Inverse function of $f(x)=x^{2019}+x$ 
Let $f: ]0,\infty[ \to]0, \infty[$ and $$f(x)=x^{2019}+x$$
Show that $f$ has an strictly increasing and differentiable inverse function $f^{-1} :]0, \infty[ \to ]0, \infty[.$ Determine the derivative $(f^{-1})'(2).$

How can I show something like this? Since the function is increasing could I just try to show that $f$ is surjective in order to prove the inverse function?
 A: 
Show that $f$ has an strictly increasing and differentiable inverse function $f^{-1} :[0, \infty[ \to [0, \infty[.$ Determine the derivative $(f^{-1})(2).$

There are a few things to do: 


*

*Show that $f$ has an inverse: We have $f'(x)=2019x^{2018}+1 >0$, so $f$ is increasing. In particular, it is injective (aka one-to-one). 
Note: We could also have said that $f$ is increasing because it is the sum of two increasing functions, $x^{2019}$ and $x$. But we will need to compute $f'$ anyway.
Moreover, we have $\lim_{x\to 0^+}f(x)=0$, $\lim_{x\to \infty}f(x)=\infty$ and $f$ is continuous, so its range is $]0,\infty[$. This means that $f$ is surjective, hence bijective

*Show that $f^{-1}$ is increasing: the inverse of an increasing function is also increasing, so that part is good :)

*Show that f is differentiable: the function $f$ is polynomial, hence it is differentiable. Here comes the important theorem:

Let $f$ be a differentiable function and $b$ an element of $\textrm{range}(f)=\textrm{dom}(f^{-1})$. Then $f^{-1}$ is differentiable at $b$ if and only if $f'(f^{-1}(b))\neq 0$. In that case,
$$\left(f^{-1}\right)'(b)=\frac{1}{f'(f^{-1}(b))}$$

Since $f'(x)\geqslant 1$ for every $x>0$, we obtain that $f^{-1}$ is differentiable on its domain.


*Compute $f^{-1}(2)$: don't try to find a formula for $f^{-1}(x)$! This is probably impossible... but it is easy to remark that $f(1)=2$ so $f^{-1}(2)=1$.

*Compute $(f^{-1})'(2)$: we just need to use the formula above:
$$\left(f^{-1}\right)'(2)=\frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(1)}=\frac{1}{2020}$$
