Question

select the correct answer.
find the inverse of the given function.
$f(x) = (x + 3)^3 - 1$
$circ$ $f^{-1}(x) = \sqrt3{x - 1} + 3$
$circ$ $f^{-1}(x) = \sqrt3{x} - 2$
$circ$ $f^{-1}(x) = \sqrt3{x + 1} - 3$
$circ$ $f^{-1}(x) = \sqrt3{x - 3} + 1$

Explanation:

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

$y = (x + 3)^3 - 1$

Step2: Swap $x$ and $y$

$x = (y + 3)^3 - 1$

Step3: Isolate the cubic term

$x + 1 = (y + 3)^3$

Step4: Take cube root of both sides

$\sqrt[3]{x + 1} = y + 3$

Step5: Solve for $y$

$y = \sqrt[3]{x + 1} - 3$

Answer:

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