If $f(x)$ is a unimodal probability density function, how can I show that its mode is at $f'(x)=0$? If the function $f(x)$ is continuous and a probability density function (PDF), how can I show that its mode is at $f'(x)=0$?
 A: Find the maximum value. The Laplace distribution is defined piece-wise, and not smooth at the mode, which is at the piece-wise common point (A.K.A., corner point), and such a common, junctional corner point in a piece-wise defined function has no general continuity guarantee of either the separate piecewise defined functions' amplitudes nor any of their derivatives. The mode for a Laplace distribution occurs at $\frac{e^{-\frac{x-\mu }{\beta }}}{2 \beta }=\frac{e^{-\frac{\mu -x}{\beta }}}{2 \beta }\to x=\mu$, because $f(x=\mu)=\frac{1}{2 \beta }$ is the maximum value of $f$. Of course there are continuous functions that are differentiable nowhere, and differentiability is not a requirement. Thus, the general answer is simply find the maximum of the pdf. This can be done in closed form as above for something as simple as the Laplace distribution. It can be done numerically for nice smooth functions using derivatives example by referring to  How to prevent newton's method from finding maxima? and altering it to avoid minima. Or, more generally for nasty functions using a canned routine like FindMaximum in Mathematica with a global search routine like RandomSearch. 
@Glen_b and @Dougal, thanks for the help, anything further @me. 
A: If it's only continuous but not differentiable at the mode, you can't. Consider the Laplace distribution.

