Question

How do you find the inverse of `f(x)=x^5-2` and graph both f and `f^-1`?

Answer 1

The answer is `f^-1(x)=root(5)(x+2)`

Explanation:

Let `y=x^5-2`

`x^5=y+2`

`x=root(5)(y+2)`

Therefore,

`f^-1(x)=root(5)(x+2)`

Verification

`f(f^-1(x))=f(root(5)(x+2))=(root(5)(x+2))^5-2`

`=x+2-2=x`

The graphs of `f(x)` and `f^-1(x)` are symmetrics wrt `y=x`

graph{(y-(x^5-2))(y-x)(y-root(5)(x+2))=0 [-10, 10, -5, 4.995]}