Is $f'(x)=0$ true at a cusp? If I am given a graph that has a cusp and I am asked to find every point $x$ where $f'(x)=0$ is satisfied, does this include the cusp? I know that when the derivative of a function equals zero this means that there is a horizontal tangent at that point, I also know that the derivative does not exist at a cusp. However, a cusp does have a horizontal tangent but is not differentiable at that point, so do we include the point $x$ where the cusp is when the question is asking us where $f'(x)=0$?

By cusp I mean when
$$\lim_{x \to a^{+}} f'(x)=+ \infty \text{ and } \lim_{x \to a^{-}} f'(x)=- \infty$$
if the derivative does not exist it is, in particular not $=0$
Regards Jotam. If a derivative does not exist at the cusp, say at point $x_o$, then it means that $f'(x_o)$ does not exist, it does not have any unique value. The limits in both direction are different :
$$  \lim_{h \rightarrow 0} \frac{ f(x_o + h) - f(x_o) }{h} \ne  lim_{h \rightarrow 0^{-}} \frac{ f(x_o + h) - f(x_o) }{h}$$

A tangent line at a point $ x $ is a straight line with gradient $f'(x)$. Since there is no derivative at the cusp, there is no tangent at the cusp. 

To be spesific for your question, $f'(x_o)$ cannot be zero because it can not have any unique value.

A tangent line represents the $\textbf{continuous} $ rate of change at the considered point. But you can observe that there is a discontinuity of change at the cusp. $ \textbf{There is a sudden change at the cusp}.$

A tangent line has a direction.  

Thanks.