Berikut ini adalah kumpulan soal mengenai Fungsi Komposisi. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Diketahui
{f(x)=2x-3} dan
{\left(g\circ f\right)(x)=2x+1}. Tentukan nilai
g(x).
\(\begin{aligned}
\left(g\circ f\right)(x)&=2x+1\\
g\left(f(x)\right)&=2x+1\\
g(2x-3)&=2x+1
\end{aligned}\)
| CARA BIASA | CARA CEPAT |
|---|
Misal 2x-3=u
\(\begin{aligned}
2x&=u+3\\
x&=\dfrac{u+3}2
\end{aligned}\)
\(\begin{aligned}
g(2x-3)&=2x+1\\
g(u)&=2\left(\dfrac{u+3}2\right)+1\\[8pt]
&=u+3+1\\
&=u+4\\
g(x)&=\boxed{\boxed{x+4}}\end{aligned}\) | \(\begin{aligned}
g(2x-3)&=2x+1\\
g(2x-3)&=2x{\color{blue}{-3+3}}+1\\
g({\color{blue}{2x-3}})&={\color{blue}{2x-3}}+4\\
g(x)&=\boxed{\boxed{x+4}}
\end{aligned}\) |
No. 2
Jika
{f(x)=5x+3} dan
{g(f(x))=4x+9}, nilai
g(13) adalah....
\(\begin{aligned}
f(x)&=13\\
5x+3&=13\\
5x&=10\\
x&=2
\end{aligned}\)
\(\begin{aligned}
g(f(x))&=4x+9\\
g(13)&=4(2)+9\\
&=17
\end{aligned}\)
No. 3
Diketahui
{f(x)=2x+3} dan
{\left(g\circ f\right)(x)=4x^2+16x+16}. Rumus fungsi
g(x) adalah ....
\(\begin{aligned}
\left(g\circ f\right)(x)&=4x^2+16x+16\\
g(f(x))&=4x^2+16x+16\\
g(2x+3)&=4x^2+16x+16
\end{aligned}\)
Misal t=2x+3
\(\begin{aligned}
t-3&=2x\\
\dfrac{t-3}2&=x\\
x&=\dfrac{t-3}2
\end{aligned}\)
\(\begin{aligned}
g(t)&=4\left(\dfrac{t-3}2\right)^2+16\left(\dfrac{t-3}2\right)+16\\
&=4\left(\dfrac{t^2-6t+9}4\right)+8(t-3)+16\\
&=t^2-6t+9+8t-24+16\\
&=t^2+2t+1\\
g(x)&=\boxed{\boxed{x^2+2x+1}}
\end{aligned}\)
No. 4
Jika
{g(x)=x-2} dan
{(g\circ f)(x)=x^2+2x+3}, maka
(f\circ g)(3) adalah ....
\(\begin{aligned}
(f\circ g)(3)&=f(g(3))\\
&=f(3-2)\\
&=f(1)
\end{aligned}\)
\(\begin{aligned}
(g\circ f)(1)&=1^2+2(1)+3\\
g(f(1))&=1+2+3\\
f(1)-2&=6\\
f(1)&=8
\end{aligned}\)
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)= ....
- {3x^2-2x+5}
- {3x^2-2x+37}
- {3x^2-2x+50}
\(\begin{aligned}
\left(f\circ g\right)(x)&=12x^2+32x+26\\
f\left(g(x)\right)&=12x^2+32x+26\\
f\left(2x+3\right)&=12x^2+32x+26
\end{aligned}\)
Misal
\(\begin{aligned}
2x+3&=u\\
2x&=u-3\\
x&=\dfrac{u-3}2
\end{aligned}\)
\(\begin{aligned}
f\left(u\right)&=12\left(\dfrac{u-3}2\right)^2+32\left(\dfrac{u-3}2\right)+26\\[8pt]
&=12\left(\dfrac{u^2-6u+9}4\right)+16\left(u-3\right)+26\\[8pt]
&=3\left(u^2-6u+9\right)+16u-48+26\\
&=3u^2-18u+27+16u-22\\
&=3u^2-2u+5\\
f(x)&=\boxed{\boxed{3x^2-2x+5}}
\end{aligned}\)
No. 6
Jika
{f(x)=\dfrac3{2x-1}} dan
{\left(f\circ g\right)(x)=\dfrac{3x+3}{x-1}}, maka
{g(x-1)=}
- \dfrac{x+2}x, x\neq0
- \dfrac{x-2}x, x\neq0
- \dfrac{x+1}x, x\neq0
- \dfrac{x-1}x, x\neq0
- \dfrac{x}{x+1}, x\neq-1
\(\begin{aligned}
\left(f\circ g\right)(x)&=\dfrac{3x+3}{x-1}\\[8pt]
f\left( g(x)\right)&=\dfrac{3x+3}{x-1}\\[8pt]
\dfrac3{2g(x)-1}&=\dfrac{3x+3}{x-1}\\[8pt]
\left(2g(x)-1\right)(3x+3)&=3(x-1)\\
2(3x+3)g(x)-3x-3&=3x-3\\
(6x+6)g(x)&=6x\\
g(x)&=\dfrac{6x}{6x+6}\color{red}\dfrac{:6}{:6}\\[8pt]
&=\dfrac{x}{x+1}\\[8pt]
g(x-1)&=\dfrac{x-1}{x-1+1}\\
&=\boxed{\boxed{\dfrac{x-1}x}}
\end{aligned}\)
No. 7
Jika tabel berikut menyatakan hasil fungsi
f dan
g
maka nilai
{(f\circ g\circ f)(0)+(g\circ f\circ g)(1)=}
\(\eqalign{
(f\circ g\circ f)(0)+(g\circ f\circ g)(1)&=f(g(f(0)))+g(f(g(1)))\\
&=f(g(1))+g(f(0))\\
&=f(0)+g(1)\\
&=1+0\\
&=\boxed{\boxed{1}}
}\)
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