For each of the following sentences, we assume $f$ is Riemann integrable on $[a, b]$ and $F$ is the function on $[a, b]$ defined by $$ F(x)=\int_a^x f(t) \, \mathrm{d} t $$ Determine whether the following statements are true or not:
(a) If $F$ is differentiable at $c \in(a, b)$, then $f$ is continuous at $c$.
(b) There exists $c \in[a, b]$ such that $$ \frac{F(b)-F(a)}{b-a}=f(c) $$
Regarding the first part, my guess was to provide the piece-wise function $f$ which could be differentiable on open subinterval but become discontinuous at some point of this interval. For instance, define a function $f$ on $[-1,1]$ by $$ f(x)=\left\{\begin{array}{ll} -1, & -1<x<0 \\ 0, & x=0 \\ 1, & 0<x<1 \end{array}\right. $$
Then, $F(x)$ would be differentiable at any point other than $0$, whereas there may exist some point besides the zero for which $f$ is discontinuous there. Are there any insightful suggestions on how to proceed further, especially with the second part?!
Both statements are false.
(a) Take$$\begin{array}{rccc}f\colon&[-1,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x\ne0\\1&\text{ if }x=0.\end{cases}\end{array}$$Then $F(x)=0$ for every $x\in[-1,1]$. So, $F$ is differentiable everywhere, but $f$ is discontinuous at $0$.
(b) Take$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}2x&\text{ if }x\ne\frac12\\0&\text{ if }x=\frac12.\end{cases}\end{array}$$Then $F(x)=x^2$ for every $x\in[-1,1]$. Therefore$$\frac{F(1)-F(0)}{1-0}=1,$$but $1$ is not in the range of $f$.