How to solve an Inverse differentiation problem

Question

If f is a one-to-one function where $f(3)=2$ and $f'(3)=6$, what is the value of $(f^{-1})'(2)$?

I am not even sure where to start with this question. I was hoping someone can help

$f$ of $3 =2$ and $f$ prime of $3 = 6$ we need to find the value of the derivative inverse of $f$ of $2$

Answer 1

first thing $f$ sends $3$ to $2$ and $f^{-1}$ sends $2$ to $3.$ the slopes of $f$ at $2$ and of $(f^{-1})'$ at $3$ are reciprocals. here is one way to see that.

since $f$ has slope $6$ at $x = 3, y = 2.$ if $f$ were a line it would be $y = f(x) = 6(x-3) + 2 = 6x -16.$ the inverse would be $f^{-1}(x)= \frac{1}{6}(x-2)+3$ which has slope $\frac16$ at the point $x = 2, y = 3.$ therefore $$(f^{-1})'(2)= \frac16. $$

Answer 2

Hint:

$$(f^{-1})'(2)=\displaystyle\frac{1}{f'(f^{-1}(2))}$$

Answer 3

This formula comes in handy from time to time, and it's good to know how to derive it on the spot. Let $f$ be such a function where $y = f^{-1}(x)$. Then $f(y)=x$ and by implicit differentiation we have $$f'(y)\cdot y' = 1 \\ \implies y' = \frac{1}{f'(y)}$$ In letting $y = f^{-1}(x)$ we may differentiate both sides to get $y' = (f^{-1}(x))'$ and then we can use these two equations to rewrite $y' = \frac{1}{f'(y)}$ as $$(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$$ The rest of your question can be answered with algebra, as you now have an equation for $(f^{-1}(2))'$ and you were already given everything you need to solve for it.