Question

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

when
`x > - 2`
find
`\frac{d {f}^( - 1) }{dx} at \: x = 16`
​

Answer 1

`(df^(1)(16))/(dx) = \pm (1)/(10)`

Explanation:

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

First we find the inverse function.

`y = x^2 + 4x - 5`

`x = y^2 + 4y - 5`

`y^2 + 4y - 5 = x`

`y^2 + 4y = x + 5`

`y^2 + 4y + 4 = x + 5 + 4`

`(y + 2)^2 = x + 9`

`y + 2 = \pm√(x + 9)`

`y = -2 \pm√(x + 9)`

`f^(-1)(x) = -2 \pm√(x + 9)`

`f^(-1)(x) = -2 \pm (x + 9)^{(1)/(2)}`

Now we find the derivative of the inverse function.

`(df^(-1)(x))/(dx) = \pm (1)/(2)(x + 9)^{-(1)/(2)}`

`(df^(-1)(x))/(dx) = \pm (1)/(2√(x + 9))`

Now we evaluate the derivative of the inverse function at x = 16.

`(df^(-1)(16))/(dx) = \pm (1)/(2√(16 + 9))`

`(df^(-1)(16))/(dx) = \pm (1)/(2√(25))`

`(df^(-1)(16))/(dx) = \pm (1)/(2 * 5 )`

`(df^(-1)(16))/(dx) = \pm (1)/(10)`