Question

for the function $f(x) = sqrt5{x} - 3$, find $f^{-1}(x)$.
answer attempt 1 out of 2
$\circ$ $f^{-1}(x) = x^5 + 3$ $\circ$ $f^{-1}(x) = (x - 3)^5$
$\circ$ $f^{-1}(x) = (x + 3)^5$ $\circ$ $f^{-1}(x) = x^5 - 3$

Explanation:

Step1: Set $y = f(x)$

$y = \sqrt[5]{x} - 3$

Step2: Swap $x$ and $y$

$x = \sqrt[5]{y} - 3$

Step3: Isolate the radical term

$x + 3 = \sqrt[5]{y}$

Step4: Eliminate the 5th root

$(x + 3)^5 = y$

Step5: Replace $y$ with $f^{-1}(x)$

$f^{-1}(x) = (x + 3)^5$

Answer:

$f^{-1}(x) = (x + 3)^5$