Let f(x)=x^3-7x^2+25x-39 and let g be the inverse function of f. what is the value of g'(0)?

Thank you for posting your question here at . I hope the answer will help you. Feel free to ask more questions here.

Below are the choices that can be found from other sources:

a) -1/25 

b) 1/25 

c) 1/10 

d) 10 

e) 25

f(x) = x^3 - 7x^2 + 25x - 39 
f'(x) = 3x^2 - 14x + 25 

Since g is the inverse function of f 
f(g(x)) = x 

Differentiate both sides of equation 
f'(g(x)) g'(x) = 1 

Now f(3) = 0, so g(0) = 3 

g'(x) = 1 / f'(g(x)) 
g'(0) = 1 / f'(g(0)) 
g'(0) = 1 / f'(3) 
g'(0) = 1 / 10

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.