Question

find $f^{-1}$ for the function $f(x)=\sqrt3{x - 2}+8$.
(1 point)
\\(\circ\\) $f^{-1}(x)=(x - 8)^3+2$
\\(\circ\\) $f^{-1}(x)=(x + 8)^3+2$
\\(\circ\\) $f^{-1}(x)=\sqrt3{x - 8}+2$
\\(\circ\\) $f^{-1}(x)=(x - 8)^3-2$

Explanation:

Step1: Set $y=f(x)$

$y = \sqrt[3]{x-2} + 8$

Step2: Isolate the cube root term

$y - 8 = \sqrt[3]{x-2}$

Step3: Cube both sides

$(y - 8)^3 = x - 2$

Step4: Solve for $x$

$x = (y - 8)^3 + 2$

Step5: Swap $x$ and $y$

$f^{-1}(x) = (x - 8)^3 + 2$

Answer:

A. $f^{-1}(x)=(x-8)^3 + 2$