Determine the greatest interval where the function is invertible

The assingment is to determine the greatest interval around $x=0$ where the function:

$$f(x)=x^5-5x+3$$

is invertible. After that, determine $(f^{-1})'(3)$

I have totally forgotten all about invertible functions, how do I do it and when is it not possible?

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Ideally you solve $y=x^5-5x+3$ for $x$ to find candidates to the inverse function. From there you choose an appropriate one, but this approach is unlikely to work here. Alternatively you can use the fact that the function is continuous and study its derivative to look for changes in its mononicity. This is what I would try. – Git Gud Aug 10 '14 at 13:03
If you sketch this, it has three real zeroes and two turning points (one each side of zero). Finding the turning points is relatively easy in this particular case and helps you answer the first part. – Henry Aug 10 '14 at 13:12

The function $f$ has the following properties:

1. $\lim_{x\to-\infty}f(x)=-\infty$
2. $\lim_{x\to\infty}f(x)=\infty$
3. $f'(x)=5x^4-5=5(x-1)(x+1)(x^2+1)$

Thus $f$ has a relative maximum at $-1$, with $f(-1)=7$ and a relative minimum at $1$, with $f(1)=-1$.

Then $f$ is invertible in $[-1,1]$ and no largest interval around $0$, because in a larger interval around $0$ it will not be monotonic.
Thus $f^{-1}$ is defined on $[-1,7]$ and, by the inverse function theorem,
$\displaystyle(f^{-1})'(3)=\frac{1}{f'(a)}$
where $a\in[-1,1]$ is the unique point such that $f(a)=3$. The equation is then $a^5-5a+3=3$ or $a=0$. Thus$$(f^{-1})'(3)=\frac{1}{f'(0)}=-\frac{1}{5}.$$

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Hint: The function has a local maximum and a local minimum near 0. Find those. That will give you the requested interval.

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If a continuous function is invertible, then it is monotonic. Since you're dealing with a differentiable function, this is equivalent to finding the greatest interval around 0 where the sign of $f'$ does not change.

Find the zeros of $f'$, splice $\mathbb{R}$ into intervals with these, and you should have your solution.

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