Solutions

a) If there is more than one $x$ in $[a,b]$ with $f(x)=0$, there is not necessarily a second smallest. Case in point, consider the continuous function

$$f(x)= \begin{cases} 0, & \text{if } 0 < x \le 2,\\ x, & \text{otherwise} \end{cases}$$

on the interval $[-5,5]$, with $f(-5)=-5$ and $f(5)=5$.

To prove that there is a largest $x$ in $[a,b]$ with $f(x)=0$, consider $g(x)=f(b-x+a)$. We have $g(a)=f(b)>0$ and $g(b)=f(a)<0$. We can then apply the same proof structure in Theorem 7-1 of finding the least upper bound to the set

$$A=\{x : a \le x \le b \text{ and } g \text{ is positive on } [a,x]\}.$$

This least upper bound corresponds to the largest $x$ in $[a,b]$ with $f(x)=0$.

b) Let $f$ be a continuous function on $[a,b]$ such that $f(a)<0$ and $f(b)>0$. Consider

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

Since $B$ is bounded above by $b$ and is not empty, it has a least upper bound $\alpha$. We can assert that $\alpha < b$. Obviously, we cannot have $\alpha > b$ because $b$ is an upper bound of $B$ by definition. Moreover, $\alpha \ne b$ because otherwise by Theorem 6-3 there would exist a $\delta > 0$ such that $f(x)>0$ for $\alpha-\delta < x < \alpha$ since $f(b)>0$, thereby invalidating $\alpha$ as $\sup B$.

We now posit that $f(\alpha)=0$. For contradiction, suppose $f(\alpha)<0$. Applying Theorem 6-3 would invalidate $\alpha$ as an upper bound since there would exist $x_0$ such that $\alpha < x_0 < b$ and $f(x_0)<0$. Likewise, suppose $f(\alpha)>0$. Applying Theorem 6-3 would invalidate $\alpha$ as the least upper bound since there would exist $x_1$ such that $x_1<\alpha$ and $f(x_1)>0$. Both options lead to contradictions, leaving $f(\alpha)=0$ as the only possible alternative.

The point $\alpha$ located by this proof corresponds to the largest $x$ in $[a,b]$ such that $f(x)=0$. The set $A$, defined by the textbook's proof of Theorem 7-1, and $B$ defined in the previous proof differ when the largest $x$ in $[a,b]$ such that $f(x)=0$ is not also the smallest in that interval.