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Bomull
Group Title
prove that the function f(x)=....has a derivative at x=0
 one year ago
 one year ago
Bomull Group Title
prove that the function f(x)=....has a derivative at x=0
 one year ago
 one year ago

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Bomull Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0\]
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.1
so 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?
 one year ago

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

Bomull Group TitleBest ResponseYou've already chosen the best response.1
you'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?
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.1
*derivative at 0
 one year ago

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

Bomull Group TitleBest ResponseYou've already chosen the best response.1
oh, it's actually x²cos(1/x)
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Ah. 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.
 one year ago

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
In 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.
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.1
L'hopital's rule comes only later in the course!
 one year ago

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

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

Bomull Group TitleBest ResponseYou've already chosen the best response.1
"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?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
But 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.
 one year ago

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

robtobey Group TitleBest ResponseYou've already chosen the best response.0
Refer 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.
 one year ago

robtobey Group TitleBest ResponseYou've already chosen the best response.0
Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.1
@robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.1
why is \[\lim_{h \rightarrow 0}\frac{ f(0+h)f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }\] ?
 one year ago

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