Answer
$f^{-1}(x)=(x-1)^{5}+2$
Work Step by Step
$f(x)=1+\sqrt[5]{x-2}$
Substitute $f(x)$ by $y$:
$y=1+\sqrt[5]{x-2}$
Solve for $x$:
$y-1=\sqrt[5]{x-2}$
$\sqrt[5]{x-2}=y-1$
$x-2=(y-1)^{5}$
$x=(y-1)^{5}+2$
Interchange $x$ and $y$:
$y=(x-1)^{5}+2$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=(x-1)^{5}+2$
