# Neighborhoods/differentiability

• December 20th 2009, 10:34 AM
zhupolongjoe
Neighborhoods/differentiability
Let f(x)=x+2x^2*sin(1/x) for x=/=0 and f(0)=0

Prove that f(x) is differentiable everywhere and that f'(0)>0, but there is no neighborhood of 0 on which f is increasing.

I differentiate using the standard rules to get:

f'(x)=-2cos(1/x)+4xsin(1/x)+1 for x=/=0

And when x=0, I look at

lim(x->0) x+2x^2*sin(1/x)/x, which upon factoring, gives that the limit is equal to 1>0, and so f'(0)>0.

The part I can't get is the last part, show there is no neighborhood of 0 on which f is increasing. I know the sin(1/x) term gives wild oscillation, and 1/x is unbounded, but I don't know how to formally show the conclusion that there is no neighborhood of 0 on which f is increasing.

Thanks.
• December 20th 2009, 10:53 AM
tonio
Quote:

Originally Posted by zhupolongjoe
Let f(x)=x+2x^2*sin(1/x) for x=/=0 and f(0)=0

Prove that f(x) is differentiable everywhere and that f'(0)>0, but there is no neighborhood of 0 on which f is increasing.

I differentiate using the standard rules to get:

f'(x)=-2cos(1/x)+4xsin(1/x)+1 for x=/=0

And when x=0, I look at

lim(x->0) x+2x^2*sin(1/x)/x, which upon factoring, gives that the limit is equal to 1>0, and so f'(0)>0.

The part I can't get is the last part, show there is no neighborhood of 0 on which f is increasing. I know the sin(1/x) term gives wild oscillation, and 1/x is unbounded, but I don't know how to formally show the conclusion that there is no neighborhood of 0 on which f is increasing.

Thanks.

For any neighborhood $I_\epsilon:=(-\epsilon,\epsilon)$ of zero there exists $n\in\mathbb{N}\,\,\,s.t.\,\,\,x_n:=\frac{1}{2\pi n}\,,\,\,y_n:=\frac{1}{(4n+1)\pi\slash 2}\in I_\epsilon$ , but:

$f'(x_n) =-2\cos(2\pi n)+8\pi n\sin(2\pi n)+1 = -1<0$

$f'(y_n)=-2\cos\left(\frac{(4n+1)\pi}{2}\right)+2(4n+1)\pi\s in\left(\frac{(4n+1)\pi}{2}\right)+1=2(4n+1)\pi+1> 0$

Thus, the derivative of the function doesn't keep the same sign in any neighborhood of zero $\Longrightarrow$ f isn't increasing (or decreasing) in any such neighborhood.

Tonio
• December 21st 2009, 03:34 PM
zhupolongjoe
Thanks. One question though. Since I found in the problem that f'(0)=1, can I just use the xn you have and that or do I need to find another yn as you did?
• December 21st 2009, 03:59 PM
tonio
Quote:

Originally Posted by zhupolongjoe
Thanks. One question though. Since I found in the problem that f'(0)=1, can I just use the xn you have and that or do I need to find another yn as you did?

I don't understand your question: the xn's and yn's we defined served to show that in any neighborhood of zero f ' has positive and negative values, so it does not keep one single sign in it ==> f isn't increasing or decreasing in no neighborhood of zero. What does this have to do with the actual value of f'(0)?

Tonio
• December 21st 2009, 05:14 PM
zhupolongjoe
Okay, nevermind...that answered my question, thanks.