Question

How do you find the inverse of `f(x)=4x^2+1, x>=0`?

Answer 1

`f(x)=4x^2+1;quadquadquadx>=0`
`f^(-1)(x)=sqrt(x-1)/2;quadquadquadx>=1`

Explanation:

`y=4x^2+1`

`x=4y^2+1`

`x-1=4y^2`

`1/4(x-1)=y^2`

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

`+-sqrt(x-1)/2=y`