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anonymous
 3 years ago
prove that the function f(x)=....has a derivative at x=0
anonymous
 3 years ago
prove that the function f(x)=....has a derivative at x=0

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so a graphing calculator shows that x²cos(1/2) fluctuates near 0, and I believe it might not be continuous at 0. So the problem is to say that the piecewise function is defined and continuous and therefore has a derivative?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Is that function typed correctly? I'm getting a very smooth looking parabola as my graph for that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're right, I must have made a mistake before. but still the problem would be that if f(x) wasn't defined to be 0 at 0, it wouldn't be continuous and have a derivate at 0?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Could it have been \[x^2\cos\left(\frac{1}{2x}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, it's actually x²cos(1/x)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Ah. Well, to show that it's differentiable, you basically have to show that it's continuous, and that the limit \[\lim_{h\to0} \frac{f(0+h)f(0)}{h}\]exists.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3To show continuity, you need to show\[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so cos (1/x) at x>0 would be undefined because of (1/0), right? but then I don't know how to go further

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3What you could do, is since \(\cos(x)\) is bounded by 1 and 1, when you multiply by \(x^2\), that term will dominate. So \[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3In the first limit, \[\lim_{h\to0} \frac{f(0+h)f(0)}{h} =\lim_{h\to0} \frac{h^2\cos\left(\frac{1}{h}\right)}{h}\]We just showed that the numerator goes to 0, so we can use L'hopital's on this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0L'hopital's rule comes only later in the course!

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3In that, case, try multiplying by \[\Large\frac{\frac{1}{h}}{\frac{1}{h}}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Then we get \[\lim_{h\to0}\frac{h\cos\left(\frac{1}{h}\right)}{1}=\lim_{h\to0}h\cos\left(\frac{1}{h}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"What you could do, is since cos(x) is bounded by 1 and 1, when you multiply by x2, that term will dominate." but does it not matter that 1/0 would be undefined?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3But you aren't dividing by 0. You're just getting very close. So you have \[\cos(\text{A very large number})\] which is bounded by 1 and 1.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3I've got to go now. Hopefully that made enough sense. To do that very last limit, use the same fact that cosine is bounded.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Refer to the attached plot of\[\left\{x^2 \cos \left(\frac{1}{x}\right),\sin \left(\frac{1}{x}\right)+2 x \cos \left(\frac{1}{x}\right)\right\} \]in blue and red respectively.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is \[\lim_{h \rightarrow 0}\frac{ f(0+h)f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }\] ?

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0Because \(f(0+h)=f(h)=h^2cos(\frac{1}{h})\) (just replace x with h) Also, \(f(0)=0\), so\[\lim_{h \rightarrow 0}\frac{ f(0+h)f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ f(h)f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})0 }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})}{ h }\]
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