Find the derivative of an inverse function

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a function, $f(x)=\frac{x^3}{3}+\frac{x^2}{2}-6x+4$.

Let $I$ be the longest closed interval such that $0 \in I$ and $f$ is invertible. And let $g$ be the inverse function of $f$ in $I$.

Find $I$ and $g'(4)$.

What I've been doing:

I found $f'(x)=x^2+x-6$ and I found the roots which are $-3$ and $2$, and then I looked where the function decreases and increases.

So $f$ is strictly decreasing in $I=[-3, 2]$ so $f$ must be bijective, and then invertible (right?), also $0 \in I$.

Now I have to find $g'(4)$:

I have that $g(x)=f^{-1}(x)$ so $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$ and now my problem is: How do I find $f^{-1}(x)$?

Answer 1

The largest interval containing $0$ where $f$ is invertible is indeed $[-3,2]$. Also $f(0)=4$.

By definition, for every $x\in[-3,2]$, $$ x=g(f(x)) $$ so we can differentiate using the chain rule, getting, for $x\in(-3,2)$, $$ 1=g'(f(x))f'(x) $$ Now substitute $x=0$.

Answer 2

In general, to find the derivative of an inverse function $f$ given $f'$, we use the method known as implicit differentiation. I'll give a different example: differentiating $y=\arcsin x$. $$\begin{split} y &= \arcsin x\\ \sin y & = x\\ \cos y \cdot y' & = 1\\ y' & = \sec y\\ y' & = \sec(\arcsin x)\\ \frac{d}{dx}\arcsin x &= \frac{1}{\sqrt{1-x^2}}. \end{split}$$ Can you apply this idea to your case?