Contoh Soal 4 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: f\: \: \textrm{dan}\: \: g\: \textrm{adalah fungsi yang }\\ &\textrm{mempunyai invers dan memenuhi}\\ & f(2x)=g(x-3),\: \textrm{maka}\: \: f^{-1}(x)\\ & \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&g^{-1}\left ( \displaystyle \frac{x}{2}- \frac{2}{3}\right )\\ \textrm{b}.&g^{-1}\left ( \displaystyle \frac{x}{2} \right )-\displaystyle \frac{2}{3}\\ \textrm{c}.&g^{-1}(2x+6)\\ \textrm{d}.&2g^{-1}(x)-6\\ \color{red}\textrm{e}.&2g^{-1}(x)+6 \end{array}\\ & (\textbf{SBMPTN 2016 Mat Das})\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned} &\textrm{Misalkan bahwa}\\ &f(2x)=g(x-3)=x,\\ & \textrm{maka}\\ &\begin{cases} f^{-1}(x) &=2x \\ g^{-1}(x) &=x-3 \end{cases} \end{aligned}\\ &\begin{array}{|c|c|}\hline \textrm{Sintak}&\textrm{Hasil}\\\hline \begin{aligned}g^{-1}(x) &=x-3\\ x&=g^{-1}(x)+3\\ & \end{aligned}&\begin{aligned}f^{-1}(x) &=2x\\ &=2\left ( g^{-1}(x)+3 \right )\\ &=2g^{-1}(x)+6 \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Jika}\: \: f^{-1}(x)=\displaystyle \frac{x-1}{5}\: \: \textrm{dan}\: \: g^{-1}(x)=\displaystyle \frac{3-x}{2},\\ & \textrm{maka}\: \: \left (f\circ g \right )^{-1}(6)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0\qquad&\color{red}\textrm{c}.&1\qquad&\textrm{e}.&3 \end{array}\\ & (\textbf{UMPTN 1995})\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui b}&\textrm{ahwa}:\\ &\begin{cases} f^{-1}(x)&=\displaystyle \frac{x-1}{5} \\ g^{-1}(x)&=\displaystyle \frac{3-x}{2} \end{cases}\\ \left (f\circ g \right )^{-1}(x)&=\left (g^{-1}\circ f^{-1} \right )(x)\\ &=\displaystyle \frac{3-\left ( \displaystyle \frac{x-1}{5} \right )}{2}\\ \left (f\circ g \right )^{-1}(6)&=\displaystyle \frac{3-\left ( \displaystyle \frac{6-1}{5} \right )}{2}\\ &=\displaystyle \frac{3-1}{2}=\frac{2}{2}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Invers dari}\: \: f(x)=125^{x}\: \: \textrm{adalah}\: \: f^{-1}(x)\\ &\textrm{Nilai dari}\: \: f^{-1}\left ( 5\sqrt{5} \right )=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{3}{5}\\ \color{red}\textrm{b}.&\displaystyle \frac{1}{2}&\textrm{c}.&\displaystyle \frac{1}{6}&\textrm{e}.&-\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned} &\textrm{Diketahui bahwa}\: \: f(x)=125^{x},\: \textrm{maka}\\ &f(x)=y=125^{x}\\ &\textrm{Kedua ruas dilogkan masing-masing}\\ &\log y=\log 125^{x}\\ &\Leftrightarrow \log y=x\log 125\\ &\Leftrightarrow x\log 125=\log y\\ &\Leftrightarrow x=\displaystyle \frac{\log y}{\log 125}\\ &\Leftrightarrow x= \, ^{125}\log y\\ &\Leftrightarrow f^{-1}(x)=\, ^{125}\log x\\ &\textrm{Selanjutnya}\\ &f^{-1}(x)=\, ^{125}\log x\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\, ^{125}\log (5\sqrt{5})\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\, ^{5^{3}}\log 5^{.^{ \frac{3}{2}}}\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\displaystyle \frac{\displaystyle \frac{3}{2}}{3}\,^{5}\log 5\\ &\Leftrightarrow f^{-1}(5\sqrt{5})= \displaystyle \frac{1}{2}.1\\ &\Leftrightarrow f^{-1}(5\sqrt{5})= \color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

Contoh Soal 3 Fungsi Komposisi dan Fungsi Invers

 $\begin{array}{ll}\\ 11.&\textrm{Diketahui beberapa fungsi memiliki }\\ &\textrm{sifat-sifat sebagaimana berikut ini}:\\ &(i)\quad \Phi (-x)=-\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &(ii)\quad \Phi (-x)=\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &\textrm{Jika diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\\ & \textrm{memiliki sifat}\: \: (i)\: \: \textrm{dan fungsi}\: \: h\: \: \textrm{dan}\: \: k\\ &\textrm{memiliki sifat}\: \: (ii),\: \: \textrm{maka pernyataan }\\ &\textrm{berikut yang salah adalah}\: ....\\ &(1)\quad (f+g)(-x)=-(f+g)(x)\\ &(2)\quad (f.k)(-x)=-(f.k)(x)\\ &(3)\quad (h-k)(-x)=(h-k)(x)\\ &(4)\quad (h-g)(-x)=(h-g)(x)\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&(1),(2)\: \textrm{dan}\: (3)\\ \textrm{b}.&(1)\: \textrm{dan}\: (3)\\ \textrm{c}.&(2)\: \textrm{dan}\: (4)\\ \textrm{d}.&(4)\: \textrm{saja}\\ \textrm{e}.&\textrm{semuanya benar} \end{array}\\ & (\textbf{SIMAK UI 2014 Mat Das})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} \Phi (-x)=-\Phi (x) \\ (\textrm{fungsi ganjil})\begin{cases} f & \text{ misal } f(x)=x \\ g & \text{ misal } g(x)=2x \end{cases} \\\\ \Phi (-x)=\Phi (x)\\ (\textrm{fungsi genap})\begin{cases} h & \text{ misal } h(x)=x^{2} \\ k & \text{ misal } k(x)=2x^{2} \end{cases} \end{cases}\\ & \end{aligned}\\ &\begin{array}{|c|}\hline \begin{aligned}(1)\quad (f+g)(-x)&=-(f+g)(x)\\ &\textrm{benar}\\ (2)\qquad (f.k)(-x)&=-(f.k)(x)\\ &\textrm{benar}\\ \end{aligned}\\\hline \begin{aligned} (3)\quad (h-k)(-x)&=(h-k)(x)\\ &\textrm{benar}\\ (4)\quad (h-g)(-x)&=(h-g)(x)\\ &\color{blue}\textrm{salah} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{2x-1}\: \: \textrm{dan}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2}\\ &\textrm{maka}\: \: g(x)=\: ....\\ &(\textbf{UMPTN 1998})\\ &\begin{array}{llllll}\\ \textrm{a}.&x+\displaystyle \frac{1}{2}&&&\textrm{d}.&1-\displaystyle \frac{2}{x}\\\\ \textrm{b}.&x-\displaystyle \frac{1}{2}&\color{red}\textrm{c}.&2-\displaystyle \frac{1}{x}&\textrm{e}.&2-\displaystyle \frac{1}{2x} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dan}\\ &(f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{1}{\left (\displaystyle \frac{3x-2}{x} \right )}\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\color{magenta}2g(x)-1=\displaystyle \frac{3x-2}{x}\\ &\Leftrightarrow 2g(x)=1+\left ( 3-\displaystyle \frac{2}{x} \right )\\ &\Leftrightarrow 2g(x)=4-\displaystyle \frac{2}{x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dengan}\\ &f^{-1}(x)=\displaystyle \frac{x+1}{2x}.........(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\displaystyle \frac{f(g(x))+1}{2\left ( f(g(x)) \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\left ( \displaystyle \frac{x}{3x-2} \right )+1}{2\left ( \displaystyle \frac{x}{3x-2} \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\displaystyle \frac{4x-2}{3x-2}}{\displaystyle \frac{2x}{3x-2}}\\ &\Leftrightarrow g(x)=\displaystyle \frac{4x-2}{2x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: f(x)=x^{2}-9\: \: \textrm{dan}\: \: (f\circ g)(x)=x(x-6)\\ &\textrm{rumus fungsi}\: \: g(x)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+3&&&\textrm{d}.&3x+1\\ \textrm{b}.&x-3&\color{red}\textrm{c}.&-x&\textrm{e}.&x \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dan}\\ &(f\circ g)(x)=x(x-6)=x^{2}-6x,\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x+9-9\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=(x-3)^{2}-9\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\qquad g(x)=x-3 \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{x+9}\: .......(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=x^{2}-6x,\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\sqrt{\left ( f(g(x)) \right )+9}\\ &\Leftrightarrow g(x)=\sqrt{x^{2}-6x+9}\\ &\Leftrightarrow g(x)=\sqrt{(x-3)^{2}}\\ &\Leftrightarrow g(x)=x-3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}}\: \: \textrm{dan}\\ &\left ( f\circ g \right )(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\\ & \textrm{maka}\: \: g(x+2)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{x+3}&&&\textrm{d}.&x+3\\ \textrm{b}.&\displaystyle \frac{1}{x-2}\qquad&\textrm{c}.&x-2\qquad&\color{red}\textrm{e}.&x+5 \end{array}\\ & (\textbf{UM UGM 2010 Mat Das})\\\\ &\textrm{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ f\left ( g(x) \right )&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \displaystyle \frac{1}{\sqrt{\left (g(x) \right )^{2}-2}}&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \left (g(x) \right )^{2}-2&=x^{2}+6x+7\\ \left (g(x) \right )^{2}&=x^{2}+6x+9\\ g(x)&=\sqrt{x^{2}+6x+9}=\sqrt{(x+3)^{2}}\\ g(x)&=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5 \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}},\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{\displaystyle \frac{1}{x^{2}}+2}\: .......(\textrm{akan ditunjukkan})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\: \: \textrm{maka}\\ &\begin{array}{|c|c|}\hline \begin{aligned}f(x) =y&= \displaystyle \frac{1}{\sqrt{x^{2}-2}}\\ y^{2}&=\displaystyle \frac{1}{x^{2}-2}\\ x^{2}-2&=\displaystyle \frac{1}{y^{2}}\\ x^{2}&=\displaystyle \frac{1}{y^{2}}+2\\ x&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(y)&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(x)&=\sqrt{\displaystyle \frac{1}{x^{2}}+2} \end{aligned}&\begin{aligned}g(x)&=\left (f^{-1}\circ f\circ g \right )(x)\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \right )^{2}}+2}\\ &=\sqrt{\left ( x^{2}+6x+7 \right )+2}\\ &=\sqrt{\left ( x^{2}+6x+9 \right )}\\ &=\sqrt{(x+3)^{2}}\\ &=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5\\ & \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: g(x)=2x+4\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=4x^{2}+8x-3,\\ & \textrm{maka}\: \: f^{-1}(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+9\\ \textrm{b}.&\sqrt{x}+2\\ \textrm{c}.&x^{2}-4x-3\\ \textrm{d}.&\sqrt{x+1}+2\\ \color{red}\textrm{e}.&\sqrt{x+7}+2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{Sintak 1}&\textrm{ Sintak 2}&\textrm{Hasil Invers}\\\hline \begin{aligned}g(x)=y&=2x+4\\ y-4&=2x\\ x&=\displaystyle \frac{y-4}{2}\\ f^{-1}(y)&=\displaystyle \frac{y-4}{2}\\ f^{-1}(x)&=\displaystyle \frac{x-4}{2}\\ & \end{aligned}&\begin{aligned}f(x)&=\left (f\circ g\circ g^{-1} \right )(x)\\ &=4\left ( g^{^{-1}}(x) \right )^{2}+8\left ( g^{-1}(x) \right )-3\\ &=4\left ( \displaystyle \frac{x-4}{2} \right )^{2}+8\left ( \displaystyle \frac{x-4}{2} \right )-3\\ &=\left ( \displaystyle x^{2}-8x+16 \right )+4x-16-3\\ &=x^{2}-4x-3\\ &=x^{2}-4x+4-7\\ &=(x-2)^{2}-7 \end{aligned}&\begin{aligned}f(x)=y&=(x-2)^{2}-7\\ y+7&=(x-2)^{2}\\ \sqrt{y+7}&=(x-2)\\ (x-2)&=\sqrt{y+7}\\ x&=\sqrt{y+7}+2\\ f^{-1}(y)&=\sqrt{y+7}+2\\ f^{-1}(x)&=\sqrt{x+7}+2 \end{aligned} \\\hline \end{array} \end{array}$

Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

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

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

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

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

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

Contoh Soal 1 Fungsi Komposisi dan Fungsi Invers

 $\begin{array}{ll}\\ 1.&\textrm{Diketahui fungsi}\: \: f(2x)=8x-9\\ & \textrm{dan}\: \: g(3x+1)=6x+3.\\ &\textrm{Rumus untuk}\: \: \left ( f+g \right )(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&6x+8&&&\textrm{d}.&14x-6\\ \color{red}\textrm{b}.&6x-8&\textrm{c}.&14x+6&\textrm{e}.&6x-6 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(2x) =8x-9\\ &\Rightarrow \quad f(x)=f\left ( 2\left ( \displaystyle \frac{x}{2} \right ) \right )\\ &=8\left ( \displaystyle \frac{x}{2} \right )-9=4x-9 \\ &\color{blue}g(3x+1) =6x+3\\ &\Rightarrow \quad g(x)=g\left ( 3\left ( \displaystyle \frac{x-1}{3} \right )+1 \right )\\ &=6\left ( \displaystyle \frac{x-1}{3} \right )+3\\ &=2x+1 \end{cases}\\ &\left ( f+ g \right )(x)=(4x-9)+(2x+1)\\ &=6x-8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui fungsi}\: \: f(x)=2x-1\\ & \textrm{dan}\: \: g(x)=x^{2}.\: \: \textrm{Fungsi}\: \: (f+g)(x^{2})=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&x^{2}+2x-1\\ \color{red}\textrm{b}.&x^{4}+2x^{2}-1\\ \textrm{c}.&x^{4}+2x-1\\ \textrm{d}.&x^{4}+(2x-1)^{2}\\ \textrm{e}.&x^{4}+2x \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =2x-1\\ &\color{blue}g(x) =x^{2} \end{cases}\\ &\left ( f+ g \right )(x)=(2x-1)+(x^{2})\\ &=x^{2}+2x-1\\ &\textrm{maka}\\ &(f+g)(x^{2})=(x^{2})^{2}+2(x^{2})-1\\ &\: \: \: \: \qquad\qquad =x^{4}+2x^{2}-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: f(x)=3-x,\: \: \textrm{maka}\\ & f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2x^{2}-6x+4\\ \textrm{b}.&2x^{2}+4x+6\\ \textrm{c}.&2x^{2}-4x-6\\ \textrm{d}.&6x+4\\ \color{red}\textrm{e}.&-4x+6 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ & f(x)=3-x,\: \textrm{sehingga}\\ &f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)\\ &=\left ( 3-x^{2} \right )+\left ( 3-x \right )^{2}-2(3-x)\\ &=\left (3-x^{2} \right )+\left ( 9-6x+x^{2} \right )-\left ( 6-2x \right )\\ &=-x^{2}+x^{2}-6x+2x+3+9-6\\ &=-4x+6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\\ & g:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dirumuskan dengan}\\ &f(x)=x-1\: \: \textrm{dan}\: \: g(x)=x^{2}+2x-3.\\ & \textrm{Fungsi komposisi}\: \: g\: \: \textrm{atas}\: \: f\\ &\textrm{dinotasikan dengan}\\ &\begin{array}{ll}\\ \color{red}\textrm{a}.&\left ( g\circ f \right )(x)=x^{2}-4\\ \textrm{b}.&\left ( g\circ f \right )(x)=x^{2}-5\\ \textrm{c}.&\left ( g\circ f \right )(x)=x^{2}-6\\ \textrm{d}.&\left ( g\circ f \right )(x)=x^{2}-4x-4\\ \textrm{e}.&\left ( g\circ f \right )(x)=x^{2}-4x-5 \end{array}\\ & (\textbf{UN 2016})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} \color{blue}f(x) &\color{blue}=x-1 \\ \color{blue}g(x) &\color{blue}=x^{2}+2x-3 \end{cases}\\ \left ( g\circ f \right )&=g\left ( f(x) \right )\\ &=\left ( f(x) \right )^{2}+2\left ( f(x) \right )-3\\ &=\left ( x-1 \right )^{2}+2\left ( x-1 \right )-3\\ &=\left (x^{2}-2x+1 \right )+\left ( 2x-2 \right )-3\\ &=x^{2}-4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui fungsi}\: \: f(x)=6x-3\\ & \textrm{dan}\: \: g(x)=5x+4 \: \: \textrm{dan}\: \: (f\circ g)(a)=81\\ &\textrm{Nilai}\: \: a\: \: \textrm{adalah}\: ........\: \: (\textbf{Ebtanas 2001})\\ &\begin{array}{llllll}\\ \textrm{a}.&-2&&&\color{red}\textrm{d}.&2\\ \textrm{b}.&-1&\textrm{c}.&1&\textrm{e}.&27 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =6x-3\\ &\color{blue}g(x) =5x+4\end{cases}\\ &\left ( f\circ g \right )(a)=81\\ &\textrm{maka}\\ &(f\circ g)(x)=f\left ( g(x) \right )\\ &\: \qquad\qquad =6(g(x))-3\\ &\: \qquad\qquad =6(5x+4)-3\\ &\: \qquad\qquad =30x+24-3\\ &\: \qquad\qquad =30x+21=81\\ &30x+21=81\\ &30x=81-21=60\\ &x=\displaystyle \frac{60}{30}=2 \end{aligned} \end{array}$.

Fungsi Komposisi dan Fungsi Invers

$\color{blue}\textrm{A. Fungsi Komposisi}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline \textrm{Syarat}&\textrm{Sifat-sifat}\\\hline \begin{aligned}&R_{_{f}}\cap D_{_{g}}\neq \left \{ \: \right \} \end{aligned}&\begin{aligned}1.\: \: &\textrm{Tidak komutatif}\\ &(f\circ g)(x)\neq (g\circ f)(x)\\ 2.\: \: &\textrm{Bersifat asosiatif}\\ & f\circ (g\circ h)(x)= (f\circ g)\circ h(x)\\ 3.\: \: &\textrm{Adanya unsur dentitas}\\ & (f\circ I)(x)=(I\circ f)(x)=f(x) \end{aligned}\\\hline \end{array}$.

$\color{blue}\textrm{B. Fungsi Invers}$

$\begin{aligned}&\bullet \quad \textrm{Suatu fungsi}\: \: f:A\rightarrow B\\ &\qquad \textrm{memiliki fungsi invers} \: \: g:B\rightarrow A\\ &\qquad \textbf{jika dan hanya jika}\: \: f\\ &\qquad \textrm{merupakan fungsi}\: \textbf{bijektif}\\ &\bullet \quad \textrm{Jika fungsi}\: \: g\: \: \textrm{ada, maka}\\ &\qquad g\: \: \textrm{dinyatakan dengan}\: \: f^{-1}\: \: (\textrm{dibaca}:\: \: f\: \: \textbf{invers})\end{aligned}$.

$\begin{array}{ll}\\ &\color{blue}\textbf{Catatan}\\ &\textrm{Perlu diingat bahwa pada invers }\\ &\textrm{fungsi komposisi berlaku }\\ &\textrm{ketentuan sebagai berikut}\\ \blacklozenge &\left ( g\circ f \right )^{-1}(x)=\left ( f^{-1}\circ g^{-1} \right )(x)\\ \blacklozenge &\left ( f\circ g \right )^{-1}(x)=\left ( g^{-1}\circ f^{-1} \right )(x)\\ \blacklozenge &f(x)=\left ( \left (f^{-1} \right )^{-1}(x) \right )\\ \blacklozenge &x=f^{-1}\left ( f(x) \right )=\left ( f^{-1}\circ f \right )(x)\\ &\quad=\left ( f\circ f^{-1} \right )(x)=f\left ( f^{-1}(x) \right ) \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah}\: \: (f\circ g)(x)\: \: \textrm{dan}\: \: (g\circ f)(x)\: \: \textrm{Jika}:\\ &\textrm{a}.\quad f(x)=2-x\: \: \textrm{dah}\: \: g(x)=5x+3\\ &\textrm{b}.\quad f(x)=2x+1\: \: \textrm{dah}\: \: g(x)=x^{2}-4\\ &\textrm{c}.\quad f(x)=\displaystyle \frac{5}{x-4}\: \: \textrm{dah}\: \: g(x)=3x^{2}\\ &\textrm{d}.\quad f(x)=\sqrt{4-x}\: \: \textrm{dah}\: \: g(x)=x^{2}+x\\ &\textrm{e}.\quad f(x)=x^{3}+1\: \: \textrm{dah}\: \: g(x)=\displaystyle \frac{x}{x-1}\\ &\textrm{f}.\quad f(x)=\displaystyle \frac{3}{x-2}\: \: \textrm{dah}\: \: g(x)=\sqrt{x-4}\\ \end{array}$.

Jawab: 

hanya no. 1 a saja yang dibahas

$\begin{aligned}1.\quad \textrm{a}.\: \: \: \: (f\circ g)(x)&=f\left ( g(x) \right )\\ &=f\left ( 5x+3 \right )\\ &=2-\left ( 5x+3 \right )\\ &=-5x-1\ \textrm{dan}\: \: \: \: \: \: \: \: \: \: &\\ (g\circ f)(x)&=g\left ( f(x) \right )\\ &=g(2-x)\\ &=5(2-x)+3\\ &=10-5x+3\\ &=13-10x \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui bahwa}\: \: g(x)=3x+2\\ & \textrm{dan}\: \: (g\circ f)(x)=4x-5.\: \: \textrm{Tentukanlah}\: \: f(x)\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}(g\circ f)(x)&=4x-5\\ g(f(x))&=4x-5\\ 3.f(x)+2&=4x-5\\ 3.f(x)&=4x-7\\ f(x)&=\displaystyle \frac{4x-7}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa}\: \: g(x)=x+4\: \: \textrm{dan}\\ & (f\circ g)(x)=2x^{2}+3.\: \: \textrm{Tentukanlah}\: \: f(x)\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}(f\circ g)(x)&=2x^{2}+3\\ f(g(x))&=2x^{2}+3\\ f(x+4)&=2x^{2}+3,\\ & \textrm{misalkan}\: \: x+4=a\Rightarrow x=a-4,\\ \textrm{sehingga}&,\\ f(a)&=2(a-4)^{2}+3\\ f(a)&=2(a^{2}-8a+16)+3\\ &=2a^{2}-16a+35\\ f(x)&=2x^{2}-16x+35 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: f(x)=3x\: \: \textrm{dan}\: \: g(x)=3^{x}.\\ &\textrm{Tentukanlah rumus untuk}\: \: ^{^{27}}\log (g\circ f)(x)\\\\ &\color{red}\textrm{Jawab}:\\ &\begin{aligned}^{^{27}}\log (g\circ f)(x)&=\: ^{^{27}}\log g(f(x))\\ &=\: ^{^{27}}\log 3^{3x}\\ &=\: ^{^{3^{3}}}\log 3^{3x}\\ &=\: ^{^{^{\left (3^{3} \right )}}}\log \left ( 3^{3} \right )^{x}\\ &=x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah invers dari}\: \: f(x)=6^{2x}\\\\ &\color{red}\textrm{Jawab}:\\ &\begin{aligned}f(x)&=6^{2x}\\ y&=6^{2x}\\ \log y&=\log 6^{2x}\\ \log y&=2x\log 6\\ \displaystyle \frac{\log y}{\log 6}&=2x\\ \displaystyle \frac{\log y}{2\log 6}&=x\\ \displaystyle \frac{\log y}{\log 6^{2}}&=x\\ \displaystyle \frac{\log y}{\log 36}&=x\\ x&=\displaystyle \frac{\log y}{\log 36}&\\ x&=\: ^{^{36}}\log y\\ f^{-1}(x)&=\: ^{^{36}}\log x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukanlah inver dari}\\ & f(x)=\displaystyle \frac{2x+3}{4x-5},\: \: \: x\neq \displaystyle \frac{5}{4}\\\\ &\color{red}\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{2x+3}{4x-5}\\ y&=\displaystyle \frac{2x+3}{4x-5}\\ (4x-5)y&=2x+3\\ 4xy-5y&=2x+3\\ 4xy-2x&=5y+3\\ x(4y-2)&=5y+3\\ x&=\displaystyle \frac{5y+3}{4y-2}\\ f^{-1}(y)&=\displaystyle \frac{5y+3}{4y-2}\\ \textrm{maka},&\\ f^{-1}(x)&=\displaystyle \frac{5x+3}{4x-2},\: \: \: x\neq \displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: f(x)=-2x-4\: \: \textrm{dan}\: \: g(x)=\displaystyle \frac{20-3x}{2},\\ & \textrm{maka nilai dari}\: \: (f\circ g)^{-1}(2)=....\\\\ &\color{blue}\textrm{Jawab}:\\ &\textbf{Perhatikan bahwa}\\ &\begin{aligned}&\\ &(f\circ g)^{-1}(x)=\left ( g^{-1}\circ f^{-1} \right )(x)\\ & \end{aligned}\\ &\begin{array}{|c|c|}\hline (f\circ g)^{-1}(x)&\left ( g^{-1}\circ f^{-1} \right )(x)\\\hline \begin{aligned}(f\circ g)(x)&=f(g(x))\\ y&=-2\left ( \displaystyle \frac{20-3x}{2} \right )-4\\ y&=3x-20-4\\ y&=3x-24\\ y+24&=3x\\ x&=\displaystyle \frac{y+24}{3}\\ (f\circ g)^{-1}y&=\displaystyle \frac{y+24}{3}\\ (f\circ g)^{-1}(2)&=\displaystyle \frac{2+24}{3}\\ &=\displaystyle \frac{26}{3} \end{aligned}&\begin{aligned}\left ( g^{-1}\circ f^{-1} \right )(x)&=g^{-1}\left ( f^{-1}(x) \right )\\ &=......\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ \left ( g^{-1}\circ f^{-1} \right )(2)&=....\\ &=.... \end{aligned}\\\hline \end{array} \end{array}$.

DAFTAR PUSTAKA

1. Soedyarto, Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
2. Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.