Homework 7

Solution.

By Exercise 17.5, polynomials are continuous functions on $\R$. (This is not hard to prove--you can prove that $x^n$ is continuous for each $n \in \N$ by induction; then every polynomial is a finite linear combination of these.) Thus, by Theorem 17.4(iii), $\frac{p}{q}$ is continuous at $x_0$ provided $q\p{x_0} \neq 0$, which is exactly the domain of $f$.

Solution.

Consider $f\p{x} = \frac{1}{x - x_0}$. Since $x_0 \notin S$, it follows that $x - x_0 \neq 0$ for all $x \in S$. In particular, by Theorem 17.4(iii) again, $f$ is a continuous function on $S$.

To show that $f$ is unbounded, let $M > 0$. Since $x_n \to x_0$, there exists $n \in \N$ such that $\abs{x_n - x_0} < \frac{1}{M}$, so for this $n$, we have

Since $M$ was arbitrary, this means $f$ is unbounded.

Solution.

We follow the hint. Note that $x \mapsto f\p{x + 1}$ is continuous since it is the composition of two continuous functions, so $g$ itself is continuous since it is the difference of two continuous functions.

By definition, we have

Thus, $0$ lies between (or is one of) $g\p{0}$ and $g\p{1}$. Thus, by the intermediate value theorem, there exists $x_0 \in \br{0, 1}$ such that $g\p{x_0} = 0$. But this is equivalent to

so we are done by setting $\p{x, y} = \p{x_0, x_0 + 1}$.