What Happens at the Cusp on the Original Graph When Viewing the Derivative?

[sarahr]

If the original graph, f, has a cusp, obviously the derivative is not defined at the x-value of the cusp (resulting in an asymptote).

but, what if you are viewing a graph of the derivative, f ', and it has a cusp.. what is going on at the x-value of the cusp on the original graph, f ?

 

[HallsofIvy]

If the derivative graph has a cusp, that means that the second derivative does not exist. Think about y= |x| which does not have a derivative at x= 0. If we let f '(x)= |x| = (x if x>= 0 and -x if x< 0) and integrate we get
f(x)= ((1/2)x^2 if x>=0 and -(1/2)x^2 if x< 0). What does its graph look like around x= 0?

 

[sarahr]

perfect example! thankss