if $f(x)=\frac{1}{9}x - 2$, what is $f^{-1}(x)$?\n$f^{-1}(x)=9x + 18$\n$f^{-1}(x)=\frac{1}{9}x+2$\n$f^{-1}(x)…

if $f(x)=\frac{1}{9}x - 2$, what is $f^{-1}(x)$?\n$f^{-1}(x)=9x + 18$\n$f^{-1}(x)=\frac{1}{9}x+2$\n$f^{-1}(x)=9x + 2$\n$f^{-1}(x)=-2x+\frac{1}{9}$

if $f(x)=\frac{1}{9}x - 2$, what is $f^{-1}(x)$?\n$f^{-1}(x)=9x + 18$\n$f^{-1}(x)=\frac{1}{9}x+2$\n$f^{-1}(x)=9x + 2$\n$f^{-1}(x)=-2x+\frac{1}{9}$

Answer

Explanation:

Step1: Let $y = f(x)$

$y=\frac{1}{9}x - 2$

Step2: Swap $x$ and $y$

$x=\frac{1}{9}y - 2$

Step3: Solve for $y$

Add 2 to both sides: $x + 2=\frac{1}{9}y$. Then multiply both sides by 9, we get $y = 9x+18$.

Answer:

A. $f^{-1}(x)=9x + 18$