Determine the greatest interval where the function is invertible

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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$$(f^{-1})'(3)=\frac{1}{f'(0)}=-\frac{1}{5}.$$