If the function fis defined by f(x) = x ^ 5 - 1 then the inverse, f ^ - 1 is defined by f ^ - 1 * (x) =

Given the function:
[tex]f(x)=x^5-1[/tex]We have that f(x) = y, so:
[tex]y=x^5-1[/tex]Substitute x with y:
[tex]x=y^5-1[/tex]And solve for y:
[tex]\begin{gathered} x+1=y^5-1+1 \\ x+1=y^5 \\ y=\sqrt[5]{x+1} \end{gathered}[/tex]Answer:
[tex]\sqrt[5]{x+1}[/tex]