Question

Diketahui f(x) = (2x)/(3 - x), x ne3 dang ^ - 1 * (x) = 6/(2 + x) ?x ne-2 . Nilai (g f)^ -1 (1) adalah​

Answer 1

[tex]\begin{aligned}&\textsf{Diketahui }f(x)=\frac{2x}{3-x},\:x\ne3\ {\sf dan}\\&g^{-1}(x)=\frac{6}{2+x},\:x\ne-2.\ \textsf{Nilai dari}\\\vphantom{\bigg|}&(g\circ f)^{-1}(1)\ {\sf adalah\ }\boxed{\,\bf\frac{3}{2}\,}\ .\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Komposisi dan Invers Fungsi

Perhatikan bahwa:

[tex]\begin{aligned}\boxed{\vphantom{\Big|}\,(g\circ f)^{-1}(x)=\left(f^{-1}\circ g^{-1}\right)(x)\,}\end{aligned}[/tex]

Maka kita cari [tex]f^{-1}(x)[/tex] terlebih dahulu.

[tex]\begin{aligned}f(x)=y=&=\frac{2x}{3-x},\:x\ne3\\y(3-x)&=2x\\3y-xy&=2x\\2x+xy&=3y\\x(2+y)&=3y\\f(y)=x&=\frac{3y}{y=2}\\\vphantom{\Bigg|}\therefore\ f^{-1}(x)&=\frac{3x}{x+2}\\\end{aligned}[/tex]

Untuk menentukan nilai dari [tex](g\circ f)^{-1}(1)[/tex], kita bisa langsung menggunakan [tex]x = 1[/tex], atau mencari [tex](g\circ f)^{-1}(x)[/tex] terlebih dahulu.

CARA I

[tex]\begin{aligned}(g\circ f)^{-1}(1)&=\left(f^{-1}\circ g^{-1}\right)(1)\\&=f^{-1}\left(g^{-1}(1)\right)\\&=f^{-1}\left(\frac{6}{2+1}\right)\\&=f^{-1}(2)\\&=\frac{3\cdot2}{2+2}=\frac{6}{4}\\(g\circ f)^{-1}(1)&=\boxed{\,\bf\frac{3}{2}\,}\end{aligned}[/tex]

CARA II

[tex]\begin{aligned}(g\circ f)^{-1}(x)&=\left(f^{-1}\circ g^{-1}\right)(x)\\&=f^{-1}\left(g^{-1}(x)\right)\\&=\frac{3g^{-1}(x)}{g^{-1}(x)+2}\\&=\frac{\left(3\cdot\dfrac{6}{2+x}\right)}{\left(\dfrac{6}{2+x}+2\right)}\\&=\frac{\left(\dfrac{18}{\cancel{2+x}}\right)}{\left(\dfrac{10+2x}{\cancel{2+x}}\right)}\\&=\frac{18}{10+2x}\\(g\circ f)^{-1}(x)&=\frac{9}{5+x}\end{aligned}[/tex]

Maka:

[tex]\begin{aligned}(g\circ f)^{-1}(1)&=\frac{9}{5+1}=\frac{9}{6}\\&=\boxed{\,\bf\frac{3}{2}\,}\end{aligned}[/tex]