Solutions

a) Let $N=f\left(\frac{a+b}{2}\right)$. By definition, there exists $\delta_1>0$ such that for all $x$,

$$f(x)>N.$$

Likewise, there exists $\delta_2>0$ such that for all $x$,

$$0<b-x<\delta_2\implies f(x)>N.$$

Thus, we have that $f(x)>N$ for $x\in(a,a+\delta_1)$ and $(b-\delta_2,b)$. Consider the interval

$$D=[a+\delta_1,b-\delta_2].$$

It is continuous, so there exists $y\in D$ such that $f(y)\le f(x)$ for all $x\in D$. Since $\frac{a+b}{2}\in D$, we have that

$$f(x)>f\left(\frac{a+b}{2}\right)=N\ge f(y)$$

for all $x$ in $(a,b)$.

b) The proof is similar to that in a) except that we let $N=f(0)$.