Soal dan Pembahasan Komposisi Fungsi

\(\begin{align}
(g ∘ f)(x)&= g(f(x))\\
&=g(x^3+2)\\
&=\frac{2}{(x^3+2)-1}\\
&=\frac{2}{x^3+1}\\
\end{align}\)

\(\begin{align}
(f∘g)(x)&=f(g(x))\\
&=f(\sqrt{2x})\\
&=\frac{2(\sqrt{2x})}{(\sqrt{2x})^2-4}\\
&=\frac{2\sqrt{2x}}{2x-4}\\
&=\frac{2\sqrt{2x}}{2(x-2)}\\
&=\frac{\sqrt{2x}}{x-2}\\
\end{align}\)

\(\begin{align}
(g∘f)(x)&=g(f(x))\\
&=g(\sqrt{x+1})\\
&=(\sqrt{x+1})^2-1\\
&=x+1-1\\
&=x\\
\end{align}\)

\((f∘g)(x)=f(g(x))\)
\((f∘g)(a)=f(g(a))=81\)
\(⇒f(5a+4)=81\)
\(⇒6(5a+4)-3=81\)
\(⇒6(5a+4)=84\)
\(⇒(5a+4)=14\)
\(⇒5a=14-4\)
\(⇒5a=10\)
\(⇒a=2\)

\(\begin{align}
(g∘f)(x)&=g(f(x))\\
(g∘f)(2)&=g(f(2))\\
&=g(4(2)+2)\\
&=g(10)\\
\end{align}\)
karena \(g(x) = 4\) maka \(g(10)=4\)

\(f(g(x))=-\frac{x}{2}+1\)
\(⇒4g(x)=-\frac{x}{2}+1\)
\(⇒g(x)=\frac{-\frac{x}{2}+1}{4}\)
\(⇒g(x)=\frac{1}{8}(-x+2)\)

\(\begin{align}
(f∘g)(x)&=f(g(x))\\
&=f(x^2-1)\\
&=2(x^2-1)-2\\
&=2x^2-2-2\\
&=2x^2-4\\
(f∘g)(x+1)&=2(x+1)^2-4\\
&=2(x^2+2x+1)-4\\
&=2x^2+4x+2-4\\
&=2x^2+4x-2\\
\end{align}\)

Cara 1: Penjabaran biasa

\((f∘g)(x)=x^2+3x+1\)
\(⇒f(g(x))=x^2+3x+1\)
\(⇒f(x+1)=x^2+3x+1\)

langkah selanjutnya persamaan bagian kanan diubah dalam bentuk \((x+1)\)

\(⇒f(x+1)=x^2+2x+1+x\)
\(⇒f(x+1)=(x+1)^2+(x+1)-1\)
\(⇒f(x)=x^2+x-1\)

Cara 2: menggunakan cara invers,
misalkan \(y=x+1⇒x=y-1\), maka

\((f∘g)(x)=x^2+3x+1\)
\(⇒f(g(x))=x^2+3x+1\)
\(⇒f(x+1)=x^2+3x+1\)
\(⇒f(y)=(y-1)^2+3(y-1)+1\)
\(⇒f(y)=y^2-2y+1+3y-3+1\)
\(⇒f(y)=y^2+y-1\)
\(⇒f(x)=x^2+x-1\)

\((f∘g)(x)=\frac{x}{3x-2}\)
\(⇒(f(g(x))=\frac{x}{3x-2}\)
\(⇒\frac{1}{2(g(x)-1}=\frac{x}{3x-2}\)
\(⇒3x-2=(2g(x)-1)x\)
\(⇒3x-2=2xg(x)-x\)
\(⇒4x-2=2xg(x)\)
\(⇒g(x)=\frac{4x-2}{2x}\)
\(⇒g(x)=2-\frac{1}{x}\)

Cara 1: Penjabaran
\((g∘f)(x)=2x+1\)
\(g(f(x))=2x+1\)
\(g(2x-3)=2x+1\)
\(g(2x-3)=2x-3+4\)
\(g(x)=x+4\)

Cara : Menggunakan invers
misalkan \(y=2x-3⇒2x=y+3\)
\((g∘f)(x)=2x+1\)
\(g(f(x))=2x+1\)
\(g(2x-3)=2x+1\)
\(g(y)=y+3+1\)
\(g(y)=y+4\)
\(g(x)=x+4\)