please help

Question

given f(x)=x^2-8x-4 For less than or equal to 4. find the value of (f^-1)'(5)

Answer 1

If f and g are inverses then
g'(X) = 1/[f'(g(X))]
you can derive this using the fact
f(g(X)) = X and take derivative with chain rule.

 

So given f(X) = x^2-8x-4

Complete the square

f(X) = x^2-8x+(8/2)^2-4-(8/2)^2 = (X-4)^2-20

 

we are given X is less than or equal to 4. This is important because the function has to be 1-1 for inverse to exist. reting the domain to X less than or equal to 4 lets it be 1-1. So inverse exists and we can find

 

Derivative f:

f'(X) = 2x-8

 

thus g'(5) = 1/f'(g(5)) = 1/f'(-1) = 1/-10 =-1/10

 

Note g(5) = -1 because if f(X) = 5 = (X-4)^2-20 then we can solve for x. We get X =-1 or X = 9. But X has to be less than 4 so -1 is answer.