Question

`f` is a function on real where `f(x)=2x^2-x` How do you find `f^-1(x)` ?

Answer 1

`f^(-1)(x)=(1+-sqrt(1+8x))/4`

Explanation:

We have `y=f(x)=2x^2-x`

or `2x^2-x-y=0`

or `x=(1+-sqrt(1+8y))/4`

Hence `f^(-1)(x)=(1+-sqrt(1+8x))/4`