Finding Critical Point of f(x) with f(0) = 0

[cateater2000]

let f(x)=sin(1/x)*x^2 for x not 0, and f(0)=0. show that x=0 is a critical point for f which is neither a local minimum, a local maximum, nor an inflection point.

well I tried differentiating this, and got f'=-cos(1/x) +2xsin(1/x). to find a critical point i make f'=0. Not sure how to do this. Do I take the limx->0?

Any hints or tips would be great

 

[arildno]

What you've done here, is to find the derivative of f at all points EXCEPT at x=0!
But you are to find f'(0)...
Use the definition of the derivative.

 

[cateater2000]

k thanks i'll try that

 

[arildno]

If you're a bit unsure what I mean, the definition of f'(0) is:
f'(0)=\lim_{h\to{0}}\frac{f(0+h)-f(0)}{h}

 

[cateater2000]

yeh i got that to work, now how do I show that it's not a local min,max or inflection. Would I look at the second derivative? If that's not defined it's not anything?

 

[arildno]

yeh i got that to work, now how do I show that it's not a local min,max or inflection. Would I look at the second derivative? If that's not defined it's not anything?

The standard second-derivative fails, since the first derivative is discontinuous at x=0 (the 2.derivative is not defined).

It remains to be shown that f(0) is not a local maximum/minimum.
This should be fairly easy to show..

Use, for example, the following definition of local maximum:
We say that a function f has a local maximum at x_{0}, iff there exists a \delta>0 so that for all x\in{D}(x_{0},\delta),f(x)\leq{f}(x_{0})
I've assumed that the x's in the open \delta-disk are in the domain of f, as is the case in your problem.

Note that this definition makes no assumption of differentiability or continuity of f.

 

[cateater2000]

ok i finished that part of the question.( this is a 4 part question)

I can't figure out these 2 parts. Any tips would be fantastic
f(x)=x^2*sin(1/x)
1.let g(x)=2x^2 +f(x) (f from the first question i asked)

Show g has a global minimum at x=0 but g'(x) changes sign infinitely often on both (0,e) and (-e,0) for any e>0.

For this question I can easily show 0 is a critical point. But when I show it's a minimum is what's difficult, when I differentiate twice I cannot see that f''(0)>0


2. Let h(x)=x+2f(x). Show h'(0)>0, but h is not monotone increasing on any interval that includes 0.

I know how to show h'(0)>0 but have no idea how to show it's monotone increasing.



Again any help would be fantastic
thanks in advance

 

[arildno]

Show that for 1., g(x)>=x^2 for ALL x.
How can that help you in showing that x=0 must be a global minimum?