Exercise Zone : Fungsi Komposisi (Composite Function)

Berikut ini adalah kumpulan soal mengenai Fungsi Komposisi. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Diketahui ${f(x)=2x-3}$ dan ${\left(g\circ f\right)(x)=2x+1}$. Tentukan nilai $g(x)$.

dari Nasywa

$\begin{aligned}\left(g\circ f\right)(x)&=2x+1\\g\left(f(x)\right)&=2x+1\\g(2x-3)&=2x+1\end{aligned}$

CARA BIASA	CARA CEPAT
Misal $2x-3=u$ $\begin{aligned}2x&=u+3\\x&=\dfrac{u+3}2\end{aligned}$ $\begin{aligned}g(2x-3)&=2x+1\\g(u)&=2\left(\dfrac{u+3}2\right)+1\\[8pt]&=u+3+1\\&=u+4\\g(x)&=\boxed{\boxed{x+4}}\end{aligned}$	$\begin{aligned}g(2x-3)&=2x+1\\g(2x-3)&=2x{\color{blue}{-3+3}}+1\\g({\color{blue}{2x-3}})&={\color{blue}{2x-3}}+4\\g(x)&=\boxed{\boxed{x+4}}\end{aligned}$

Jadi, $g(x)=x+4$.

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No. 2

Jika ${f(x)=5x+3}$ dan ${g(f(x))=4x+9}$, nilai $g(13)$ adalah....

1. $13$
2. $14$
3. $15$

4. $16$
5. $17$

$\begin{aligned}f(x)&=13\\5x+3&=13\\5x&=10\\x&=2\end{aligned}$

$\begin{aligned}g(f(x))&=4x+9\\g(13)&=4(2)+9\\&=17\end{aligned}$

Jadi, $g(13)=17$.
JAWAB: E

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No. 3

Diketahui ${f(x)=2x+3}$ dan ${\left(g\circ f\right)(x)=4x^2+16x+16}$. Rumus fungsi $g(x)$ adalah ....

1. $x^2-2x$
2. $x^2-2x-1$
3. $x^2-2x+1$

4. $x^2+2x+1$
5. $x^2+2x-1$

$\begin{aligned}\left(g\circ f\right)(x)&=4x^2+16x+16\\g(f(x))&=4x^2+16x+16\\g(2x+3)&=4x^2+16x+16\end{aligned}$

Misal $t=2x+3$
$\begin{aligned}t-3&=2x\\\dfrac{t-3}2&=x\\x&=\dfrac{t-3}2\end{aligned}$

$\begin{aligned}g(t)&=4\left(\dfrac{t-3}2\right)^2+16\left(\dfrac{t-3}2\right)+16\\&=4\left(\dfrac{t^2-6t+9}4\right)+8(t-3)+16\\&=t^2-6t+9+8t-24+16\\&=t^2+2t+1\\g(x)&=\boxed{\boxed{x^2+2x+1}}\end{aligned}$

Jadi, $g(x)=x^2+2x+1$.
JAWAB: D

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No. 4

Jika ${g(x)=x-2}$ dan ${(g\circ f)(x)=x^2+2x+3}$, maka $(f\circ g)(3)$ adalah ....

1. $5$
2. $6$
3. $7$

4. $8$
5. $9$

$\begin{aligned}(f\circ g)(3)&=f(g(3))\\&=f(3-2)\\&=f(1)\end{aligned}$

$\begin{aligned}(g\circ f)(1)&=1^2+2(1)+3\\g(f(1))&=1+2+3\\f(1)-2&=6\\f(1)&=8\end{aligned}$

Jadi,

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No. 5

Diketahui ${f:R\to R}$, ${g:R\to R}$, ${g(x)=2x+3}$ dan ${\left(f\circ g\right)(x)=12x^2+32x+26}$. Rumus $f(x)=$ ....

1. ${3x^2-2x+5}$
2. ${3x^2-2x+37}$
3. ${3x^2-2x+50}$

4. ${3x^2+2x-5}$
5. ${3x^2+2x-50}$

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$\begin{aligned}\left(f\circ g\right)(x)&=12x^2+32x+26\\f\left(g(x)\right)&=12x^2+32x+26\\f\left(2x+3\right)&=12x^2+32x+26\end{aligned}$

Misal
$\begin{aligned}2x+3&=u\\2x&=u-3\\x&=\dfrac{u-3}2\end{aligned}$

$\begin{aligned}f\left(u\right)&=12\left(\dfrac{u-3}2\right)^2+32\left(\dfrac{u-3}2\right)+26\\[8pt]&=12\left(\dfrac{u^2-6u+9}4\right)+16\left(u-3\right)+26\\[8pt]&=3\left(u^2-6u+9\right)+16u-48+26\\&=3u^2-18u+27+16u-22\\&=3u^2-2u+5\\f(x)&=\boxed{\boxed{3x^2-2x+5}}\end{aligned}$

Jadi, $f(x)=3x^2-2x+5$.
JAWAB: A

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No. 6

Jika ${f(x)=\dfrac3{2x-1}}$ dan ${\left(f\circ g\right)(x)=\dfrac{3x+3}{x-1}}$, maka ${g(x-1)=}$

1. $\dfrac{x+2}x$, $x\neq0$
2. $\dfrac{x-2}x$, $x\neq0$
3. $\dfrac{x+1}x$, $x\neq0$

4. $\dfrac{x-1}x$, $x\neq0$
5. $\dfrac{x}{x+1}$, $x\neq-1$

$\begin{aligned}\left(f\circ g\right)(x)&=\dfrac{3x+3}{x-1}\\[8pt]f\left( g(x)\right)&=\dfrac{3x+3}{x-1}\\[8pt]\dfrac3{2g(x)-1}&=\dfrac{3x+3}{x-1}\\[8pt]\left(2g(x)-1\right)(3x+3)&=3(x-1)\\2(3x+3)g(x)-3x-3&=3x-3\\(6x+6)g(x)&=6x\\g(x)&=\dfrac{6x}{6x+6}\color{red}\dfrac{:6}{:6}\\[8pt]&=\dfrac{x}{x+1}\\[8pt]g(x-1)&=\dfrac{x-1}{x-1+1}\\&=\boxed{\boxed{\dfrac{x-1}x}}\end{aligned}$

Jadi, $g(x-1)=\dfrac{x-1}x$.
JAWAB: D

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No. 7

Jika tabel berikut menyatakan hasil fungsi $f$ dan $g$

$x$	$0$	$1$	$2$	$3$
$f(x)$	$1$	$3$	$1$	$-1$
$g(x)$	$2$	$0$	$1$	$2$

maka nilai ${(f\circ g\circ f)(0)+(g\circ f\circ g)(1)=}$

1. $4$
2. $3$
3. $2$

4. $1$
5. $0$

$\begin{array}{rl}(f\circ g\circ f)(0)+(g\circ f\circ g)(1)& =f(g(f(0)))+g(f(g(1)))\\ & =f(g(1))+g(f(0))\\ & =f(0)+g(1)\\ & =1+0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, ${(f\circ g\circ f)(0)+(g\circ f\circ g)(1)=1}$.
JAWAB: D

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