Combinations of Continuous Functions

**shadow_2145** · June 5th 2008, 08:44 AM

(1) Let $g(x) = \sqrt[3]{x}$ .
(a) Prove that $g$ is continuous at $c=0$ .
(b) Prove that $g$ is continuous at a point $c \not=0$ . (The identity $a^3-b^3=(a-b)(a^2+ab+b^2$ ) will be helpful)

(2) Using the $\epsilon-\delta$ characterization of continuity (and thus using no previous results about sequences), show that the linear function $f(x)= ax+b$ is continuous at every point of $\mathbb{R}$ .

(3)
(a) Using the definition: ( A function $f: A \rightarrow \mathbb{R}$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$ , there exists a $\delta > 0$ such that whenever $|x-c| < \delta$ (and $x \in A$ ) it follows that $|f(x)-f(c)| < \epsilon$ . If $f$ is continuous at every point in the domain $A$ , then we say that $f$ is continuous on $A$ .) show that any function $f$ with domain $\mathbb{Z}$ will necessarily be continuous at every point in its domain.

(b) Show in general that if $c$ is an isolated point of $A$ $\subseteq \mathbb{R}$ , then $f: A\rightarrow R$ is continuous at $c$ .

(4) Assume $h: \mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and let $K = {x: h(x) = 0}$ . Show that $K$ is a closed set.

If anyone could figure out any of these, I would greatly appreciate it! Thanks!

**o_O** · June 5th 2008, 11:20 AM

$\forall \epsilon > 0, \exists \delta > 0$ such that $|x - c| < \delta \: \: \Rightarrow \: \: |f(x) - f(c)| < \epsilon$

1.(a) c = 0:
$|f(x) - f(c)| = |\sqrt[3]{x} - \sqrt[3]{0}| < \epsilon$
$|\sqrt[3]{x}| < \epsilon$
$|x| < \epsilon^{3}$

We know that |x - c| = |x - 0| = |x| < delta and we know that |x| < epsilon cubed. What should we choose delta to be?

(b) c not 0:
$|f(x) - f(c)| = |\sqrt[3]{x} - \sqrt[3]{c}| < \epsilon$
$\iff |x^{\frac{1}{3}}-c^{\frac{1}{3}}| \cdot \left|\frac{ x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}} }{x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}} }\right| < \epsilon$
(Think of x - c as a difference of cubes. So, the $x^{\frac{1}{3}} - c^{\frac{1}{3}}$ corresponds to the $(a-b)$ part of $a^{3} - b^{3} = (a-b)(a^2 + ab +b^2)$ )

$\iff \frac{|x-c|}{|x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}}|} < \epsilon$

etc.

2. $|f(x) - f(c)| < \epsilon$
$|ax + b - (ac + b)| < \epsilon$
$|ax - ac| < \epsilon$
$|a| \: |x - c| < \epsilon$
$|x-c| < \frac{\epsilon}{|a|}$

and the conclusion follows.