Assuming that the original equation is
f(x)= [tex] \sqrt[3]{x-5} [/tex]
the inverse of f(x) is f(y)
f(x)= [tex] \sqrt[3]{x-5} [/tex]
y=[tex] \sqrt[3]{x-5} [/tex]
f₋₁(y)= [tex] \sqrt[3]{y-5} [/tex]
x=[tex] \sqrt[3]{y-5} [/tex]
x³= y-5
y=x³+5
Thus f₋₁=y=x³+5
