Answer
$f^{-1}(x)=2+(x-1)^{5}$
Work Step by Step
We are given:
$f(x)=1+\sqrt[5]{x-2}$
To find the inverse function, we switch $x$ and $y$ and solve for $y$:
$y=1+\sqrt[5]{x-2}$
$x=1+\sqrt[5]{y-2}$
$x-1=\sqrt[5]{y-2}$
$(x-1)^5=y-2$
$(x-1)^5+2=y$
Thus the inverse is:
$f^{-1}(x)=2+(x-1)^{5}$
