Solutions

a) Suppose the minimum of $f$ on $[a,b]$ is at $a$. Let

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

By definition, $f(a)\le f(x)$ for $x\in(a,b]$. Write $x=a+h$ for $h>0$. Then,

$$f(a)\le f(a+h)\implies 0\le f(a+h)-f(a)\implies 0\le \frac{f(a+h)-f(a)}{h}=g(h).$$

Hence,

$$f'(a)=\lim_{h\to 0^+}g(h)\ge \lim_{h\to 0^+}0=0.$$

We can prove that $f'(b)\le 0$ if the minimum is at $b$ similarly.

b) Since $f$ is differentiable on $[a,b]$, we know it is continuous on the same. Hence, by the Extreme Value Theorem, we know it has a minimum on $[a,b]$. Now, suppose that $f'(a)>0$ and $f'(b)<0$. We know that the minimum cannot be at either $a$ or $b$. Otherwise, by a), $f'(a)\ge 0$ or $f'(b)\le 0$ which contradicts our premise above.

Thus, the minimum of $f$ is in $(a,b)$. By Theorem 1), if $x$ is the minimum point then $f'(x)=0$ because $f$ is differentiable on $(a,b)$.

c) Let $g(x)=f(x)-cx$. We have $g'(x)=f'(x)-c$. Since $f'(a)<c<f'(b)$, then $g'(a)<0$ and $g'(b)>0$. Thus, by b), $g'(x)=0$ and equivalently, $f'(x)=c$ for some $x\in(a,b)$.