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No. Diketahui
{f(x)=2x-3} dan
{\left(g\circ f\right)(x)=2x+1} . Tentukan nilai
g(x) .
Penyelesaian \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. Jika
{f(x)=5x+3} dan
{g(f(x))=4x+9} , nilai
g(13) adalah....
Penyelesaian \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. Diketahui
{f(x)=2x+3} dan
{\left(g\circ f\right)(x)=4x^2+16x+16} . Rumus fungsi
g(x) adalah ....
Penyelesaian \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} 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. Jika
{g(x)=x-2} dan
{(g\circ f)(x)=x^2+2x+3} , maka
(f\circ g)(3) adalah ....
Penyelesaian
\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\\
(f\circ g)(3)&=\boxed{\boxed{8}}
\end{aligned}
No. 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} Penyelesaian \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} \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.
Jika
g(x)=\dfrac{ax+2}{x+3} dan
h(x)=\dfrac{5x-4}{-x+a} , nilai
(g\circ h)(1)=2 , maka nilai dari
3a adalah
Penyelesaian \begin{aligned}
(g\circ h)(1)&=2\\
g(h(1))&=2\\
g\left(\dfrac{5(1)-4}{-1+a}\right)&=2\\[8pt]
g\left(\dfrac1{a-1}\right)&=2\\[8pt]
\dfrac{a\left(\dfrac1{a-1}\right)+2}{\dfrac1{a-1}+3}&=2\\[20pt]
\dfrac{\dfrac{a+2(a-1)}{a-1}}{\dfrac{1+3(a-1)}{a-1}}&=2\\[20pt]
\dfrac{\dfrac{a+2a-2}{a-1}}{\dfrac{1+3a-3}{a-1}}&=2\\[20pt]
\dfrac{\dfrac{3a-2}{a-1}}{\dfrac{3a-2}{a-1}}&=2\\[20pt]
\dfrac{3a-2}{3a-2}&=2\\[8pt]
3a-2&=6a-4\\
3a&=\boxed{\boxed{2}}
\end{aligned}
No.
Diketahui fungsi
f(x)=5x+3 dan
g(x)=x^2+ax+b . Jika
\left(g\circ f\right)(1)=53 dan
\left(g\circ f\right)(0)=8 , maka nilai
a+b adalah
Penyelesaian \begin{aligned}
\left(g\circ f\right)(1)&=53\\
g\left(f(1)\right)&=53\\
g\left(5(1)+3\right)&=53\\
g(8)&=53\\
8^2+a(8)+b&=53\\
64+8a+b&=53\\
8a+b&=-11
\end{aligned} \begin{aligned}
\left(g\circ f\right)(1)&=53\\
g\left(f(0)\right)&=8\\
g\left(5(0)+3\right)&=8\\
g(3)&=8\\
3^2+a(3)+b&=8\\
9+3a+b&=8\\
3a+b&=-1
\end{aligned} \begin{aligned}
8a+b&=-11\\
3a+b&=-1\qquad-\\\hline
5a&=-10\\
a&=-2
\end{aligned} \begin{aligned}
3a+b&=-1\\
3(-2)+b&=-1\\
-6+b&=-1\\
b&=5
\end{aligned} \begin{aligned}
a+b&=-2+5\\
&=\boxed{\boxed{3}}
\end{aligned}
No.
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 Penyelesaian \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.
Diketahui
{f(x)=x^2-5x+1} dan
{g(x)=x-4} . Jika
{\left(g\circ f\right)(a)=21} , maka
a= ...
Alternatif Penyelesaian
\begin{aligned} \left(g\circ f\right)(a)&=21\\
g\left(f(a)\right)&=21\\
g\left(a^2-5a+1\right)&=21\\
a^2-5a+1-4&=21\\
a^2-5a-3&=21\\
a^2-5a-24&=0\\
(a-8)(a+3)&=0\end{aligned}
No.
Diketahui
{f (x) = 3x + p} dan
{g (x) = 4x - 120} dengan
(f\circ g) (x) = (g\circ f) (x),} maka
f (20) =
Alternatif Penyelesaian
\begin{aligned}
\left(f\circ g\right)(x)&=\left(g\circ f\right)(x)\\
f\left(g(x)\right)&=g\left(f(x)\right)\\
f\left(4x-120\right)&=g\left(3x+p\right)\\
3\left(4x-120\right)+p&=4\left(3x+p\right)-120\\
12x-360+p&=12x+4p-120\\
-360+p&=4p-120\\
-3p&=240\\
p&=\dfrac{240}{-3}\\
&=-80
\end{aligned}
\begin{aligned}
f(x)&=3x-80\\
f(20)&=3(20)-80\\
&=60-80\\
&=\boxed{\boxed{-20}}
\end{aligned}
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