[tex]\begin{aligned}\sf a.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-7}{2}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x-5}{2}\,}\\\sf b.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-2}{6}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x-11}{6}\,}\\\sf c.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-4}{15}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x+8}{15}\,}\\\end{aligned}[/tex]
Komposisi dan Invers Fungsi
Untuk menentukan [tex](g\circ f)^{-1}(x)[/tex] dan [tex](f\circ g)^{-1}(x)[/tex], setidaknya kita punya dua cara.
__________________
[tex]f(x)=x+2,\ g(x)=2x+3[/tex]
CARA I
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=2f(x)+3\\&=2(x+2)+3\\(g\circ f)(x)&=2x+7\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)=y&=2x+7\\2x&=y-7\\x&=\frac{y-7}{2}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-7}{2}\,}\end{aligned}[/tex]
Kemudian,
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=g(x)+2\\&=2x+3+2\\(f\circ g)(x)&=2x+5\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)(x)=y&=2x+5\\2x&=y-5\\x&=\frac{y-5}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-5}{2}\,}\end{aligned}[/tex]
CARA II
[tex]\begin{aligned}f(x)&=x+2&&\Rightarrow f^{-1}(x)=x-2\\g(x)&=2x+3&&\Rightarrow g^{-1}(x)=\frac{x-3}{2}\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=g^{-1}(x)-2\\&=\frac{x-3}{2}-2\\&=\frac{x-3-4}{2}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-7}{2}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)-3}{2}\\&=\frac{x-2-3}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-5}{2}\,}\end{aligned}[/tex]__________________
[tex]f(x)=3x-1,\ g(x)=2x+4[/tex]
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=2f(x)+4\\&=2(3x-1)+4\\(g\circ f)(x)&=6x+2\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)(x)=y&=6x+2\\6x&=y-2\\x&=\frac{y-2}{6}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-2}{6}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=3g(x)-1\\&=3(2x+4)-1\\(f\circ g)(x)&=6x+11\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)(x)=y&=6x+11\\6x&=y-11\\x&=\frac{y-11}{6}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-11}{6}\,}\end{aligned}[/tex]
[tex]\begin{aligned}f(x)&=3x-1&&\Rightarrow f^{-1}(x)=\frac{x+1}{3}\\g(x)&=2x+4&&\Rightarrow g^{-1}(x)=\frac{x-4}{2}\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=\frac{g^{-1}(x)+1}{3}\\&=\frac{\dfrac{x-4}{2}+1}{3}\\&=\frac{\dfrac{x-4+2}{2}}{3}\\&=\dfrac{x-2}{2}\cdot\frac{1}{3}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-2}{6}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)-4}{2}\\&=\frac{\dfrac{x+1}{3}-4}{2}\\&=\frac{\dfrac{x+1-12}{3}}{2}\\&=\frac{x-11}{3}\cdot\frac{1}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-11}{6}\,}\end{aligned}[/tex]__________________
[tex]f(x)=5x+2,\ g(x)=3x-2[/tex]
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=3f(x)-2\\&=3(5x+2)-2\\(g\circ f)(x)&=15x+4\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)(x)=y&=15x+4\\15x&=y-4\\x&=\frac{y-4}{15}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-4}{15}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=5g(x)+2\\&=5(3x-2)+2\\(f\circ g)(x)&=15x-8\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)(x)=y&=15x-8\\15x&=y+8\\x&=\frac{y+8}{15}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x+8}{15}\,}\end{aligned}[/tex]
[tex]\begin{aligned}f(x)&=5x+2&&\Rightarrow f^{-1}(x)=\frac{x-2}{5}\\g(x)&=3x-2&&\Rightarrow g^{-1}(x)=\frac{x+2}{3}\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=\frac{g^{-1}(x)-2}{5}\\&=\frac{\dfrac{x+2}{3}-2}{5}\\&=\frac{\dfrac{x+2-6}{3}}{5}\\&=\dfrac{x-4}{3}\cdot\frac{1}{5}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-4}{15}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)+2}{3}\\&=\frac{\dfrac{x-2}{5}+2}{3}\\&=\frac{\dfrac{x-2+10}{5}}{3}\\&=\dfrac{x+8}{5}\cdot\frac{1}{3}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x+8}{15}\,}\end{aligned}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
[tex]\begin{aligned}\sf a.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-7}{2}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x-5}{2}\,}\\\sf b.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-2}{6}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x-11}{6}\,}\\\sf c.\ \ &(g\circ f)^{-1}(x)=\boxed{\,\frac{x-4}{15}\,}\\&(f\circ g)^{-1}(x)=\boxed{\,\frac{x+8}{15}\,}\\\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
Komposisi dan Invers Fungsi
Untuk menentukan [tex](g\circ f)^{-1}(x)[/tex] dan [tex](f\circ g)^{-1}(x)[/tex], setidaknya kita punya dua cara.
Tentukan [tex](g\circ f)(x)[/tex] dan [tex](f\circ g)(x)[/tex] terlebih dahulu, lalu tentukan inversnya.
Kita gunakan identitas:
[tex]\begin{aligned}\bullet\ &(g\circ f)^{-1}(x)=\left(f^{-1}\circ g^{-1}\right)(x)\\\bullet\ &(f\circ g)^{-1}(x)=\left(g^{-1}\circ f^{-1}\right)(x)\\\end{aligned}[/tex]
__________________
Soal a.
[tex]f(x)=x+2,\ g(x)=2x+3[/tex]
CARA I
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=2f(x)+3\\&=2(x+2)+3\\(g\circ f)(x)&=2x+7\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)=y&=2x+7\\2x&=y-7\\x&=\frac{y-7}{2}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-7}{2}\,}\end{aligned}[/tex]
Kemudian,
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=g(x)+2\\&=2x+3+2\\(f\circ g)(x)&=2x+5\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(f\circ g)(x)=y&=2x+5\\2x&=y-5\\x&=\frac{y-5}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-5}{2}\,}\end{aligned}[/tex]
CARA II
[tex]\begin{aligned}f(x)&=x+2&&\Rightarrow f^{-1}(x)=x-2\\g(x)&=2x+3&&\Rightarrow g^{-1}(x)=\frac{x-3}{2}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=g^{-1}(x)-2\\&=\frac{x-3}{2}-2\\&=\frac{x-3-4}{2}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-7}{2}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)-3}{2}\\&=\frac{x-2-3}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-5}{2}\,}\end{aligned}[/tex]
__________________
Soal b.
[tex]f(x)=3x-1,\ g(x)=2x+4[/tex]
CARA I
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=2f(x)+4\\&=2(3x-1)+4\\(g\circ f)(x)&=6x+2\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)=y&=6x+2\\6x&=y-2\\x&=\frac{y-2}{6}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-2}{6}\,}\end{aligned}[/tex]
Kemudian,
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=3g(x)-1\\&=3(2x+4)-1\\(f\circ g)(x)&=6x+11\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(f\circ g)(x)=y&=6x+11\\6x&=y-11\\x&=\frac{y-11}{6}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-11}{6}\,}\end{aligned}[/tex]
CARA II
[tex]\begin{aligned}f(x)&=3x-1&&\Rightarrow f^{-1}(x)=\frac{x+1}{3}\\g(x)&=2x+4&&\Rightarrow g^{-1}(x)=\frac{x-4}{2}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=\frac{g^{-1}(x)+1}{3}\\&=\frac{\dfrac{x-4}{2}+1}{3}\\&=\frac{\dfrac{x-4+2}{2}}{3}\\&=\dfrac{x-2}{2}\cdot\frac{1}{3}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-2}{6}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)-4}{2}\\&=\frac{\dfrac{x+1}{3}-4}{2}\\&=\frac{\dfrac{x+1-12}{3}}{2}\\&=\frac{x-11}{3}\cdot\frac{1}{2}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x-11}{6}\,}\end{aligned}[/tex]
__________________
Soal c.
[tex]f(x)=5x+2,\ g(x)=3x-2[/tex]
CARA I
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=3f(x)-2\\&=3(5x+2)-2\\(g\circ f)(x)&=15x+4\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)=y&=15x+4\\15x&=y-4\\x&=\frac{y-4}{15}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-4}{15}\,}\end{aligned}[/tex]
Kemudian,
[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=5g(x)+2\\&=5(3x-2)+2\\(f\circ g)(x)&=15x-8\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(f\circ g)(x)=y&=15x-8\\15x&=y+8\\x&=\frac{y+8}{15}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x+8}{15}\,}\end{aligned}[/tex]
CARA II
[tex]\begin{aligned}f(x)&=5x+2&&\Rightarrow f^{-1}(x)=\frac{x-2}{5}\\g(x)&=3x-2&&\Rightarrow g^{-1}(x)=\frac{x+2}{3}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=\frac{g^{-1}(x)-2}{5}\\&=\frac{\dfrac{x+2}{3}-2}{5}\\&=\frac{\dfrac{x+2-6}{3}}{5}\\&=\dfrac{x-4}{3}\cdot\frac{1}{5}\\\therefore\ (g\circ f)^{-1}(x)&=\boxed{\,\frac{x-4}{15}\,}\end{aligned}[/tex]
[tex]\begin{aligned}(f\circ g)^{-1}(x)&=\left(g^{-1}\circ f^{-1}\right)(x)\\&=g^{-1}\left(f^{-1}(x)\right)\\&=\frac{f^{-1}(x)+2}{3}\\&=\frac{\dfrac{x-2}{5}+2}{3}\\&=\frac{\dfrac{x-2+10}{5}}{3}\\&=\dfrac{x+8}{5}\cdot\frac{1}{3}\\\therefore\ (f\circ g)^{-1}(x)&=\boxed{\,\frac{x+8}{15}\,}\end{aligned}[/tex]