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)?

Question

nottim · Answer

I guess you should do the inverse of the equation first?

anonymous · Answer

can someone pleaseeee help me????????

anonymous · Answer

How would I do that?

anonymous · Answer

I think I have the same question...hold on

nottim · Answer

you didnt leanr inverse?

anonymous · Answer

Inverse of the equation would take too much work. Consider the following: let f and g be as stated. Then since they are inverses, clearly f(g(x))=x. Taking the derivative of each side, you get $$f'(g(x))g'(x)=1$$ Thus, $$g'(x)=\frac{1}{f'(g(x))}$$ Thus, $$g'(0)=\frac{1}{f'(g(0))}$$ Since g(0) is just the point x such that f(x)=0. A little algebra shows the only real answer to this is x=3. Since $$f'(x)=3x^2-14x+25$$ you clearly have that $$f'(3)=27-42+25=10$$ Hence, you have the final answer that $$g'(0)=\frac{1}{10}$$

anonymous · Answer

not completely

anonymous · Answer

Thank you imperialist