Misal f : R→R dan f : R→R dengan f(x)=4x+3 dan g(x)=x−1x, x=1. Tentukanlah: a. (f∘g)−1(x) b. (f∘g)−1(2)

Pertanyaan

Misal $f\;:\;R\rightarrow R$ dan $f\;:\;R\rightarrow R$ dengan $f\left(x\right)=4x+3$ dan $g open parentheses x close parentheses equals fraction numerator x over denominator x minus 1 end fraction comma space x not equal to 1.$

Tentukanlah:

a. $\left(f\circ g\right)^{-1}\left(x\right)$

b. $\left(f\circ g\right)^{-1}\left(2\right)$

H. Nur

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Jawaban

bentuk dari ${\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol=$ $fraction numerator bold italic x bold minus bold 3 over denominator bold italic x bold minus bold 7 end fraction$ dan $\begin{array}{rcl}{\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\mathbf2\boldsymbol)}&\boldsymbol=&\end{array}$ $\begin{array}{rcl}&&\frac{\mathbf1}{\mathbf5}\end{array}$ .

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah ${\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol=$ $fraction numerator bold italic x bold minus bold 3 over denominator bold italic x bold minus bold 7 end fraction$ dan $\begin{array}{rcl}{\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\mathbf2\boldsymbol)}&\boldsymbol=&\end{array}$ $\begin{array}{rcl}&&\frac{\mathbf1}{\mathbf5}\end{array}$ .

Ingat!

Ada 2 cara menentukan formula invers fungsi komposisi, yaitu:

Mula-mula menentukan fungsi komposisinya, kemudian diinverskan.
Mula-mula menentukan invers masing-masing fungsi kemudian dikomposisikan.

Ingat pula $\left(f\circ g\right)^{-1}\left(x\right)=\left(g^{-1}\circ f^{-1}\right)\left(x\right)$ .

Diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 4 x plus 3 end cell row cell g open parentheses x close parentheses end cell equals cell fraction numerator x over denominator x minus 1 end fraction comma space x not equal to 1 end cell end table$

Ditanya:

a. $\left(f\circ g\right)^{-1}\left(x\right)$

b. $\left(f\circ g\right)^{-1}\left(2\right)$

Jawab:

Cari $f^{-1}\left(x\right)$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 4 x plus 3 end cell row y equals cell 4 x plus 3 end cell row cell 4 x end cell equals cell y minus 3 end cell row x equals cell fraction numerator y minus 3 over denominator 4 end fraction end cell row cell f to the power of negative 1 end exponent open parentheses y close parentheses end cell equals cell fraction numerator y minus 3 over denominator 4 end fraction end cell row cell f to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x minus 3 over denominator 4 end fraction end cell end table$

Cari $g^{-1}\left(x\right)$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses x close parentheses end cell equals cell fraction numerator x over denominator x minus 1 end fraction end cell row y equals cell fraction numerator x over denominator x minus 1 end fraction end cell row cell y open parentheses x minus 1 close parentheses end cell equals x row cell x y minus y end cell equals x row cell x y minus x end cell equals y row cell x open parentheses y minus 1 close parentheses end cell equals y row x equals cell fraction numerator y over denominator y minus 1 end fraction end cell row cell g to the power of negative 1 end exponent open parentheses y close parentheses end cell equals cell fraction numerator y over denominator y minus 1 end fraction end cell row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x over denominator x minus 1 end fraction end cell end table$

Mencari $\left(f\circ g\right)^{-1}\left(x\right)$ dengan sifat invers $\left(f\circ g\right)^{-1}\left(x\right)=\left(g^{-1}\circ f^{-1}\right)\left(x\right)$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses fraction numerator x minus 3 over denominator 4 end fraction close parentheses end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses minus begin display style 4 over 4 end style end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator x minus 3 minus 4 over denominator 4 end fraction end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator x minus 3 over denominator 4 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator x minus 7 over denominator 4 end fraction end style close parentheses end fraction end cell row blank equals cell fraction numerator x minus 3 over denominator up diagonal strike 4 end fraction times fraction numerator up diagonal strike 4 over denominator x minus 7 end fraction end cell row blank equals cell fraction numerator x minus 3 over denominator x minus 7 end fraction end cell end table$

Mencari nilai $\left(f\circ g\right)^{-1}\left(2\right)$ dengan substitusi $x=2$ pada $\left(f\circ g\right)^{-1}\left(x\right)$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x minus 3 over denominator x minus 7 end fraction end cell row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 2 minus 3 over denominator 2 minus 7 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator negative 5 end fraction end cell row blank equals cell 1 fifth end cell end table$

Dengan demikian, bentuk dari ${\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol=$ $fraction numerator bold italic x bold minus bold 3 over denominator bold italic x bold minus bold 7 end fraction$ dan $\begin{array}{rcl}{\boldsymbol(\boldsymbol f\boldsymbol\circ\boldsymbol g\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\mathbf2\boldsymbol)}&\boldsymbol=&\end{array}$ $\begin{array}{rcl}&&\frac{\mathbf1}{\mathbf5}\end{array}$ .