Calculus inverse function solving for $f^{-1}$ 
*

*Show that the function $f(x) = x^5 -x^3 +2x$ is invertible. Compute the derivative of $f^{-1}$
at 2.


To find $f^{-1}$ I switched $x$ and $y$ which gave me $x = y^5 - y^3 + 2y$
this is where i got stuck because I am not sure how to solve for y after that step.
Can the $y^5 - y^3$ become $y^2$?
 A: A function $f:\mathbb R\to\text{range(f)}$ is invertible, if its derivative, which is here $f'(x)=5x^4-3x^2+2$, is positive. The positivity can be easily checked, since $f'(x)$ has no zeros, is continuous and has one positive value. In particular $f(1)=2$, so that $(f^{-1}(2))'=1/f'(1)=1/4$.
A: Just to answer the last question for a start:  No, you certainly can't replace $y^5 - y^3$ by $y^2$.  (Replace $y$ by almost any number you like and you will see this.)
You have probably learnt that a function is invertible if and only if it is a bijection.  It is going to be much easier to show that your $f$ is a bijection (i.e. both one-to-one and onto) than to find an explicit inverse function (and note that the question doesn't ask you to find the inverse!).
A: *

*$f'(x)$ is a quadratic polynomial in $x^2$: factor it. if this is positive everywhere or negative everywhere, then $f(x)$ is invertible

*plug some simple arguments into $f(x)$ until the expression matches $2$, and you'll find $f^{-1}(2)$

A: We claim $f(x)=f(y)\iff x=y$.
Consider $f'(x)=5x^4-3x^2+2 = 5(x^2-\frac{3}{10})^2+\frac{31}{20}>0$. So $f(x)$ is strictly monotonically increasing, which implies $f(x)=f(y)\iff x=y$.
Let $g=f^{-1}$. Then $g(2)=1$ as $f(1)=2$. Hence $g'(2)=\cfrac{1}{f'(g(2))}=\cfrac{1}{f'(1)}=\cfrac{1}{4}$
A: $f(x) = x^5 -x^3 +2x$, $f(1)=2.$
$$(f^{-1})'(2)=\frac{1}{f'(1)}=\frac{1}{4}$$
