Question

question 4 let $f(x) = 6 + \sqrt{3x - 9}$. find $f^{-1}(x)$. $f^{-1}(x) = $

Explanation:

Step1: Set $y = f(x)$

$y = 6 + \sqrt{3x - 9}$

Step2: Isolate the square root

$y - 6 = \sqrt{3x - 9}$

Step3: Square both sides

$(y - 6)^2 = 3x - 9$

Step4: Solve for $x$

$3x = (y - 6)^2 + 9$
$x = \frac{(y - 6)^2 + 9}{3}$

Step5: Swap $x$ and $y$

$y = \frac{(x - 6)^2 + 9}{3}$

Answer:

$\frac{(x - 6)^2 + 9}{3}$ (or simplified as $\frac{1}{3}(x-6)^2 + 3$)