Solutions

a) Consider the sets

$$A=\{x : a \le x \le x_0,\ f(x)<0\}$$

and

$$B=\{x : x_0 \le x \le b,\ f(x)<0\}.$$

If $A=\varnothing$, we can let $c=a$. Otherwise, by the proof in Exercise 3a, we can let $c$ be the largest $x$ in $A$ with $f(x)=0$. In either case, we have $f(c)=0$, $a \le c < x_0$, and $f(x)>0$ for all $x \in (c,x_0]$.

Likewise, if $B=\varnothing$, we can let $d=b$. Otherwise, we can let $d$ be the smallest $x$ in $B$ with $f(x)=0$, by the proof of Theorem 7-1. In either case, we have $f(d)=0$, $x_0<d\le b$, and $f(x)>0$ for all $x \in [x_0,d)$.

Thus, we found $c$ and $d$ with $a\le c < x_0 < d \le b$ such that $f(c)=f(d)=0$, but $f(x)>0$ for all $x$ in $(c,d)$.

b) Let

$$c=\sup\{x : a \le x \le b,\ f(x)=f(a)\}.$$

We must have $f(c)=f(a)$. If $f(c)<f(a)$, then by the Intermediate Value Theorem there would exist $x \in [c,b]$ such that $f(x)=f(a)$, thereby invalidating $c$ as an upper bound. If $f(c)>f(a)$, we could find a lower upper bound. Hence, only $f(c)=f(a)$ is possible.

In addition, $f(x)>f(c)$ for $c<x\le b$. This can also be proven by contradiction.

Now, let

$$d=\inf\{x : c \le x \le b,\ f(x)=f(b)\}.$$

We can similarly prove that $f(d)=f(b)$ by contradiction.

In addition, $f(x)<f(d)$ for $c<x<d$. This can also be proven by contradiction.

Hence, $c$ and $d$ satisfy $a\le c<d\le b$ such that $f(c)=f(a)$ and $f(d)=f(b)$ and

$$f(a)<f(x)<f(d)$$

for all $x$ in $(c,d)$.