Jika f(x)=x−3, g(x)=x−12 dan h(x)=4+x2. Tunjukkan bahwa (h∘g∘f)(x)=(h∘g∘f)−1(x).

Pertanyaan

Jika $f open parentheses x close parentheses equals x minus 3 comma space g open parentheses x close parentheses equals fraction numerator 2 over denominator x minus 1 end fraction$ dan $h\left(x\right)=4+\frac2x$ . Tunjukkan bahwa $\left(h\circ g\circ f\right)\left(x\right)=\left(h\circ g\circ f\right)^{-1}\left(x\right).$

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

nilai ${\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol={\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}$ adalah sama-sama $\boldsymbol x$ .

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut ${\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol={\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}$ adalah sama-sama $\boldsymbol x$ .

Ingat!

Misalkan fungsi $f:A\rightarrow B$ , fungsi $g:B\rightarrow C$ , dan fungsi $h:C\rightarrow D$ , maka terdapat komposisi dari tiga fungsi. yaitu $\left(h\circ g\circ f\right):A\rightarrow D$ .
$\left(h\circ g\circ f\right)\left(x\right)=h\left(g\left(f\left(x\right)\right)\right)$

Diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell x minus 3 end cell row cell g open parentheses x close parentheses end cell equals cell fraction numerator 2 over denominator x minus 1 end fraction end cell row cell h open parentheses x close parentheses end cell equals cell 4 plus 2 over x end cell end table$

Ditanya: Tunjukkan bahwa $\left(h\circ g\circ f\right)\left(x\right)=\left(h\circ g\circ f\right)^{-1}\left(x\right).$

Jawab:

Dari rumus $\left(h\circ g\circ f\right)\left(x\right)=h\left(g\left(f\left(x\right)\right)\right)$ , maka harus mencari $g\left(f\left(x\right)\right)$ dahulu.

$table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses f open parentheses x close parentheses close parentheses end cell equals cell g open parentheses x minus 3 close parentheses end cell row blank equals cell fraction numerator 2 over denominator open parentheses x minus 3 close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 2 over denominator x minus 4 end fraction end cell end table$

Mencari $\left(h\circ g\circ f\right)\left(x\right)=h\left(g\left(f\left(x\right)\right)\right)$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses h ring operator g ring operator f close parentheses open parentheses x close parentheses end cell equals cell h open parentheses g open parentheses f open parentheses x close parentheses close parentheses close parentheses end cell row blank equals cell h open parentheses fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell row blank equals cell 4 plus fraction numerator 2 over denominator begin display style fraction numerator 2 over denominator x minus 4 end fraction end style end fraction end cell row blank equals cell 4 plus up diagonal strike 2 times fraction numerator x minus 4 over denominator up diagonal strike 2 end fraction end cell row blank equals cell 4 plus x minus 4 end cell row blank equals x end table$

Mencari $\begin{array}{rcl}&&\left(h\circ g\circ f\right)^{-1}\left(x\right)\end{array}$ :

$\begin{array}{rcl}y&=&\left(h\circ g\circ f\right)\left(x\right)=x\\y&=&x\\x&=&y\\\left(h\circ g\circ f\right)^{-1}\left(y\right)&=&y\\\left(h\circ g\circ f\right)^{-1}\left(x\right)&=&x\end{array}$

Dengan demikian, nilai ${\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}{\boldsymbol(\boldsymbol x\boldsymbol)}\boldsymbol={\boldsymbol(\boldsymbol h\boldsymbol\circ\boldsymbol g\boldsymbol\circ\boldsymbol f\boldsymbol)}^{\boldsymbol-\mathbf1}{\boldsymbol(\boldsymbol x\boldsymbol)}$ adalah sama-sama $\boldsymbol x$ .