F'(x)=2|x|. What is the graph of f(x).

Question

I have been told that f(x) would look like this:

y+0.5=x^2 for x>0

y-0.5=-x^2 for x<0

With two holes at x=0 (no point exists there.)

https://i.stack.imgur.com/nmBxY.png

I thought because dy/dx has a value at x=0, and f(x) is not differentiable at x=0, this should be false. Would this be disproven by the logic that because dy/dx is not differentiable at x=0, this is the correct f(x) for dy/dx? It would be optimal if this could be answered / confirmed without the use of integrals.

Answer 1

You have $F(x) = F(0)+\int_0^x F'(t) dt = F(0)+\int_0^x |t| dt = \begin{cases} F(0)+{x^2 \over 2}, & x \ge 0 \\ F(0)-{x^2 \over 2}, & x < 0 \end{cases}$.

Simplifying gives $F(x) = F(0) + {1 \over 2} x |x|$.