Let be a differentiable function such that $f\left(3\right)=15$, $f\left(6\right)=3$ ,$f^{\prime}\left(6\right) = -2$ , and $f^{\prime}\left(3\right) = -8$. The function $g\left(x\right)$ is differentiable and $g\left(x\right) = f^{-1}\left(x\right)$ for all $x$. What is the value of $g^{\prime}\left(3\right)$?
My answer:
Because $g^{\prime}\left(x\right) = \frac{1}{f^{\prime}\left(x\right)}$
Thus $g^{\prime}\left(3\right) = \frac{1}{-8}$
I have realized my mistake.
I incorrectly understood the derivative of an inverse function. I should have stated
$ \frac{d}{dx} f^{-1} = \frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)} $