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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.

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is that g = f on (-1/2, 1/2) ? – Rustyn Jan 28 '13 at 1:14
@Rustyn: The domain given in the restriction doesn’t match the one in the description. It should probably be closed in both places, but emka will have to tell us which was intended. – Brian M. Scott Jan 28 '13 at 1:15
@Brian Yes, that was an oversight on my part. $f$ restricted to $[-\frac{1}{2},\frac{1}{2}]$ is $g$. – emka Jan 28 '13 at 1:18
As long as you need only a reasonable sketch of the graph, the easiest way to graph $g^{-1}$ is to use the fact that the graph of $y=g^{-1}(x)$ is the reflection of the graph of $y=g(x)$ in the line $y=x$. – Brian M. Scott Jan 28 '13 at 1:19
So $g^{-1}(x)=-x^3+x$? – emka Jan 28 '13 at 1:25
up vote 1 down vote accepted

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.

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On my first attempt of sketching things I noticed that $g^{-1}$ is just a skinnier $g$ – emka Jan 28 '13 at 12:56
@emka: $g^{-1}$ has a vertical tangent at $(0,0)$ and $g$ has a horizontal tangent there. This is the only inflection point. This forces $g$ to go from concave down to concave up at the origin and $g^{-1}$ to go from up to down. The (one-sided) tangents at the end points are also different. – Barbara Osofsky Jan 28 '13 at 16:12

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.

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