Question

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

Explanation:

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

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

Step2: Swap $x$ and $y$

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

Step3: Isolate the term with $y$

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

Step4: Solve for $y$

$y = 9(x - 2) = 9x - 18$

Answer:

B. $h(x) = 9x - 18$