# Continuity.

• Aug 11th 2008, 04:54 AM
pearlyc
Continuity.
Question 1.
Prove using $\epsilon - \delta$ arguments that

(i) $f(x) = x^4$ is continuous at x=a, where a not equal to 0.
(ii) Deduce (quoting an appropriate theorem) that $g(y) = y^{1/4}$ is continuous on the interval $[0,\infty)$.

Question 2.
Consider the function

$f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{ } x\leq 2\\x-1, &\mbox{ } 2

(i) At which points is f(x) continuous?
(ii) Are there points at which f is continuous on the left or on the right but not continuous?
(iii) Give the smallest value of c for which the function f is an injection if its domain is restricted to the interval [c,3]

Thanks a lot for your guidance guys!
• Aug 11th 2008, 08:09 AM
kalagota
Quote:

Originally Posted by pearlyc
Question 1.
Prove using $\epsilon - \delta$ arguments that

(i) $f(x) = x^4$ is continuous at x=a, where a not equal to 0.
(ii) Deduce (quoting an appropriate theorem) that $g(y) = y^{1/4}$ is continuous on the interval $[0,\infty)$.

why don't you show us what you have done..

Quote:

Originally Posted by pearlyc
Question 2.
Consider the function

$f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{ } x\leq 2\\x-1, &\mbox{ } 2

(i) At which points is f(x) continuous?
(ii) Are there points at which f is continuous on the left or on the right but not continuous?
(iii) Give the smallest value of c for which the function f is an injection if its domain is restricted to the interval [c,3]

Thanks a lot for your guidance guys!

$\cos$ is continuous, polynomials and constants are continuous..

so check whether $f(a) = \lim_{x\rightarrow a^+} f(x) = \lim_{x\rightarrow a^-} f(x)$ where $a$ are the possible values of x which can give discontinuities, in short, $2$ and $4$.
• Aug 11th 2008, 09:20 AM
ThePerfectHacker
Quote:

Originally Posted by pearlyc
(i) $f(x) = x^4$ is continuous at x=a, where a not equal to 0.

Say $|x-a|<\delta$ then $|x| \leq |x-a|+|a| < \delta + |a|$.
This means,
$|f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3|$ $< \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|) +|a|^3]$.

Then for $\epsilon > 0$ choose $\delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a| )+|a|^3} \right\}$
• Aug 12th 2008, 12:15 AM
pearlyc
Quote:

Originally Posted by ThePerfectHacker
Say $|x-a|<\delta$ then $|x| \leq |x-a|+|a| < \delta + |a|$.
This means,
$|f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3|$ $< \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|) +|a|^3]$.

Then for $\epsilon > 0$ choose $\delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a| )+|a|^3} \right\}$

is that all we have to do for proving this question? they gave me a hint for this question saying that it may help to commence with the restriction that | x - a | < |a|/2. what does that mean?

can you please explain how did $\delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a| )+|a|^3} \right\}$ came about?
• Aug 13th 2008, 05:40 AM
pearlyc
anyone else that is willing to give me a light in this? (Doh)