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Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$.
A problem from Mark Krusemeyer. Should one be using the symmetry of the graphs along $x=y$ due the functions being inverses of one another? Just a thought.
Here's a brute force approach. There's probably a more elegant way of doing it.
The two curves can be written as $y=x^c$ and $y=x^{1/c}$. They always intersect at $x=0$ and $x=1$ (for $c=1$ the curves coincide).
The derivatives of the two curves are $c x^{c-1}$ and $c^{-1}x^{1/c-1}$ respectively. For $0<c\neq1$ the angle between them at $x=0$ is clearly always $\pi/2$. At $x=1$ we want the angle between them to be $\pi/4$, so $$ \tan^{-1}(c) - \tan^{-1}(1/c) = \pm\frac\pi4 \implies c = \pm 1+\sqrt2 $$
The two solutions just reflect the fact that you can swap the two curves, or equivalently, set $c\to1/c$.
There are also two negative solutions to the above equation which are excluded by the question. These are also valid curves and solutions, which can be obtained by the transformation $x \to 1/x$.
It is possible to use the symmetry as a slight shortcut to Simon's answer: Given the graph and conditions as laid out there, the relevant point of intersection must be at $x=y=1$. Because of the symmetry, the angles formed by the tangent lines to the curves at that point and the line $y=x$ must be $\frac{\pi}{8}$, so the slope of the tangent line to the blue curve (lower for $0< x<1$) at $x=1$ must be $\tan\frac{3\pi}{8}=1+\sqrt{2}$. Now, either $cx^{c-1}|_{x=1}=c=1+\sqrt{2}$ or $c^{-1}x^{1/c-1}|_{x=1}=\frac{1}{c}=1+\sqrt{2}$, so $c=1+\sqrt{2}$ or $c=\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$.
