Give an example of a function who is nondifferentiable on (0, 2) but has an antiderivative on (0, 2)

Question

Originally when I was playing around with this problem, I tried to first find a function who was differentiable, but whose derivative was not differentiable at a specific point. So I figured out the piecewise function $x^2$ for x > 0 and 0 for x $\leq$ 0 satisfied my conditions. However, I need this to be on the interval (0, 2) so I thought, I could just shift the graph over by 1 and change where the graph switches from 0 to 1. So I have $(x-1)^2$ for x > 1 and 0 for x $\leq$ 1. But now I need to change this to be a function who is nondifferentiable on (0, 2) and still has an antiderivative on (0, 2). So I just took the derivatives of the two pieces of my function to finally get $2(x-1)$ for x > 1 and 0 for x $\leq$ 1. However, I was trying to check this by using the limit definition of the derivative, and I'm not sure my example works. Can someone clarify?

Answer 1

Your approach is admirable, but I think that you're overthinking this. The function $$ f(x) = |x-1| = \begin{cases} 1-x& x < 1\\ x - 1 & x \geq 1 \end{cases} $$ Is not differentiable in $(0,2)$ (since it is not differentiable at $x=1$), but has an antiderivative (since it is continuous on $(0,2)$).

Answer 2

Your example of $(x-1)^2$ for $x \gt 1$ and zero otherwise does not work as its derivative at $x=1$ is zero and so it has a derivative everywhere.

But $2(x-1)$ for $x \gt 1$ and zero otherwise does work as it has no derivative when $x=1$ but it has an anti-derivative the previous function.

If you want a continuous function that has no derivative anywhere but has an antiderivative, try the Weierstrass function $$f(x)=\sum_{n=0} ^\infty a^n \cos(b^n \pi x)$$ where $0\lt a \lt 1$, $b$ is a positive odd integer, and $ab \gt 1+\frac{3}{2} \pi$.