Solutions

a) Since $f$ is differentiable on $[a,b]$, it is also continuous on $[a,b]$. By the mean value theorem, there exists $y \in (a,b)$ such that

$$\frac{f(b)-f(a)}{b-a} = f'(y) \ge M.$$

b) The line of reasoning is identical to that in a) but with an opposite inequality.

c) If $|f'(x)| \le M$ for all $x$ in $[a,b]$ then

$$\frac{|f(b)|-|f(a)|}{b-a} \le \left|\frac{f(b)-f(a)}{b-a}\right| \le M.$$