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
$$\lim_{x \to a^{+}} f'(x)=+ \infty \text{ and } \lim_{x \to a^{-}} f'(x)=- \infty$$
 A: if the derivative does not exist it is, in particular not $=0$
A: For graphs with an explicit equation
$$y=f(x)$$
(you are not mentioning a parametric curve), the only possibility for a cusp is a point where $f(x)$ is finite and
$$f'(x^-)=-f'(x^+)=\pm\infty,$$ i.e. a vertical tangent with a change of direction.
For example
$$y=\sqrt[3]{|x|}$$ has a cusp, while 
$$y=\sqrt[3]{x}$$ not.
A: The prime examples of a cusp are given by  equations  of the form  (a): $\>x^2=y^3\>,\ $ or (b): $\>x^3=y^2$.
Converting these equations to functions $x\mapsto f(x)$ results in case (a) to $$y=f(x):=|x|^{2/3}\qquad(-\infty<x<\infty)\ .$$
This function $f$ is continuous on all of ${\mathbb R}$, but not differentiable at $x=0$. One has
$$\lim_{x\to0-}{f(x)-f(0)\over x}=-\infty,\qquad \lim_{x\to0+}{f(x)-f(0)\over x}=+\infty$$
as well as 
$$\lim_{x\to0-}f'(x)=-\infty,\qquad \lim_{x\to0+}f'(x)=+\infty\ .$$
In case (b) solving $x^3=y^2$ for $y$ as a function of $x$ gives no solution when $x<0$ and two solutions when $x>0$. The function
$$y=g(x):=x^{3/2}\geq0\qquad(x\geq0)$$ obtained by choosing the nonnegative branch has
$$g'(0)=\lim_{x\to0}{g(x)-g(0)\over x}=0$$
as well as $$\lim_{x\to0+}g'(x)=\lim_{x\to0+}{3\over2}\sqrt{x}=0\ .$$
A: 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.
