1
$\begingroup$

Let $f:[0,\infty)\to\mathbb{R}$ be defined by: $$\begin{cases}&x\sin(\frac{1}{x}) \, \, &\text{if}\, \, x > 0\\ & 0, &\text{if} \, \, x = 0\end{cases}$$

Show that $f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.

I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.

$\endgroup$
  • $\begingroup$ What does the phrase "local maximum or minimum in the endpoint" mean? $\endgroup$ – BigbearZzz Dec 8 '18 at 10:10
  • $\begingroup$ According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :) $\endgroup$ – kareem bokai Dec 8 '18 at 10:19
  • $\begingroup$ I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum? $\endgroup$ – BigbearZzz Dec 8 '18 at 10:21
  • $\begingroup$ Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min $\endgroup$ – kareem bokai Dec 8 '18 at 10:25
2
$\begingroup$

An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with $$ -|x| \le f(x) \le |x|, $$ where $\pm|x|\to 0 $ as $x\to 0$. Differentiability of $f$ on $(0,\infty)$ is obvious.

For the second part, we want to show that for any $\varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,\varepsilon)$. Choose $n$ large enough so that $\frac 1{2\pi n+\pi/2}<\varepsilon$, then $$ f\left(\frac 1{2\pi n+\pi/2}\right) = \frac 1{2\pi n+\pi/2}>0 = f(0) $$ so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $\sin(1/x)=-1$)

$\endgroup$
0
$\begingroup$

It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=x\cdot g(x)$ is continuous at $x=0$.

For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $\lim_{x\to0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.