Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll}\\ 6.&\textrm{Fungsi}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{ditentukan oleh}\\ &g(x)=x^{2}-x+3\: \: \textrm{dan}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{sehingga}\: \: (f\circ g)(x)=3x^{2}-3x+4 \:, \\ &\textrm{maka fungsi}\: \: f(x-2)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&2x-11&&&\textrm{d}.&3x-7\\ \textrm{b}.&2x-7&\textrm{c}.&3x+1&\color{red}\textrm{e}.&3x-11 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =f(x)\\ &\color{blue}g(x) =x^{2}-x+3\end{cases}\\ &\left ( f\circ g \right )(x)=3x^{2}-3x+4\\ &\begin{aligned}f(g(x))&=3x^{2}-3x+4\\ f(x^{2}-&x+3)=3(x^{2}-x+3)-5\\ \textrm{Sehing}&\textrm{ga}\\ f(x)&=3x-5\\ f(x-2)&=3(x-2)-5\\ &=3x-11 \end{aligned} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{dengan}\: \: f(x)=x-3\: \: \textrm{dan}\\ & g(x)=x^{2}+5.\: \textrm{Jika}\: \: (f\circ g)(x)=(g\circ f)(x)\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&4\\ \color{red}\textrm{b}.&2&\textrm{c}.&3&\textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =x-3\\ &\color{blue}g(x) =x^{2}+5\end{cases}\\ &\left ( f\circ g \right )(x)=(g\circ f)(x)\\ &\begin{aligned}f(g(x))&=g(f(x))\\ (x^{2}+5)-3&=(x-3)^{2}+5\\ x^{2}+2&=x^{2}-6x+14\\ 6x&=14-2\\ x&=\displaystyle \frac{12}{6}\\ x&=2 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{dengan}\: \: f(x)=3x-10\: \: \textrm{dan}\\ & g(x)=4x+n.\: \textrm{Jika}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ &\textrm{maka nilai}\: \: n\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&15&&&\textrm{d}.&-10\\ \textrm{b}.&10&\textrm{c}.&5&\color{red}\textrm{e}.&-15 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =3x-10\\ &\color{blue}g(x) =4x+n\end{cases}\\ &\left ( g\circ f \right )(x)-(f\circ g)(x)=0\\ &\begin{aligned}g(f(x))&=f(g(x))\\ 3(4x+n)-10&=4(3x-10)+n\\ 12x+3n-10&=12x-40+n\\ 2n&=-30\\ x&=\displaystyle \frac{-30}{2}\\ x&=-15 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: f(x)=\sqrt{2x+3}\\ & \textrm{dan}\: \: g(x)=x^{2}+1,\: \textrm{maka}\: \: \left ( f\circ g \right )(2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&2,24&&&\textrm{d}.&6\\ \textrm{b}.&3\qquad&\color{red}\textrm{c}.&3,61\qquad&\textrm{e}.&6,16 \end{array}\\ & (\textbf{SAT Subject Test})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=f\left ( g(x) \right )\\ &=\sqrt{2g(x)+3}\\ &=\sqrt{2\left ( x^{2}+1 \right )+3}\\ \left ( f\circ g \right )(2)&=\sqrt{2\left ( 2^{2}+1 \right )+3}\\ &=\sqrt{2(5)+3}\\ &=\sqrt{13}\\ &\color{blue}\approx 3,61 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Misalkan}\: \: f(x)=x^{2},\: \: g(x)=2x\\ &\textrm{dan}\: \: h(x)=1-x.\\ & \textrm{Fungsi}\: \: (f\circ g\circ h)(x)=\: ....\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&4x^{2}-8x+4\\ \textrm{b}.&4x^{2}+8x-4\\ \textrm{c}.&2x^{2}-4x+1\\ \textrm{d}.&x^{2}-2x+1\\ \textrm{e}.&4-2x+x^{2} \end{array}\\\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \bullet \quad&f(x)=x^{2}\\ \bullet \quad &g(x)=2x\\ \bullet \quad&h(x)=1-x\\ \textrm{mak}&\textrm{a}\\ &(f\circ g\circ h)=f\left ( g(h(x)) \right )\\ &\: =\left ( 2(1-x) \right )^{2}\\ &\: =(2-2x)^{2}\\ &\: =4x^{2}-8x+4 \end{aligned} \end{array}$