Solutions

a) We have

$$f(a+k)\le g(a+k)\le h(a+k).$$

Hence,

$$\frac{f(a+k)-f(a)}{k}\le \frac{g(a+k)-f(a)}{k}\le \frac{h(a+k)-f(a)}{k}.$$

Since $f'(a)=h'(a)$, $g'(a)=f'(a)$ by the Squeeze Theorem.

b) This is obvious with $g(x)=0$ and $f(x)=h(x)=x$. We have

$$-x\le 0\le x$$

for all $x$ and $f'(x)=h'(x)=1$. However, $g'(x)=0$.