AP Calculus BC - Derivative of inverse problem

Question

Let $g(x)$ be the inverse of the function $f(x)$. Given the following values on the table below, at which value $x=a$ will $g'(a)=1/6$? (No calculator allowed)

x     -1   2   4   6   7
f(x)   3   5  -2   8  -6
f'(x)  2   6   9   7   5

I'm confused with derivatives of inverses and have no idea proceed with this problem. A step-by-step explanation would be super awesome! Thank you.

Answer 1

Hint: Since $g$ is the inverse of $f$, $g(f(y))=y$. Taking the implicit derivative of both sides, it follows that $g'(f(y))f'(y)=1$. Using the substitution $a=f(y)$ and $g'(a)=\frac{1}{6}$, you need $f'(y)=6$.

Answer 2

By the inverse function theorem, if the function f is differentiable, the inverse $g(x)$ is differentiable as long as $f'(x) \neq 0$, and if we set $y=f(x)$, $\frac{dx}{dy}=\frac{1}{dy/dx}$.

So, in this case, we have $f'(x)=\frac{1}{g'(a)}=\frac{1}{1/6}=6$. According to the chart, at $x=2$, we have $f'(x)=6$. Therefore the answer would be $x=2$.

Hope that helps!