Question

Question #b727a

Answer 1

Given:

`f(x) = 9x^2,x>=0`

Substitute `f^-1(x)` every x within `f(x)`:

`f(f^-1(x)) = 9(f^-1(x))^2,f^-1(x)>=0`

The left side becomes x, because one of the two parts of the relationship of a function to it inverse is `f(f^-1(x)) = x`:

`x = 9(f^-1(x))^2,f^-1(x)>=0`

Flip the equation:

`9(f^-1(x))^2= x,f^-1(x)>=0`

Divide both sides by 9:

`(f^-1(x))^2= x/9,f^-1(x)>=0`

Normally, when we use the square root on both sides, we must use the `+-` on the right side but the restriction `f^-1(x)>=0` tells us to use only the positive value:

`f^-1(x)= sqrtx/3`

I will leave it to you to verify that `f(f^-1(x)) = x` and `f^-1(f(x))= x`