Solutions

a) Let $h(x) = f(x) - g(x)$. Then $h'(x) = f'(x) - g'(x)$ and for $x > a$

$$\frac{h(x)-h(a)}{x-a} = h'(y)$$

for some $y \in (a,x)$. But $h'(y) = f'(y) - g'(y) > 0$ so

$$\frac{f(x)-g(x)-f(a)+g(a)}{x-a} > 0 \implies f(x) > g(x)$$

since $f(a) = g(a)$. Likewise, for $x < a$ we have

$$\frac{h(a)-h(x)}{a-x} = h'(z)$$

for some $z \in (x,a)$ so

$$\frac{f(a)-g(a)-f(x)+g(x)}{a-x} > 0 \implies g(x) > f(x).$$

b) Let $f(x)=2x-1$ and $g(x)=x$. Then $f'(x)=2>1=g'(x)$ for all $x$. Clearly $f(0)=-1 \ne 0=g(0)$. However,

$$f\left(\frac{1}{4}\right) = -\frac{1}{2} < \frac{1}{4} = g\left(\frac{1}{4}\right).$$

c) Suppose for contradiction that $f(y)\le g(y)$ for some $y=x_0>a$. Let $h(x)=f(x)-g(x)$. By the mean value theorem, for any $x>a$ we have

$$\frac{h(x)-h(a)}{x-a}=h'(z)=f'(z)-g'(z)\ge 0$$

for some $z \in (a,x)$. But this implies that $f(x)\ge g(x)$ for any $x>a$. Hence, we can trivially eliminate the possibility that $f(y)<g(y)$ and that leaves us with $f(y)=g(y)$, equivalently, $h(y)=0$.

We now assert that $h(x)=0$ for all $x\in[a,y]$. Suppose for contradiction this were not so: $h(t)\ne 0$ for some $t\in(a,y)$. We trivially discount $h(t)<0$ as we did above. If $h(t)>0$, then

$$\begin{align*} -h(t) < 0 &\implies h(y)-h(t)<0 \\ &\implies \frac{h(y)-h(t)}{y-t}<0, \end{align*}$$

and by the mean value theorem $h'(z)<0$ for some $z\in(t,y)$ which contradicts the premise that $h'\ge 0$.

Hence, $h(x)=0$ for all $x\in[a,y]$. But this implies that $h'(x)=0$ for all $x\in[a,x_0]\subseteq[a,y]$ and therefore $f'(x_0)=g'(x_0)$.