1. Continuity.

Question 1.
Prove using $\displaystyle \epsilon - \delta$ arguments that

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

Question 2.
Consider the function

$\displaystyle f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{ } x\leq 2\\x-1, &\mbox{ } 2<x<4\\4, &\mbox{ } x\geq 4\end{array}\right.$

(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!

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

(i) $\displaystyle f(x) = x^4$ is continuous at x=a, where a not equal to 0.
(ii) Deduce (quoting an appropriate theorem) that $\displaystyle g(y) = y^{1/4}$ is continuous on the interval $\displaystyle [0,\infty)$.
why don't you show us what you have done..

Originally Posted by pearlyc
Question 2.
Consider the function

$\displaystyle f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{ } x\leq 2\\x-1, &\mbox{ } 2<x<4\\4, &\mbox{ } x\geq 4\end{array}\right.$

(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!
$\displaystyle \cos$ is continuous, polynomials and constants are continuous..

so check whether $\displaystyle f(a) = \lim_{x\rightarrow a^+} f(x) = \lim_{x\rightarrow a^-} f(x)$ where $\displaystyle a$ are the possible values of x which can give discontinuities, in short, $\displaystyle 2$ and $\displaystyle 4$.

3. Originally Posted by pearlyc
(i) $\displaystyle f(x) = x^4$ is continuous at x=a, where a not equal to 0.
Say $\displaystyle |x-a|<\delta$ then $\displaystyle |x| \leq |x-a|+|a| < \delta + |a|$.
This means,
$\displaystyle |f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3|$$\displaystyle < \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|) +|a|^3]. Then for \displaystyle \epsilon > 0 choose \displaystyle \delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a| )+|a|^3} \right\} 4. Originally Posted by ThePerfectHacker Say \displaystyle |x-a|<\delta then \displaystyle |x| \leq |x-a|+|a| < \delta + |a|. This means, \displaystyle |f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3|$$\displaystyle < \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|) +|a|^3]$.

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

5. anyone else that is willing to give me a light in this?