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 have the function \( f(x)=\frac{1}{9}x + 2 \), so we write it as \( y=\frac{1}{9}x + 2 \).

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

After swapping, we get \( x=\frac{1}{9}y + 2 \).

Step3: Solve for \( y \)

First, subtract 2 from both sides: \( x - 2=\frac{1}{9}y \).
Then, multiply both sides by 9: \( y = 9(x - 2)=9x-18 \).
So the inverse function \( h(x)=9x - 18 \).

Answer:

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