Critical point

by cateater2000
Tags: critical, point
 P: 35 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
 Sci Advisor HW Helper PF Gold P: 12,016 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.
 P: 35 k thanks i'll try that
 Sci Advisor HW Helper PF Gold P: 12,016 Critical point If you're a bit unsure what I mean, the definition of f'(0) is: $$f'(0)=\lim_{h\to{0}}\frac{f(0+h)-f(0)}{h}$$
 P: 35 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?
HW Helper
PF Gold
P: 12,016
 Quote by 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?
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.
 P: 35 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
 Sci Advisor HW Helper PF Gold P: 12,016 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?

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