
Originally Posted by
wingless
$\displaystyle f(x) = 2^{\sin x} + 2^{\cos x}$
$\displaystyle f'(x) = 0 ~\rightarrow~ 2^{\sin x}\cos x = 2^{\cos x}\sin x$
One obvious attempt to solve this equation is $\displaystyle \sin x = \cos x$. The solution set to this equation is $\displaystyle x = \left \{ \frac{\pi}{4}, \frac{\pi}{4}+\pi\right \}$. One of these is maximum and the other is minimum. Note that there are infinitely many maximums and minimums, as the function has a period of $\displaystyle 2\pi$.