Suppose f^−1 is the inverse function of a differentiable function f and f(3)=7,f′(3)=7/4, then (f^−1)′(7)=?

Question

Suppose f^−1 is the inverse function of a differentiable function f and f(3)=7,f′(3)=7/4,  then (f^−1)′(7)=?

anonymous · Answer

ok the 12 didnt work

angela210793 · Answer

Wht makes u think tht the answer's 12???

anonymous · Answer

$$f'(x)\approx \frac{f(x+h)-f(x)}{x}$$

angela210793 · Answer

I thought that too...but I'm not sure if that it goes this way :S

anonymous · Answer

Oh, duh - it's 3

anonymous · Answer

Lol, I'm not reading the question right.

jim_thompson5910 · Answer

$$\Large (f^{-1})^{\prime}(x) = \frac{1}{f^{\prime}(f^{-1}(x))}$$


$$\Large (f^{-1})^{\prime}(7) = \frac{1}{f^{\prime}(f^{-1}(7))}$$


$$\Large (f^{-1})^{\prime}(7) = \frac{1}{f^{\prime}(3)}$$


$$\Large (f^{-1})^{\prime}(7) = \frac{1}{\frac{7}{4}}$$


$$\Large (f^{-1})^{\prime}(7) = \frac{4}{7}$$

anonymous · Answer

4/7 works thanks

anonymous · Answer

Props to jim_thompson5910. I learned something new today.

angela210793 · Answer

me too xD

jim_thompson5910 · Answer

Yes it's a very slick and useful formula...unfortunately the tick marks are a bit squished (next to the f), but oh well

anonymous · Answer

OH, that's why I couldn't figure out where that first line was coming from.  Nice.

jim_thompson5910 · Answer

Yeah the right side of the formula reads out to be: "1 over (f prime of f inverse of x)"

anonymous · Answer

Awesome, thanks!

jim_thompson5910 · Answer

you're welcome