Question

prove that the function f(x)=....has a derivative at x=0

Answer 1

\[f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0\]

Answer 2

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?

Answer 3

Is that function typed correctly? I'm getting a very smooth looking parabola as my graph for that.

Answer 4

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?

Answer 5

*derivative at 0

Answer 6

Could it have been \[x^2\cos\left(\frac{1}{2x}\right)\]

Answer 7

oh, it's actually x²cos(1/x)

Answer 8

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.

Answer 9

To show continuity, you need to show\[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

Answer 10

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

Answer 11

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 \[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

Answer 12

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.

Answer 13

L'hopital's rule comes only later in the course!

Answer 14

In that, case, try multiplying by \[\Large\frac{\frac{1}{h}}{\frac{1}{h}}\]

Answer 15

Then we get \[\lim_{h\to0}\frac{h\cos\left(\frac{1}{h}\right)}{1}=\lim_{h\to0}h\cos\left(\frac{1}{h}\right)\]

Answer 16

"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?

Answer 17

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.

Answer 18

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.

Answer 19

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.

Answer 20

Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.

Answer 21

@robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)

Answer 22

why is \[\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }\] ?

Answer 23

Because \(f(0+h)=f(h)=h^2cos(\frac{1}{h})\) (just replace x with h)
Also, \(f(0)=0\), so\[\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ f(h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})-0 }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})}{ h }\]