Question

what is the inverse of the function $f(x)=\frac{1}{9}x + 2$?
$\bigcirc$ $h(x)=18x - 2$
$\bigcirc$ $h(x)=9x - 18$
$\bigcirc$ $h(x)=9x + 18$
$\bigcirc$ $h(x)=18x + 2$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=\frac{1}{9}x + 2 \). Replace \( f(x) \) with \( y \), so we have \( y=\frac{1}{9}x + 2 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x=\frac{1}{9}y + 2 \).

Step3: Solve for \( y \)

First, subtract 2 from both sides of the equation: \( x - 2=\frac{1}{9}y \).
Then, multiply both sides by 9 to isolate \( y \): \( y = 9(x - 2) \).
Simplify the right - hand side: \( y=9x-18 \).
Since the inverse function of \( f(x) \) is denoted as \( f^{-1}(x) \) (or in this case \( h(x) \) as per the options), the inverse function \( h(x)=9x - 18 \).

Answer:

\( h(x)=9x - 18 \) (corresponding to the option: \( h(x)=9x - 18 \))