Fungsi Komposisi dan Fungsi Invers

$\text{A. Fungsi Komposisi}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|cc|}\hline \text{Syarat}& \text{Sifat-sifat}\\ \begin{array}{rl}& R_{{}_f}\cap D_{{}_g}\ne \left\{\right\}\end{array}& \begin{array}{rl}1.& \text{Tidak komutatif}\\ & (f\circ g)(x)\ne (g\circ f)(x)\\ 2.& \text{Bersifat asosiatif}\\ & f\circ (g\circ h)(x)=(f\circ g)\circ h(x)\\ 3.& \text{Adanya unsur dentitas}\\ & (f\circ I)(x)=(I\circ f)(x)=f(x)\end{array}\\ \hline\end{array}$.

$\text{B. Fungsi Invers}$

$\begin{array}{rl}& \bullet \text{Suatu fungsi}f:A\to B\\ & \text{memiliki fungsi invers}g:B\to A\\ & \mathbf{\text{jika dan hanya jika}}f\\ & \text{merupakan fungsi}\mathbf{\text{bijektif}}\\ & \bullet \text{Jika fungsi}g\text{ada, maka}\\ & g\text{dinyatakan dengan}{f}^{-1}(\text{dibaca}:f\mathbf{\text{invers}})\end{array}$.

$\begin{array}{l}\\ & \mathbf{\text{Catatan}}\\ & \text{Perlu diingat bahwa pada invers }\\ & \text{fungsi komposisi berlaku }\\ & \text{ketentuan sebagai berikut}\\ ⧫& {(g\circ f)}^{-1}(x)=(f^{-1}\circ g^{-1})(x)\\ ⧫& {(f\circ g)}^{-1}(x)=(g^{-1}\circ f^{-1})(x)\\ ⧫& f(x)=({\left(f^{-1}\right)}^{-1}(x))\\ ⧫& x=f^{-1}(f(x))=(f^{-1}\circ f)(x)\\ & =(f\circ f^{-1})(x)=f(f^{-1}(x))\end{array}$.

$\text{CONTOH SOAL}$

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

Jawab: 

hanya no. 1 a saja yang dibahas

$\begin{array}{rl}1.\text{a}.(f\circ g)(x)& =f(g(x))\\ & =f(5x+3)\\ & =2-(5x+3)\\ & =-5x-1\text{dan}& \\ (g\circ f)(x)& =g(f(x))\\ & =g(2-x)\\ & =5(2-x)+3\\ & =10-5x+3\\ & =13-10x\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui bahwa}g(x)=3x+2\\ & \text{dan}(g\circ f)(x)=4x-5.\text{Tentukanlah}f(x)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}(g\circ f)(x)& =4x-5\\ g(f(x))& =4x-5\\ 3.f(x)+2& =4x-5\\ 3.f(x)& =4x-7\\ f(x)& ={\displaystyle \frac{4x-7}{3}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 4.& \text{Diketahui}f(x)=3x\text{dan}g(x)=3^x.\\ & \text{Tentukanlah rumus untuk}{}^{{}^{27}}\log (g\circ f)(x)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^{{}^{27}}\log (g\circ f)(x)& ={}^{{}^{27}}\log g(f(x))\\ & ={}^{{}^{27}}\log 3^{3x}\\ & ={}^{{}^{3^3}}\log 3^{3x}\\ & ={}^{{}^{{}^{\left(3^3\right)}}}\log {\left(3^3\right)}^x\\ & =x\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah invers dari}f(x)=6^{2x}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =6^{2x}\\ y& =6^{2x}\\ \log y& =\log 6^{2x}\\ \log y& =2x\log 6\\ {\displaystyle \frac{\log y}{\log 6}}& =2x\\ {\displaystyle \frac{\log y}{2\log 6}}& =x\\ {\displaystyle \frac{\log y}{\log 6^2}}& =x\\ {\displaystyle \frac{\log y}{\log 36}}& =x\\ x& ={\displaystyle \frac{\log y}{\log 36}}& \\ x& ={}^{{}^{36}}\log y\\ f^{-1}(x)& ={}^{{}^{36}}\log x\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah inver dari}\\ & f(x)={\displaystyle \frac{2x+3}{4x-5},x\ne {\displaystyle \frac{5}{4}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{2x+3}{4x-5}}\\ y& ={\displaystyle \frac{2x+3}{4x-5}}\\ (4x-5)y& =2x+3\\ 4xy-5y& =2x+3\\ 4xy-2x& =5y+3\\ x(4y-2)& =5y+3\\ x& ={\displaystyle \frac{5y+3}{4y-2}}\\ f^{-1}(y)& ={\displaystyle \frac{5y+3}{4y-2}}\\ \text{maka},& \\ f^{-1}(x)& ={\displaystyle \frac{5x+3}{4x-2},x\ne {\displaystyle \frac{1}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}f(x)=-2x-4\text{dan}g(x)={\displaystyle \frac{20-3x}{2},}\\ & \text{maka nilai dari}(f\circ g{)}^{-1}(2)=....\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Perhatikan bahwa}}\\ & \begin{array}{rl}& \\ & (f\circ g{)}^{-1}(x)=(g^{-1}\circ f^{-1})(x)\\ & \end{array}\\ & \begin{array}{|cc|}\hline (f\circ g{)}^{-1}(x)& (g^{-1}\circ f^{-1})(x)\\ \begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ y& =-2\left({\displaystyle \frac{20-3x}{2}}\right)-4\\ y& =3x-20-4\\ y& =3x-24\\ y+24& =3x\\ x& ={\displaystyle \frac{y+24}{3}}\\ (f\circ g{)}^{-1}y& ={\displaystyle \frac{y+24}{3}}\\ (f\circ g{)}^{-1}(2)& ={\displaystyle \frac{2+24}{3}}\\ & ={\displaystyle \frac{26}{3}}\end{array}& \begin{array}{rl}(g^{-1}\circ f^{-1})(x)& =g^{-1}(f^{-1}(x))\\ & =......\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ (g^{-1}\circ f^{-1})(2)& =....\\ & =....\end{array}\\ \hline\end{array}\end{array}$.

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.