tag:blogger.com,1999:blog-31575763163250344652024-03-12T21:51:03.453-07:00matematikaSegala hal mengenai matematikaJanuar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-3157576316325034465.post-47696606586649138852012-03-19T02:48:00.002-07:002012-03-19T02:48:46.032-07:00Pertidaksamaan<h3>Sifat-Sifat Pertidaksamaan</h3><ol start="1"><li>tanda pertidaksamaan <span style="text-decoration: underline;">tidak berubah</span> jika kedua ruas <span style="text-decoration: underline;">ditambah atau dikurangi</span> dengan bilangan yang sama</li>
</ol><div style="padding-left: 90px;">Jika a < b maka:</div><div style="padding-left: 90px;">a + c < b + c</div><div style="padding-left: 90px;">a – c < b – c</div><ol start="2"><li>tanda pertidaksamaan <span style="text-decoration: underline;">tidak berubah</span> jika kedua ruas <span style="text-decoration: underline;">dikali atau dibagi</span> dengan <span style="text-decoration: underline;">bilangan positif</span> yang sama</li>
</ol><div style="padding-left: 90px;">Jika a < b, dan c adalah bilangan positif, maka:</div><div style="padding-left: 90px;">a.c < b.c</div><div style="padding-left: 90px;">a/b < b/c</div><ol start="3"><li>tanda pertidaksamaan akan <span style="text-decoration: underline;">berubah</span> jika kedua ruas pertidaksamaan <span style="text-decoration: underline;">dikali atau dibagi</span> dengan <span style="text-decoration: underline;">bilangan negatif</span> yang sama</li>
</ol><div style="padding-left: 90px;">Jika a < b, dan c adalah bilangan negatif, maka:</div><div style="padding-left: 90px;">a.c > b.c</div><div style="padding-left: 90px;">a/c > b/c</div><ol start="4"><li>tanda pertidaksamaan <span style="text-decoration: underline;">tidak berubah</span> jika kedua ruas <span style="text-decoration: underline;">positif</span> masing-masing <span style="text-decoration: underline;">dikuadratkan</span></li>
</ol><div style="padding-left: 90px;">Jika a < b; a dan b sama-sama positif, maka: a<sup>2</sup> < b<sup>2</sup></div><br />
<h3>Pertidaksamaan Linear</h3>→ Variabelnya berpangkat 1<br />
<em><strong><span style="text-decoration: underline;">Penyelesaian:</span></strong></em><br />
Suku-suku yang mengandung variabel dikumpulkan di ruas kiri, dan konstanta diletakkan di ruas kanan<br />
<span style="text-decoration: underline;">Contoh:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture1.gif"><img alt="" class="alignnone size-full wp-image-103" src="http://learnwithalice.files.wordpress.com/2011/07/picture1.gif?w=500" title="Picture1" /></a><br />
<br />
<h3>Pertidaksamaan Kuadrat</h3>→ Variabelnya berpangkat 2<br />
<em><span style="text-decoration: underline;"><strong>Penyelesaian:</strong></span></em><br />
<ol start="1"><li>Ruas kanan dibuat menjadi nol</li>
<li>Faktorkan</li>
<li>Tentukan harga nol, yaitu nilai variabel yang menyebabkan nilai faktor sama dengan nol</li>
<li>Gambar garis bilangannya</li>
</ol><div style="padding-left: 60px;">Jika tanda pertidaksamaan ≥ atau ≤, maka harga nol ditandai dengan titik hitam •</div><div style="padding-left: 60px;">Jika tanda pertidaksamaan > atau <, maka harga nol ditandai dengan titik putih °</div><ol start="5"><li>Tentukan tanda (+) atau (–) pada masing-masing interval di garis bilangan. Caranya adalah dengan memasukkan salah satu bilangan pada interval tersebut pada persamaan di ruas kiri.</li>
</ol><div style="padding-left: 60px;">Tanda pada garis bilangan berselang-seling, kecuali jika ada batas rangkap (harga nol yang muncul 2 kali atau sebanyak bilangan genap untuk pertidaksamaan tingkat tinggi), batas rangkap tidak merubah tanda</div><ol start="6"><li>Tentukan himpunan penyelesaian</li>
</ol><div style="padding-left: 60px;">→ jika tanda pertidaksamaan > 0 berarti daerah pada garis bilangan yang diarsir adalah yang bertanda (+)</div><div style="padding-left: 60px;">→ jika tanda pertidaksamaan < 0 berarti daerah pada garis bilangan yang diarsir adalah yang bertanda (–)</div><span style="text-decoration: underline;">Contoh:</span><br />
(2x – 1)<sup>2</sup> ≥ (5x – 3).(x – 1) – 7<br />
4x<sup>2</sup> – 4x + 1 ≥ 5x<sup>2</sup> – 5x – 3x + 3 – 7<br />
4x<sup>2</sup> – 4x + 1 – 5x<sup>2</sup> + 5x + 3x – 3 + 7 ≥ 0<br />
–x<sup>2</sup> + 4x + 5 ≥ 0<br />
–(x<sup>2</sup> – 4x – 5) ≥ 0<br />
–(x – 5).(x + 1) ≥ 0<br />
Harga nol: x – 5 = 0 atau x + 1 = 0<br />
x = 5 atau x = –1<br />
Garis bilangan:<br />
<ul><li>menggunakan titik hitam karena tanda pertidaksamaan ≥</li>
<li>jika dimasukkan x = 0 hasilnya positif</li>
<li>karena 0 berada di antara –1 dan 5, maka daerah tersebut bernilai positif, di kiri dan kanannya bernilai negatif</li>
<li>karena tanda pertidaksamaan ≥ 0, maka yang diarsir adalah yang positif</li>
</ul><a href="http://learnwithalice.files.wordpress.com/2011/07/garis_01.jpg"><img alt="" class="alignnone size-full wp-image-104" src="http://learnwithalice.files.wordpress.com/2011/07/garis_01.jpg?w=500" title="garis_01" /></a><br />
Jadi penyelesaiannya: {x | –1 ≤ x ≤ 5}<br />
<br />
<h3>Pertidaksamaan Tingkat Tinggi</h3>→ Variabel berpangkat lebih dari 2<br />
<em><strong>Penyelesaian sama dengan pertidaksamaan kuadrat</strong></em><br />
<span style="text-decoration: underline;">Contoh:</span><br />
(2x + 1)<sup>2</sup>.(x<sup>2</sup> – 5x + 6) < 0<br />
(2x + 1)<sup>2</sup>.(x – 2).(x – 3) < 0<br />
Harga nol: 2x + 1 = 0 atau x – 2 = 0 atau x – 3 = 0<br />
x = –1/2 atau x = 2 atau x = 3<br />
Garis bilangan:<br />
<ul><li>menggunakan titik putih karena tanda pertidaksamaan <</li>
<li>jika dimasukkan x = 0 hasilnya positif</li>
<li>karena 0 berada di antara –1/2 dan 2, maka daerah tersebut bernilai positif</li>
<li>karena –1/2 adalah batas rangkap (–1/2 muncul sebanyak 2 kali sebagai harga nol, jadi –1/2 merupakan batas rangkap), maka di sebelah kiri –1/2 juga bernilai positif</li>
<li>selain daerah yang dibatasi oleh batas rangkap, tanda positif dan negatif berselang-seling</li>
<li>karena tanda pertidaksamaan ³ 0, maka yang diarsir adalah yang positif</li>
</ul><a href="http://learnwithalice.files.wordpress.com/2011/07/garis_02.jpg"><img alt="" class="alignnone size-full wp-image-105" src="http://learnwithalice.files.wordpress.com/2011/07/garis_02.jpg?w=500" title="garis_02" /></a><br />
Jadi penyelesaiannya: {x | 2 < x < 3}<br />
<br />
<h3>Pertidaksamaan Pecahan</h3>→ ada pembilang dan penyebut<br />
<em><span style="text-decoration: underline;"><strong>Penyelesaian:</strong></span></em><br />
<ol start="1"><li>Ruas kanan dijadikan nol</li>
<li>Samakan penyebut di ruas kiri</li>
<li>Faktorkan pembilang dan penyebut (jika bisa)</li>
<li>Cari nilai-nilai variabel yang menyebabkan pembilang dan penyebutnya sama dengan nol (harga nol untuk pembilang dan penyebut)</li>
<li>Gambar garis bilangan yang memuat semua nilai yang didapatkan pada langkah 4</li>
</ol><div style="padding-left: 60px;">Apapun tanda pertidaksamaannya, harga nol untuk penyebut selalu digambar dengan titik putih (penyebut suatu pecahan tidak boleh sama dengan 0 agar pecahan tersebut mempunyai nilai)</div><ol start="6"><li>Tentukan tanda (+) atau (–) pada masing-masing interval</li>
</ol><span style="text-decoration: underline;">Contoh 1:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture2.gif"><img alt="" class="alignnone size-full wp-image-107" src="http://learnwithalice.files.wordpress.com/2011/07/picture2.gif?w=500" title="Picture2" /></a><br />
Harga nol pembilang: –5x + 20 = 0<br />
–5x = –20 → x = 4<br />
Harga nol penyebut: x – 3 = 0 → x = 3<br />
Garis bilangan:<br />
→ x = 3 digambar menggunakan titik putih karena merupakan harga nol untuk penyebut<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_03.jpg"><img alt="" class="alignnone size-full wp-image-109" src="http://learnwithalice.files.wordpress.com/2011/07/garis_03.jpg?w=500" title="garis_03" /></a><br />
Jadi penyelesaiannya: {x | 3 < x ≤ 4}<br />
<br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture3.gif"><img alt="" class="alignnone size-full wp-image-108" src="http://learnwithalice.files.wordpress.com/2011/07/picture3.gif?w=500" title="Picture3" /></a><br />
Harga nol pembilang: x – 2 = 0 atau x + 1 = 0<br />
x = 2 atau x = –1<br />
Harga nol penyebut: tidak ada, karena penyebut tidak dapat difaktorkan dan jika dihitung nilai diskriminannya:<br />
D = b<sup>2</sup> – 4.a.c = 1<sup>2</sup> – 4.1.1 = 1 – 4 = –3<br />
Nilai D-nya negatif, sehingga persamaan tersebut tidak mempunyai akar real<br />
(Catatan: jika nilai D-nya tidak negatif, gunakan rumus abc untuk mendapat harga nol-nya)<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_04.jpg"><img alt="" class="alignnone size-full wp-image-106" src="http://learnwithalice.files.wordpress.com/2011/07/garis_04.jpg?w=500" title="garis_04" /></a><br />
Jadi penyelesaiannya: {x | x ≤ –1 atau x ≥ 2}<br />
<br />
<h3>Pertidaksamaan Irasional/Pertidaksamaan Bentuk Akar</h3>→ variabelnya berada dalam tanda akar<br />
<em><span style="text-decoration: underline;"><strong>Penyelesaian:</strong></span></em><br />
<ol start="1"><li>Kuadratkan kedua ruas</li>
<li>Jadikan ruas kanan sama dengan nol</li>
<li>Selesaikan seperti menyelesaikan pertidaksamaan linear/kuadrat</li>
<li>Syarat tambahan: yang berada di dalam setiap tanda akar harus ≥ 0</li>
</ol><span style="text-decoration: underline;">Contoh 1:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture4.gif"><img alt="" class="alignnone size-full wp-image-110" src="http://learnwithalice.files.wordpress.com/2011/07/picture4.gif?w=500" title="Picture4" /></a><br />
Kuadratkan kedua ruas:<br />
x<sup>2</sup> – 5x – 6 < x<sup>2</sup> – 3x + 2<br />
x<sup>2</sup> – 5x – 6 – x<sup>2</sup> + 3x – 2 < 0<br />
–2x – 8 < 0<br />
Semua dikali –1:<br />
2x + 8 > 0<br />
2x > –8<br />
x > –4<br />
<em>Syarat 1:</em><br />
x<sup>2</sup> – 5x – 6 ≥ 0<br />
(x – 6).(x + 1) ≥ 0<br />
Harga nol: x – 6 = 0 atau x + 1 = 0<br />
x = 6 atau x = –1<br />
<em>Syarat 2:</em><br />
x<sup>2</sup> – 3x + 2 ≥ 0<br />
(x – 2).(x – 1) ≥ 0<br />
Harga nol: x – 2 = 0 atau x – 1 = 0<br />
x = 2 atau x = 1<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_05.jpg"><img alt="" class="alignnone size-full wp-image-111" src="http://learnwithalice.files.wordpress.com/2011/07/garis_05.jpg?w=500" title="garis_05" /></a><br />
Jadi penyelesaiannya: {x | –4 < x ≤ –1 atau x ≥ 6}<br />
<br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture5.gif"><img alt="" class="alignnone size-full wp-image-112" src="http://learnwithalice.files.wordpress.com/2011/07/picture5.gif?w=500" title="Picture5" /></a><br />
Kuadratkan kedua ruas:<br />
x<sup>2</sup> – 6x + 8 < x<sup>2</sup> – 4x + 4<br />
x<sup>2</sup> – 6x + 8 – x<sup>2</sup> + 4x – 4 < 0<br />
–2x + 4 < 0<br />
–2x < –4<br />
Semua dikalikan –1<br />
2x > 4<br />
x > 2<br />
Syarat:<br />
x<sup>2</sup> – 6x + 8 ≥ 0<br />
(x – 4).(x – 2) ≥ 0<br />
Harga nol: x – 4 = 0 atau x – 2 = 0<br />
x = 4 atau x = 2<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_06.jpg"><img alt="" class="alignnone size-full wp-image-113" src="http://learnwithalice.files.wordpress.com/2011/07/garis_06.jpg?w=500" title="garis_06" /></a><br />
Jadi penyelesaiannya: {x | x ≥ 4}<br />
<br />
<h3>Pertidaksamaan Nilai Mutlak</h3>→ variabelnya berada di dalam tanda mutlak | ….. |<br />
(tanda mutlak selalu menghasilkan hasil yang positif, contoh: |3| = 3; |–3| = 3)<br />
Pengertian nilai mutlak:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture6.gif"><img alt="" class="alignnone size-full wp-image-114" src="http://learnwithalice.files.wordpress.com/2011/07/picture6.gif?w=500" title="Picture6" /></a><br />
<em><span style="text-decoration: underline;"><strong>Penyelesaian:</strong></span></em><br />
Jika |x| < a berarti: –a < x < a, dimana a ≥ 0<br />
Jika |x| > a berarti: x < –a atau x > a, dimana a ≥ 0<br />
<br />
<span style="text-decoration: underline;">Contoh 1:</span><br />
|2x – 3| ≤ 5<br />
berarti:<br />
–5 ≤ 2x – 3 ≤ 5<br />
–5 + 3 ≤ 2x ≤ 5 + 3<br />
–2 ≤ 2x ≤ 8<br />
Semua dibagi 2:<br />
–1 ≤ x ≤ 4<br />
<br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
|3x + 7| > 2<br />
berarti:<br />
3x + 7 < –2 atau 3x + 7 > 2<br />
3x < –2 – 7 atau 3x > 2 – 7<br />
x < –3 atau x > –5/3<br />
<br />
<span style="text-decoration: underline;">Contoh 3:</span><br />
|2x – 5| < |x + 4|<br />
Kedua ruas dikuadratkan:<br />
(2x – 5)<sup>2</sup> < (x + 4)<sup>2</sup><br />
(2x – 5)<sup>2</sup> – (x + 4)<sup>2</sup> < 0<br />
(2x – 5 + x + 4).(2x – 5 – x – 4) < 0 <em>(Ingat! a<sup>2</sup> – b<sup>2</sup> = (a + b).(a – b))</em><br />
(3x – 1).(x – 9) < 0<br />
Harga nol: 3x – 1 = 0 atau x – 9 = 0<br />
x = 1/3 atau x = 9<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_07.jpg"><img alt="" class="alignnone size-full wp-image-115" src="http://learnwithalice.files.wordpress.com/2011/07/garis_07.jpg?w=500" title="garis_07" /></a><br />
Jadi penyelesaiannya: {x | 1/3 < x < 4}<br />
<br />
<span style="text-decoration: underline;">Contoh 4:</span><br />
|4x – 3| ≥ x + 1<br />
Kedua ruas dikuadratkan:<br />
(4x – 3)<sup>2</sup> ≥ (x + 1)<sup>2</sup><br />
(4x – 3)<sup>2</sup> – (x + 1)<sup>2</sup> ≥ 0<br />
(4x – 3 + x + 1).(4x – 3 – x – 1) ≥ 0<br />
(5x – 2).(3x – 4) ≥ 0<br />
Harga nol: 5x – 2 = 0 atau 3x – 4 = 0<br />
x = 2/5 atau x = 4/3<br />
<em>Syarat:</em><br />
x + 1 ≥ 0<br />
x ≥ –1<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_08.jpg"><img alt="" class="alignnone size-full wp-image-117" src="http://learnwithalice.files.wordpress.com/2011/07/garis_08.jpg?w=500" title="garis_08" /></a><br />
Jadi penyelesaiannya: {x | –1 ≤ x ≤ 2/5 atau x ≥ 4/3}<br />
<br />
<span style="text-decoration: underline;">Contoh 5:</span><br />
|x – 2|<sup>2</sup> – |x – 2| < 2<br />
Misalkan |x – 2| = y<br />
y<sup>2</sup> – y < 2<br />
y<sup>2</sup> – y – 2 < 0<br />
(y – 2).(y + 1) < 0<br />
Harga nol: y – 2 = 0 atau y + 1 = 0<br />
y = 2 atau y = –1<br />
Garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/garis_09.jpg"><img alt="" class="alignnone size-full wp-image-116" src="http://learnwithalice.files.wordpress.com/2011/07/garis_09.jpg?w=500" title="garis_09" /></a><br />
Artinya:<br />
–1 < y < 2<br />
–1 < |x – 2| < 2<br />
Karena nilai mutlak pasti bernilai positif, maka batas kiri tidak berlaku<br />
|x – 2| < 2<br />
Sehingga:<br />
–2 < x – 2 < 2<br />
–2 + 2 < x < 2 + 2<br />
0 < x < 4Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com68tag:blogger.com,1999:blog-3157576316325034465.post-50855110378163650782012-03-19T02:47:00.002-07:002012-03-19T02:47:49.159-07:00Sistem Persamaan (Linear dan Kuadrat)<h3>Sistem Persamaan Linear Dua Variabel (SPLDV)</h3>→ mengandung 2 variabel berpangkat 1<br />
<span style="text-decoration: underline;"><strong>Bentuk umum:</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture11.gif"><img alt="" class="alignnone size-full wp-image-123" src="http://learnwithalice.files.wordpress.com/2011/07/picture11.gif?w=500" title="Picture1" /></a><br />
dimana a<sub>1</sub>, a<sub>2</sub>, b<sub>1</sub>, b<sub>2</sub>, c<sub>1</sub>, dan c<sub>2</sub> adalah bilangan real<br />
<em><strong>Catatan:</strong></em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture21.gif"><img alt="" class="alignnone size-full wp-image-124" src="http://learnwithalice.files.wordpress.com/2011/07/picture21.gif?w=500" title="Picture2" /></a><br />
<em><strong>Penyelesaian:</strong></em><br />
<ol start="1"><li>Metode grafik</li>
<li>Metode substitusi</li>
<li>Metode eliminasi</li>
<li>Metode gabungan substitusi-eliminasi</li>
</ol><span style="text-decoration: underline;">Contoh:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture31.gif"><img alt="" class="alignnone size-full wp-image-125" src="http://learnwithalice.files.wordpress.com/2011/07/picture31.gif?w=500" title="Picture3" /></a><br />
<em>Metode grafik:</em><br />
→ gambar grafik untuk tiap persamaan, cara paling mudah: masukkan x = 0, hitung nilai y untuk mendapatkan titik pertama; lalu masukkan y = 0, hitung nilai x untuk mendapatkan titik kedua<br />
→ jika saat dimasukkan x = 0, didapatkan nilai y = 0, untuk mendapatkan titik kedua masukkan nilai x selain 0<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/tabel.jpg"><img alt="" class="alignnone size-full wp-image-122" src="http://learnwithalice.files.wordpress.com/2011/07/tabel.jpg?w=500" title="tabel" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/grafik_01.jpg"><img alt="" class="alignnone size-full wp-image-133" height="273" src="http://learnwithalice.files.wordpress.com/2011/07/grafik_01.jpg?w=500&h=273" title="grafik_01" width="500" /></a><br />
<em>Metode substitusi:</em><br />
Dari persamaan 1: 2x – y = 8 → 2x – 8 = y<br />
Masukkan ke persamaan 2:<br />
x + 2y = 14<br />
x + 2.(2x – 8 ) = 14<br />
x + 4x – 16 = 14<br />
5x = 14 + 16<br />
5x = 30<br />
x = 30/5 = 6<br />
y = 2x – 8 = 2.6 – 8 = 12 – 8 = 4<br />
Jadi penyelesaiannya: {(6, 4)}<br />
<em>Metode eliminasi:</em><br />
Eliminasi x: (Persamaan 2 dikali 2)<br />
2x – y = 8<br />
<span style="text-decoration: underline;">2x + 4y = 28 – </span> (dikurangi karena nilai x-nya sama-sama positif)<br />
–5y = –20<br />
y = –20/–5 = 4<br />
Eliminasi y: (Persamaan 1 dikali 2)<br />
4x – 2y = 16<br />
<span style="text-decoration: underline;"> x + 2y = 14 +</span> (ditambah karena nilai y-nya positif dan negatif)<br />
5x = 30<br />
x = 30/5 = 6<br />
Jadi penyelesaiannya: {(6, 4)}<br />
<em>Metode gabungan (eliminasi-substitusi)</em><br />
Eliminasi x: (Persamaan 2 dikali 2)<br />
2x – y = 8<br />
<span style="text-decoration: underline;">2x + 4y = 28 – </span> (dikurangi karena nilai x-nya sama-sama positif)<br />
–5y = –20<br />
y = –20/–5 = 4<br />
Masukkan ke salah satu persamaan, misalnya persamaan 1:<br />
2x – y = 8<br />
2x – 4 = 8<br />
2x = 8 + 4<br />
2x = 12<br />
x = 12/2 = 6<br />
Jadi penyelesaiannya: {(6, 4)}<br />
<h3>Sistem Persamaan Linear Tiga Variabel (SPLTV)</h3><span style="text-decoration: underline;"><strong>Bentuk umum:</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture41.gif"><img alt="" class="alignnone size-full wp-image-126" src="http://learnwithalice.files.wordpress.com/2011/07/picture41.gif?w=500" title="Picture4" /></a><br />
dimana a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub>, c<sub>1</sub>, c<sub>2</sub>, c<sub>3</sub>, d<sub>1</sub>, d<sub>2</sub> dan d<sub>3</sub> adalah bilangan real<br />
<em><strong>Penyelesaian:</strong></em><br />
→ Eliminasi salah satu variabel dari sistem sehingga mernjadi SPLDV (misal: dari persamaan 1 dan 2 eliminasi x, persamaan 1 dan 3 atau 2 dan 3 juga eliminasi x)<br />
<span style="text-decoration: underline;">Contoh:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture51.gif"><img alt="" class="alignnone size-full wp-image-127" src="http://learnwithalice.files.wordpress.com/2011/07/picture51.gif?w=500" title="Picture5" /></a><br />
Eliminasi z dari persamaan 1 dan 2 (persamaan 1 dikali 2):<br />
2x + 2y + 2z = 12<br />
<span style="text-decoration: underline;">2x + 3y – 2z = 2 (+)</span><br />
4x + 5y = 14 …… Persamaan 4<br />
Eliminasi z dari persamaan 1 dan 3:<br />
x + y + z = 6<br />
<span style="text-decoration: underline;">3x – 2y + z = 2 (–)</span><br />
–2x + 3y = 4 …… Persamaan 5<br />
Eliminasi x dari persamaan 4 dan 5 (persamaan 5 dikali 2):<br />
4x + 5y = 14<br />
<span style="text-decoration: underline;">–4x + 6y = 8 (+)</span><br />
11y = 22<br />
y = 22/11 = 2<br />
Masukkan y ke persamaan 5:<br />
–2x + 3y = 4<br />
–2x + 3.2 = 4<br />
–2x + 6 = 4<br />
–2x = 4 – 6<br />
–2x = –2<br />
x = –2/–2 = 1<br />
Masukkan x dan y ke persamaan 1:<br />
x + y + z = 6<br />
1 + 2 + z = 6<br />
z = 6 – 1 – 2 = 3<br />
Jadi penyelesaiannya: {(1, 2, 3)}<br />
<h3>Sistem Persamaan Linear Kuadrat Dua Variabel (SPLKDV)</h3><span style="text-decoration: underline;"><strong>Bentuk Umum:</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture61.gif"><img alt="" class="alignnone size-full wp-image-128" src="http://learnwithalice.files.wordpress.com/2011/07/picture61.gif?w=500" title="Picture6" /></a><br />
<em><strong>Penyelesaian:</strong></em><br />
→ Substitusi persamaan 1 ke 2 diperoleh:<br />
mx + n = ax<sup>2</sup> + bx + c<br />
ax<sup>2</sup> + (b –m)x + (c – n) = 0<br />
Nilai diskriminannya: D = b<sup>2</sup> – 4.a.c = (b – m)<sup>2</sup> – 4.a.(c – n)<br />
<ul><li>D > 0 → SPLKV mempunyai 2 akar (penyelesaian) nyata</li>
<li>D = 0 → SPLKV mempunyai 1 akar (penyelesaian) nyata</li>
<li>D < 0 → SPLKV tidak mempunyai akar (penyelesaian) nyata</li>
</ul>→ Dapat juga diselesaikan dengan grafik<br />
<span style="text-decoration: underline;">Contoh:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture7.gif"><img alt="" class="alignnone size-full wp-image-129" src="http://learnwithalice.files.wordpress.com/2011/07/picture7.gif?w=500" title="Picture7" /></a><br />
Substitusi persamaan 1 ke 2<br />
2 – x = x<sup>2</sup><br />
x<sup>2</sup> + x – 2 = 0<br />
(x + 2).(x – 1) = 0<br />
x + 2 = 0 atau x – 1 = 0<br />
x = –2 atau x = 1<br />
untuk x = –2 → y = 2 – (–2) = 2 + 2 = 4 (nilai x juga dapat dimasukkan ke persamaan 2)<br />
untuk x = 1 → y = 2 – 1 = 1<br />
Jadi penyelesaiannya: {(–2, 4), (1, 1)}<br />
<em>Grafik:</em><br />
→ cara menggambar grafik fungsi kuadrat: lihat di bab <a href="http://learnwithalice.wordpress.com/2011/06/29/fungsi-kuadrat/">FUNGSI KUADRAT</a><br />
→ cara menggambar garis: lihat di bagian SPLDV<br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/grafik_02.jpg"><img alt="" class="alignnone size-full wp-image-134" height="273" src="http://learnwithalice.files.wordpress.com/2011/07/grafik_02.jpg?w=500&h=273" title="grafik_02" width="500" /></a><br />
<h3>Sistem Persamaan Kuadrat (SPK)</h3><span style="text-decoration: underline;"><strong>Bentuk umum:</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture8.gif"><img alt="" class="alignnone size-full wp-image-130" src="http://learnwithalice.files.wordpress.com/2011/07/picture8.gif?w=500" title="Picture8" /></a><br />
<em><strong>Penyelesaian:</strong></em><br />
→ Jika persamaan 1 = persamaan 2, maka SPK mempunyai banyak penyelesaian<br />
→ Jika persamaan 1 ≠ persamaan 2, maka substitusi persamaan 1 ke 2, sehingga diperoleh:<br />
ax<sup>2</sup> + bx + c = px<sup>2</sup> + qx + r<br />
(a – p)x<sup>2</sup> + (b – q)x + (c – r) = 0<br />
Hitung nilai Diskriminan: D = (b – q)<sup>2</sup> – 4.(a – p).(c – r)<br />
<ul><li>D > 0 → SPK mempunyai 2 akar (penyelesaian) real</li>
<li>D = 0 → SPK mempunyai 1 akar (penyelesaian) real</li>
<li>D < 0 → SPK tidak mempunyai akar (penyelesaian) real</li>
</ul>→ dapat juga diselesaikan dengan cara grafik<br />
<span style="text-decoration: underline;">Contoh 1:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture9.gif"><img alt="" class="alignnone size-full wp-image-131" src="http://learnwithalice.files.wordpress.com/2011/07/picture9.gif?w=500" title="Picture9" /></a><br />
Substitusi persamaan1 ke 2:<br />
x<sup>2</sup> – 2x – 3 = –x<sup>2</sup> – 2x – 5<br />
x<sup>2</sup> – 2x – 3 + x<sup>2</sup> + 2x + 5 = 0<br />
2x<sup>2</sup> + 2 = 0<br />
Semua dibagi 2:<br />
x<sup>2</sup> + 1 = 0<br />
Karena persamaan tidak dapat difaktorkan, hitung nilai D:<br />
D = b<sup>2</sup> – 4.a.c = 0<sup>2</sup> – 4.1.1 = a – 4<br />
Karena D < 0 maka SPK tidak mempunya penyelesaian real<br />
<em>Grafik:</em><br />
→ Cara menggambar grafik fungsi kuadrat: lihat di bab <a href="http://learnwithalice.wordpress.com/2011/06/29/fungsi-kuadrat/">FUNGSI KUADRAT</a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/grafik_03.jpg"><img alt="" class="alignnone size-full wp-image-135" height="273" src="http://learnwithalice.files.wordpress.com/2011/07/grafik_03.jpg?w=500&h=273" title="grafik_03" width="500" /></a><br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/picture10.gif"><img alt="" class="alignnone size-full wp-image-132" src="http://learnwithalice.files.wordpress.com/2011/07/picture10.gif?w=500" title="Picture10" /></a><br />
Substitusi persamaan 1 ke 2:<br />
x<sup>2</sup> – 2x = –1/2 x<sup>2</sup> + 4x – 6<br />
Semua dikalikan 2:<br />
2x<sup>2</sup> – 4x = –x<sup>2</sup> + 8x – 12<br />
2x<sup>2</sup> – 4x + x<sup>2</sup> – 8x + 12 = 0<br />
3x<sup>2</sup> – 12x + 12 = 0<br />
Semua dibagi 3:<br />
x<sup>2</sup> – 4x + 4 = 0<br />
(x – 2).(x – 2) = 0<br />
x = 2 → y = x<sup>2</sup> – 2x = 2<sup>2</sup> – 2.2 = 4 – 4 = 0<br />
Jadi penyelesaiannya: {(2, 0)}<br />
<em>Grafik:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/07/grafik_04.jpg"><img alt="" class="alignnone size-full wp-image-136" height="273" src="http://learnwithalice.files.wordpress.com/2011/07/grafik_04.jpg?w=500&h=273" title="grafik_04" width="500" /></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com17tag:blogger.com,1999:blog-3157576316325034465.post-62393505915783099012012-03-19T02:46:00.003-07:002012-03-19T02:46:51.335-07:00Persamaan dan Pertidaksamaan Trigonometri<h3>Persamaan Dasar</h3><b>sin x = sin a </b><br />
<div style="text-align: center;">x = a + k.360° atau x = (180 – a) + k.360° (kuadran I atau II)</div><b>cos x = cos a</b><br />
<div style="text-align: center;">x = a + k.360° atau x = –a + k.360° (kuadran I atau IV)</div><b>tan x = tan a</b><br />
<div style="text-align: center;">x = a + k.180</div><i>*k = bilangan bulat</i><br />
<span style="text-decoration: underline;">Catatan:</span><br />
Jika ada persamaan cos x = sin a, cot x = tan a, sec x = cosec a, dan sebaliknya, salah satu diubah menjadi (90 – a)°,<br />
contoh: cos x = sin a → cos x = cos (90 – a)°<br />
<i><b>Contoh:</b></i><br />
<ul><li>Tentukan HP (Himpunan Penyelesaian) dari 2 cos x – √3 = 0 untuk 0 ≤ x ≤ 360°</li>
</ul><div style="padding-left: 60px;">2 cos x = √3</div><div style="padding-left: 60px;">cos x = ½ √3</div><div style="padding-left: 60px;">cos x = cos 30°</div><div style="padding-left: 60px;">x = 30° + k.360° atau x = (180 – 30)° + k.360°</div><div style="padding-left: 60px;">k = 0 → x = 30° x = 150° + k.360°</div><div style="padding-left: 60px;">k = 1 → x = 390° (tidak memenuhi) k = 0 → x = 150°</div><div style="padding-left: 60px;">Jadi HP = {30°, 150°}</div><ul><li>Tentukan HP dari tan (60 – ½ x)° = cot (x + 120)° untuk 0 ≤ x ≤ 360°</li>
</ul><div style="padding-left: 60px;">tan (60 – ½ x)° = tan (90 – (x + 120))°</div><div style="padding-left: 60px;">tan (60 – ½ x)° = tan (–x – 30)°</div><div style="padding-left: 60px;">60° – ½ x = –x – 30° + k.180°</div><div style="padding-left: 60px;">x – ½ x = –30° – 60° + k.180°</div><div style="padding-left: 60px;">½ x = –90° + k.180°</div><div style="padding-left: 60px;">x = –180° + k.360°</div><div style="padding-left: 60px;">k = 1 → x = 180°</div><div style="padding-left: 60px;">Jadi HP = {180°}</div><br />
<h3>Persamaan bentuk a cos nx + b sin nx</h3>a cos nx + b sin nx diubah menjadi k cos(nx – α)<br />
dimana<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture14.gif"><img alt="" class="size-full wp-image-190 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture14.gif?w=500" title="Picture1" /></a>Selanjutnya diselesaikan seperti menyelesaikan persamaan dasar cos x = cos a<br />
Penentuan letak α:<br />
<ul><li>Jika a +, b + → α di kuadran I</li>
<li>Jika a –, b + → α di kuadran II</li>
<li>Jika a –, b – → α di kuadran III</li>
<li>Jika a +, b – → α di kuadran IV</li>
</ul>Untuk persamaan a cos nx + b sin nx = c, syarat agar persamaan ini dapat diselesaikan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture22.gif"><img alt="" class="size-full wp-image-191 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture22.gif?w=500" title="Picture2" /></a>Dan agar persamaan ini tidak dapat diselesaikan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture32.gif"><img alt="" class="size-full wp-image-192 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture32.gif?w=500" title="Picture3" /></a><br />
<h3>Persamaan bentuk a cos<sup>2</sup>x + b sin x.cos x + c sin<sup>2</sup>x = d</h3>Caranya, lakukan dengan mengubah unsur-unsurnya seperti berikut ini:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture42.gif"><img alt="" class="size-full wp-image-193 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture42.gif?w=500" title="Picture4" /></a>Selanjutnya persamaan diselesaikan seperti menyelesaikan persamaan a cos nx + b sin nx = c<br />
<br />
<h3>Persamaan bentuk a(cos x ± sin x) + b sin x.cos x + c = 0</h3>Caranya:<br />
Misalkan (cos x ± sin x) = p<br />
maka<br />
(cos x ± sin x)<sup>2</sup> = p<sup>2</sup><br />
cos<sup>2</sup>x ± 2 sin x.cos x + sin<sup>2</sup>x = p<sup>2</sup><br />
1 ± 2 sin x.cos x = p<sup>2</sup><br />
± 2 sin x.cos x = p<sup>2</sup> – 1<br />
Sehingga 2 sin x.cos x = ± ½ (p<sup>2</sup> – 1)<br />
Sehingga persamaan di atas akan menjadi persamaan kuadrat:<br />
a.p ± ½ b(p<sup>2</sup> – 1) + c = 0<br />
Selesaikan dengan cara pemfaktoran atau rumus abc untuk mendapatkan nilai p, kemudian persamaan cos x ± sin x = p dapat diselesaikan dengan cara seperti menyelesaikan persamaan a cos nx + b sin nx = c<br />
<br />
<h3>Nilai ekstrim y = a cos nx + b sin nx + c</h3><a href="http://learnwithalice.files.wordpress.com/2011/10/picture52.gif"><img alt="" class="size-full wp-image-188 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture52.gif?w=500" title="Picture5" /></a><br />
<h3>Pertidaksamaan Trigonometri</h3>→ mencari harga nol sama dengan cara menyelesaikan persamaan trigonometri<br />
→ diselesaikan dengan menggunakan garis bilangan<br />
<i><b>Contoh:</b></i><br />
Selesaikan sin 2x < cos x untuk 0 ≤ x ≤ 360°<br />
Cara:<br />
sin 2x – cos x < 0<br />
2 sin x.cos x – cos x < 0<br />
cos x.(2 sin x – 1) < 0<br />
harga nol:<br />
<ul><li>cos x = 0</li>
</ul><div style="padding-left: 60px;">cos x = cos 90°</div><div style="padding-left: 60px;">x = 90° + k.360° atau x = –90° + k.360°</div><div style="padding-left: 60px;">k = 0 → x = 90° k = 1 → x = 270°</div><ul><li>2 sin x – 1 = 0</li>
</ul><div style="padding-left: 60px;">2 sin x = 1</div><div style="padding-left: 60px;">sin x = ½</div><div style="padding-left: 60px;">sin x = sin 30°</div><div style="padding-left: 60px;">x = 30° + k.360° atau x = (180 – 30)° + k.360°</div><div style="padding-left: 60px;">k = 0 → x = 30° x = 150° + k.360°</div><div style="padding-left: 60px;"> k = 0 → x = 150°</div>Memberi tanda (+) dan (-) pada garis bilangan:<br />
Jika x = 180° maka sin 2.180° – cos 180° = sin 360° – cos 180° = 0 – (–1) = 1 (+)<br />
Jadi garis bilangannya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/grs_bil.jpg"><img alt="" class="alignnone size-full wp-image-189" src="http://learnwithalice.files.wordpress.com/2011/10/grs_bil.jpg?w=500" title="Grs_Bil" /></a><br />
karena yang diminta kurang dari (<) 0, maka yang diarsir adalah bagian-bagian yang bertanda (-)<br />
Sehingga HP-nya: {0° ≤ x < 30° atau 90° < x < 150° atau 270° < x ≤ 360°}Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com2tag:blogger.com,1999:blog-3157576316325034465.post-21278043813382553992012-03-19T02:37:00.000-07:002012-03-19T02:37:31.813-07:00Trigonometri<h3><strong>Ukuran Sudut</strong></h3>1 putaran = 360 derajat (360°) = 2π radian<br />
<h3>Perbandingan trigonometri</h3><a href="http://learnwithalice.files.wordpress.com/2011/10/segitiga.jpg"><img alt="" class="alignnone size-full wp-image-162" height="126" src="http://learnwithalice.files.wordpress.com/2011/10/segitiga.jpg?w=500&h=126" title="segitiga" width="500" /></a><br />
Catatan:<br />
<ul><li>Sin = sinus</li>
<li>Cos = cosinus</li>
<li>Tan/Tg = tangens</li>
<li>Sec = secans</li>
<li>Cosec/Csc = cosecans</li>
<li>Cot/Ctg = cotangens</li>
</ul>Dari gambar tersebut dapat diperoleh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture1.gif"><img alt="" class="alignnone size-full wp-image-153" src="http://learnwithalice.files.wordpress.com/2011/10/picture1.gif?w=500" title="Picture1" /></a><br />
(sec merupakan kebalikan dari cos,<br />
csc merupakan kebalikan dari sin, dan<br />
cot merupakan kebalikan dari tan)<br />
<em><strong>Contoh:</strong></em><br />
Dari segitiga berikut ini:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/segi3soal.jpg"><img alt="" class="alignnone size-full wp-image-164" src="http://learnwithalice.files.wordpress.com/2011/10/segi3soal.jpg?w=500" title="segi3soal" /></a><br />
Diketahui panjang AB = 12 cm, AC = 13 cm. Hitung semua nilai perbandingan trigonometri untuk sudut A!<br />
Pertama, hitung dulu panjang BC dengan menggunakan rumus Phytagoras:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture2.gif"><img alt="" class="alignnone size-full wp-image-154" src="http://learnwithalice.files.wordpress.com/2011/10/picture2.gif?w=500" title="Picture2" /></a><br />
<h3><strong>Nilai perbandingan trigonometri beberapa sudut istimewa</strong></h3><h3><a href="http://learnwithalice.files.wordpress.com/2011/10/tabel.jpg"><img alt="" class="alignnone size-full wp-image-139" src="http://learnwithalice.files.wordpress.com/2011/10/tabel.jpg?w=500" title="tabel" /></a></h3>* tambahan: sin 37° = cos 53° = 0,6<br />
<h3>Kuadran</h3>Kuadran adalah pembagian daerah pada sistem koordinat kartesius → dibagi dalam 4 daerah<br />
Nilai perbandingan trigonometri untuk sudut-sudut di berbagai kuadran memenuhi aturan seperti pada gambar:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/4kuadran.jpg"><img alt="" class="size-full wp-image-140 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/4kuadran.jpg?w=500" title="4kuadran" /></a><br />
Untuk sudut b > 360° → b = (k . 360 + a) → b = a<br />
(k = bilangan bulat > 0)<br />
<strong>Mengubah fungsi trigonometri suatu sudut ke sudut lancip</strong><br />
<ul><li>Jika menggunakan 90 ± a atau 270 ± a maka fungsi berubah:</li>
</ul><div style="padding-left: 60px;">sin ↔ cos</div><div style="padding-left: 60px;">tan ↔ cot</div><div style="padding-left: 60px;">sec ↔ csc</div><ul><li>Jika menggunakan 180 ± a atau 360 ± a maka fungsi tetap</li>
</ul><strong>Sudut dengan nilai negatif</strong><br />
Nilai negatif diperoleh karena sudut dibuat dari sumbu x, diputar searah jarum jam<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/sudut_negatif.jpg"><img alt="" class="alignnone size-full wp-image-163 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/sudut_negatif.jpg?w=500" title="sudut_negatif" /></a>Untuk sudut dengan nilai negatif, sama artinya dengan sudut yang berada di kuadran IV<br />
<em><strong>Contoh:</strong></em><br />
<ul><li>Cos 120º = cos (180 – 60)º = – cos 60º = – 1/2 (120º ada di kuadran II sehingga nilai cos-nya negatif)</li>
<li>Cos 120º = cos (90 + 30)º = – sin 30º = – 1/2</li>
<li>Tan 1305º = tan (3.360 + 225)º = tan 225º = tan (180 + 45)º = tan 45º = 1 (225º ada di kuadran III sehingga nilai tan-nya positif)</li>
<li>Sin –315º = – sin 315º = – sin (360 – 45)º = –(– sin 45)º = sin 45º = 1/2 √2</li>
</ul><h3>Identitas Trigonometri</h3><div style="text-align: left;"><a href="http://learnwithalice.files.wordpress.com/2011/10/picture9.gif"><img alt="" height="248" src="http://learnwithalice.files.wordpress.com/2011/10/picture9.gif?w=321&h=248" title="Picture9" width="321" /></a></div><div style="text-align: left;">Sehingga, secara umum, berlaku:</div><div style="text-align: center;">sin<sup>2</sup>a + cos<sup>2</sup>a = 1</div><div style="text-align: center;">1 + tan<sup>2</sup>a = sec<sup>2</sup>a</div><div style="text-align: center;">1 + cot<sup>2</sup>a = csc<sup>2</sup>a</div><h3>Grafik fungsi trigonometri</h3><strong>y = sin x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_01.jpg"><img alt="" class="size-full wp-image-141 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_01.jpg?w=500&h=213" title="graph_01" width="500" /></a><br />
<strong>y = cos x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_02.jpg"><img alt="" class="size-full wp-image-142 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_02.jpg?w=500&h=213" title="graph_02" width="500" /></a><br />
<strong>y = tan x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_03.jpg"><img alt="" class="size-full wp-image-143 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_03.jpg?w=500&h=213" title="graph_03" width="500" /></a><br />
<strong>y = cot x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_04.jpg"><img alt="" class="size-full wp-image-144 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_04.jpg?w=500&h=213" title="graph_04" width="500" /></a><br />
<strong>y = sec x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_05.jpg"><img alt="" class="size-full wp-image-145 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_05.jpg?w=500&h=213" title="graph_05" width="500" /></a><br />
<strong>y = csc x</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_06.jpg"><img alt="" class="size-full wp-image-146 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_06.jpg?w=500&h=213" title="graph_06" width="500" /></a><br />
<h3><strong>Menggambar Grafik fungsi y = A sin/cos/tan/cot/sec/csc (kx ± b) ± c</strong></h3><ol start="1"><li>Periode fungsi untuk sin/cos/sec/csc = 2π/k → artinya: grafik akan berulang setiap kelipatan 2π/k</li>
</ol><div style="padding-left: 30px; text-align: left;"> Periode fungsi untuk tan/cot = π/k → artinya: grafik akan berulang setiap kelipatan π/k</div><ol start="2"><li>Nilai maksimum = c + |A|, nilai minimum = c – |A|</li>
<li>Amplitudo = ½ (y<sub>max</sub> – y<sub>min</sub>)</li>
<li>Cara menggambar:</li>
<ol start="1"><li>Gambar grafik fungsi dasarnya seperti pada gambar di atas</li>
<li>Hitung periode fungsi, dan gambarkan grafik sesuai dengan periode fungsinya</li>
<li>Jika A ≠ 1, kalikan semua nilai y pada grafik fungsi dasar dengan A</li>
<li>Untuk kx + b → grafik digeser ke kiri sejauh b/k</li>
</ol></ol><div style="padding-left: 60px;"> Untuk kx – b → grafik digeser ke kanan sejauh b/k</div><ol start="4"><ol start="5"><li>Untuk + c → grafik digeser ke atas sejauh c</li>
</ol></ol><div style="padding-left: 60px;"> Untuk – c → grafik digeser ke bawah sejauh c</div><em><strong>Contoh:</strong></em> y = 2 sin (3x + 90)° + 3<br />
→ periode fungsi = 2p/3 = 120°<br />
<span style="text-decoration: underline;"><em><strong>Langkah-Langkah:</strong></em></span><br />
Grafik fungsi y = sin x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_07.jpg"><img alt="" class="size-full wp-image-147 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_07.jpg?w=500&h=213" title="graph_07" width="500" /></a><br />
Karena periode fungsinya 2π/3, maka dalam selang 0 hingga 2π, terjadi 3 gelombang sinus → y = sin 3x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_08.jpg"><img alt="" class="size-full wp-image-148 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_08.jpg?w=500&h=213" title="graph_08" width="500" /></a><br />
Ampitudo dikali 2 → y = 2 sin 3x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_09.jpg"><img alt="" class="size-full wp-image-149 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_09.jpg?w=500&h=213" title="graph_09" width="500" /></a><br />
Grafik digeser ke kiri sejauh 90°/3 = 30° = π/6 → y = 2 sin (3x + 90)°<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_10.jpg"><img alt="" class="size-full wp-image-150 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_10.jpg?w=500&h=213" title="graph_10" width="500" /></a><br />
Grafik digeser ke atas sejauh 3 satuan → y = 2 sin (3x + 90)° + 3<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/graph_11.jpg"><img alt="" class="size-full wp-image-151 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/10/graph_11.jpg?w=500&h=213" title="graph_11" width="500" /></a><br />
<h3>Aturan-Aturan pada Segitiga ABC</h3><a href="http://learnwithalice.files.wordpress.com/2011/10/segi3_abc.jpg"><img alt="" class="size-full wp-image-161 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/segi3_abc.jpg?w=500" title="segi3_ABC" /></a><strong>Aturan Sinus</strong><br />
Dari segitiga ABC di atas:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture3.gif"><img alt="" class="alignnone size-full wp-image-155" src="http://learnwithalice.files.wordpress.com/2011/10/picture3.gif?w=500" title="Picture3" /></a><br />
Sehingga, secara umum, dalam segitiga ABC berlaku rumus:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture4.gif"><img alt="" class="size-full wp-image-156 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture4.gif?w=500" title="Picture4" /></a><br />
<strong>Aturan Cosinus</strong><br />
Dari segitiga ABC di atas:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture5.gif"><img alt="" class="alignnone size-full wp-image-157" src="http://learnwithalice.files.wordpress.com/2011/10/picture5.gif?w=500" title="Picture5" /></a><br />
Sehingga, secara umum:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture6.gif"><img alt="" class="size-full wp-image-158 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture6.gif?w=500" title="Picture6" /></a><br />
<strong>Luas Segitiga</strong><br />
Dari segitiga ABC di atas diperoleh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture7.gif"><img alt="" class="alignnone size-full wp-image-159" src="http://learnwithalice.files.wordpress.com/2011/10/picture7.gif?w=500" title="Picture7" /></a><br />
Sehingga, secara umum:<br />
<br />
<br />
<br />
<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture8.gif"><img alt="" class="size-full wp-image-160 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture8.gif?w=500" title="Picture8" /></a> <br />
<br />
<h3>Rumus Jumlah dan Selisih Sudut</h3>Dari gambar segitiga ABC berikut:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/segitigaabcd.jpg"><img alt="" class="alignnone size-full wp-image-171" src="http://learnwithalice.files.wordpress.com/2011/10/segitigaabcd.jpg?w=500" title="segitigaABCD" /></a><br />
AD = b.sin α<br />
BD = a.sin β<br />
CD = a.cos β = b.cos α<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture11.gif"><img alt="" class="alignnone size-full wp-image-170" src="http://learnwithalice.files.wordpress.com/2011/10/picture11.gif?w=500" title="Picture1" /></a><br />
Untuk mencari cos(α+β) = sin (90 – (α+β))°<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture21.gif"><img alt="" class="alignnone size-full wp-image-172" src="http://learnwithalice.files.wordpress.com/2011/10/picture21.gif?w=500" title="Picture2" /></a><br />
<br />
Untuk fungsi tangens:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture31.gif"><img alt="" class="alignnone size-full wp-image-173" src="http://learnwithalice.files.wordpress.com/2011/10/picture31.gif?w=500" title="Picture3" /></a><br />
Sehingga, rumus-rumus yang diperoleh adalah:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture41.gif"><img alt="" class="size-full wp-image-174 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture41.gif?w=500" title="Picture4" /></a><br />
<h3> Rumus Sudut Rangkap</h3><a href="http://learnwithalice.files.wordpress.com/2011/10/picture51.gif"><img alt="" class="alignnone size-full wp-image-175" src="http://learnwithalice.files.wordpress.com/2011/10/picture51.gif?w=500" title="Picture5" /></a><br />
Sehingga, rumus-rumus yang diperoleh adalah:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture61.gif"><img alt="" class="size-full wp-image-176 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture61.gif?w=500" title="Picture6" /></a> Penurunan dari rumus cos2α:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture71.gif"><img alt="" class="size-full wp-image-177 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture71.gif?w=500" title="Picture7" /></a><br />
<h3>Rumus Perkalian Fungsi Sinus dan Kosinus</h3>Dari rumus-rumus jumlah dan selisih dua sudut dapat diturunkan rumus-rumus baru sebagai berikut:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture81.gif"><img alt="" class="alignnone size-full wp-image-178" src="http://learnwithalice.files.wordpress.com/2011/10/picture81.gif?w=500" title="Picture8" /></a><br />
Sehingga, rumus-rumus yang diperoleh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture91.gif"><img alt="" class="size-full wp-image-179 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture91.gif?w=500" title="Picture9" /></a><br />
<h3>Rumus Jumlah dan Selisih Fungsi Sinus dan Kosinus</h3>Dari rumus perkalian fungsi sinus dan kosinus dapat diturunkan rumus jumlah dan selisih fungsi sinus dan kosinus.<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture10.gif"><img alt="" class="alignnone size-full wp-image-180" src="http://learnwithalice.files.wordpress.com/2011/10/picture10.gif?w=500" title="Picture10" /></a><br />
Maka akan diperoleh rumus-rumus:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture111.gif"><img alt="" class="size-full wp-image-181 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/10/picture111.gif?w=500" title="Picture11" /></a><br />
<strong>Contoh-contoh soal:</strong><br />
(1) Tanpa menggunakan daftar, buktikan bahwa:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture12.gif"><img alt="" class="alignnone size-full wp-image-182" src="http://learnwithalice.files.wordpress.com/2011/10/picture12.gif?w=500" title="Picture12" /></a><br />
<br />
(2) Buktikan bahwa dalam segitiga ABC berlaku:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/10/picture13.gif"><img alt="" class="alignnone size-full wp-image-183" src="http://learnwithalice.files.wordpress.com/2011/10/picture13.gif?w=500" title="Picture13" /></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com19tag:blogger.com,1999:blog-3157576316325034465.post-12724669944674945942012-03-19T02:25:00.002-07:002012-03-19T02:25:17.221-07:00Irisan KerucutTerdapat 4 macam irisan kerucut: lingkaran, parabola,elips, hiperbola<br />
<h3><a href="http://learnwithalice.files.wordpress.com/2011/11/irisan_kerucut.jpg"><img alt="" class="size-full wp-image-206 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/11/irisan_kerucut.jpg?w=500" title="irisan_kerucut" /></a></h3><h3>Definisi</h3><span style="text-decoration: underline;"><strong>Lingkaran</strong></span><br />
Lingkaran adalah tempat kedudukan titik-titik yang berjarak sama terhadap suatu titik tertentu.<br />
<ul><li>Titik tertentu itu disebut <strong>pusat</strong> lingkaran</li>
<li>Jarak yang sama itu disebut <strong>jari-jari/radius (r)</strong></li>
</ul>Luas lingkaran = π.r<sup>2</sup> (r = jari-jari)<br />
<em><strong>Contoh gambar:</strong></em><br />
Lingkaran dengan pusat (0, 0) dan jari-jari 2<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/lingkaran.jpg"><img alt="" class="size-full wp-image-207 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/lingkaran.jpg?w=500&h=213" title="lingkaran" width="500" /></a><br />
<span style="text-decoration: underline;"><strong>Parabola</strong></span><br />
Parabola adalah tempat kedudukan titik-titik yang berjarak sama terhadap sebuah titik dan sebuah garis tertentu.<br />
<ul><li>Titik itu disebut <strong>fokus/titik api (F)</strong></li>
<li>Garis tertentu itu disebut <strong>garis direktris/garis arah</strong></li>
<li>Garis yang melalui F dan tegak lurus dengan garis arah disebut <strong>sumbu simetri</strong> parabola</li>
<li>Titik potong parabola dengan sumbu simetri disebut <strong>puncak</strong> parabola</li>
<li>Tali busur terpendek yang melalui F disebut <strong>Latus Rectum</strong> → tegak lurus dengan sumbu simetri</li>
</ul><em><strong>Contoh gambar:</strong></em><br />
Parabola horisontal dengan puncak (0,0), fokus (1, 0), dan garis arah x = –1<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/parabola2.jpg"><img alt="" class="size-full wp-image-208 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/parabola2.jpg?w=500&h=213" title="parabola2" width="500" /></a><br />
Parabola vertikal dengan puncak (0,0), fokus (0, 1), dan garis arah y = –1<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/parabola.jpg"><img alt="" class="size-full wp-image-209 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/parabola.jpg?w=500&h=213" title="parabola" width="500" /></a><br />
<span style="text-decoration: underline;"><strong>Elips</strong></span><br />
(1) Elips adalah tempat kedudukan titik-titik yang <span style="text-decoration: underline;">jumlah</span> jaraknya terhadap 2 titik tertentu tetap.<br />
<ul><li>Jumlah jarak itu = 2a (untuk elips horisontal) atau 2b (untuk elips vertikal)</li>
<li>Kedua titik tetap itu disebut <strong>fokus (F)</strong> → jarak antara F<sub>1</sub> dan F<sub>2</sub> adalah 2c</li>
</ul>(2) Elips adalah tempat kedudukan semua titik yang perbandingan jaraknya terhadap sebuah titik dan sebuah garis tetap = e (<strong>eksentrisitet</strong>), dimana 0 < e < 1<br />
<ul><li>Titik itu adalah <strong>fokus (F)</strong>, dan garis itu adalah <strong>garis arah</strong>.</li>
<li>Ruas garis yang melalui kedua fokus dan memotong elips disebut <strong>sumbu mayor</strong></li>
<li><strong>Pusat</strong> elips adalah titik tengah F<sub>1</sub> dan F<sub>2</sub></li>
<li>Ruas garis yang melalui pusat, tegak lurus sumbu mayor dan memotong elips disebut <strong>sumbu minor</strong></li>
</ul>Luas Elips = π.a.b (a = ½ panjang horisontal; b = ½ panjang vertikal)<br />
<em><strong>Contoh gambar:</strong></em><br />
Elips horisontal dengan pusat (0, 0), puncak-puncak (5, 0), (–5, 0), (0, 4), (0, –4), fokus (3, 0), (–3, 0), dan garis arah x = ±25/3<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/elips.jpg"><img alt="" class="size-full wp-image-201 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/elips.jpg?w=500&h=213" title="elips" width="500" /></a><br />
Elips vertikal dengan pusat (0, 0), puncak-puncak (√2, 0), (–√2, 0), (0, 2), (0, –2), fokus (0,√2), (0, –√2), dan garis arah y = ±2√2/3<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/elips2.jpg"><img alt="" class="size-full wp-image-200 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/elips2.jpg?w=500&h=213" title="elips2" width="500" /></a><br />
<span style="text-decoration: underline;"><strong>Hiperbola</strong></span><br />
(1) Hiperbola adalah tempat kedudukan titik-titik yang <span style="text-decoration: underline;">selisih</span> jaraknya terhadap 2 titik tertentu tetap<br />
<ul><li>Selisih jarak itu = 2a (untuk elips horisontal) atau 2b (untuk elips vertikal)</li>
<li>Kedua titik tetap itu disebut <strong>fokus (F)</strong> → jarak antara F<sub>1</sub> dan F<sub>2</sub> adalah 2c</li>
</ul>(2) Hiperbola adalah tempat kedudukan semua titik yang perbandingan jaraknya terhadap sebuah titik dan sebuah garis tetap = e , dimana e > 1<br />
<ul><li>Titik-titik tertentu itu disebut <strong>fokus (F<sub>1</sub> dan F<sub>2</sub>)</strong></li>
<li>Garis yang melalui titik-titik F<sub>1</sub> dan F<sub>2</sub> disebut <strong>sumbu transvers (sumbu utama)/ sumbu nyata</strong></li>
<li>Titik tengah F<sub>1</sub> dan F<sub>2</sub> disebut <strong>pusat hiperbola (P)</strong></li>
<li>Garis yang melalui P dan tegak lurus sumbu transvers disebut <strong>sumbu konjugasi (sumbu sekawan)/ sumbu imajiner</strong></li>
<li>Titik-titik potong hiperbola dan sumbu transvers disebut <strong>puncak hiperbola</strong></li>
<li>Garis yang melalui fokus dan tegak lurus pada sumbu nyata dan memotong hiperbola di 2 titik → ruas garis penghubung kedua titik tersebut = <strong>Latus Rectum</strong></li>
</ul><em><strong>Contoh gambar:</strong></em><br />
Hiperbola horisontal dengan pusat (0, 0), puncak (2, 0), (–2, 0), fokus (√6, 0), (–√6, 0), dan asimtot y = ± ½√2 x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/hiperbola.jpg"><img alt="" class="size-full wp-image-205 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/hiperbola.jpg?w=500&h=213" title="hiperbola" width="500" /></a><br />
Hiperbola vertikal dengan pusat (0, 0), puncak (√2, 0), (–√2, 0), fokus (0, √6), (0, –√6), dan asimtot y = ± ½√2 x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/hiperbola2.jpg"><img alt="" class="size-full wp-image-204 aligncenter" height="213" src="http://learnwithalice.files.wordpress.com/2011/11/hiperbola2.jpg?w=500&h=213" title="hiperbola2" width="500" /></a><br />
<h3>Persamaan</h3><a href="http://learnwithalice.files.wordpress.com/2011/11/tabel_persamaan1.jpg"><img alt="" class="size-full wp-image-213 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/11/tabel_persamaan1.jpg?w=500" title="tabel_persamaan" /></a><br />
<strong>Tips!</strong><br />
Cara membedakan persamaan-persamaan irisan kerucut:<br />
<ul><li>Pada persamaan <strong>Lingkaran</strong>: koefisien x<sup>2</sup> dan y<sup>2</sup> sama</li>
<li>Pada persamaan <strong>Parabola</strong>: hanya salah satu yang bentuknya kuadrat (x<sup>2</sup> saja atau y<sup>2</sup> saja)</li>
<li>Pada persamaan <strong>Elips</strong>: koefisien x<sup>2</sup> dan y<sup>2</sup> bertanda sama (sama-sama positif atau sama-sama negatif)</li>
<li>Pada persamaan <strong>Hiperbola</strong>: koefisien x<sup>2</sup> dan y<sup>2</sup> berbeda tanda (salah satu positif, yang lain negatif)</li>
</ul><em><strong>Contoh:</strong></em><br />
<ul><li><strong>3</strong>x<sup>2</sup> <strong>+ 3</strong>y<sup>2</sup> + 6x + y = 5 → Persamaan Lingkaran</li>
<li>3<strong>x<sup>2</sup></strong> + 3<strong>y</strong> + 6x = 5 → Persamaan Parabola</li>
<li><strong>3</strong>x<sup>2</sup> <strong>+</strong> y<sup>2</sup> + 6x + y = 5 → Persamaan Elips</li>
<li><strong>3</strong>x<sup>2</sup> <strong>– 3</strong>y<sup>2</sup> + 6x + y = 5 → Persamaan Hiperbola</li>
</ul><br />
<h3><strong>Kedudukan Titik terhadap Irisan Kerucut</strong></h3>Cara mencari kedudukan titik terhadap kerucut:<br />
<ol><li>Jadikan ruas kanan pada persamaan irisan kerucut = 0</li>
<li>Masukkan koordinat titik pada persamaan:</li>
</ol><div style="padding-left: 60px;">→ Jika hasil ruas kiri < 0 → titik berada di dalam irisan kerucut</div><div style="padding-left: 60px;">→ Jika hasil ruas kiri = 0 → titik berada tepat pada irisan kerucut tersebut</div><div style="padding-left: 60px;">→ Jika hasil ruas kanan > 0 → titik berada di luar irisan kerucut</div><strong><em>Contoh:</em></strong><br />
Tentukan kedudukan titik (5, –1) terhadap elips dengan persamaan 3x<sup>2</sup> + y<sup>2</sup> + 6x + y = 5<br />
Cara:<br />
3x<sup>2</sup> + y<sup>2</sup> + 6x + y – 5 = 0<br />
Ruas kiri: 3.5<sup>2</sup> + (–1)<sup>2</sup> + 6.5 + (–1) – 5 = 75 + 1 + 30 – 1 – 5 =100<br />
→ 100 > 0, jadi titik (5, –1) berada di luar elips tersebut<br />
<br />
<h3><strong>Kedudukan Garis terhadap Irisan Kerucut</strong></h3>Cara mencari kedudukan garis terhadap irisan kerucut:<br />
<ol><li>Persamaan garis dijadikan persamaan x = … atau y = …</li>
<li>Substitusikan persamaan garis itu pada persamaan irisan kerucut, sehingga menghasilkan suatu persamaan kuadrat.</li>
<li>Hitung nilai Diskriminan (D) dari persamaan kuadrat tersebut (Ingat! D = b<sup>2</sup> – 4.a.c)</li>
</ol><div style="padding-left: 60px;">→ Jika D < 0 → garis berada di luar irisan kerucut</div><div style="padding-left: 60px;">→ Jika D = 0 → garis menyinggung irisan kerucut di 1 titik</div><div style="padding-left: 60px;">→ Jika D > 0 → garis memotong irisan kerucut di 2 titik</div><strong><em>Contoh:</em></strong><br />
Tentukan kedudukan garis x + 2y = 4 terhadap parabola dengan persamaan 3x<sup>2</sup> + 3y + 6x = 5<br />
Cara:<br />
Garis: x = 4 – 2y<br />
3(4 – 2y)<sup>2</sup> + 3y + 6(4 – 2y) – 5 = 0<br />
3(16 – 16y + 4y<sup>2</sup>) + 3y + 24 – 12y – 5 = 0<br />
48 – 48y + 12y<sup>2</sup> + 3y + 24 – 12y – 5 = 0<br />
12y<sup>2</sup> – 57y + 67 = 0<br />
D = b<sup>2</sup> – 4.a.c = (–57)<sup>2</sup> – 4.12.67 = 33<br />
Karena D > 0 maka garis x + 2y = 4 memotong parabola tersebut<br />
<br />
<h3>Persamaan Garis Singgung</h3><span style="text-decoration: underline;"><strong>Persamaan garis singgung dengan gradien m</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/tabel_pgs.jpg"><img alt="" class="size-full wp-image-199 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/11/tabel_pgs.jpg?w=500" title="tabel_pgs" /></a><br />
<span style="text-decoration: underline;"><strong>Persamaan garis singgung pada titik (x<sub>1</sub>, y<sub>1</sub>)</strong></span><br />
→ selalu gunakan sistem bagi adil:<br />
<div style="padding-left: 30px;">(…)<sup>2</sup> menjadi (…).(…)</div><div style="padding-left: 30px;">(…) menjadi ½ (…) + ½ (…)</div><div style="padding-left: 30px;">Pada salah satu (…) akan dimasukkan koordinat titik yang diketahui</div>→ masukkan titik ke persamaan hasil bagi adil<br />
<ol start="1"><li>Jika titik terletak pada irisan kerucut, akan menghasilkan persamaan garis singgung</li>
<li>Jika titik terletak di luar irisan kerucut, akan menghasilkan persamaan garis polar</li>
</ol>Potongkan garis polar dengan irisan kerucut untuk mendapatkan 2 titik potong<br />
Masukkan kedua titik potong itu ke dalam persamaan hasil bagi adil untuk mendapatkan 2 buah persamaan garis singgung<br />
<br />
<strong><em>Contoh 1:</em></strong><br />
Tentukan persamaan garis singgung pada lingkaran x<sup>2</sup> + y<sup>2</sup> + 4x = 13 pada titik (2, 1)<br />
Cara:<br />
(2, 1) terletak pada lingkaran (2<sup>2</sup> + 1<sup>2</sup> + 4.2 = 13)<br />
Persamaan bagi adil:<br />
x<sub>1</sub>.x + y<sub>1</sub>.y + 2.x<sub>1</sub> + 2.x = 9<br />
Masukkan (2, 1) sebagai x<sub>1</sub> dan y<sub>1</sub>:<br />
2.x + 1.y + 2.2 + 2.x = 9<br />
4x + y – 5 = 0 → persamaan garis singgung<br />
<br />
<strong><em>Contoh 2:</em></strong><br />
Tentukan persamaan garis singgung pada lingkaran x<sup>2</sup> + y<sup>2</sup> + 4x = 13 pada titik (4, 1)<br />
Cara:<br />
(4, 1) terletak di luar lingkaran (4<sup>2</sup> + 1<sup>2</sup> + 4.4 = 33 > 16)<br />
Persamaan bagi adil:<br />
x<sub>1</sub>.x + y<sub>1</sub>.y + 2.x<sub>1</sub> + 2.x = 9<br />
Masukkan (4, 1) sebagai x<sub>1</sub> dan y<sub>1</sub>:<br />
4.x + 1.y + 2.4 + 2.x = 9<br />
6x + y – 1 = 0 → persamaan garis polar<br />
y = 1 – 6x<br />
Substitusikan persamaan garis polar ke dalam persamaan lingkaran:<br />
x<sup>2</sup> + (1 – 6x)<sup>2</sup> + 4x – 13 = 0<br />
x<sup>2</sup> + 1 – 12x + 36x<sup>2</sup> + 4x – 13 = 0<br />
37x<sup>2</sup> – 8x – 12 = 0<br />
Gunakan rumus abc:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/picture1.gif"><img alt="" class="alignnone size-full wp-image-210" src="http://learnwithalice.files.wordpress.com/2011/11/picture1.gif?w=500" title="Picture1" /></a><br />
Masukkan (x<sub>1</sub>, y<sub>1</sub>) dan (x<sub>2</sub>, y<sub>2</sub>) ke dalam persamaan hasil bagi adil<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/picture2.gif"><img alt="" class="alignnone size-full wp-image-211" src="http://learnwithalice.files.wordpress.com/2011/11/picture2.gif?w=500" title="Picture2" /></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com19tag:blogger.com,1999:blog-3157576316325034465.post-59112963547200531052012-03-19T02:03:00.002-07:002012-03-19T02:03:36.696-07:00Suku Banyak<div class="entry"> <h3>Bentuk Umum:</h3>a<sub>n </sub>x<sup>n</sup> + a<sub>n – 1</sub> x<sup>n – 1</sup> + a<sub>n – 2</sub> x<sup>n – 2</sup> + … + … a<sub>2</sub>x<sup>2</sup> + a<sub>1</sub>x + a<sub>0</sub><br />
n = derajat suku banyak<br />
a<sub>0</sub> = konstanta<br />
a<sub>n</sub>, a<sub>n – 1</sub>, a<sub>n – 2</sub>, … = koefisien dari x<sup>n</sup>, x<sup>n – 1</sup>, x<sup>n – 2</sup>, …<br />
<h3>Pembagian Suku Banyak</h3><span style="text-decoration: underline;"><strong>Bentuk Umum</strong></span><br />
<div style="text-align: center;">F(x) = P(x).H(x) + S(x)</div>F(x) = suku banyak<br />
P(x) = pembagi<br />
H(x) = hasil bagi<br />
S(x) = sisa<br />
<span style="text-decoration: underline;"><strong>Teorema Sisa:</strong></span><br />
Jika suatu suku banyak F(x) dibagi oleh (x – k) maka sisanya adalah F(k)<br />
Jika <em><strong>pembagi berderajat n</strong></em> maka <em><strong>sisanya berderajat n – 1</strong></em><br />
Jika <em><strong>suku banyak berderajat m</strong></em> dan pembagi berderajat n, maka <em><strong>hasil baginya berderajat m – n</strong></em><br />
<span style="text-decoration: underline;"><strong>Cara Pembagian Suku Banyak</strong></span><br />
<span style="text-decoration: underline;">Contoh:</span><br />
F(x) = 2x<sup>3</sup> – 3x<sup>2</sup> + x + 5 dibagi dengan P(x) = 2x<sup>2</sup> – x – 1<br />
<strong>1. Pembagian biasa</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/pembagian.jpg"><img alt="" class="alignnone size-full wp-image-220" src="http://learnwithalice.files.wordpress.com/2011/11/pembagian.jpg?w=500" title="pembagian" /></a><br />
Jadi hasil baginya: H(X) = x – 1, sisanya S(x) = x + 4<br />
<strong>2. Cara Horner/Skema</strong><br />
bisa digunakan untuk pembagi berderajat 1 atau pembagi yang dapat difaktorkan menjadi pembagi-pembagi berderajat 1<br />
Cara:<br />
<ul><li>Tulis koefisiennya saja → harus runtut dari koefisien x<sup>n</sup>, x<sup>n – 1</sup>, … hingga konstanta (jika ada variabel yang tidak ada, maka koefisiennya ditulis 0)</li>
</ul><div style="padding-left: 60px;">Contoh: untuk 4x<sup>3</sup> – 1, koefisien-koefisiennya adalah 4, 0, 0, dan -1 (untuk x<sup>3</sup>, x<sup>2</sup>, x, dan konstanta)</div><ul><li>Jika koefisien derajat tertinggi P(x) ≠ 1, maka hasil baginya harus dibagi dengan koefisien derajat tertinggi P(x)</li>
<li>Jika pembagi dapat difaktorkan, maka:</li>
</ul><div style="padding-left: 60px;">Jika pembagi dapat difaktorkan menjadi P<sub>1</sub> dan P<sub>2</sub>, maka S(x) = P<sub>1</sub>.S<sub>2</sub> + S<sub>1</sub></div><div style="padding-left: 60px;">Jika pembagi dapat difaktorkan menjadi P<sub>1</sub>, P<sub>2</sub>, P<sub>3</sub>, maka S(x) = P<sub>1</sub>.P<sub>2</sub>.S<sub>3</sub> + P<sub>1</sub>.S<sub>2</sub> + S<sub>1</sub></div><div style="padding-left: 60px;">Jika pembagi dapat difaktorkan menjadi P<sub>1</sub>, P<sub>2</sub>, P<sub>3</sub>, P<sub>4</sub>, maka S(x) = P<sub>1</sub>.P<sub>2</sub>.P<sub>3</sub>.S<sub>4</sub> + P<sub>1</sub>.P<sub>2</sub>.S<sub>3</sub> + P<sub>1</sub>.S<sub>2</sub> + S<sub>1</sub></div><div style="padding-left: 60px;">dan seterusnya</div>Untuk soal di atas,<br />
P(x) = 2x<sup>2</sup> – x – 1 = (2x + 1)(x – 1)<br />
P<sub>1</sub>: 2x + 1 = 0 → x = –½<br />
P<sub>2</sub>: x – 1 = 0 → x = 1<br />
Cara Hornernya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/horner.jpg"><img alt="" class="alignnone size-full wp-image-222" src="http://learnwithalice.files.wordpress.com/2011/11/horner.jpg?w=500" title="Horner" /></a><br />
H(x) = 1.x – 1 = x – 1<br />
S(x) = P<sub>1</sub>.S<sub>2</sub> + S<sub>1</sub> = (2x + 1).1/2 + 7/2 = x + ½ + 7/2 = x + 4<br />
<strong>3. Cara koefisien tak tentu</strong><br />
F(x) = P(x).H(x) + S(x)<br />
Untuk soal di atas, karena F(x) berderajat 3 dan P(x) berderajat 2, maka<br />
H(x) berderajat 3 – 2 = 1<br />
S(x) berderajat 2 – 1 = 1<br />
Jadi, misalkan H(x) = ax + b dan S(x) = cx + d<br />
Maka:<br />
2x<sup>3</sup> – 3x<sup>2</sup> + x + 5 = (2x<sup>2</sup> – x – 1).(ax + b) + (cx + d)<br />
Ruas kanan:<br />
= 2ax<sup>3</sup> + 2bx<sup>2</sup> – ax<sup>2</sup> – bx – ax – b + cx + d<br />
= 2ax<sup>3</sup> + (2b – a)x<sup>2</sup> + (–b – a + c)x + (–b + d)<br />
Samakan koefisien ruas kiri dan ruas kanan:<br />
x<sup>3</sup> → 2 = 2a → a = 2/2 = 1<br />
x<sup>2</sup> → –3 = 2b – a → 2b = –3 + a = –3 + 1 = –2 → b = –2/2 = –1<br />
x → 1 = –b – a + c → c = 1 + b + a = 1 – 1 + 1 → c = 1<br />
Konstanta → 5 = –b + d → d = 5 + b = 5 – 1 → d = 4<br />
Jadi:<br />
H(x) = ax + b = 1.x – 1 = x – 1<br />
S(x) = cx + d = 1.x + 4 = x + 4<br />
<h3>Teorema Faktor</h3>Suatu suku banyak F(x) mempunyai faktor (x – k) jika F(k) = 0 (sisanya jika dibagi dengan (x – k) adalah 0)<br />
Catatan: jika (x – k) adalah faktor dari F(x) maka k dikatakan sebagai akar dari F(x)<br />
<strong>Tips:</strong><br />
<ol><li>Untuk mencari akar suatu suku banyak dengan cara Horner, dapat dilakukan dengan mencoba-coba dengan angka dari faktor-faktor konstantanya ang akan memberikan sisa = 0</li>
<li>Jika jumlah koefisien suku banyak = 0, maka pasti salah satu akarnya adalah x = 1</li>
<li>Jika jumlah koefisien suku di posisi genap = jumlah koefisien suku di posisi ganjil, maka pasti salah satu akarnya adalah x = –1</li>
</ol><span style="text-decoration: underline;">Contoh:</span><br />
Tentukan penyelesaian dari x<sup>3</sup> – 2x<sup>2</sup> – x + 2 = 0<br />
Faktor-faktor dari konstantanya, yaitu 2, adalah ±1 dan ±2<br />
Karena jumlah seluruh koefisien + konstantanya = 0 (1 – 2 – 1 + 2 = 0), maka, pasti x = 1 adalah salah satu faktornya, jadi:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/11/horner2.jpg"><img alt="" class="alignnone size-full wp-image-221" src="http://learnwithalice.files.wordpress.com/2011/11/horner2.jpg?w=500" title="horner2" /></a><br />
Jadi x<sup>3</sup> – 2x<sup>2</sup> – x + 2 = (x – 1)(x<sup>2</sup> – x – 2)<br />
= (x – 1)(x – 2)(x + 1)<br />
x = 1 x = 2 x = –1<br />
Jadi himpunan penyelesaiannya: {–1, 1, 2}<br />
<h3>Sifat Akar-Akar Suku Banyak</h3>Pada persamaan berderajat 3:<br />
ax<sup>3</sup> + bx<sup>2</sup> + cx + d = 0 akan mempunyai akar-akar x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub><br />
dengan sifat-sifat:<br />
<ul><li>Jumlah 1 akar: x<sub>1</sub> + x<sub>2</sub> + x<sub>3</sub> = – b/a</li>
<li>Jumlah 2 akar: x<sub>1</sub>.x<sub>2</sub> + x<sub>1</sub>.x<sub>3</sub> + x<sub>2</sub>.x<sub>3</sub> = c/a</li>
<li>Hasil kali 3 akar: x<sub>1</sub>.x<sub>2</sub>.x<sub>3</sub> = – d/a</li>
</ul>Pada persamaan berderajat 4:<br />
ax<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e = 0 akan mempunyai akar-akar x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub><br />
dengan sifat-sifat:<br />
<ul><li>Jumlah 1 akar: x<sub>1</sub> + x<sub>2</sub> + x<sub>3</sub> + x<sub>4 </sub>= – b/a</li>
<li>Jumlah 2 akar: x<sub>1</sub>.x<sub>2</sub> + x<sub>1</sub>.x<sub>3</sub> + x<sub>1</sub>.x<sub>4</sub> + x<sub>2</sub>.x<sub>3</sub> + x<sub>2</sub>.x<sub>4</sub> + x<sub>3</sub>.x<sub>4</sub> = c/a</li>
<li>Jumlah 3 akar: x<sub>1</sub>.x<sub>2</sub>.x<sub>3</sub> + x<sub>1</sub>.x<sub>2</sub>.x<sub>4</sub> + x<sub>2</sub>.x<sub>3</sub>.x<sub>4</sub> = – d/a</li>
<li>Hasil kali 4 akar: x<sub>1</sub>.x<sub>2</sub>.x<sub>3</sub>.x<sub>4</sub> = e/a</li>
</ul>Dari kedua persamaan tersebut, kita dapat menurunkan rumus yang sama untuk persamaan berderajat 5 dan seterusnya<br />
(amati pola: –b/a, c/a, –d/a , e/a, …)<br />
<h3>Pembagian Istimewa</h3><a href="http://learnwithalice.files.wordpress.com/2011/11/picture11.gif"><img alt="" class="size-full wp-image-223 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/11/picture11.gif?w=500" title="Picture1" /></a><br />
</div>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com6tag:blogger.com,1999:blog-3157576316325034465.post-55662540957484863412012-03-19T02:02:00.002-07:002012-03-19T02:02:53.585-07:00Fungsi<h3><strong>Pasangan terurut</strong></h3><em>Contoh:</em><br />
A = {1, 2, 3}, B = {4, 5}<br />
Himpunan semua pasangan terurut dari A dan B adalah:<br />
{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}<br />
<br />
<strong>Relasi</strong><br />
Relasi adalah himpunan dari pasangan terurut ang memenuhi aturan tertentu<br />
<em>Contoh:</em><br />
A = {1, 2, 3, 4}, B = {2, 4}<br />
Jika ada relasi R dari A ke B dengan aturan ”faktor dari”, maka himpunan pasangan terurut untuk relasi tersebut adalah:<br />
R = {(1, 2), (1, 4), (2, 2), (2, 4), (4, 4)}<br />
Diagram panahnya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_1.jpg"><img alt="" class="alignnone size-full wp-image-229" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_1.jpg?w=500" title="diagram_panah_1" /></a><br />
<br />
<strong>Fungsi</strong><br />
Fungsi dari A ke B adalah relasi yang memasangkan <span style="text-decoration: underline;">setiap anggota himpunan A</span> ke <span style="text-decoration: underline;">hanya satu anggota himpunan B</span><br />
Notasi fungsi f dari A ke B ditulis f : A → B<br />
A disebut <strong>domain (daerah asal)</strong><br />
B disebut <strong>kodomain (daerah kawan)</strong><br />
Himpunan bagian dari B yang merupakan hasil dari fungsi A ke B disebut <strong>range (daerah hasil)</strong><br />
Fungsi juga dapat dinyatakan dengan lambang f : x → y = f(x)<br />
dimana y = f(x) adalah rumus fungsi dengan x sebagai variabel bebas dan y sebagai variabel terikat (tak bebas)<br />
Contoh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_2.jpg"><img alt="" class="size-full wp-image-230 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_2.jpg?w=500" title="diagram_panah_2" /></a>Untuk fungsi yang digambarkan dalam diagram panah di atas:<br />
Domain = D<sub>f</sub> = {1, 2, 3, 4}<br />
Range = R<sub>f</sub> = {2, 4}<br />
<h3>Menentukan Daerah Asal Fungsi</h3>Agar suatu fungsi terdefinisi (mempunyai daerah hasil di himpunan bilangan real), maka ada beberapa syarat yang harus dipenuhi.<br />
1. Fungsi di dalam akar<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture1.gif"><img alt="" class="alignnone size-full wp-image-234" src="http://learnwithalice.files.wordpress.com/2011/12/picture1.gif?w=500" title="Picture1" /></a></div>2. Fungsi pecahan<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture2.gif"><img alt="" class="alignnone size-full wp-image-235" src="http://learnwithalice.files.wordpress.com/2011/12/picture2.gif?w=500" title="Picture2" /></a></div>3. Fungsi dimana penyebutnya adalah fungsi lain dalam bentuk akar<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture3.gif"><img alt="" class="alignnone size-full wp-image-236" src="http://learnwithalice.files.wordpress.com/2011/12/picture3.gif?w=500" title="Picture3" /></a></div>4. Fungsi logaritma<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture4.gif"><img alt="" class="alignnone size-full wp-image-237" src="http://learnwithalice.files.wordpress.com/2011/12/picture4.gif?w=500" title="Picture4" /></a></div><em>Contoh:</em><br />
Daerah asal untuk fungsi<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture5.gif"><img alt="" class="alignnone size-full wp-image-238" src="http://learnwithalice.files.wordpress.com/2011/12/picture5.gif?w=500" title="Picture5" /></a><br />
adalah:<br />
x<sup>2</sup> + 3x – 4 > 0<br />
(x + 4)(x – 1) > 0<br />
Pembuat nol: x = –4 dan x = 1<br />
Jika x = 0 maka hasilnya 0<sup>2</sup> + 3.0 – 4 = –4 (negatif)<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/garis_bil.jpg"><img alt="" class="alignnone size-full wp-image-233" src="http://learnwithalice.files.wordpress.com/2011/12/garis_bil.jpg?w=500" title="garis_bil" /></a><br />
Jadi D<sub>f</sub> = {x | x < –4 atau x > 1}<br />
<h3>Aljabar Fungsi</h3>Jika f : x → f(x) dan g : x → g(x) maka:<br />
<ol><li>(f + g)(x) = f(x) + g(x)</li>
<li>(f – g)(x) = f(x) – g(x)</li>
<li>(f × g)(x) = f(x) × g(x)</li>
<li><a href="http://learnwithalice.files.wordpress.com/2011/12/picture6.gif"><img alt="" class="alignnone size-full wp-image-239" src="http://learnwithalice.files.wordpress.com/2011/12/picture6.gif?w=500" title="Picture6" /></a></li>
</ol>Daerah asalnya:<br />
D<sub>f+g</sub>, D<sub>f–g</sub>, D<sub>f</sub>×<sub>g</sub> = D<sub>f</sub> ∩ D<sub>g</sub> (irisan dari D<sub>f</sub> dan D<sub>g</sub>)<br />
D<sub>f/g</sub> = D<sub>f</sub> ∩ D<sub>g</sub> dan g(x) ≠ 0<br />
<h3>Komposisi fungsi</h3><strong>Notasi:</strong><br />
f komposisi g dapat dinyatakan dengan f o g (dapat juga dibaca ”f bundaran g”)<br />
(f o g)(x) = f(g(x)) (g dimasukkan ke f)<br />
<strong>Ilustrasi:</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_gof.jpg"><img alt="" class="size-full wp-image-231 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_gof.jpg?w=500" title="diagram_panah_gof" /></a> <em>Contoh:</em> f(1) = 2, g(2) = 0, maka (g o f )(1) = g(f(1)) = g(2) = 0<br />
<strong>Sifat-Sifat Komposisi Fungsi</strong><br />
1. Tidak bersifat komutatif<br />
<div style="padding-left: 30px;">(f o g)(x) ≠ (g o f)(x)</div>2. Asosiatif<br />
<div style="padding-left: 30px;">(f o (g o h))(x) = ((f o g) o h)(x)</div>3. Terdapat fungsi identitas I(x) = x<br />
<div style="padding-left: 30px;">(f o I)(x) = (I o f)(x) = f(x)</div><br />
<em>Contoh 1:</em><br />
f(x) = 3x + 2<br />
g(x) = 2x + 5<br />
h(x) = x<sup>2</sup> – 1<br />
Cari (f o g)(x), (g o f)(x), dan (f o g o h)(x)!<br />
(f o g)(x) = f(g(x)) = f(2x + 5)<br />
= 3(2x + 5) + 2<br />
= 6x + 15 + 2 = 6x + 17<br />
(g o f)(x) = g(f(x)) = g(3x + 2)<br />
= 2(3x + 2) + 5<br />
= 6x + 4 + 5 = 6x + 9<br />
(f o g o h)(x) = f(g(h(x))) = f(g(x<sup>2</sup> – 1))<br />
= f(2(x<sup>2</sup> – 1) + 5)<br />
= f(2x<sup>2</sup> – 2 + 5)<br />
= f(2x<sup>2</sup> + 3)<br />
= 3(2x<sup>2</sup> + 3) + 2<br />
= 6x<sup>2</sup> + 9 + 2 = 6x<sup>2</sup> + 11<br />
atau dengan menggunakan rumus (f o g)(x) yang sudah diperoleh sebelumnya,<br />
(f o g o h)(x) = (f o g)(h(x)) = (f o g)(x<sup>2</sup> – 1)<br />
= 6(x<sup>2</sup> – 1) + 17<br />
= 6x<sup>2</sup> – 6 + 17<br />
= 6x<sup>2</sup> + 11<br />
<br />
<em>Contoh 2:</em><br />
f(x) = 3x + 2<br />
(f o g)(x) = 6x + 17<br />
Cari g(x)!<br />
(f (g(x)) = 6x + 17<br />
3.g(x) + 2 = 6x + 17<br />
3.g(x) = 6x + 17 – 2<br />
3.g(x) = 6x + 15<br />
g(x) = 2x + 5<br />
<br />
<em>Contoh 3:</em><br />
g(x) = 2x + 5<br />
(f o g)(x) = 6x + 17<br />
Cari f(x)!<br />
f(2x + 5) = 6x + 17<br />
misalkan: 2x + 5 = a → 2x = a – 5<br />
f(a) = 3(a – 5) + 17<br />
f(a) = 3a – 15 + 17<br />
f(a) = 3a + 2<br />
f(x) = 3x + 2<br />
<br />
<em>Contoh 4:</em><br />
f(x) = x<sup>2</sup> + 2x + 5<br />
(f o g)(x) = 4x<sup>2</sup> – 8x + 8<br />
Cari g(x)!<br />
f(g(x)) = 4x<sup>2</sup> – 8x + 8<br />
(g(x))<sup>2</sup> + 2g(x) + 5 = 4x<sup>2</sup> – 8x + 8<br />
Gunakan cara melengkapkan kuadrat sempurna<br />
(g(x) + 1)<sup>2</sup> – 1 + 5 = 4x<sup>2</sup> – 8x + 8<br />
(g(x) + 1)<sup>2</sup> = 4x<sup>2</sup> – 8x + 8 – 4<br />
(g(x) + 1)<sup>2</sup> = 4x<sup>2</sup> – 8x + 4<br />
(g(x) + 1)<sup>2</sup> = (2x – 2)<sup>2</sup><br />
g(x) + 1 = 2x – 2 atau g(x) + 1 = –(2x – 2)<br />
g(x) = 2x – 3 atau g(x) = –2x + 3<br />
atau<br />
f(g(x)) = 4x<sup>2</sup> – 8x + 8<br />
(g(x))<sup>2</sup> + 2g(x) + 5 = 4x<sup>2</sup> – 8x + 8<br />
Karena pangkat tertinggi di ruas kanan = 2, maka misalkan g(x) = ax + b<br />
(ax + b)<sup>2</sup> + 2(ax + b) + 5 = 4x<sup>2</sup> – 8x + 8<br />
a<sup>2</sup>x<sup>2</sup> + 2abx + b<sup>2</sup> + 2ax + 2ab + 5 = 4x<sup>2</sup> – 8x + 8<br />
a<sup>2</sup>x<sup>2</sup> + (2ab + 2a)x + (b<sup>2</sup> + 2ab + 5) = 4x<sup>2</sup> – 8x + 8<br />
Samakan koefisien x<sup>2</sup> di ruas kiri dan kanan:<br />
a<sup>2</sup> = 4 → a = 2 atau a = –2<br />
samakan koefisien x di ruas kiri dan kanan:<br />
untuk a = 2 → 2ab + 2a = –8<br />
4b + 4 = –8<br />
4b = –12 → b = –3<br />
untuk a = –2 → 2ab + 2a = –8<br />
–4b + 4 = –8<br />
–4b = –12 → b = 3<br />
Jadi g(x) = 2x – 3 atau g(x) = –2x + 3<br />
<br />
<h3>Invers Fungsi</h3><strong>Notasi</strong><br />
Invers dari fungsi f(x) dilambangkan dengan f<sup>–1</sup> (x)<br />
<br />
<strong>Ilustrasi</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_invers.jpg"><img alt="" class="size-full wp-image-232 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_invers.jpg?w=500" title="diagram_panah_invers" /></a><em>Contoh:</em> Jika f(2) = 1 maka f<sup>–1</sup>(1) =2<br />
Jika digambar dalam koordinat cartesius, grafik invers fungsi merupakan pencerminan dari grafik fungsinya terhadap garis y = x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/cartesius.jpg"><img alt="" class="size-full wp-image-228 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/cartesius.jpg?w=500" title="cartesius" /></a><strong>Sifat-Sifat Invers Fungsi:</strong><br />
<ol><li>(f<sup>–1</sup>)<sup>–1</sup>(x) = f(x)</li>
<li>(f o f<sup>–1</sup>)(x) = (f<sup>–1</sup> o f)(x) = I(x) = x, I = fungsi identitas</li>
<li>(f o g)<sup>–1</sup>(x) = (g<sup>–1</sup> o f<sup>–1</sup>)(x)</li>
</ol><span style="text-decoration: underline;">Ingat:</span> (f o g<sup>–1</sup>)(x) ¹ (f o g)<sup>–1</sup>(x)<br />
<br />
<strong>Mencari invers fungsi</strong><br />
<ol><li>Nyatakan persamaan fungsinya y = f(x)</li>
<li>Carilah x dalam y, namai persamaan ini dengan x = f<sup>–1</sup>(y)</li>
<li>Ganti x dengan y dan y dengan x, sehingga menjadi y = f<sup>–1</sup>(x), yang merupakan invers fungsi dari f</li>
</ol><em>Contoh 1:</em><br />
f(x) = 3x – 2<br />
invers fungsinya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture7.gif"><img alt="" class="alignnone size-full wp-image-240" src="http://learnwithalice.files.wordpress.com/2011/12/picture7.gif?w=500" title="Picture7" /></a><br />
<br />
<em>Contoh 2:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture8.gif"><img alt="" class="alignnone size-full wp-image-241" src="http://learnwithalice.files.wordpress.com/2011/12/picture8.gif?w=500" title="Picture8" /></a><br />
<span style="text-decoration: underline;">Cara Cepat!</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture9.gif"><img alt="" class="alignnone size-full wp-image-242" src="http://learnwithalice.files.wordpress.com/2011/12/picture9.gif?w=500" title="Picture9" /></a><br />
<br />
<em>Contoh 3:</em><br />
f(x) = x<sup>2</sup> – 3x + 4<br />
Invers fungsinya<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture10.gif"><img alt="" class="alignnone size-full wp-image-243" src="http://learnwithalice.files.wordpress.com/2011/12/picture10.gif?w=500" title="Picture10" /></a><br />
<br />
<em>Contoh 4:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture11.gif"><img alt="" class="alignnone size-full wp-image-244" src="http://learnwithalice.files.wordpress.com/2011/12/picture11.gif?w=500" title="Picture11" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture12.gif"><img alt="" class="alignnone size-full wp-image-245" src="http://learnwithalice.files.wordpress.com/2011/12/picture12.gif?w=500" title="Picture12" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture13.gif"><img alt="" class="alignnone size-full wp-image-227" src="http://learnwithalice.files.wordpress.com/2011/12/picture13.gif?w=500" title="Picture13" /></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com24tag:blogger.com,1999:blog-3157576316325034465.post-2274901339308928252012-03-19T02:00:00.000-07:002012-03-19T02:00:12.587-07:00Limit Fungsi<div class="post-248 post type-post status-publish format-standard hentry category-matematika-indonesia tag-limit-fungsi tag-rangkuman-matematika"> <div class="entry"> Pengertian tentang limit dapat diperoleh dengan melihat contoh berikut ini.<br />
Contoh: Perhatikan fungsi<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture14.gif"><img alt="" class="alignnone size-full wp-image-254" src="http://learnwithalice.files.wordpress.com/2011/12/picture14.gif?w=500" title="Picture1" /></a><br />
untuk nilai x yang mendekati 1<br />
<table border="1" cellpadding="0" cellspacing="0"><tbody>
<tr> <td valign="top" width="38">x</td> <td valign="top" width="44">0</td> <td valign="top" width="47">0,9</td> <td valign="top" width="57">0,95</td> <td valign="top" width="57">0,98</td> <td valign="top" width="38">…</td> <td valign="top" width="66">1,0001</td> <td valign="top" width="66">1,0005</td> <td valign="top" width="47">1,05</td> <td valign="top" width="38">1,1</td> </tr>
<tr> <td valign="top" width="38">f(x)</td> <td valign="top" width="44">1</td> <td valign="top" width="47">1,9</td> <td valign="top" width="57">1,95</td> <td valign="top" width="57">1,98</td> <td valign="top" width="38">…</td> <td valign="top" width="66">2,0001</td> <td valign="top" width="66">2,0005</td> <td valign="top" width="47">2,05</td> <td valign="top" width="38">2,1</td> </tr>
</tbody> </table>Gambar grafiknya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/grafik.jpg"><img alt="" class="alignnone size-full wp-image-251" src="http://learnwithalice.files.wordpress.com/2011/12/grafik.jpg?w=500" title="grafik" /></a><br />
Dari gambar dan tabel dapat disimpulkan:<br />
→ Jika x mendekati 1 dari kiri, maka nilai f(x) mendekati 2<br />
<div style="padding-left: 60px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture21.gif"><img alt="" class="alignnone size-full wp-image-255" src="http://learnwithalice.files.wordpress.com/2011/12/picture21.gif?w=500" title="Picture2" /></a></div>→ Jika x mendekati 1 dari kanan, maka nilai f(x) mendekati 2<br />
<div style="padding-left: 60px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture31.gif"><img alt="" class="alignnone size-full wp-image-256" src="http://learnwithalice.files.wordpress.com/2011/12/picture31.gif?w=500" title="Picture3" /></a></div>→ Jadi, jika x mendekati 1, maka nilai f(x) mendekati 2<br />
<div style="padding-left: 60px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture41.gif"><img alt="" class="alignnone size-full wp-image-257" src="http://learnwithalice.files.wordpress.com/2011/12/picture41.gif?w=500" title="Picture4" /></a></div><span style="text-decoration: underline;"><strong>Teorema:</strong></span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture51.gif"><img alt="" class="alignnone size-full wp-image-258" src="http://learnwithalice.files.wordpress.com/2011/12/picture51.gif?w=500" title="Picture5" /></a><br />
Jika limit kiri dan limit kanan tidak sama, maka nilai limitnya tidak ada<br />
<br />
Hasil limit tidak boleh bentuk tak tentu:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture61.gif"><img alt="" class="alignnone size-full wp-image-259" src="http://learnwithalice.files.wordpress.com/2011/12/picture61.gif?w=500" title="Picture6" /></a><br />
<br />
<h3>Sifat-Sifat Limit<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/sifat.jpg"><img alt="" class="alignnone size-full wp-image-274" src="http://learnwithalice.files.wordpress.com/2011/12/sifat.jpg?w=500" title="sifat" /></a></h3><br />
<h3><strong>Cara Penyelesaian Limit dengan Perhitungan:</strong></h3><strong>1. </strong><strong><span style="text-decoration: underline;">Substitusi langsung</span></strong><br />
<em>Contoh:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture71.gif"><img alt="" class="alignnone size-full wp-image-260" src="http://learnwithalice.files.wordpress.com/2011/12/picture71.gif?w=500" title="Picture7" /></a><br />
<br />
<strong>2. </strong><strong><span style="text-decoration: underline;">Pemfaktoran (biasanya untuk bentuk 0/0)</span></strong><br />
<em> Contoh:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture81.gif"><img alt="" class="alignnone size-full wp-image-261" src="http://learnwithalice.files.wordpress.com/2011/12/picture81.gif?w=500" title="Picture8" /></a><br />
Ingat:<br />
<div style="padding-left: 60px;">(a<sup>2</sup> – b<sup>2</sup>) = (a – b)(a + b)</div><div style="padding-left: 60px;">(a<sup>3</sup> + b<sup>3</sup>) = (a + b)(a<sup>2</sup> – ab + b<sup>2</sup>)</div><div style="padding-left: 60px;">(a<sup>3</sup> – b<sup>3</sup>) = (a – b)(a<sup>2</sup> + ab + b<sup>2</sup>)</div><div style="padding-left: 60px;"> </div> <strong>3. </strong><strong><span style="text-decoration: underline;">Dikali sekawan (jika ada bentuk akar)</span></strong><br />
Contoh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture91.gif"><img alt="" class="alignnone size-full wp-image-262" height="42" src="http://learnwithalice.files.wordpress.com/2011/12/picture91.gif?w=500&h=42" title="Picture9" width="500" /></a><br />
<br />
<strong>4. </strong><strong><span style="text-decoration: underline;">Untuk limit tak terhingga: </span></strong><br />
→ Jika bentuknya sudah pecahan: dibagi pangkat tertinggi<br />
→ Jika bentuknya belum pecahan: dikali sekawan, baru dibagi pangkat tertinggi<br />
Sifat operasi dengan ∞:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/takhingga.jpg"><img alt="" class="alignnone size-full wp-image-275" src="http://learnwithalice.files.wordpress.com/2011/12/takhingga.jpg?w=500" title="takhingga" /></a><br />
<em>Contoh:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture101.gif"><img alt="" class="alignnone size-full wp-image-263" src="http://learnwithalice.files.wordpress.com/2011/12/picture101.gif?w=500" title="Picture10" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture111.gif"><img alt="" class="alignnone size-full wp-image-264" height="192" src="http://learnwithalice.files.wordpress.com/2011/12/picture111.gif?w=500&h=192" title="Picture11" width="500" /></a><br />
<strong>Cara cepat!</strong><br />
→ Untuk bentuk pecahan:<br />
<ul><li>Jika pangkat pembilang (atas) > penyebut (bawah), hasil =∞</li>
<li>Jika pangkat pembilang (atas) < penyebut (bawah), hasil =0</li>
<li>Jika pangkat pembilang (atas) = penyebut (bawah), hasil =koefisien pangkat tertinggi atas : koefisien pangkat tertinggi bawah</li>
</ul><em>Contoh 1:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture121.gif"><img alt="" class="alignnone size-full wp-image-265" src="http://learnwithalice.files.wordpress.com/2011/12/picture121.gif?w=500" title="Picture12" /></a><br />
<em>Contoh 2:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture131.gif"><img alt="" class="alignnone size-full wp-image-266" src="http://learnwithalice.files.wordpress.com/2011/12/picture131.gif?w=500" title="Picture13" /></a><br />
<em>Contoh 3</em>:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture141.gif"><img alt="" class="alignnone size-full wp-image-267" src="http://learnwithalice.files.wordpress.com/2011/12/picture141.gif?w=500" title="Picture14" /></a><br />
<br />
→ Untuk bentuk <a href="http://learnwithalice.files.wordpress.com/2011/12/picture15.gif"><img alt="" class="alignnone size-full wp-image-268" src="http://learnwithalice.files.wordpress.com/2011/12/picture15.gif?w=500" title="Picture15" /></a><br />
<em>Contoh:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture16.gif"><img alt="" class="alignnone size-full wp-image-269" height="69" src="http://learnwithalice.files.wordpress.com/2011/12/picture16.gif?w=500&h=69" title="Picture16" width="500" /></a><br />
<strong></strong><br />
<strong>5. </strong><strong><span style="text-decoration: underline;">Limit trigonometri:</span></strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/trigono.jpg"><img alt="" class="alignnone size-full wp-image-249" src="http://learnwithalice.files.wordpress.com/2011/12/trigono.jpg?w=500" title="trigono" /></a><br />
Untuk cosinus:<br />
1 – cos ax = 2 sin<sup>2 </sup>½ ax (dari rumus cos 2x)<br />
cos ax – 1 = –2 sin<sup>2</sup> ½ ax (dari rumus cos 2x)<br />
1 – cos<sup>2</sup>ax = sin<sup>2</sup>ax (dari sin<sup>2</sup>x + cos<sup>2</sup>x = 1)<br />
<br />
<h3>Bilangan e</h3>Bilangan e didapat dari:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture17.gif"><img alt="" class="alignnone size-full wp-image-270" src="http://learnwithalice.files.wordpress.com/2011/12/picture17.gif?w=500" title="Picture17" /></a><br />
e = 2,718281828…<br />
<br />
Rumus-rumus pengembangannya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture18.gif"><img alt="" class="alignnone size-full wp-image-271" src="http://learnwithalice.files.wordpress.com/2011/12/picture18.gif?w=500" title="Picture18" /></a><br />
<br />
<h3><strong>Kontinuitas</strong></h3>Suatu fungsi kontinu di x = a jika:<br />
1. f(a) ada (dapat dihitung/real)<br />
2. <a href="http://learnwithalice.files.wordpress.com/2011/12/picture19.gif"><img alt="" class="alignnone size-full wp-image-272" src="http://learnwithalice.files.wordpress.com/2011/12/picture19.gif?w=500" title="Picture19" /></a><br />
3. <a href="http://learnwithalice.files.wordpress.com/2011/12/picture20.gif"><img alt="" class="alignnone size-full wp-image-273" src="http://learnwithalice.files.wordpress.com/2011/12/picture20.gif?w=500" title="Picture20" /></a><br />
<br />
<strong>Ilustrasi:</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/kontinu.jpg"><img alt="" class="size-full wp-image-252 alignleft" src="http://learnwithalice.files.wordpress.com/2011/12/kontinu.jpg?w=500" title="kontinu" /></a><br />
<br />
<br />
</div><div class="postinfo"> <div class="com"><a href="http://learnwithalice.wordpress.com/2011/12/20/limit-fungsi/#respond" title="Comment on Limit Fungsi">Leave a Comment</a></div>Posted in <a href="http://learnwithalice.wordpress.com/category/matematika-indonesia/" rel="category tag" title="View all posts in Matematika (Indonesia)">Matematika (Indonesia)</a> | Tags: <a href="http://learnwithalice.wordpress.com/tag/limit-fungsi/" rel="tag">Limit Fungsi</a>, <a href="http://learnwithalice.wordpress.com/tag/rangkuman-matematika/" rel="tag">Rangkuman Matematika</a><br />
</div></div><div class="title"> <small>Posted by: <strong>alicealc</strong> | December 18, 2011 </small> <h2><a href="http://learnwithalice.wordpress.com/2011/12/18/fungsi/" rel="bookmark">Fungsi</a></h2></div><h3>Pengertian</h3><strong>Pasangan terurut</strong><br />
<em>Contoh:</em><br />
A = {1, 2, 3}, B = {4, 5}<br />
Himpunan semua pasangan terurut dari A dan B adalah:<br />
{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}<br />
<br />
<strong>Relasi</strong><br />
Relasi adalah himpunan dari pasangan terurut ang memenuhi aturan tertentu<br />
<em>Contoh:</em><br />
A = {1, 2, 3, 4}, B = {2, 4}<br />
Jika ada relasi R dari A ke B dengan aturan ”faktor dari”, maka himpunan pasangan terurut untuk relasi tersebut adalah:<br />
R = {(1, 2), (1, 4), (2, 2), (2, 4), (4, 4)}<br />
Diagram panahnya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_1.jpg"><img alt="" class="alignnone size-full wp-image-229" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_1.jpg?w=500" title="diagram_panah_1" /></a><br />
<br />
<strong>Fungsi</strong><br />
Fungsi dari A ke B adalah relasi yang memasangkan <span style="text-decoration: underline;">setiap anggota himpunan A</span> ke <span style="text-decoration: underline;">hanya satu anggota himpunan B</span><br />
Notasi fungsi f dari A ke B ditulis f : A → B<br />
A disebut <strong>domain (daerah asal)</strong><br />
B disebut <strong>kodomain (daerah kawan)</strong><br />
Himpunan bagian dari B yang merupakan hasil dari fungsi A ke B disebut <strong>range (daerah hasil)</strong><br />
Fungsi juga dapat dinyatakan dengan lambang f : x → y = f(x)<br />
dimana y = f(x) adalah rumus fungsi dengan x sebagai variabel bebas dan y sebagai variabel terikat (tak bebas)<br />
Contoh:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_2.jpg"><img alt="" class="size-full wp-image-230 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_2.jpg?w=500" title="diagram_panah_2" /></a>Untuk fungsi yang digambarkan dalam diagram panah di atas:<br />
Domain = D<sub>f</sub> = {1, 2, 3, 4}<br />
Range = R<sub>f</sub> = {2, 4}<br />
<h3>Menentukan Daerah Asal Fungsi</h3>Agar suatu fungsi terdefinisi (mempunyai daerah hasil di himpunan bilangan real), maka ada beberapa syarat yang harus dipenuhi.<br />
1. Fungsi di dalam akar<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture1.gif"><img alt="" class="alignnone size-full wp-image-234" src="http://learnwithalice.files.wordpress.com/2011/12/picture1.gif?w=500" title="Picture1" /></a></div>2. Fungsi pecahan<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture2.gif"><img alt="" class="alignnone size-full wp-image-235" src="http://learnwithalice.files.wordpress.com/2011/12/picture2.gif?w=500" title="Picture2" /></a></div>3. Fungsi dimana penyebutnya adalah fungsi lain dalam bentuk akar<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture3.gif"><img alt="" class="alignnone size-full wp-image-236" src="http://learnwithalice.files.wordpress.com/2011/12/picture3.gif?w=500" title="Picture3" /></a></div>4. Fungsi logaritma<br />
<div style="padding-left: 30px;"><a href="http://learnwithalice.files.wordpress.com/2011/12/picture4.gif"><img alt="" class="alignnone size-full wp-image-237" src="http://learnwithalice.files.wordpress.com/2011/12/picture4.gif?w=500" title="Picture4" /></a></div><em>Contoh:</em><br />
Daerah asal untuk fungsi<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture5.gif"><img alt="" class="alignnone size-full wp-image-238" src="http://learnwithalice.files.wordpress.com/2011/12/picture5.gif?w=500" title="Picture5" /></a><br />
adalah:<br />
x<sup>2</sup> + 3x – 4 > 0<br />
(x + 4)(x – 1) > 0<br />
Pembuat nol: x = –4 dan x = 1<br />
Jika x = 0 maka hasilnya 0<sup>2</sup> + 3.0 – 4 = –4 (negatif)<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/garis_bil.jpg"><img alt="" class="alignnone size-full wp-image-233" src="http://learnwithalice.files.wordpress.com/2011/12/garis_bil.jpg?w=500" title="garis_bil" /></a><br />
Jadi D<sub>f</sub> = {x | x < –4 atau x > 1}<br />
<h3>Aljabar Fungsi</h3>Jika f : x → f(x) dan g : x → g(x) maka:<br />
<ol><li>(f + g)(x) = f(x) + g(x)</li>
<li>(f – g)(x) = f(x) – g(x)</li>
<li>(f × g)(x) = f(x) × g(x)</li>
<li><a href="http://learnwithalice.files.wordpress.com/2011/12/picture6.gif"><img alt="" class="alignnone size-full wp-image-239" src="http://learnwithalice.files.wordpress.com/2011/12/picture6.gif?w=500" title="Picture6" /></a></li>
</ol>Daerah asalnya:<br />
D<sub>f+g</sub>, D<sub>f–g</sub>, D<sub>f</sub>×<sub>g</sub> = D<sub>f</sub> ∩ D<sub>g</sub> (irisan dari D<sub>f</sub> dan D<sub>g</sub>)<br />
D<sub>f/g</sub> = D<sub>f</sub> ∩ D<sub>g</sub> dan g(x) ≠ 0<br />
<h3>Komposisi fungsi</h3><strong>Notasi:</strong><br />
f komposisi g dapat dinyatakan dengan f o g (dapat juga dibaca ”f bundaran g”)<br />
(f o g)(x) = f(g(x)) (g dimasukkan ke f)<br />
<strong>Ilustrasi:</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_gof.jpg"><img alt="" class="size-full wp-image-231 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_gof.jpg?w=500" title="diagram_panah_gof" /></a> <em>Contoh:</em> f(1) = 2, g(2) = 0, maka (g o f )(1) = g(f(1)) = g(2) = 0<br />
<strong>Sifat-Sifat Komposisi Fungsi</strong><br />
1. Tidak bersifat komutatif<br />
<div style="padding-left: 30px;">(f o g)(x) ≠ (g o f)(x)</div>2. Asosiatif<br />
<div style="padding-left: 30px;">(f o (g o h))(x) = ((f o g) o h)(x)</div>3. Terdapat fungsi identitas I(x) = x<br />
<div style="padding-left: 30px;">(f o I)(x) = (I o f)(x) = f(x)</div><br />
<em>Contoh 1:</em><br />
f(x) = 3x + 2<br />
g(x) = 2x + 5<br />
h(x) = x<sup>2</sup> – 1<br />
Cari (f o g)(x), (g o f)(x), dan (f o g o h)(x)!<br />
(f o g)(x) = f(g(x)) = f(2x + 5)<br />
= 3(2x + 5) + 2<br />
= 6x + 15 + 2 = 6x + 17<br />
(g o f)(x) = g(f(x)) = g(3x + 2)<br />
= 2(3x + 2) + 5<br />
= 6x + 4 + 5 = 6x + 9<br />
(f o g o h)(x) = f(g(h(x))) = f(g(x<sup>2</sup> – 1))<br />
= f(2(x<sup>2</sup> – 1) + 5)<br />
= f(2x<sup>2</sup> – 2 + 5)<br />
= f(2x<sup>2</sup> + 3)<br />
= 3(2x<sup>2</sup> + 3) + 2<br />
= 6x<sup>2</sup> + 9 + 2 = 6x<sup>2</sup> + 11<br />
atau dengan menggunakan rumus (f o g)(x) yang sudah diperoleh sebelumnya,<br />
(f o g o h)(x) = (f o g)(h(x)) = (f o g)(x<sup>2</sup> – 1)<br />
= 6(x<sup>2</sup> – 1) + 17<br />
= 6x<sup>2</sup> – 6 + 17<br />
= 6x<sup>2</sup> + 11<br />
<br />
<em>Contoh 2:</em><br />
f(x) = 3x + 2<br />
(f o g)(x) = 6x + 17<br />
Cari g(x)!<br />
(f (g(x)) = 6x + 17<br />
3.g(x) + 2 = 6x + 17<br />
3.g(x) = 6x + 17 – 2<br />
3.g(x) = 6x + 15<br />
g(x) = 2x + 5<br />
<br />
<em>Contoh 3:</em><br />
g(x) = 2x + 5<br />
(f o g)(x) = 6x + 17<br />
Cari f(x)!<br />
f(2x + 5) = 6x + 17<br />
misalkan: 2x + 5 = a → 2x = a – 5<br />
f(a) = 3(a – 5) + 17<br />
f(a) = 3a – 15 + 17<br />
f(a) = 3a + 2<br />
f(x) = 3x + 2<br />
<br />
<em>Contoh 4:</em><br />
f(x) = x<sup>2</sup> + 2x + 5<br />
(f o g)(x) = 4x<sup>2</sup> – 8x + 8<br />
Cari g(x)!<br />
f(g(x)) = 4x<sup>2</sup> – 8x + 8<br />
(g(x))<sup>2</sup> + 2g(x) + 5 = 4x<sup>2</sup> – 8x + 8<br />
Gunakan cara melengkapkan kuadrat sempurna<br />
(g(x) + 1)<sup>2</sup> – 1 + 5 = 4x<sup>2</sup> – 8x + 8<br />
(g(x) + 1)<sup>2</sup> = 4x<sup>2</sup> – 8x + 8 – 4<br />
(g(x) + 1)<sup>2</sup> = 4x<sup>2</sup> – 8x + 4<br />
(g(x) + 1)<sup>2</sup> = (2x – 2)<sup>2</sup><br />
g(x) + 1 = 2x – 2 atau g(x) + 1 = –(2x – 2)<br />
g(x) = 2x – 3 atau g(x) = –2x + 3<br />
atau<br />
f(g(x)) = 4x<sup>2</sup> – 8x + 8<br />
(g(x))<sup>2</sup> + 2g(x) + 5 = 4x<sup>2</sup> – 8x + 8<br />
Karena pangkat tertinggi di ruas kanan = 2, maka misalkan g(x) = ax + b<br />
(ax + b)<sup>2</sup> + 2(ax + b) + 5 = 4x<sup>2</sup> – 8x + 8<br />
a<sup>2</sup>x<sup>2</sup> + 2abx + b<sup>2</sup> + 2ax + 2ab + 5 = 4x<sup>2</sup> – 8x + 8<br />
a<sup>2</sup>x<sup>2</sup> + (2ab + 2a)x + (b<sup>2</sup> + 2ab + 5) = 4x<sup>2</sup> – 8x + 8<br />
Samakan koefisien x<sup>2</sup> di ruas kiri dan kanan:<br />
a<sup>2</sup> = 4 → a = 2 atau a = –2<br />
samakan koefisien x di ruas kiri dan kanan:<br />
untuk a = 2 → 2ab + 2a = –8<br />
4b + 4 = –8<br />
4b = –12 → b = –3<br />
untuk a = –2 → 2ab + 2a = –8<br />
–4b + 4 = –8<br />
–4b = –12 → b = 3<br />
Jadi g(x) = 2x – 3 atau g(x) = –2x + 3<br />
<br />
<h3>Invers Fungsi</h3><strong>Notasi</strong><br />
Invers dari fungsi f(x) dilambangkan dengan f<sup>–1</sup> (x)<br />
<br />
<strong>Ilustrasi</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_invers.jpg"><img alt="" class="size-full wp-image-232 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/diagram_panah_invers.jpg?w=500" title="diagram_panah_invers" /></a><em>Contoh:</em> Jika f(2) = 1 maka f<sup>–1</sup>(1) =2<br />
Jika digambar dalam koordinat cartesius, grafik invers fungsi merupakan pencerminan dari grafik fungsinya terhadap garis y = x<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/cartesius.jpg"><img alt="" class="size-full wp-image-228 aligncenter" src="http://learnwithalice.files.wordpress.com/2011/12/cartesius.jpg?w=500" title="cartesius" /></a><strong>Sifat-Sifat Invers Fungsi:</strong><br />
<ol><li>(f<sup>–1</sup>)<sup>–1</sup>(x) = f(x)</li>
<li>(f o f<sup>–1</sup>)(x) = (f<sup>–1</sup> o f)(x) = I(x) = x, I = fungsi identitas</li>
<li>(f o g)<sup>–1</sup>(x) = (g<sup>–1</sup> o f<sup>–1</sup>)(x)</li>
</ol><span style="text-decoration: underline;">Ingat:</span> (f o g<sup>–1</sup>)(x) ¹ (f o g)<sup>–1</sup>(x)<br />
<br />
<strong>Mencari invers fungsi</strong><br />
<ol><li>Nyatakan persamaan fungsinya y = f(x)</li>
<li>Carilah x dalam y, namai persamaan ini dengan x = f<sup>–1</sup>(y)</li>
<li>Ganti x dengan y dan y dengan x, sehingga menjadi y = f<sup>–1</sup>(x), yang merupakan invers fungsi dari f</li>
</ol><em>Contoh 1:</em><br />
f(x) = 3x – 2<br />
invers fungsinya:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture7.gif"><img alt="" class="alignnone size-full wp-image-240" src="http://learnwithalice.files.wordpress.com/2011/12/picture7.gif?w=500" title="Picture7" /></a><br />
<br />
<em>Contoh 2:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture8.gif"><img alt="" class="alignnone size-full wp-image-241" src="http://learnwithalice.files.wordpress.com/2011/12/picture8.gif?w=500" title="Picture8" /></a><br />
<span style="text-decoration: underline;">Cara Cepat!</span><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture9.gif"><img alt="" class="alignnone size-full wp-image-242" src="http://learnwithalice.files.wordpress.com/2011/12/picture9.gif?w=500" title="Picture9" /></a><br />
<br />
<em>Contoh 3:</em><br />
f(x) = x<sup>2</sup> – 3x + 4<br />
Invers fungsinya<br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture10.gif"><img alt="" class="alignnone size-full wp-image-243" src="http://learnwithalice.files.wordpress.com/2011/12/picture10.gif?w=500" title="Picture10" /></a><br />
<br />
<em>Contoh 4:</em><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture11.gif"><img alt="" class="alignnone size-full wp-image-244" src="http://learnwithalice.files.wordpress.com/2011/12/picture11.gif?w=500" title="Picture11" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture12.gif"><img alt="" class="alignnone size-full wp-image-245" src="http://learnwithalice.files.wordpress.com/2011/12/picture12.gif?w=500" title="Picture12" /></a><br />
<a href="http://learnwithalice.files.wordpress.com/2011/12/picture13.gif"><img alt="" class="alignnone size-full wp-image-227" src="http://learnwithalice.files.wordpress.com/2011/12/picture13.gif?w=500" title="Picture13" /></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com25tag:blogger.com,1999:blog-3157576316325034465.post-18376558575986314162012-03-19T01:54:00.000-07:002012-03-19T01:54:53.976-07:00Sifat Akar Persamaan Kuadrat<div class="content"><h3>Sifat akar-akar persamaan kuadrat</h3>Jika x<sub>1</sub> dan x<sub>2</sub> adalah akar-akar persamaan kuadrat ax<sup>2</sup> + bx + c = 0, maka:<br />
<div style="text-align: center;"><strong>x<sub>1</sub> + x<sub>2</sub> = –b/a</strong></div><div style="text-align: center;"><strong>x<sub>1</sub>.x<sub>2</sub> = c/a</strong></div><div style="text-align: center;"><strong>|x<sub>1</sub> – x<sub>2</sub>| = –D/a</strong></div><div style="text-align: center;">(Ingat! D = b<sup>2</sup> – 4.a.c)</div><h3>Bentuk simetri akar-akar persamaan kuadrat</h3><strong>Jumlah kuadrat akar-akar:</strong><br />
x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup> = (x<sub>1</sub> + x<sub>2</sub>)<sup>2</sup> – 2.x<sub>1</sub>.x<sub>2</sub><br />
<strong>Jumlah pangkat tiga akar-akar:</strong><br />
x<sub>1</sub><sup>3</sup> + x<sub>2</sub><sup>3</sup> = (x<sub>1</sub> + x<sub>2</sub>)<sup>3</sup> – 3.x<sub>1</sub>.x<sub>2</sub>.(x<sub>1</sub> + x<sub>2</sub>)<br />
<strong>Jumlah pangkat empat akar-akar:</strong><br />
x<sub>1</sub><sup>4</sup> + x<sub>2</sub><sup>4</sup> = (x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup>)<sup>2</sup> – 2.x<sub>1</sub><sup>2</sup>.x<sub>2</sub><sup>2</sup><br />
<strong>Jumlah kebalikan akar-akar:</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/06/picture39.gif"><img alt="" class="alignnone size-full wp-image-83" src="http://learnwithalice.files.wordpress.com/2011/06/picture39.gif?w=500" title="Picture3" /></a><br />
<strong>Jumlah kuadrat kebalikan akar-akar:</strong><br />
<a href="http://learnwithalice.files.wordpress.com/2011/06/picture41.gif"><img alt="" class="alignnone size-full wp-image-82" src="http://learnwithalice.files.wordpress.com/2011/06/picture41.gif?w=500" title="Picture4" /></a><br />
<strong>Selisih kuadrat akar-akar:</strong><br />
x<sub>1</sub><sup>2</sup> – x<sub>2</sub><sup>2</sup> = (x<sub>1</sub> + x<sub>2</sub>).(x<sub>1</sub> – x<sub>2</sub>) dimana x<sub>1</sub> > x<sub>2</sub><br />
<h3>Hubungan Jenis Akar-akar PK dengan Nilai Diskriminan (D)</h3><ul><li>Jika D > 0 maka PK mempunyai 2 akar real yang berlainan</li>
</ul><div style="padding-left: 60px;">→ D = bilangan kuadrat berarti akar-akarnya rasional</div><div style="padding-left: 60px;">→ D bukan bilangan kuadrat berarti akar-akarnya irasional</div><ul><li>Jika D = 0 maka PK m,empunyai 1 akar real atau akar-akarnya kembar</li>
<li>Jika D ≥ 0 maka PK mempunyai 2 akar real/nyata</li>
<li>Jika D < 0 maka PK tidak mempuyai akar real / akar-akarnya imajiner</li>
<li>Jika kedua akar positif (x<sub>1</sub> > 0, x<sub>2</sub> > 0)</li>
</ul><div style="padding-left: 60px;">D ≥ 0</div><div style="padding-left: 60px;">x<sub>1</sub> + x<sub>2</sub> > 0</div><div style="padding-left: 60px;">x<sub>1</sub>.x<sub>2</sub> > 0</div><ul><li>Jika kedua akar negatif (x<sub>1</sub> < 0 dan x<sub>2</sub> < 0)</li>
</ul><div style="padding-left: 60px;">D ≥ 0</div><div style="padding-left: 60px;">x<sub>1</sub> + x<sub>2</sub> < 0</div><div style="padding-left: 60px;">x<sub>1</sub>.x<sub>2</sub> > 0</div><ul><li>Jika kedua akar berlainan tanda (1 positif, 1 negatif)</li>
</ul><div style="padding-left: 60px;">D > 0</div><div style="padding-left: 60px;">x<sub>1</sub>.x<sub>2</sub> < 0</div><ul><li>Jika kedua akar bertanda sama (sama-sama positif/sama-sama negatif)</li>
</ul><div style="padding-left: 60px;">D ≥ 0</div><div style="padding-left: 60px;">x<sub>1</sub>.x<sub>2</sub> > 0</div><ul><li>Jika kedua akar saling berlawanan (x<sub>1</sub> = –x<sub>2</sub>)</li>
</ul><div style="padding-left: 60px;">D > 0</div><div style="padding-left: 60px;">b = 0 (diperoleh dari x<sub>1</sub> + x<sub>2</sub> = 0)</div><div style="padding-left: 60px;">x<sub>1</sub>.x<sub>2</sub> < 0</div><ul><li>Jika kedua akar saling berkebalikan (x<sub>1</sub> = 1/x<sub>2</sub>)</li>
</ul><div style="padding-left: 60px;">D > 0</div><div style="padding-left: 60px;">c = a</div><span style="text-decoration: underline;">Contoh 1:</span><br />
Tentukan nilai m agar x<sup>2</sup> + 4x + (m – 4) = 0 mempunyai 2 akar real<br />
D ≥ 0<br />
b<sup>2</sup> – 4ac ≥ 0<br />
4<sup>2</sup> – 4.1.(m – 4) ≥ 0<br />
16 – 4m + 16 ≥ 0<br />
–4m ≥ –16 – 16<br />
Semua dibagi –4<br />
<em><strong>(Ingat! Jika dibagi atau dikali bilangan negatif tanda pertidaksamaan dibalik)</strong></em><br />
m ≤ 4 + 4<br />
m ≤ 8<br />
<br />
<br />
<br />
<br />
<br />
<br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
Tentukan nilai n agar akar-akar PK x<sup>2</sup> + (2n + 2)x + 5 – n = 0 bertanda sama<br />
<em>Syarat 1</em><br />
D ≥ 0<br />
b<sup>2</sup> – 4ac ≥ 0<br />
(2n + 2)<sup>2</sup> – 4.1.(5 – n) ≥ 0<br />
4n<sup>2</sup> + 8n + 4 – 20 + 4n ≥ 0<br />
4n<sup>2</sup> + 12n – 16 ≥ 0<br />
Semua dibagi 4:<br />
n<sup>2</sup> + 3n – 4 ³ 0<br />
(n + 4).(n – 1) ³ 0<br />
Pembuat nol: n = –4 atau n = 1<br />
<em>Syarat 2:</em><br />
x<sub>1</sub>.x<sub>2</sub> > 0<br />
<a href="http://learnwithalice.files.wordpress.com/2011/06/picture51.gif"><img alt="" class="alignnone size-full wp-image-85" src="http://learnwithalice.files.wordpress.com/2011/06/picture51.gif?w=500" title="Picture5" /></a><br />
Gambar garis bilangan:<br />
<a href="http://learnwithalice.files.wordpress.com/2011/06/garis_bil.jpg"><img alt="" class="alignnone size-full wp-image-84" src="http://learnwithalice.files.wordpress.com/2011/06/garis_bil.jpg?w=500" title="garis_bil" /></a><br />
Jadi: HP = {n | n ≤ –4 atau 1 ≤ n < 5}<br />
<h3>Menyusun PK</h3>PK dengan akar-akar x<sub>1</sub> dan x<sub>2</sub> adalah:<br />
<div style="text-align: center;"><strong>x<sup>2</sup> – (x<sub>1</sub> + x<sub>2</sub>)x + (x<sub>1</sub>.x<sub>2</sub>) = 0</strong></div>dengan kata lain:<br />
<div style="text-align: center;"><strong>x<sup>2</sup> – (jumlah akar-akar)x + (hasil kali akar-akar) = 0</strong></div><span style="text-decoration: underline;">Contoh 1:</span><br />
Tentukan PK yang mempunyai akar-akar 2 dan –5:<br />
x<sup>2</sup> – (2 + (–5))x + (2.(–5)) = 0<br />
x<sup>2</sup> + 3x – 10 = 0<br />
<br />
<span style="text-decoration: underline;">Contoh 2:</span><br />
Jika x<sub>1</sub> dan x<sub>2</sub> adalah akar-akar PK: x<sup>2</sup> – 3x – 1 = 0, susun PK baru yang akar-akarnya 3x<sub>1</sub> + 2 dan 3x<sub>2</sub> + 2!<br />
Karena PK tersebut tidak dapat difaktorkan,<br />
x<sub>1</sub> + x<sub>2</sub> = –b/a = –(– 3) /1 = 3<br />
x<sub>1</sub>.x<sub>2</sub> = c/a = –1/1 = –1<br />
Misal akar-akar PK baru adalah y<sub>1</sub> dan y<sub>2</sub>:<br />
y<sub>1</sub> + y<sub>2</sub> = 3.x<sub>1</sub> + 2 + 3.x<sub>2</sub> + 2<br />
= 3(x<sub>1</sub> + x<sub>2</sub>) + 4 = 9 + 4 = 13<br />
y<sub>1</sub>.y<sub>2</sub> = (3x<sub>1</sub> + 2).(3x<sub>2</sub> + 2)<br />
= 9.x<sub>1</sub>.x<sub>2</sub> + 6.x<sub>1</sub> + 6.x<sub>2</sub> + 4<br />
= 9.(–1) + 6.3 + 4 = –9 + 18 + 4 = 13<br />
Jadi PK barunya:<br />
x<sup>2</sup> – (y<sub>1</sub> + y<sub>2</sub>)x + (y<sub>1</sub>.y<sub>2</sub>) = 0<br />
x<sup>2</sup> – 13x + 13 = 0<br />
<b> </b></div><div class="content"><b>Soal</b> </div><div class="content">Tentukan nilai k agar persamaan² kuadrat berikut memiliki akar kembar <br />
<br />
a. x²-2x+k=0<br />
b. 2x²-4x+k=0<br />
c. kx²-6x+1/2=0<br />
d. 3x²-kx+5=0<br />
e. 2kx²+3x+2=0</div><br />
<b>Jawab</b> <br />
suatu persamaan kuadrat akan memiliki akar kembar jika D = 0<br />
D = b² - 4ac<br />
<br />
1.] x² - 2x + k = 0<br />
D = 0<br />
4 - 4 . 1 . k = 0<br />
4 - 4k = 0<br />
4k = 4<br />
k = 1<br />
<br />
2.] 2x² - 4x + k = 0<br />
D = 0<br />
16 - 4 . 2 . k = 0<br />
16 - 8k = 0<br />
8k = 16<br />
k = 2<br />
<br />
3.] kx² - 6x + 1/2 = 0<br />
36 - 4 . k . 1/2 = 0<br />
36 - 2k = 0<br />
2k = 36<br />
k = 18<br />
<br />
4.] 3x² - kx + 5 = 0<br />
D = 0<br />
k² - 4 . 3 . 5 = 0<br />
k² - 60 = 0<br />
k = ± √60<br />
<br />
5.] 2kx² + 3x + 2 = 0<br />
D = 0<br />
9 - 4 . 2k . 2 = 0<br />
9 - 16k = 0<br />
16k = 9<br />
k = 9/16Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com26tag:blogger.com,1999:blog-3157576316325034465.post-87623059449018600162012-03-19T01:46:00.000-07:002012-03-19T01:46:39.384-07:00Turunan dan Integral<span style="font-size: small;"><strong>Turunan fungsi f ‘ (x) didefinisikan sebagai</strong> :</span><br />
<br />
<div style="padding-left: 60px;"><img alt="f' (x) = \underset{h\rightarrow 0}{lim}\:\frac{f(x + h) - f(x)}{h}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%20%28x%29%20%3D%20%5Cunderset%7Bh%5Crightarrow%200%7D%7Blim%7D%5C%3A%5Cfrac%7Bf%28x%20%2B%20h%29%20-%20f%28x%29%7D%7Bh%7D&bg=ffffff&fg=000000&s=2" title="f' (x) = \underset{h\rightarrow 0}{lim}\:\frac{f(x + h) - f(x)}{h}" /></div><br />
<span style="font-size: small;"><strong>Rumus-rumus Turunan</strong> :</span><br />
untuk <em>a</em> = konstanta<br />
<ul><li><img alt="f(x) = ax^n" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20ax%5En&bg=ffffff&fg=000000&s=0" title="f(x) = ax^n" /> maka <img alt="f'(x) = an.x^{n-1}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20an.x%5E%7Bn-1%7D&bg=ffffff&fg=000000&s=1" title="f'(x) = an.x^{n-1}" /></li>
<li><img alt="f(x) = a" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20a&bg=ffffff&fg=000000&s=0" title="f(x) = a" /> maka <img alt="f'(x) = 0" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%200&bg=ffffff&fg=000000&s=1" title="f'(x) = 0" /></li>
<li><img alt="f(x) = x" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20x&bg=ffffff&fg=000000&s=0" title="f(x) = x" /> maka <img alt="f'(x) = 1" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%201&bg=ffffff&fg=000000&s=1" title="f'(x) = 1" /></li>
</ul><span id="more-22"></span><br />
jika U = u(x) dan V = v(x) adalah suatu fungsi<br />
<ul><li><img alt="f(x) = U + V" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20U%20%2B%20V&bg=ffffff&fg=000000&s=0" title="f(x) = U + V" /> maka <img alt="f'(x) = U' + V'" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20U%27%20%2B%20V%27&bg=ffffff&fg=000000&s=0" title="f'(x) = U' + V'" /></li>
<li><img alt="f(x) = U - V" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20U%20-%20V&bg=ffffff&fg=000000&s=0" title="f(x) = U - V" /> maka <img alt="f'(x) = U' - V'" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20U%27%20-%20V%27&bg=ffffff&fg=000000&s=0" title="f'(x) = U' - V'" /></li>
<li><img alt="f(x) = U\times V" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20U%5Ctimes%20V&bg=ffffff&fg=000000&s=0" title="f(x) = U\times V" /> maka <img alt="f'(x) = U'.V + V'.U" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20U%27.V%20%2B%20V%27.U&bg=ffffff&fg=000000&s=0" title="f'(x) = U'.V + V'.U" /></li>
<li><img align="absmiddle" alt="f(x) = \frac UV" src="http://latex.codecogs.com/gif.latex?f%28x%29&space;=&space;%5Cfrac&space;UV" /> maka <img align="absmiddle" alt="f'(x) = \frac {U'.V - V'.U}{V^2}" src="http://latex.codecogs.com/gif.latex?f%27%28x%29&space;=&space;%5Cfrac&space;%7BU%27.V&space;-&space;V%27.U%7D%7BV%5E2%7D" /></li>
<li><img alt="f(x) = U^n" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20U%5En&bg=ffffff&fg=000000&s=0" title="f(x) = U^n" /> maka <img alt="f'(x) = n.U^{n-1}.U'" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20n.U%5E%7Bn-1%7D.U%27&bg=ffffff&fg=000000&s=0" title="f'(x) = n.U^{n-1}.U'" /> dinamakan <span style="color: red;">aturan rantai</span></li>
</ul><br />
Jangan sampai lupa yah, setiap fungsi yang hendak diturunkan, pastikan dinyatakan dalam bentuk perpangkatan terlebih dulu, let’s cekidot …<br />
Contoh dan pembahasan turunan fungsi:<br />
<br />
Tentukan turunan pertama dari :<br />
<ol><li> <img alt="f(x) = 2x^5" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%202x%5E5&bg=ffffff&fg=000000&s=0" title="f(x) = 2x^5" /><br />
Jawab : <div style="padding-left: 30px;"><img alt="\begin{array}{rcl} f'(x) & = & 2.5.x^{5-1}\\ & = & 10x^4\end{array}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D%20f%27%28x%29%20%26%20%3D%20%26%202.5.x%5E%7B5-1%7D%5C%5C%20%26%20%3D%20%26%2010x%5E4%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=1" title="\begin{array}{rcl} f'(x) & = & 2.5.x^{5-1}\\ & = & 10x^4\end{array}" /></div><div style="padding-left: 30px;"><br />
</div></li>
<li> <img align="absmiddle" alt="f(x) = \frac 3x" src="http://latex.codecogs.com/gif.latex?f%28x%29&space;=&space;%5Cfrac&space;3x" /><br />
Jawab : <div style="padding-left: 30px;">* nyatakan dalam bentuk pangkat terlebih dulu menjadi <img alt="f(x) = 3.x^{-1}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%203.x%5E%7B-1%7D&bg=ffffff&fg=000000&s=0" title="f(x) = 3.x^{-1}" /></div><div style="padding-left: 30px;">* maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 3.(-1).x^{-1-1}\\ & = & (-3).x^{-2}\\ & = & -\frac{3}{x^2}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;3.%28-1%29.x%5E%7B-1-1%7D%5C%5C&space;&&space;=&space;&&space;%28-3%29.x%5E%7B-2%7D%5C%5C&space;&&space;=&space;&&space;-%5Cfrac%7B3%7D%7Bx%5E2%7D%5Cend%7Balign*%7D" /></div></li>
<li> <img alt="f(x) = \sqrt{7x}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20%5Csqrt%7B7x%7D&bg=ffffff&fg=000000&s=0" title="f(x) = \sqrt{7x}" /><br />
Jawab : <div style="padding-left: 30px;">* nyatakan dalam bentuk pangkat terlebih dulu menjadi <img align="absmiddle" alt="f(x) = \sqrt7 .\;x^{\frac{1}{2}}" src="http://latex.codecogs.com/gif.latex?f%28x%29&space;=&space;%5Csqrt7&space;.%5C;x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D" /></div><div style="padding-left: 30px;">* maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & \sqrt 7. \frac 12.x^{\frac 12-1}\\ & = & \frac 12. \sqrt 7.x^{-\frac{1}{2}}\\ & = & \frac 12. \sqrt 7. \frac{1}{\sqrt x}\\ & = & \frac{\sqrt 7}{2\sqrt x}.\frac{\sqrt x}{\sqrt x}\\ & = & \frac{\sqrt{7x}}{2x}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;%5Csqrt&space;7.&space;%5Cfrac&space;12.x%5E%7B%5Cfrac&space;12-1%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;12.&space;%5Csqrt&space;7.x%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;12.&space;%5Csqrt&space;7.&space;%5Cfrac%7B1%7D%7B%5Csqrt&space;x%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B%5Csqrt&space;7%7D%7B2%5Csqrt&space;x%7D.%5Cfrac%7B%5Csqrt&space;x%7D%7B%5Csqrt&space;x%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B%5Csqrt%7B7x%7D%7D%7B2x%7D%5Cend%7Balign*%7D" /></div></li>
<li> <img align="absmiddle" alt="f(x) = \frac{3x-2}{x+1}" src="http://latex.codecogs.com/gif.latex?f%28x%29&space;=&space;%5Cfrac%7B3x-2%7D%7Bx+1%7D" /><br />
Jawab : <div style="padding-left: 30px;">* kita misalkan <img alt="\begin{array}{rcl} U=3x-2 & maka & U'=3\\ V=x+1 & maka & V'=1\end{array}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D%20U%3D3x-2%20%26%20maka%20%26%20U%27%3D3%5C%5C%20V%3Dx%2B1%20%26%20maka%20%26%20V%27%3D1%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0" title="\begin{array}{rcl} U=3x-2 & maka & U'=3\\ V=x+1 & maka & V'=1\end{array}" /></div><div style="padding-left: 30px;">* maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & \frac{U'.V-V'.U}{V^2}\\ & = & \frac{(3)(x+1)-(1)(3x-2)}{(x+1)^2}\\ & = & \frac{3x+3-3x+2}{(x+1)^2}\\ & = & \frac{5}{(x+1)^2}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;%5Cfrac%7BU%27.V-V%27.U%7D%7BV%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B%283%29%28x+1%29-%281%29%283x-2%29%7D%7B%28x+1%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B3x+3-3x+2%7D%7B%28x+1%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B5%7D%7B%28x+1%29%5E2%7D%5Cend%7Balign*%7D" /></div></li>
<li> <img alt="f(x) = (3x^2 -5)^4" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%20%3D%20%283x%5E2%20-5%29%5E4&bg=ffffff&fg=000000&s=0" title="f(x) = (3x^2 -5)^4" /><br />
Jawab : <div style="padding-left: 30px;">* kita misalkan <img alt="U = 3x^2 -5\: maka U'=6x" class="latex" src="http://s.wordpress.com/latex.php?latex=U%20%3D%203x%5E2%20-5%5C%3A%20maka%20U%27%3D6x&bg=ffffff&fg=000000&s=0" title="U = 3x^2 -5\: maka U'=6x" /> dan <img alt="n = 4" class="latex" src="http://s.wordpress.com/latex.php?latex=n%20%3D%204&bg=ffffff&fg=000000&s=0" title="n = 4" /></div><div style="padding-left: 30px;">* lalu kita pakai <img alt="f'(x) = n.U^{n-1}.U'" class="latex" src="http://s.wordpress.com/latex.php?latex=f%27%28x%29%20%3D%20n.U%5E%7Bn-1%7D.U%27&bg=ffffff&fg=000000&s=0" title="f'(x) = n.U^{n-1}.U'" /> ( aturan rantai )</div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 4.(3x^2-5)^{4-1}.6x\\ & = & 24x(3x^2-5)^3\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;4.%283x%5E2-5%29%5E%7B4-1%7D.6x%5C%5C&space;&&space;=&space;&&space;24x%283x%5E2-5%29%5E3%5Cend%7Balign*%7D" /></div></li>
</ol><br />
<br />
Soal2<br />
1. Fungsi f ditentukan oleh <img alt="f(x)=\frac{x^2+8x+12}{x+4}\: ;\:x\neq-4" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%5Cfrac%7Bx%5E2%2B8x%2B12%7D%7Bx%2B4%7D%5C%3A%20%3B%5C%3Ax%5Cneq-4&bg=ffffff&fg=000000&s=1" title="f(x)=\frac{x^2+8x+12}{x+4}\: ;\:x\neq-4" /> dan f ‘ adalah turunan pertama dari f. Maka nilai dari f ‘(1) = ….<br />
<div style="padding-left: 30px;">a. <img alt="x^2+2x+3" class="latex" src="http://s.wordpress.com/latex.php?latex=x%5E2%2B2x%2B3&bg=ffffff&fg=000000&s=1" title="x^2+2x+3" /></div><div style="padding-left: 30px;">b. <img alt="\frac 12x^2+2x-3" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%2012x%5E2%2B2x-3&bg=ffffff&fg=000000&s=1" title="\frac 12x^2+2x-3" /></div><div style="padding-left: 30px;">c. <img alt="\frac 12x^2+2x+3" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%2012x%5E2%2B2x%2B3&bg=ffffff&fg=000000&s=1" title="\frac 12x^2+2x+3" /></div><div style="padding-left: 30px;">d. <img alt="\frac 14x^2+2x-3" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%2014x%5E2%2B2x-3&bg=ffffff&fg=000000&s=1" title="\frac 14x^2+2x-3" /></div><div style="padding-left: 30px;">e. <img alt="\frac 12x^2+2x+3" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%2012x%5E2%2B2x%2B3&bg=ffffff&fg=000000&s=1" title="\frac 12x^2+2x+3" /></div>jawab:<br />
<div style="padding-left: 30px;"><img alt="f(x)=\frac{x^2+8x+12}{x+4}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%5Cfrac%7Bx%5E2%2B8x%2B12%7D%7Bx%2B4%7D&bg=ffffff&fg=000000&s=1" title="f(x)=\frac{x^2+8x+12}{x+4}" /></div><div style="padding-left: 30px;"><img alt="\begin{array}{rcl} misalkan:u=x^2+8x+12 & maka & u'=2x+8 \\ v=x+4 & maka & v'=1 \end{array}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D%20misalkan%3Au%3Dx%5E2%2B8x%2B12%20%26%20maka%20%26%20u%27%3D2x%2B8%20%5C%5C%20v%3Dx%2B4%20%26%20maka%20%26%20v%27%3D1%20%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0" title="\begin{array}{rcl} misalkan:u=x^2+8x+12 & maka & u'=2x+8 \\ v=x+4 & maka & v'=1 \end{array}" /></div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*} maka f'(x) & = & \frac {u'.v - v'.u}{v^2}\\ & = & \frac {(2x + 8)(x + 4) - (1)(x^2 + 8x + 12)}{(x + 4)^2}\\ & = & \frac {2x^2 + 8x + 8x + 32 - x^2 - 8x - 12}{(x + 4)^2}\\ & = & \frac {x^2 + 8x + 20}{(x + 4)^2}\\ f'(1) & = & \frac {(1)^2+8(1)+20}{(1+4)^2}\\ & = & \frac {29}{25}\:jawaban\;(D)\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7D&space;maka&space;f%27%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;%7Bu%27.v&space;-&space;v%27.u%7D%7Bv%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B%282x&space;+&space;8%29%28x&space;+&space;4%29&space;-&space;%281%29%28x%5E2&space;+&space;8x&space;+&space;12%29%7D%7B%28x&space;+&space;4%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B2x%5E2&space;+&space;8x&space;+&space;8x&space;+&space;32&space;-&space;x%5E2&space;-&space;8x&space;-&space;12%7D%7B%28x&space;+&space;4%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7Bx%5E2&space;+&space;8x&space;+&space;20%7D%7B%28x&space;+&space;4%29%5E2%7D%5C%5C&space;f%27%281%29&space;&&space;=&space;&&space;%5Cfrac&space;%7B%281%29%5E2+8%281%29+20%7D%7B%281+4%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B29%7D%7B25%7D%5C:jawaban%5C;%28D%29%5Cend%7Balign*%7D" />$</div><span id="more-342"></span><br />
2. Turunan pertama fungsi <img alt="f(x)= \frac{4x-3}{-x-1}\:;\:x\neq-1" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%20%5Cfrac%7B4x-3%7D%7B-x-1%7D%5C%3A%3B%5C%3Ax%5Cneq-1&bg=ffffff&fg=000000&s=1" title="f(x)= \frac{4x-3}{-x-1}\:;\:x\neq-1" /> adalah f ‘(x) = ….<br />
<div style="padding-left: 30px;">a. <img alt="\frac{1}{(-x-1)^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%7B1%7D%7B%28-x-1%29%5E2%7D&bg=ffffff&fg=000000&s=1" title="\frac{1}{(-x-1)^2}" /></div><div style="padding-left: 30px;">b. <img alt="\frac{5}{(-x-1)^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%7B5%7D%7B%28-x-1%29%5E2%7D&bg=ffffff&fg=000000&s=1" title="\frac{5}{(-x-1)^2}" /></div><div style="padding-left: 30px;">c. <img alt="\frac{-7}{(-x-1)^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%7B-7%7D%7B%28-x-1%29%5E2%7D&bg=ffffff&fg=000000&s=1" title="\frac{-7}{(-x-1)^2}" /></div><div style="padding-left: 30px;">d. <img alt="\frac{1}{(4x-3)^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%7B1%7D%7B%284x-3%29%5E2%7D&bg=ffffff&fg=000000&s=1" title="\frac{1}{(4x-3)^2}" /></div><div style="padding-left: 30px;">e. <img alt="\frac{7}{(4x-3)^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cfrac%7B7%7D%7B%284x-3%29%5E2%7D&bg=ffffff&fg=000000&s=1" title="\frac{7}{(4x-3)^2}" /></div>jawab:<br />
<div style="padding-left: 30px;"><img alt="f(x)= \frac{4x-3}{-x-1}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%20%5Cfrac%7B4x-3%7D%7B-x-1%7D&bg=ffffff&fg=000000&s=1" title="f(x)= \frac{4x-3}{-x-1}" /></div><div style="padding-left: 30px;"><img alt="\begin{array}{rcl} misalkan:u=4x-3 & maka & u'=4 \\ v=-x-1 & maka & v'= -1 \end{array}" class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D%20misalkan%3Au%3D4x-3%20%26%20maka%20%26%20u%27%3D4%20%5C%5C%20v%3D-x-1%20%26%20maka%20%26%20v%27%3D%20-1%20%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0" title="\begin{array}{rcl} misalkan:u=4x-3 & maka & u'=4 \\ v=-x-1 & maka & v'= -1 \end{array}" /></div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*} maka f'(x) & = & \frac {u'.v - v'.u}{v^2}\\ & = & \frac {(4)(-x-1) - (-1)(4x-3)}{(-x-1)^2}\\ & = & \frac {-4x-4+4x-3}{(-x-1)^2}\\ & = & \frac {-7}{(-x -1)^2}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7D&space;maka&space;f%27%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;%7Bu%27.v&space;-&space;v%27.u%7D%7Bv%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B%284%29%28-x-1%29&space;-&space;%28-1%29%284x-3%29%7D%7B%28-x-1%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B-4x-4+4x-3%7D%7B%28-x-1%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;%7B-7%7D%7B%28-x&space;-1%29%5E2%7D%5Cend%7Balign*%7D" /></div>3. Diketahui <img alt="f(x)=(3x+4)^4" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%283x%2B4%29%5E4&bg=ffffff&fg=000000&s=0" title="f(x)=(3x+4)^4" /> dan f ‘(x) adalah turunan pertama dari f(x). Maka nilai dari f ‘(-1) = ….<br />
<div style="padding-left: 30px;">a. 4</div><div style="padding-left: 30px;">b. 12</div><div style="padding-left: 30px;">c. 16</div><div style="padding-left: 30px;">d. 84</div><div style="padding-left: 30px;">e. 112</div>jawab:<br />
<div style="padding-left: 30px;"><img alt="f(x)=(3x+4)^4" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D%283x%2B4%29%5E4&bg=ffffff&fg=000000&s=0" title="f(x)=(3x+4)^4" /></div><div style="padding-left: 30px;">misalkan u = 3x + 4 maka u’ = 3 dan n = 4</div><div style="padding-left: 30px;">gunakan aturan rantai, maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & n.u^{n-1}. u'\\ & = & 4.(3x + 4)^{4-1}.3\\ & = & 12(3x+4)^3\\ f'(-1) & = & 12(3(-1)+4)^3\\ & = & 12(-3+4)^3\\ & = & 12\:jawaban(B)\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;n.u%5E%7Bn-1%7D.&space;u%27%5C%5C&space;&&space;=&space;&&space;4.%283x&space;+&space;4%29%5E%7B4-1%7D.3%5C%5C&space;&&space;=&space;&&space;12%283x+4%29%5E3%5C%5C&space;f%27%28-1%29&space;&&space;=&space;&&space;12%283%28-1%29+4%29%5E3%5C%5C&space;&&space;=&space;&&space;12%28-3+4%29%5E3%5C%5C&space;&&space;=&space;&&space;12%5C:jawaban%28B%29%5Cend%7Balign*%7D" /></div><br />
4. Turunan pertama fungsi <img alt="f(x)=x^2-3x+\frac{4}{x^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3Dx%5E2-3x%2B%5Cfrac%7B4%7D%7Bx%5E2%7D&bg=ffffff&fg=000000&s=1" title="f(x)=x^2-3x+\frac{4}{x^2}" /> adalah f ‘(x) = ….<br />
<div style="padding-left: 30px;">a. <img alt="f(x)=x-3+\frac{4}{x^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3Dx-3%2B%5Cfrac%7B4%7D%7Bx%5E2%7D&bg=ffffff&fg=000000&s=1" title="f(x)=x-3+\frac{4}{x^2}" /></div><div style="padding-left: 30px;">b. <img alt="f(x)=x-3+\frac{4}{x^3}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3Dx-3%2B%5Cfrac%7B4%7D%7Bx%5E3%7D&bg=ffffff&fg=000000&s=1" title="f(x)=x-3+\frac{4}{x^3}" /></div><div style="padding-left: 30px;">c. <img alt="f(x)=2x-3-\frac{8}{x}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D2x-3-%5Cfrac%7B8%7D%7Bx%7D&bg=ffffff&fg=000000&s=1" title="f(x)=2x-3-\frac{8}{x}" /></div><div style="padding-left: 30px;">d. <img alt="f(x)=2x-3-\frac{4}{x^3}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D2x-3-%5Cfrac%7B4%7D%7Bx%5E3%7D&bg=ffffff&fg=000000&s=1" title="f(x)=2x-3-\frac{4}{x^3}" /></div><div style="padding-left: 30px;">e. <img alt="f(x)=2x-3-\frac{8}{x^3}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D2x-3-%5Cfrac%7B8%7D%7Bx%5E3%7D&bg=ffffff&fg=000000&s=1" title="f(x)=2x-3-\frac{8}{x^3}" /></div>jawab:<br />
<div style="padding-left: 30px;"><img alt="f(x)=x^2-3x+\frac{4}{x^2}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3Dx%5E2-3x%2B%5Cfrac%7B4%7D%7Bx%5E2%7D&bg=ffffff&fg=000000&s=1" title="f(x)=x^2-3x+\frac{4}{x^2}" /> nyatakan dalam bentuk pangkat</div><div style="padding-left: 30px;"><img alt="f(x)=x^2-3x+4.x^{-2}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3Dx%5E2-3x%2B4.x%5E%7B-2%7D&bg=ffffff&fg=000000&s=1" title="f(x)=x^2-3x+4.x^{-2}" /></div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 2x - 3+4(-2).x^{-2-1}\\ & = & 2x-3-8x^{-3}\\ & = & 2x-3-\frac{8}{x^3}\:jawaban(E)\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;2x&space;-&space;3+4%28-2%29.x%5E%7B-2-1%7D%5C%5C&space;&&space;=&space;&&space;2x-3-8x%5E%7B-3%7D%5C%5C&space;&&space;=&space;&&space;2x-3-%5Cfrac%7B8%7D%7Bx%5E3%7D%5C:jawaban%28E%29%5Cend%7Balign*%7D" /></div>5. Turunan pertama dari <img alt="f(x)=6x\sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D6x%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="f(x)=6x\sqrt x" /> adalah f ‘(x) = …<br />
<div style="padding-left: 30px;">a.<img alt="3 \sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=3%20%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="3 \sqrt x" /></div><div style="padding-left: 30px;">b. <img alt="5 \sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=5%20%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="5 \sqrt x" /></div><div style="padding-left: 30px;">c. <img alt="6\sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=6%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="6\sqrt x" /></div><div style="padding-left: 30px;">d. <img alt="9\sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=9%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="9\sqrt x" /></div><div style="padding-left: 30px;">e. <img alt="12\sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=12%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="12\sqrt x" /></div>jawab:<br />
<div style="padding-left: 30px;"><img alt="f(x)=6x \sqrt x" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D6x%20%5Csqrt%20x&bg=ffffff&fg=000000&s=0" title="f(x)=6x \sqrt x" /> nyatakan dalam bentuk pangkat</div><div style="padding-left: 30px;"><img alt="f(x)=6x^{\frac 32}" class="latex" src="http://s.wordpress.com/latex.php?latex=f%28x%29%3D6x%5E%7B%5Cfrac%2032%7D&bg=ffffff&fg=000000&s=0" title="f(x)=6x^{\frac 32}" /></div><div style="padding-left: 30px;">maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 6.(\frac 32).x^{\frac 32 -1}\\ & = & 9.x^{\frac 12}\\ & = & 9\sqrt x\:jawaban(D)\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;6.%28%5Cfrac&space;32%29.x%5E%7B%5Cfrac&space;32&space;-1%7D%5C%5C&space;&&space;=&space;&&space;9.x%5E%7B%5Cfrac&space;12%7D%5C%5C&space;&&space;=&space;&&space;9%5Csqrt&space;x%5C:jawaban%28D%29%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div><ol start="6"><li> Jika <img align="absmiddle" alt="g(x)=\left ( 5-3x \right )^{10}" src="http://latex.codecogs.com/gif.latex?g%28x%29=%5Cleft&space;%28&space;5-3x&space;%5Cright&space;%29%5E%7B10%7D" /> maka g ‘(2) = …. A. -30<br />
B. -10<br />
C. 30<br />
D. 60<br />
E. 90<br />
<br />
Jawab :<br />
<div style="padding-left: 30px;">* misal <img align="absmiddle" alt="u=5-3x" src="http://latex.codecogs.com/gif.latex?u=5-3x" /> maka <img align="absmiddle" alt="u'=-3" src="http://latex.codecogs.com/gif.latex?u%27=-3" /></div><div style="padding-left: 30px;"><img align="absmiddle" alt="n=10" src="http://latex.codecogs.com/gif.latex?n=10" /></div><div style="padding-left: 30px;">* kita pakai aturan rantai sehingga :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}g'(x) & = & {\color{Red} n.u^{n-1}.u'}\\ & = & 10.(5-3x)^{10-1}.(-3)\\ & = & (-30)(5-3x)^9\\g'(2) & = & (-30)(5-3.2)^9\\ & = & (-30)(-1)^9\\ & = & 30 \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Dg%27%28x%29&space;&&space;=&space;&&space;%7B%5Ccolor%7BRed%7D&space;n.u%5E%7Bn-1%7D.u%27%7D%5C%5C&space;&&space;=&space;&&space;10.%285-3x%29%5E%7B10-1%7D.%28-3%29%5C%5C&space;&&space;=&space;&&space;%28-30%29%285-3x%29%5E9%5C%5Cg%27%282%29&space;&&space;=&space;&&space;%28-30%29%285-3.2%29%5E9%5C%5C&space;&&space;=&space;&&space;%28-30%29%28-1%29%5E9%5C%5C&space;&&space;=&space;&&space;30&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><span id="more-999"></span></div></li>
<li> Jika <img align="absmiddle" alt="f(x)=x^3-\frac{x}{x^2-1}" src="http://latex.codecogs.com/gif.latex?f%28x%29=x%5E3-%5Cfrac%7Bx%7D%7Bx%5E2-1%7D" /> maka f ‘(x) = … A. <img align="absmiddle" alt="3x^2+\frac{x^2+1}{(x^2-1)^2}" src="http://latex.codecogs.com/gif.latex?3x%5E2+%5Cfrac%7Bx%5E2+1%7D%7B%28x%5E2-1%29%5E2%7D" /><br />
B. <img align="absmiddle" alt="3x^2-\frac{x^2-1}{(x^2-1)^2}" src="http://latex.codecogs.com/gif.latex?3x%5E2-%5Cfrac%7Bx%5E2-1%7D%7B%28x%5E2-1%29%5E2%7D" /><br />
C. <img align="absmiddle" alt="x^2+\frac{3x+1}{(x^2-1)^2}" src="http://latex.codecogs.com/gif.latex?x%5E2+%5Cfrac%7B3x+1%7D%7B%28x%5E2-1%29%5E2%7D" /><br />
D. <img align="absmiddle" alt="x^2-\frac{3x+1}{(x^2-1)^2}" src="http://latex.codecogs.com/gif.latex?x%5E2-%5Cfrac%7B3x+1%7D%7B%28x%5E2-1%29%5E2%7D" /><br />
E. <img align="absmiddle" alt="3x^2-\frac{3x+1}{(x^2-1)^2}" src="http://latex.codecogs.com/gif.latex?3x%5E2-%5Cfrac%7B3x+1%7D%7B%28x%5E2-1%29%5E2%7D" /><br />
<br />
Jawab :<br />
<div style="padding-left: 30px;">* terdapat dua suku yang harus diturunkan, kita turunkan suku yang pertama secara langsung dan suku yang kedua menggunakan rumus <img align="absmiddle" alt="{\color{Red} y=\frac uv}\;\;\;maka\;\;\;{\color{Red} y'=\frac{u'v-v'u}{v^2}}" src="http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7BRed%7D&space;y=%5Cfrac&space;uv%7D%5C;%5C;%5C;maka%5C;%5C;%5C;%7B%5Ccolor%7BRed%7D&space;y%27=%5Cfrac%7Bu%27v-v%27u%7D%7Bv%5E2%7D%7D" /></div><div style="padding-left: 30px;">* perhatikan suku kedua misalkan :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{array}{lcl}u=x & \Leftrightarrow & u'=1\\v=x^2-1 & \Leftrightarrow & v'=2x \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Blcl%7Du=x&space;&&space;%5CLeftrightarrow&space;&&space;u%27=1%5C%5Cv=x%5E2-1&space;&&space;%5CLeftrightarrow&space;&&space;v%27=2x&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;">maka</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f(x) & = & x^3-\frac{x}{x^2-1}\\f'(x) & = & 3x^2-\left [ \frac{u'.v-v'.u}{v^2} \right ]\\ & = & 3x^2-\left [ \frac{1.(x^2-1)-2x(x)}{(x^2-1)^2} \right ]\\ & = & 3x^2-\left [ \frac{x^2-1-2x^2}{(x^2-1)^2} \right ]\\ & = & 3x^2-\left [ \frac{-x^2-1}{(x^2-1)^2} \right ]\\ & = & 3x^2+\frac{x^2+1}{(x^2-1)^2} \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28x%29&space;&&space;=&space;&&space;x%5E3-%5Cfrac%7Bx%7D%7Bx%5E2-1%7D%5C%5Cf%27%28x%29&space;&&space;=&space;&&space;3x%5E2-%5Cleft&space;[&space;%5Cfrac%7Bu%27.v-v%27.u%7D%7Bv%5E2%7D&space;%5Cright&space;]%5C%5C&space;&&space;=&space;&&space;3x%5E2-%5Cleft&space;[&space;%5Cfrac%7B1.%28x%5E2-1%29-2x%28x%29%7D%7B%28x%5E2-1%29%5E2%7D&space;%5Cright&space;]%5C%5C&space;&&space;=&space;&&space;3x%5E2-%5Cleft&space;[&space;%5Cfrac%7Bx%5E2-1-2x%5E2%7D%7B%28x%5E2-1%29%5E2%7D&space;%5Cright&space;]%5C%5C&space;&&space;=&space;&&space;3x%5E2-%5Cleft&space;[&space;%5Cfrac%7B-x%5E2-1%7D%7B%28x%5E2-1%29%5E2%7D&space;%5Cright&space;]%5C%5C&space;&&space;=&space;&&space;3x%5E2+%5Cfrac%7Bx%5E2+1%7D%7B%28x%5E2-1%29%5E2%7D&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><br />
</div></li>
<li> Turunan pertama dari <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?y=%5Cfrac%7B%28x+2%29%28x+1%29%7D%7B%28x+3%29%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /> adalah ….. A. <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?%7By%7D%27=%5Cfrac%7Bx%5E2+9x+7%7D%7Bx%5E2+9%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br />
B. <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?%7By%7D%27=%5Cfrac%7Bx%5E2+6x+11%7D%7Bx%5E2+6x+9%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br />
C. <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?%7By%7D%27=%5Cfrac%7Bx%5E2+6x+7%7D%7Bx%5E2+6x+9%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br />
D. <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?%7By%7D%27=%5Cfrac%7Bx%5E2+9x+11%7D%7Bx%5E2+6x+9%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br />
E. <img alt="" id="equationview" src="http://latex.codecogs.com/gif.latex?%7By%7D%27=%5Cfrac%7Bx%5E2+6x+11%7D%7Bx%5E2+9%7D" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /><br />
<br />
Jawab :<br />
<div style="padding-left: 30px;">* untuk model soal yang seperti ini kita kalikan pembilangnya sehingga menjadi bentuk kuadrat, didapat <img align="absmiddle" alt="y=\frac{x^2+3x+2}{x+3}" src="http://latex.codecogs.com/gif.latex?y=%5Cfrac%7Bx%5E2+3x+2%7D%7Bx+3%7D" /> baru kita gunakan <img align="absmiddle" alt="{\color{Red} y=\frac uv}\;\;\;maka\;\;\;{\color{Red} y'=\frac{u'v-v'u}{v^2}}" src="http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7BRed%7D&space;y=%5Cfrac&space;uv%7D%5C;%5C;%5C;maka%5C;%5C;%5C;%7B%5Ccolor%7BRed%7D&space;y%27=%5Cfrac%7Bu%27v-v%27u%7D%7Bv%5E2%7D%7D" /></div><div style="padding-left: 30px;">* misalkan</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{array}{lcl}u=x^2+3x+2 & \Leftrightarrow & u'=2x+3\\v=x+3 & \Leftrightarrow & v'=1 \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Blcl%7Du=x%5E2+3x+2&space;&&space;%5CLeftrightarrow&space;&&space;u%27=2x+3%5C%5Cv=x+3&space;&&space;%5CLeftrightarrow&space;&&space;v%27=1&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;">* maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}y' & = & \frac{u'.v-v'u}{v^2}\\ & = & \frac{(2x+3)(x+3)-(1)(x^2+3x+2)}{(x+3)^2}\\ & = & \frac{2x^2+9x+9-x^2-3x-2}{(x+3)^2}\\ & = & \frac{x^2+6x+7}{x^2+6x+9} \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Dy%27&space;&&space;=&space;&&space;%5Cfrac%7Bu%27.v-v%27u%7D%7Bv%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B%282x+3%29%28x+3%29-%281%29%28x%5E2+3x+2%29%7D%7B%28x+3%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B2x%5E2+9x+9-x%5E2-3x-2%7D%7B%28x+3%29%5E2%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7Bx%5E2+6x+7%7D%7Bx%5E2+6x+9%7D&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><br />
</div></li>
<li> Diketahui <img align="absmiddle" alt="y=\sqrt{3-4x}" src="http://latex.codecogs.com/gif.latex?y=%5Csqrt%7B3-4x%7D" /> maka <img align="absmiddle" alt="\frac{\partial y}{\partial x}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cpartial&space;y%7D%7B%5Cpartial&space;x%7D" /> = …. A. <img align="absmiddle" alt="\frac{1}{2\sqrt{3-4x}}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B2%5Csqrt%7B3-4x%7D%7D" /><br />
B. <img align="absmiddle" alt="\frac{1}{\sqrt{3-4x}}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B%5Csqrt%7B3-4x%7D%7D" /><br />
C. <img align="absmiddle" alt="\frac{2}{\sqrt{3-4x}}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B2%7D%7B%5Csqrt%7B3-4x%7D%7D" /><br />
D. <img align="absmiddle" alt="\frac{-1}{\sqrt{3-4x}}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B-1%7D%7B%5Csqrt%7B3-4x%7D%7D" /><br />
E. <img align="absmiddle" alt="\frac{-2}{\sqrt{3-4x}}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B-2%7D%7B%5Csqrt%7B3-4x%7D%7D" /><br />
<br />
Jawab :<br />
<div style="padding-left: 30px;">* nyatakan y dalam bentuk pangkat menjadi <img align="absmiddle" alt="y=\left ( 3-4x \right )^{\frac 12}" src="http://latex.codecogs.com/gif.latex?y=%5Cleft&space;%28&space;3-4x&space;%5Cright&space;%29%5E%7B%5Cfrac&space;12%7D" /></div><div style="padding-left: 30px;">* nah…ingat kita pakai aturan rantai</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}y' & = & n.u^{n-1}.u'\\ & = & \frac 12.(3-4x)^{\frac{1}{2}-1}.(-4)\\ & = & (-2)(3-4x)^{-\frac 12}\\ & = & \frac{-2}{\sqrt{3-4x}}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Dy%27&space;&&space;=&space;&&space;n.u%5E%7Bn-1%7D.u%27%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;12.%283-4x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D-1%7D.%28-4%29%5C%5C&space;&&space;=&space;&&space;%28-2%29%283-4x%29%5E%7B-%5Cfrac&space;12%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B-2%7D%7B%5Csqrt%7B3-4x%7D%7D%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div></li>
<li> Jika <img align="absmiddle" alt="f(3+2x)=4-2x+x^2" src="http://latex.codecogs.com/gif.latex?f%283+2x%29=4-2x+x%5E2" /> maka f ‘ (1) = … A. -4<br />
B. -2<br />
C. -1<br />
D. 0<br />
E. <img align="absmiddle" alt="\frac{1}{2}" src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B2%7D" /><br />
<br />
Jawab :<br />
<div style="padding-left: 30px;">* masih ingatkah materi komposisi fungsi ….???</div>* kita misalkan<br />
<div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}{\color{Blue} 3+2x} & = & {\color{Blue} y}\\ {\color{Red} x} & = & {\color{Red} \frac{y-3}{2}}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7D%7B%5Ccolor%7BBlue%7D&space;3+2x%7D&space;&&space;=&space;&&space;%7B%5Ccolor%7BBlue%7D&space;y%7D%5C%5C&space;%7B%5Ccolor%7BRed%7D&space;x%7D&space;&&space;=&space;&&space;%7B%5Ccolor%7BRed%7D&space;%5Cfrac%7By-3%7D%7B2%7D%7D%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;">*subitusikan ke <img align="absmiddle" alt="f(3+2x)=4-2x+x^2" src="http://latex.codecogs.com/gif.latex?f%283+2x%29=4-2x+x%5E2" /> menjadi :</div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f({\color{Blue} 3+2x}) & = & 4-2{\color{Red} x}+{\color{Red} x}^2 \\f({\color{Blue} y}) & = & 4-2\left ( {\color{Red} \frac{y-3}{2}} \right )+\left ( {\color{Red} \frac{y-3}{2}} \right )^2\\f(y) & = & 4-y+6+\left ( \frac{y^2-6y+9}{4} \right )\\ & = & \frac{16-4y+12+y^2-6y+9}{4}\\f(y) & = & \frac{y^2-10y+37}{4}\\f(x) & = & \frac 14x^2-\frac{10}{4}x+\frac{37}{4}\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28%7B%5Ccolor%7BBlue%7D&space;3+2x%7D%29&space;&&space;=&space;&&space;4-2%7B%5Ccolor%7BRed%7D&space;x%7D+%7B%5Ccolor%7BRed%7D&space;x%7D%5E2&space;%5C%5Cf%28%7B%5Ccolor%7BBlue%7D&space;y%7D%29&space;&&space;=&space;&&space;4-2%5Cleft&space;%28&space;%7B%5Ccolor%7BRed%7D&space;%5Cfrac%7By-3%7D%7B2%7D%7D&space;%5Cright&space;%29+%5Cleft&space;%28&space;%7B%5Ccolor%7BRed%7D&space;%5Cfrac%7By-3%7D%7B2%7D%7D&space;%5Cright&space;%29%5E2%5C%5Cf%28y%29&space;&&space;=&space;&&space;4-y+6+%5Cleft&space;%28&space;%5Cfrac%7By%5E2-6y+9%7D%7B4%7D&space;%5Cright&space;%29%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B16-4y+12+y%5E2-6y+9%7D%7B4%7D%5C%5Cf%28y%29&space;&&space;=&space;&&space;%5Cfrac%7By%5E2-10y+37%7D%7B4%7D%5C%5Cf%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;14x%5E2-%5Cfrac%7B10%7D%7B4%7Dx+%5Cfrac%7B37%7D%7B4%7D%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;">* baru kita turunkan tiap sukunya</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f(x) & = & \frac 14x^2-\frac{10}{4}x+\frac{37}{4}\\f'(x) & = & \frac 12x^2-\frac{10}{4}\\f'(1) & = & \frac 12-\frac 52\\f'(1) & = & -\frac 42\\ & = & -2\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;14x%5E2-%5Cfrac%7B10%7D%7B4%7Dx+%5Cfrac%7B37%7D%7B4%7D%5C%5Cf%27%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;12x%5E2-%5Cfrac%7B10%7D%7B4%7D%5C%5Cf%27%281%29&space;&&space;=&space;&&space;%5Cfrac&space;12-%5Cfrac&space;52%5C%5Cf%27%281%29&space;&&space;=&space;&&space;-%5Cfrac&space;42%5C%5C&space;&&space;=&space;&&space;-2%5Cend%7Balign*%7D" /></div></li>
</ol><div style="padding-left: 30px;"><br />
</div><div style="padding-left: 30px;"><br />
</div><br />
<ol start="11"><li> Turunan pertama dari <em>f(x) = 7 cos (5 – 3x)</em> adalah <em>f ‘ (x) </em>= …..</li>
A. <em>35 sin (5 – 3x)</em>
B. <em>- 15 sin (5 – 3x)</em>
C. 21<em> sin (5 – 3x)</em>
D. <em>- 21 sin (5 – 3x)</em>
E. <em>- 35 sin (5 – 3x)</em>
<span id="more-1086"></span>
Jawab :
<div style="padding-left: 30px;">* ingat <img align="absmiddle" alt="f(x) = {\color{Red} a.cos\:(bx+c)}\;\;\;\;maka \;\;\;\;f'(x)= {\color{Red} -ab.sin\:(bx+c)}" src="http://latex.codecogs.com/gif.latex?f%28x%29&space;=&space;%7B%5Ccolor%7BRed%7D&space;a.cos%5C:%28bx+c%29%7D%5C;%5C;%5C;%5C;maka&space;%5C;%5C;%5C;%5C;f%27%28x%29=&space;%7B%5Ccolor%7BRed%7D&space;-ab.sin%5C:%28bx+c%29%7D" /></div><div style="padding-left: 30px;">* maka:</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{align*}f(x) & = & 7 cos (5 - 3x)\\f'(x) & = & -7.(-3).sin(5-3x)\\ & = & 21\;sin(5-3x) \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28x%29&space;&&space;=&space;&&space;7&space;cos&space;%285&space;-&space;3x%29%5C%5Cf%27%28x%29&space;&&space;=&space;&&space;-7.%28-3%29.sin%285-3x%29%5C%5C&space;&&space;=&space;&&space;21%5C;sin%285-3x%29&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 60px;">
</div><div style="padding-left: 60px;">
</div><li> Jika <em>f </em>‘(<em>x</em>) adalah turunan dari <em>f</em>(<em>x</em>) dan jika <em>f(x) = ( 3x – 2 ) sin (2x + 1)</em> maka <em>f ‘ </em>(<em>x</em>) adalah …</li>
A. 3 cos ( 2<em>x </em>+ 1 )
B. 6 cos ( 2<em>x </em>+ 1 )
C. 3 sin ( 2<em>x </em>+ 1 ) + (6<em>x </em>– 4) cos (2<em>x </em>+ 1)
D. (6<em>x </em>– 4) sin ( 2<em>x </em>+ 1 ) + 3 cos ( 2<em>x </em>+ 1 )
E. 3 sin ( 2<em>x </em>+ 1) + ( 3<em>x </em>– 2 ) cos( 2<em>x </em>+ 1 )
Jawab :
<div style="padding-left: 30px;">* <img align="absmiddle" alt="f (x) = {\color{Red} (3x-2)}\;{\color{DarkGreen} sin( 2x + 1 )}" src="http://latex.codecogs.com/gif.latex?f&space;%28x%29&space;=&space;%7B%5Ccolor%7BRed%7D&space;%283x-2%29%7D%5C;%7B%5Ccolor%7BDarkGreen%7D&space;sin%28&space;2x&space;+&space;1&space;%29%7D" /> kita misalkan terlebih dulu</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{array}{lcl}{\color{Red} u}={\color{Red} 3x-2} & maka & u'=3 \\v={\color{DarkGreen} sin(2x+1)} & maka & v'=2\;cos(2x+1) \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Blcl%7D%7B%5Ccolor%7BRed%7D&space;u%7D=%7B%5Ccolor%7BRed%7D&space;3x-2%7D&space;&&space;maka&space;&&space;u%27=3&space;%5C%5Cv=%7B%5Ccolor%7BDarkGreen%7D&space;sin%282x+1%29%7D&space;&&space;maka&space;&&space;v%27=2%5C;cos%282x+1%29&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;">* ingat rumus turunan perkalian dua fungsi :</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{array}{rcl}f'(x) & = & u'.v+v'.u\\ & = & 3.{\color{DarkGreen} sin(2x+1)}+2cos(2x+1).({\color{Red} 3x-2})\\ & = & 3\;sin(2x+1)+(6x-4)\;cos(2x+1) \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Brcl%7Df%27%28x%29&space;&&space;=&space;&&space;u%27.v+v%27.u%5C%5C&space;&&space;=&space;&&space;3.%7B%5Ccolor%7BDarkGreen%7D&space;sin%282x+1%29%7D+2cos%282x+1%29.%28%7B%5Ccolor%7BRed%7D&space;3x-2%7D%29%5C%5C&space;&&space;=&space;&&space;3%5C;sin%282x+1%29+%286x-4%29%5C;cos%282x+1%29&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 30px;">
</div><li> Turunan pertama fungsi <em>f </em>(<em>x</em>) = 5 sin <em>x </em>cos <em>x </em>adalah <em>f ‘ </em>(<em>x</em>) = …</li>
A. 5 sin 2<em>x </em>
<em> </em>B. 5 cos 2<em>x</em>
C. 5 sin<sup>2 </sup><em>x </em>cos <em>x </em>
<em> </em>D. 5 sin <em>x </em>cos<span style="font-size: 11px;"><sup>2</sup></span> <em>x </em>
<em> </em>E. 5 sin 2<em>x </em>cos <em>x</em>
Jawab :
<div style="padding-left: 30px;">* <img align="absmiddle" alt="f (x) = {\color{Red} 5\;sin\;x}\;{\color{DarkGreen} cos\;x}" src="http://latex.codecogs.com/gif.latex?f&space;%28x%29&space;=&space;%7B%5Ccolor%7BRed%7D&space;5%5C;sin%5C;x%7D%5C;%7B%5Ccolor%7BDarkGreen%7D&space;cos%5C;x%7D" /> kita misalkan terlebih dulu</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{array}{lcl}{\color{Red} u}={\color{Red} 5sin\;x} & maka & u'=5cos\;x\\v={\color{DarkGreen} cos\;x} & maka & v'=-sin\;x \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Blcl%7D%7B%5Ccolor%7BRed%7D&space;u%7D=%7B%5Ccolor%7BRed%7D&space;5sin%5C;x%7D&space;&&space;maka&space;&&space;u%27=5cos%5C;x%5C%5Cv=%7B%5Ccolor%7BDarkGreen%7D&space;cos%5C;x%7D&space;&&space;maka&space;&&space;v%27=-sin%5C;x&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 30px;">* ingat rumus turunan</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{array}{rcl}f'(x) & = & u'.v+v'.u\\ & = & 5cos\;x.{\color{DarkGreen} cos\;x}+(-sin\;x).({\color{Red} 5sin\;x})\\ & = & 5\;cos^2x-5\;sin^2x\\ & = & 5(cos^2x-sin^2x)\\ & = & 5.cos\;2x \end{array}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Barray%7D%7Brcl%7Df%27%28x%29&space;&&space;=&space;&&space;u%27.v+v%27.u%5C%5C&space;&&space;=&space;&&space;5cos%5C;x.%7B%5Ccolor%7BDarkGreen%7D&space;cos%5C;x%7D+%28-sin%5C;x%29.%28%7B%5Ccolor%7BRed%7D&space;5sin%5C;x%7D%29%5C%5C&space;&&space;=&space;&&space;5%5C;cos%5E2x-5%5C;sin%5E2x%5C%5C&space;&&space;=&space;&&space;5%28cos%5E2x-sin%5E2x%29%5C%5C&space;&&space;=&space;&&space;5.cos%5C;2x&space;%5Cend%7Barray%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 30px;">eitttts…..tapi cara yang satu ini lebih simple…kita bisa pakai neh,cekidot…</div><div style="padding-left: 30px;">* ingat bahwa <img align="absmiddle" alt="sin\;2x=2\;sin\;x.cos\;x" src="http://latex.codecogs.com/gif.latex?sin%5C;2x=2%5C;sin%5C;x.cos%5C;x" /></div><div style="padding-left: 30px;">* sehingga :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f(x) & = & 5\;sin\;x\;cos\;x\\ & = & \frac{5}{2}.{\color{DarkBlue} 2.sin\;x.cos\;x}\\ & = & \frac 52.sin\;2x \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28x%29&space;&&space;=&space;&&space;5%5C;sin%5C;x%5C;cos%5C;x%5C%5C&space;&&space;=&space;&&space;%5Cfrac%7B5%7D%7B2%7D.%7B%5Ccolor%7BDarkBlue%7D&space;2.sin%5C;x.cos%5C;x%7D%5C%5C&space;&&space;=&space;&&space;%5Cfrac&space;52.sin%5C;2x&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;">* maka :</div><div style="padding-left: 30px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & \frac 52.2.cos\;2x\\ & = & 5\;cos\;2x\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;%5Cfrac&space;52.2.cos%5C;2x%5C%5C&space;&&space;=&space;&&space;5%5C;cos%5C;2x%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 30px;"><em>Dengan hasil yang sama namun lebih cepat dalam pengerjaannya…silahkan pilih cara yang lebih disukai…</em></div><div style="padding-left: 30px;">
</div>
<li> Jika<img align="absmiddle" alt="f(x)=sin^2 \left ( 2x+\frac{\pi}{6} \right )" src="http://latex.codecogs.com/gif.latex?f%28x%29=sin%5E2&space;%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29" /> , maka nilai dari <em>f </em>‘ (0) = …..</li>
A . <img align="absmiddle" alt="2\sqrt{3}" src="http://latex.codecogs.com/gif.latex?2%5Csqrt%7B3%7D" />
B. 2
C. <img align="absmiddle" alt="\sqrt{3}" src="http://latex.codecogs.com/gif.latex?%5Csqrt%7B3%7D" />
D. <img align="absmiddle" alt="12\sqrt{3}" src="http://latex.codecogs.com/gif.latex?12%5Csqrt%7B3%7D" />
E. <img align="absmiddle" alt="\sqrt{2}" src="http://latex.codecogs.com/gif.latex?%5Csqrt%7B2%7D" />
Jawab :
<div style="padding-left: 30px;">* perlu diingat bahwa :</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{align*}f(x) & = & sin^2\left ( 2x+\frac{\pi}{6} \right )\\ & = & \left ( {\color{Red} sin\left ( 2x+\frac{\pi}{6} \right )} \right )^2 \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%28x%29&space;&&space;=&space;&&space;sin%5E2%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%5C%5C&space;&&space;=&space;&&space;%5Cleft&space;%28&space;%7B%5Ccolor%7BRed%7D&space;sin%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%7D&space;%5Cright&space;%29%5E2&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;">* nah, baru kita misalkan <img align="absmiddle" alt="{\color{Red} u}={\color{Red} sin\left ( 2x+\frac{\pi}{6} \right )}\;\;maka\;\;u'=2\;cos\left ( 2x+\frac{\pi}{6} \right )" src="http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7BRed%7D&space;u%7D=%7B%5Ccolor%7BRed%7D&space;sin%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%7D%5C;%5C;maka%5C;%5C;u%27=2%5C;cos%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29" /></div><div style="padding-left: 30px;">* fungsi menjadi <img align="absmiddle" alt="f(x)=u^2" src="http://latex.codecogs.com/gif.latex?f%28x%29=u%5E2" /> baru pakai aturan rantai <img align="absmiddle" alt="f'(x) & = & n.u^{n-1}.u'" src="http://latex.codecogs.com/gif.latex?f%27%28x%29&space;&&space;=&space;&&space;n.u%5E%7Bn-1%7D.u%27" /></div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 2.u.u'\\ & = & 2.{\color{Red} sin\left ( 2x+\frac{\pi}{6} \right )}.2cos\left ( 2x+\frac{\pi}{6} \right )\\f'(0) & = & 4.sin\left ( 2.0+\frac{\pi}{6} \right ).cos\left ( 2.0+\frac{\pi}{6} \right )\\ & = & 4.sin\left ( \frac{\pi}{6} \right ).cos\left ( \frac{\pi}{6} \right )\\ & = & 4.\frac 12.\frac 12\sqrt3\\ & = & \sqrt3\end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;2.u.u%27%5C%5C&space;&&space;=&space;&&space;2.%7B%5Ccolor%7BRed%7D&space;sin%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%7D.2cos%5Cleft&space;%28&space;2x+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%5C%5Cf%27%280%29&space;&&space;=&space;&&space;4.sin%5Cleft&space;%28&space;2.0+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29.cos%5Cleft&space;%28&space;2.0+%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%5C%5C&space;&&space;=&space;&&space;4.sin%5Cleft&space;%28&space;%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29.cos%5Cleft&space;%28&space;%5Cfrac%7B%5Cpi%7D%7B6%7D&space;%5Cright&space;%29%5C%5C&space;&&space;=&space;&&space;4.%5Cfrac&space;12.%5Cfrac&space;12%5Csqrt3%5C%5C&space;&&space;=&space;&&space;%5Csqrt3%5Cend%7Balign*%7D" /></div><div style="padding-left: 60px;">
</div><div style="padding-left: 60px;">
</div><li> Turunan pertama dari <img align="absmiddle" alt="f(x)=sin^4(3-2x)" src="http://latex.codecogs.com/gif.latex?f%28x%29=sin%5E4%283-2x%29" /> adalah <em>f ’ (x) </em>=……</li>
A. - <img align="absmiddle" alt="8\;sin^3(3-2x)\;cos(6-4x)" src="http://latex.codecogs.com/gif.latex?8%5C;sin%5E3%283-2x%29%5C;cos%286-4x%29" />
B. – <img align="absmiddle" alt="8\;sin(3-2x)\;sin(6-4x)" src="http://latex.codecogs.com/gif.latex?8%5C;sin%283-2x%29%5C;sin%286-4x%29" />
C. - <img align="absmiddle" alt="4\;sin^3(3-2x)\;cos(3-2x)" src="http://latex.codecogs.com/gif.latex?4%5C;sin%5E3%283-2x%29%5C;cos%283-2x%29" />
D. - <img align="absmiddle" alt="4\;sin^2(3-2x)\;sin(6-4x)" src="http://latex.codecogs.com/gif.latex?4%5C;sin%5E2%283-2x%29%5C;sin%286-4x%29" />
E. - <img align="absmiddle" alt="8\;sin(3-2x)\;sin(6-4x)" src="http://latex.codecogs.com/gif.latex?8%5C;sin%283-2x%29%5C;sin%286-4x%29" />
Jawab :
<div style="padding-left: 30px;">* pengerjaannya hampir sama dengan soal no.4 kita misalkan terlebih dulu</div><div style="padding-left: 60px;"><img align="absmiddle" alt="u={\color{Red} sin(3-2x)}\;\;\;maka\;\;\;\;u'=-2.cos(3-2x)" src="http://latex.codecogs.com/gif.latex?u=%7B%5Ccolor%7BRed%7D&space;sin%283-2x%29%7D%5C;%5C;%5C;maka%5C;%5C;%5C;%5C;u%27=-2.cos%283-2x%29" /></div><div style="padding-left: 30px;">* didapat <img align="absmiddle" alt="f(x)=u^4" src="http://latex.codecogs.com/gif.latex?f%28x%29=u%5E4" /> kita pakai aturan rantai <img align="absmiddle" alt="f'(x)=n.u^{n-1}.u'" src="http://latex.codecogs.com/gif.latex?f%27%28x%29=n.u%5E%7Bn-1%7D.u%27" /> maka :</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = & 4.u^3.u'\\ & = & 4.{\color{Red} sin}^3{\color{Red} (3-2x)}.(-2)cos(3-2x)\\ & = & -8.sin^3(3-2x).cos(3-2x) \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&&space;4.u%5E3.u%27%5C%5C&space;&&space;=&space;&&space;4.%7B%5Ccolor%7BRed%7D&space;sin%7D%5E3%7B%5Ccolor%7BRed%7D&space;%283-2x%29%7D.%28-2%29cos%283-2x%29%5C%5C&space;&&space;=&space;&&space;-8.sin%5E3%283-2x%29.cos%283-2x%29&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 30px;">ups….saat kita cek di pilgan ternyata jawaban tersebut tidak ada pilihannya, so lanjut ke next step ….</div><div style="padding-left: 30px;">* ingat bahwa <img align="absmiddle" alt="{\color{DarkBlue} 2.sin\;x.cos\;x}={\color{DarkBlue} sin\;2x}" src="http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7BDarkBlue%7D&space;2.sin%5C;x.cos%5C;x%7D=%7B%5Ccolor%7BDarkBlue%7D&space;sin%5C;2x%7D" /></div><div style="padding-left: 30px;">
</div><div style="padding-left: 60px;"><img align="absmiddle" alt="\begin{align*}f'(x) & = &-8.sin^3(3-2x).cos(3-2x)\\ & = & -4.{\color{DarkBlue} 2.sin(3-2x).cos(3-2x)}.sin^2(3-2x)\\ & = & -4.{\color{DarkBlue} sin\;2(3-2x)}.sin^2(3-2x)\\ & = & -4.sin(6-4x).sin^2(3-2x)\\ & = & -4\;sin^2(3-2x)\;sin(3-4x) \end{align*}" src="http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7Df%27%28x%29&space;&&space;=&space;&-8.sin%5E3%283-2x%29.cos%283-2x%29%5C%5C&space;&&space;=&space;&&space;-4.%7B%5Ccolor%7BDarkBlue%7D&space;2.sin%283-2x%29.cos%283-2x%29%7D.sin%5E2%283-2x%29%5C%5C&space;&&space;=&space;&&space;-4.%7B%5Ccolor%7BDarkBlue%7D&space;sin%5C;2%283-2x%29%7D.sin%5E2%283-2x%29%5C%5C&space;&&space;=&space;&&space;-4.sin%286-4x%29.sin%5E2%283-2x%29%5C%5C&space;&&space;=&space;&&space;-4%5C;sin%5E2%283-2x%29%5C;sin%283-4x%29&space;%5Cend%7Balign*%7D" /></div><div style="padding-left: 60px;">
</div><div style="padding-left: 30px;">
</div></ol><div style="padding-left: 30px;"><strong><em>taraaaaa…..selesai sudah latihan soal dan pembahasan turunan trigonometri kita…</em></strong><br />
<strong><em> semoga bermanfaat yah….</em></strong></div><div style="padding-left: 30px;"></div><div style="padding-left: 30px;"></div>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com11tag:blogger.com,1999:blog-3157576316325034465.post-69856442639250843002012-03-18T23:43:00.002-07:002012-03-18T23:43:46.749-07:00Persamaan Kuadrat1. persamaan kuadrat yang akar-akarnya 3 kurang dari akar-akar persamaan kuadrat <br />
6X^2 - 11X + 10 =0<br />
<br />
2. persamaan kuadrat yg akar-akarnya kebalikan dari akar-akar persamaan 2X^2 + 4X - 3 = 0<br />
<br />
Jawab:<br />
<br />
1.] 6x² - 11x + 10 = 0<br />
misal akar-akarnya adalah p dan q<br />
p + q = -b/a = 11/6<br />
pq = c/a = 10/6<br />
<br />
p' = p - 3 ; p = p' + 3<br />
q' = q - 3 ; q = q' + 3<br />
<br />
substitusi :<br />
-> p + q = 11/6<br />
p' + 3 + q' + 3 = 11/6<br />
p' + q' = 11/6 - 36/6<br />
p' + q' = -25/6<br />
<br />
-> pq = 10/6<br />
(p' + 3) (q' + 3) = 10/6<br />
p'q' + 3 (p' + q') + 9 = 10/6<br />
p'q' + 3 . -25/6 = 10/6 - 54/6<br />
p'q' = -44/6 + 25/2<br />
p'q' = 31/6<br />
<br />
masukkan :<br />
x² - (p' + q') x + p'q' = 0<br />
x² + 25/6 x + 31/6 = 0<br />
____________________ . 6<br />
6x² + 25x + 31 = 0<br />
<br />
2.] 2x² + 4x - 3 = 0<br />
misal akar-akarnya adalah m dan n<br />
m + n = -b/a = -4/2 = -2<br />
mn = c/a = -3/2<br />
<br />
m' = 1/m ; m = 1/m'<br />
n' = 1/n ; n = 1/n'<br />
<br />
substitusi :<br />
-> mn = -3/2<br />
1 / m'n' = -3/2<br />
m'n' = -2/3<br />
<br />
-> m + n = -2<br />
1/m' + 1/n' = -2<br />
[m' + n'] / m'n' = -2<br />
m' + n' = -2 . -2/3<br />
m' + n' = 4/3<br />
<br />
masukkan :<br />
x² - (m' + n') x + m'n' = 0<br />
x² - 4/3 x - 2/3 = 0<br />
_____________________ . 3<br />
3x² - 4x - 2 = 0Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com0tag:blogger.com,1999:blog-3157576316325034465.post-1034640099763310192011-07-25T01:59:00.001-07:002011-07-25T01:59:40.310-07:00TrigonometriDownload materi Trigonometri disini: <br />
<a class="normal12blue" href="http://www.ziddu.com/download/15812418/TrigonometriuntukSMA.pdf.html"><b>http://www.ziddu.com/download/15812418/TrigonometriuntukSMA.pdf.html</b></a>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com0tag:blogger.com,1999:blog-3157576316325034465.post-27873009324369720032010-04-07T03:12:00.001-07:002010-04-07T03:12:40.296-07:00Konvers, Invers, dan Kontraposisi<span id="fullpost">Dari pernyataan yang berupa implikasi p <code>⇒</code> q dapat dibuat pernyataan implikasi baru sbagai brikut:<br />
(a) Pernyataan q <code>⇒</code> p disebut Konvers dari p <code>⇒</code> q <br />
(b) Pernyataan ~p <code>⇒</code> ~q disebut Invers dari p <code>⇒</code> q <br />
(c) Pernyataan ~q <code>⇒</code> ~p disebut Kontraposisi dari p <code>⇒</code> q.<br />
<br />
Untuk melihat hubungan nilai kebenaran antara implikasi, konvers, invers dan kontraposisi perhatikanlah tabel kebenaran berikut :<br />
<br />
<span xmlns=""><br />
<br />
<div style="margin-left: 18pt;"><table border="0" style="border-collapse: collapse;"><colgroup><col style="width: 26px;"></col><col style="width: 26px;"></col><col style="width: 91px;"></col><col style="width: 84px;"></col><col style="width: 78px;"></col><col style="width: 98px;"></col></colgroup><tbody valign="top">
<tr><td style="border: 0.5pt solid; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">p</span> </div></td><td style="border-color: -moz-use-text-color; border-style: solid solid solid none; border-width: 0.5pt 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">q</span> </div></td><td style="border-color: -moz-use-text-color; border-style: solid solid solid none; border-width: 0.5pt 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">Implikasi</span></div><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">p <code>⇒</code> q</span></div></td><td style="border-color: -moz-use-text-color; border-style: solid solid solid none; border-width: 0.5pt 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">Konvers</span></div><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">q <code>⇒</code> p</span></div></td><td style="border-color: -moz-use-text-color; border-style: solid solid solid none; border-width: 0.5pt 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">Invers</span></div><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">~p <code>⇒</code> ~q</span></div></td><td style="border-color: -moz-use-text-color; border-style: solid solid solid none; border-width: 0.5pt 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">Kontraposisi</span></div><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">~q <code>⇒</code> ~p</span></div></td></tr>
<tr><td style="border-color: -moz-use-text-color; border-style: none solid solid; border-width: medium 0.5pt 0.5pt; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td></tr>
<tr><td style="border-color: -moz-use-text-color; border-style: none solid solid; border-width: medium 0.5pt 0.5pt; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td></tr>
<tr><td style="border-color: -moz-use-text-color; border-style: none solid solid; border-width: medium 0.5pt 0.5pt; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td></tr>
<tr><td style="border-color: -moz-use-text-color; border-style: none solid solid; border-width: medium 0.5pt 0.5pt; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">S</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td><td style="border-color: -moz-use-text-color; border-style: none solid solid none; border-width: medium 0.5pt 0.5pt medium; padding-left: 7px; padding-right: 7px;"><div style="text-align: center;"><span style="font-family: Times New Roman; font-size: 12pt;">B</span> </div></td></tr>
</tbody></table></div><span style="font-family: Times New Roman; font-size: 12pt;"><br />
</span> </span><br />
<br />
Dari tabel di atas ternyata:<br />
Implikasi ekuivalen dengan kontraposisinya atau ditulis <br />
<br />
p <code>⇒</code> q <code>≡</code> ~q <code>⇒</code> ~p <br />
<br />
dengan kata lain jika implikasi bernilai benar maka kontraposi-sinya juga bernilai benar atau jika implikasi bernilai salah maka kontraposisinya juga bernilai salah.<br />
<br />
Konvers suatu implikasi ekuivalen dengan inversnya atau ditulis <br />
<br />
q <code>⇒</code> p <code>≡</code> ~p <code>⇒</code> ~q .<br />
<br />
Contoh:<br />
Tentukanlah konvers, invers dan kontraposisi dari pernyataan:<br />
(1) Jika harga bahan bakar minyak naik maka harga beras naik.<br />
(2) Jika x > 6 maka x<code>²</code> <code>≥</code> 36<br />
<br />
Penyelesaian: <br />
<br />
Soal (1)<br />
Konvers : Jika harga beras naik maka harga bahan bakar minyak naik.<br />
Invers : Jika harga bahan bakar minyak tidak naik maka harga beras tidak naik.<br />
Kontraposisi: Jika harga beras tidak naik maka harga bahan bakar minyak tidak naik.<br />
<br />
Soal (2) <br />
Tulis <br />
p: jika x<code>²</code> <code>&re;</code> 36<br />
q: x > 6.<br />
Jadi ~p: x<code>²</code> < 36<br />
~q: x <code>≤</code> 6.<br />
Jadi konvers p <code>⇒</code> q <code>≡</code> q <code>⇒</code> p <code>≡</code> “jika x > 6 maka x<code>²</code> <code>&re;</code> 36”,<br />
<br />
invers p <code>⇒</code> q <code>≡</code> ~p <code>⇒</code> ~q <code>≡</code> ”jika x<code>²</code> < 36 maka x <code>≤</code> 6”,<br />
<br />
kontraposisi p <code>⇒</code> q <code>≡</code> ~q <code>⇒</code> ~p <code>≡</code> “jika x <code>≤</code> 6 maka x<code>²</code> < 36”. <br />
<br />
Soal (3) <br />
Jika (p <code>∧</code> q) <code>⇒</code> r <br />
Jelas konvers (p <code>∧</code> q) <code>⇒</code> r <code>≡</code> r <code>⇒</code> (p <code>∧</code> q),<br />
invers (p <code>∧</code> q) <code>⇒</code> r <code>≡</code> ~(p <code>∧</code> q) <code>⇒</code> r <code>≡</code> (p <code>∨</code> q) <code>⇒</code> r,<br />
kontraposisi (p <code>∧</code> q) <code>⇒</code> r <code>≡</code> r <code>⇒</code> ~(p <code>∧</code> q) <code>≡</code> r <code>⇒</code> (~p <code>∨</code> q).<br />
<br />
<br />
<br />
Tugas 4<br />
<br />
(Soal nomor 1)<br />
Tentukan invers, konves dan kontraposisi dari proposisi <br />
berikut ini:<br />
<br />
(a) (p <code>∧</code> q) <code>⇒</code> r<br />
(b) p <code>⇒</code> (q <code>∧</code> r)<br />
(c) ~p <code>⇒</code> (q <code>∧</code> ~r)<br />
(d) (p <code>∨</code> ~q) <code>⇒</code> (q <code>∧</code> r)<br />
(e) (~q <code>∧</code> ~r) <code>⇒</code> (~p <code>∨</code> q)<br />
(f) (q <code>∨</code> ~r) <code>⇒</code> (p <code>∧</code> r)<br />
<br />
(Soal nomor 2)<br />
Tentukan invers, konvers, dan kontraposisi pernyataan:<br />
<br />
(a) Jika hasil produksi melimpah maka harganya turun.<br />
(b) Jika lapangan pekerjaan tidak banyak maka pengangguran meningkat.<br />
(c) Jika ABCD bujur sangkar maka ABCD segi empat.<br />
(d) Jika x > 10 maka x<code>²</code> > 100<br />
(e) Jika x<code>²</code> – 16 = 0 , maka x = 4 atau x = – 4.<br />
(f) Jika sin x = 90<code>°</code> – cos x, maka x merupakan sudut lancip.<br />
(g) Jika tan x = -1, maka x = 135<code>°</code> dan x = 315<code>°</code></span>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com2tag:blogger.com,1999:blog-3157576316325034465.post-84218186785389255642009-12-02T05:36:00.000-08:002009-12-02T05:36:00.871-08:00Rumus Bangun Datar - MatematikaRumus Bujur Sangkar<br />
<div class="content"> Bujur sangkar adalah bangun datar yang memiliki empat buah sisi sama panjang<br />
- Keliling : Panjang salah satu sisi dikali 4 (4S) (AB + BC + CD + DA)<br />
- Luas : Sisi dikali sisi (S x S)<br />
Rumus Persegi Panjang<br />
Persegi panjang adalah bangun datar mirip bujur sangkar namun dua sisi yang berhadapan lebih pendek atau lebih panjang dari <br />
dua sisi yang lain. Dua sisi yang panjang disebut panjang, sedangkan yang pendek disebut lebar.<br />
- Keliling : Panjang tambah lebar kali 2 ((p+l)x2) (AB + BC + CD + DA)<br />
- Luas : Panjang dikali lebar (pl)<br />
Rumus Segitiga<br />
- Keliling : Sisi pertama + sisi kedua + sisi ketiga (AB + BC + CA)<br />
- Luas : Panjang alas dikali pangjang tinggi dibagi dua (a x t / 2)<br />
Rumus Lingkaran<br />
- Keliling : diameter dikali phi (d x phi) atau phi dikali 2 jari-jari (phi x (r + r)<br />
- Luas : phi dikali jari-jari dikali jari-jari (phi x r x r)<br />
- phi = 22/7 = 3,14<br />
Rumus Jajar Genjang atau Jajaran Genjang<br />
- Keliling : Penjumlahan dari keempat sisi yang ada (AB + BC + CD + DA)<br />
- Luas : alas dikali tinggi (a x t)<br />
Rumus Belah Ketupat<br />
- Keliling : Penjumlahan dari keempat sisi yang ada (AB + BC + CD + DA)<br />
- Luas : alas dikali panjang diagonal dibagi 2 (a x diagonal / 2)<br />
- Diagonal : Garis tengah dua sisi berlawanan<br />
Rumus Trapesium<br />
- Keliling : Penjumlahan dari keempat sisi yang ada (AB + BC + CD + DA)<br />
- Luas : Jumlah sisi sejajar dikali tinggi dibagi 2 ((AB + CD) / 2)<br />
</div>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com0tag:blogger.com,1999:blog-3157576316325034465.post-59210403475541006262009-12-02T05:34:00.000-08:002009-12-02T05:34:32.790-08:00Rumus Bangun Ruang - Matematika<div class="content"> Rumus Kubus<br />
- Volume : Sisi pertama dikali sisi kedua dikali sisi ketiga (S pangkat 3)<br />
Rumus Balok<br />
- Volume : Panjang dikali lebar dikali tinggi (p x l x t)<br />
Rumus Bola<br />
- Volume : phi dikali jari-jari dikali tinggi pangkat tiga kali 4/3 (4/3 x phi x r x t x t x t)<br />
- Luas : phi dikali jari-jari kuadrat dikali empat (4 x phi x r x r)<br />
Rumus Limas Segi Empat<br />
- Volume : Panjang dikali lebar dikali tinggi dibagi tiga (p x l x t x 1/3)<br />
- Luas : ((p + l) t) + (p x l)<br />
Rumus Tabung<br />
- Volume : phi dikali jari-jari dikali jari-jari dikali tinggi (phi x r2 x t)<br />
- Luas : (phi x r x 2) x (t x r)<br />
Rumus Kerucut<br />
- Volume : phi dikali jari-jari dikali jari-jari dikali tinggi dibagi tiga (phi x r2 x t x 1/3)<br />
- Luas : (phi x r) x (S x r)<br />
- S : Sisi miring kerucut dari alas ke puncak (bukan tingi)<br />
Rumus Prisma Segitiga Siku-siku<br />
- Volume : alas segitiga kali tinggi segitiga kali tinggi prisma bagi dua (as x ts x tp/2)<br />
</div>Januar Ivanhttp://www.blogger.com/profile/16471371764439671083noreply@blogger.com0