Question

`f(x) = {x}^(2) + 4x - 5` ; >-2

Find
`\frac{d {f}^( - 1) }{dx}` at x=16​

Please show solving

Answer 1

The inverse function theorem says

`(\mathrm df^(-1))/(\mathrm dx)(16)=\frac1{(\mathrm df)/(\mathrm dx)(f^(-1)(16))}`

We have

`f(x)=x^2+4x-5`

defined on
`x>-2`, for which we get

`f^(-1)(x)=-2+√(x+9)`

and

`f^(-1)(16)=-2+√(16+9)=3`

The derivative of
`f(x)` is

`f'(x)=2x+4`

So we end up with

`(\mathrm df^(-1))/(\mathrm dx)(16)=\frac1{(\mathrm df)/(\mathrm dx)(3)}=\frac1{10}`