Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{l}\\ 6.& \text{Fungsi}g:\mathbb{R}\to \mathbb{R}\text{ditentukan oleh}\\ & g(x)=x^2-x+3\text{dan}f:\mathbb{R}\to \mathbb{R}\\ & \text{sehingga}(f\circ g)(x)=3x^2-3x+4,\\ & \text{maka fungsi}f(x-2)=....\\ & \begin{array}{l}\\ \text{a}.& 2x-11& & & \text{d}.& 3x-7\\ \text{b}.& 2x-7& \text{c}.& 3x+1& \text{e}.& 3x-11\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=f(x)\\ & g(x)=x^2-x+3\end{array}\\ & (f\circ g)(x)=3x^2-3x+4\\ & \begin{array}{rl}f(g(x))& =3x^2-3x+4\\ f(x^2-& x+3)=3(x^2-x+3)-5\\ \text{Sehing}& \text{ga}\\ f(x)& =3x-5\\ f(x-2)& =3(x-2)-5\\ & =3x-11\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}g:\mathbb{R}\to \mathbb{R}\\ & \text{dengan}f(x)=x-3\text{dan}\\ & g(x)=x^2+5.\text{Jika}(f\circ g)(x)=(g\circ f)(x)\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& 4\\ \text{b}.& 2& \text{c}.& 3& \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=x-3\\ & g(x)=x^2+5\end{array}\\ & (f\circ g)(x)=(g\circ f)(x)\\ & \begin{array}{rl}f(g(x))& =g(f(x))\\ (x^2+5)-3& =(x-3{)}^2+5\\ x^2+2& =x^2-6x+14\\ 6x& =14-2\\ x& ={\displaystyle \frac{12}{6}}\\ x& =2\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}g:\mathbb{R}\to \mathbb{R}\\ & \text{dengan}f(x)=3x-10\text{dan}\\ & g(x)=4x+n.\text{Jika}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ & \text{maka nilai}n\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 15& & & \text{d}.& -10\\ \text{b}.& 10& \text{c}.& 5& \text{e}.& -15\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=3x-10\\ & g(x)=4x+n\end{array}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ & \begin{array}{rl}g(f(x))& =f(g(x))\\ 3(4x+n)-10& =4(3x-10)+n\\ 12x+3n-10& =12x-40+n\\ 2n& =-30\\ x& ={\displaystyle \frac{-30}{2}}\\ x& =-15\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}f(x)=\sqrt{2x+3}\\ & \text{dan}g(x)=x^2+1,\text{maka}(f\circ g)(2)=....\\ \\ & \begin{array}{l}\\ \text{a}.& 2,24& & & \text{d}.& 6\\ \text{b}.& 3& \text{c}.& 3,61& \text{e}.& 6,16\end{array}\\ & (\mathbf{\text{SAT Subject Test}})\\ \\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ & =\sqrt{2g(x)+3}\\ & =\sqrt{2(x^2+1)+3}\\ (f\circ g)(2)& =\sqrt{2(2^2+1)+3}\\ & =\sqrt{2(5)+3}\\ & =\sqrt{13}\\ & \approx 3,61\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Misalkan}f(x)=x^2,g(x)=2x\\ & \text{dan}h(x)=1-x.\\ & \text{Fungsi}(f\circ g\circ h)(x)=....\\ & \begin{array}{l}\\ \text{a}.& 4x^2-8x+4\\ \text{b}.& 4x^2+8x-4\\ \text{c}.& 2x^2-4x+1\\ \text{d}.& x^2-2x+1\\ \text{e}.& 4-2x+x^2\end{array}\\ \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}\\ \bullet & f(x)=x^2\\ \bullet & g(x)=2x\\ \bullet & h(x)=1-x\\ \text{mak}& \text{a}\\ & (f\circ g\circ h)=f(g(h(x)))\\ & ={(2(1-x))}^2\\ & =(2-2x{)}^2\\ & =4x^2-8x+4\end{array}\end{array}$