Question

find the inverse of $f$.
$f(x)=x^3 - 5$
$\bigcirc \\ f^{-1}(x)=\sqrt3{x - 5}$
$\bigcirc \\ f^{-1}(x)=\sqrt3{x} - 5$
$\bigcirc \\ f^{-1}(x)=-\sqrt3{x + 3}$
$\bigcirc \\ f^{-1}(x)=\sqrt3{x} + 5$
$\bigcirc \\ f^{-1}(x)=\sqrt3{x + 5}$

Explanation:

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

$y = x^3 - 5$

Step2: Swap $x$ and $y$

$x = y^3 - 5$

Step3: Solve for $y$

Add 5 to both sides: $x + 5 = y^3$
Take cube root: $y = \sqrt[3]{x + 5}$

Step4: Replace $y$ with $f^{-1}(x)$

$f^{-1}(x) = \sqrt[3]{x + 5}$

Answer:

$f^{-1}(x) = \sqrt[3]{x + 5}$ (the last option)