Suppose we have a real function $f(x)$ which is differentiable everywhere, except possibly at $x=2$.
If $\lim_{x\to2^-}f^\prime(x)=\lim_{x\to2^+}f^\prime(x)$, is that enough to conclude that $f(x)$ is differentiable at $x=2$. The 'tangent slopes' are equal on the left and the right.
I believe the answer is no, it's not enough to guarantee differentiability at $x=2$, right?
However, if we add in the condition that $f(x)$ is continuous everywhere, including $x=2$, then are the above limits enough to guarantee differentiability at $x=2?$
I believe that if we add in the continuity assumption at $x=2$, then checking the left/right limits of $f^\prime(x)$ is enough to guarantee $f(x)$ is differentiable at $x=2$. But why? How is the continuity of $f(x)$ related to the continuity of $f^\prime(x)?$
Best Answer
Consider the function $f:\mathbb{R}\to\mathbb{R}$ which is defined by $f(x)=0$ for $x<2$ and $f(x)=1$ for $x\geq 2$. This is an explicit example of a function which is differentiable away from $2$, and the limits of slopes agree, but it is not differentiable at $2$.
Generally speaking, since being differentiable at a point implies being continuous at that point, continuity is a necessary condition for being differentiable. On the other hand, there are functions which are everywhere differentiable but whose derivative is not continuous, a concrete example is $x^2\sin(1/x)$.
Finally, if $f$ is continuous, and differentiable away from 2, and the limits of derivatives agree, then $f$ is also differentiable at $2$, and its derivative is the limit of $f'$ at $2$.
Edit: Okay I thought I had a prove, but it relies on the assumption that $f'$ is continious in a neighborhood of $2$. Then you can consider $g(x)=\int_2^xf'(t)dt+f(2)$, which is differentiable by the fundamental theorem of calculus. But one has $f=g$: first of all $f(2)=g(2)$ by definition. For $x>2$ we have $f(x)=\lim_{\epsilon\to 0^+}\int_{2+\epsilon}^xf'(t)dt+f(2+\epsilon)=\int_2^xf'(t)dt+f(2)=g(x)$, analog for $x<2$.
Edit 2: One can weaken the assumptions a bit, one only needs $f'$ be integrable in a suitable sense, Riemann works, and maybe even Lebesgue suffices.