Piecewise continuous differentiable if I have a piecewise continuously differentiable function. How do I see that on each open interval, where the derivative is continuous, there is a continous extension on the larger closed interval?
 A: This is true when $f$ satisfies the condition: the lateral limits exist. And false in other cases.
Let $f:[a,b]\to\mathbb{R}$ be a piecewise continuously differentiable function. Then there is a partition $P=\{x_i\}_{i=1}^n$ of $[0,1]$ (i.e. $a=x_0<x_1<\ldots<x_n=b$) such that each $I_i=(x_{i-1},x_i)$ is a maximal interval where $f$ is continuously differentiable.
We want to extend some $g_i=f|{I_i}$ to the whole interval $\overline{I_i}=[x_{i-1},x_i]$. Define $g_i':\overline{I_i}\to\mathbb{R}$ as $g_i'|{I_i}=g$ and
$$g_i'(x_{i-1})=\lim_{x\to x_{i-1}^-}g(x),\quad g_i'(x_i)=\lim_{x\to x_i^+}g(x).$$
exists since lateral limit exists.
Is easy to see that $g'$ is a continuous extension of $g$. Note that you can't always extend $f$ in the whole $[a,b]$ since $\lim_{x\to x_i^-}g(x)$ is not necessary the same value as $\lim_{x\to x_{i-1}^+}g(x)$, so here we have $g'_i(x_{i})\neq g'_{i+1}(x_i)$. 
If $f$ is not bounded this is false, note that $f(x)=\tan{x}$ where $\tan$ is definied and $f(x)=0$ where $tan$ is not definied is a piecewise differentiable function but if $P$ is a partition that maximices intervals where $f$ is continuously differentiable, $\lim_{x\to x_i^-}g(x)$ and $\lim_{x\to x_i^+}g(x)$ doesn't exists ($g$ defined as before).
If $f$ is discontinuous in some point of a maximal differentiable partition but bounded, this is also false, consider the function $f(x)=\sin{\frac{1}{x}}$ defined on $[-1,1]\setminus\{0\}$ and $f(0)=0$ then $x_0=-1$, $x_1=0$, $x_2=1$. Then  $\lim_{x\to 0^-}g(x)$ and $\lim_{x\to 0^+}g(x)$ doesn't exists ($g$ defined as before). In other cases it is easy to give examples.
