Derivative of an accumulation function with absolute value integrand

By FTC Part 1, since the integrand $f(t) = |t-3|$ is continuous everywhere, $F'(x) = f(x) = |x-3|$ for every real $x$. This holds directly from the theorem: differentiating an accumulation function just evaluates the integrand at the upper limit, with no chain rule needed here since the upper limit is simply $x$.

Now consider where $F'(x) = |x-3|$ itself fails to be differentiable, since $F$ is differentiable wherever $F'$ exists as a genuine derivative — this requires $F'(x)=|x-3|$ to be smooth. The function $|x-3|$ has a sharp corner at $x=3$ (it is not differentiable there, since the left and right derivatives are $-1$ and $1$ respectively). Therefore $F(x)$ is not differentiable at $x=3$, even though $F$ itself is continuous there.

The answer is x equals three. F prime of x equals the absolute value of x minus three, and this function has a sharp corner exactly where the expression inside the absolute value is zero. That occurs at x equals three, so this is the one point where F fails to be differentiable.