Question about bijective function when restricting domain Let $f(x):x^3-x$.  By an appropriate restriction of the domain and range to find a bijective function $g$.  Then graph $g$ and $g^{-1}$.
The function I found is $g=f|_{\left[-\frac{1}{2},\frac{1}{2}\right]}$.  So $g:\left[-\frac{1}{2},\frac{1}{2}\right] \to \left[-\frac{3}{8}, \frac{3}{8}\right]$.  This function is both one to one and onto.
My problem is trying to explicitly find $g^{-1}$ so I can graph it.  
 A: The function $f$ has critical values at $\pm \sqrt(3)/3$ and these are local extrema. The restriction $g$ of $f$ to the interval $[\ -\sqrt(3)/3\ ,\ \sqrt(3)/3\ ]$ has a positive derivative on this interval except for a $0$ at $0$ and the endpoints, so it is monotone increasing with horizontal tangent at those three points and a change in concavity at $0$. The range of $g$ is $[\ -f\left(\sqrt(3)/3\right)\ ,\ \sqrt(3)/3\ ]$  so that interval is the domain of $g^{-1}$. Also, $g^{-1}$ is monotone increasing on this domain, has a vertical tangent at $0$ with a change of concavity there. The endpoints of its range are $\pm 1$. You do not need a formula to sketch the graph of either $g$ or $g^{-1}$ if you have decent graph paper. 
A: If $f(x) = x^3 - x$ then $f'(x) = 3x^2 - 1$.  This function is 1-1 on three subintervals of the line: $(\infty, 1/\sqrt{3}]$, $[-1/\sqrt{3}, 1/\sqrt{3}]$ and $[1/\sqrt{3}, \infty)$
The inverse on any of these intervals can be explicitly found using Cardano's formulae.  You can then restrict the ranges appropriately to get bijections.
