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f(x) = x^p cos(1/x) if x>0, 0 if x = 0

domain [0,inf), p real number

*f*,

*f'*are continuous in regards to changing values of p when...

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- Thread starter savanna91
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- #1

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- 0

f(x) = x^p cos(1/x) if x>0, 0 if x = 0

domain [0,inf), p real number

Thanks!

- #2

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- 4

f(x) = x^p cos(1/x) if x>0, 0 if x = 0

domain [0,inf), p real number

f,f'are continuous in regards to changing values of p when...

Thanks!

I don't see any questions asked here?

- #3

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What values of p make f and f' continuous?

- #4

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You do know that f( x ) = cos(1/x) when x >0 and 0 when x = 0 is not continuous at 0 right?

well, suppose I had xcos(1/x), can you see that this is continuous at 0?

Okay, let's say you can see that it is; then (xcos(1/x) ) ' = cos(1/x) + [stuff] = cos(1/x) + sin(1/x)/x. This is not continuous at 0 right? especially not with the cos(1/x) term in there!

But, let's see what happens with x^2 cos(1/x):

( x^2cos(1/x) ) ' = 2xcos(1/x) + sin(1/x). This is almost good, since 2xcos(1/x) is continous at 0, but sin(1/x) isn't; if only we had an xsin(1/x) there instead ( which is continuous at 0 ).

But you can notice a pattern: (xcos(1/x)) ' had a sin(1/x)/x term, (x^2cos(1/x)) ' had a sin(1/x) term, hopefully (x^3 cos(1/x) ) ' will have a xsin(1/x) term in it, and indeed:

(x^3 cos(1/x) ) ' = 3x^2cos(1/x) + xsin(1/x) and indeed, this is continuous at 0 ( you should know that xcos(1/x) is continuous at 0, x^2cos(1/x) is as well, cos(1/x) is bounded between -1 and 1 while x or x^2 goes to 0, so the whole thing goes to 0 ).

Working out the derivatives on your own will show you that the whole thing relies on the "power rule" ( and perhaps the periodic and bounded properties of cos and sin )

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