Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$f(x) = \begin{cases}x + 2x^2\sin\left(\frac1x\right),& x\ne 0\\0,& x = 0\;.\end{cases}$$

I am having a tough time answering this question in a rigorous mathematical way, here is what I have tried:

I have proved in a previous part of this question that $f\,'(0) = 1$.

The derivative when $x\ne 0$ is,

$$f\,'(x) = 1+4x\sin\left(\frac1x\right)−2\cos\left(\frac1x\right)\;.$$

I have $$\lim\limits_{x\to 0} f\,'(x) = 1 - 2\cos\left(\frac1x\right)\;,$$ which oscillates between $3$ & $-1$.

Since $f\,'(x)$ containing $0$, is not continuous, it cannot be increasing on the interval. Am I on the right track here?

Thanks in advance!

share|improve this question
    
the function seems to be increasing check this out –  Santosh Linkha Jun 14 '12 at 7:18
1  
@experimentX Your picture shows that every interval containing $0$ contains infinitely many subintervals where the function is decreasing for a little bit. These decreasing subintervals are easier to see farther away from $0$, but they are still there when you get close to $0$. So this picture does indeed illustrate what OP's problem states. –  alex.jordan Jun 14 '12 at 7:29

3 Answers 3

up vote 3 down vote accepted

You're on the right track, but the equation $$\lim\limits_{x\to 0} f\,'(x) = 1 - 2\cos\left(\frac1x\right)$$ doesn't make sense as it stands. It almost makes sens, though, in that the omitted term goes to zero as $x\to0$. So it is definitely true that the derivative oscillates infinitely much near the origin, with upper and lower bounds that approach $3$ and $-1$. For your purposes, though, you just need a sequence of points approaching $0$ where the derivative is negative. Good candidates for such a sequence are points $x$ where $\cos(1/x)=-1$. I trust you can take it from there on your own.

share|improve this answer
    
Thanks for your help Harald. So if I let x = 1/nPi, then as x->0, then I can show that there can always be a value for x such that f'(x) will be -1. Therefore the function is not increasing. Is that correct? –  JackReacher Jun 15 '12 at 5:45
    
Yup. Though I spotted a small mistake in my answer just now: You are looking for points where $\cos(1/x)=+1$ (I had missed the minus sign in the formula for $f'$), i.e., $x=1/(2n\pi)$. Incidentally, $\sin(1/x)=0$ at these points, which takes care of a minor complication. –  Harald Hanche-Olsen Jun 15 '12 at 7:12

(Note that the second term in your derivative for non-zero $x$ should be $2x\sin(1/x)$, and the third should be $-\cos(1/x)$, unless the second term in the function is supposed to be $2x^2\sin(1/x)$.)

You’re working too hard: in order to show that $f$ is not increasing in any interval containing $0$, it suffices to show that in any interval containing $0$ there is a point $x$ such that $f\,'(x)<0$.

share|improve this answer
    
Methinks your last sentence might be misleading. –  Harald Hanche-Olsen Jun 14 '12 at 7:10
    
@Harald: Argh. Forgot that the baseline was $y=x$. –  Brian M. Scott Jun 14 '12 at 7:21

You are correct till the part you compute your first derivative. I assume that the function is $$ f(x) = \begin{cases} x + 2x^2\sin(1/x) & \text{ for }x \neq 0 \\ 0 & \text{ for } x = 0\end{cases} $$so that the derivative is given by $$ f'(x) = \begin{cases} 1 + 4x \sin(1/x) - 2 \cos(1/x) & x \neq 0\\ 1 & x = 0\end{cases}$$ Consider any interval containing $0$ i.e say $(-a,b)$ where $a,b \in \mathbb{R}^+$. By archimedean property, you can find $n \in \mathbb{N}$, such that $$\dfrac1{n \pi} \in (-a,b).$$ Note that since $\dfrac1{(n+1) \pi} < \dfrac1{n \pi}$, we have that $$\dfrac1{(n+1) \pi} \in (-a,b).$$

Now look at the derivatives at $\dfrac1{n \pi}$ and $\dfrac1{(n+1) \pi}$ and conclude what you want.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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