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Math Help - continuity / differentiability problem

  1. #1
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    continuity / differentiability problem

    this problem i know how to do ("</=" means less than or equal to)

    f(x) = x^2 , x </= 1
    x^(1/2) , x > 1

    f(1) = 1^2 = 1
    lim f(x) = 1
    x~> 1+
    lim f(x) = 1
    x~> 1-
    since f(1) = lim f(x) = 1
    x~> 1
    the piecewise function is continuous at x = 1. then to check differentiability i took the derivative:
    f'(x) = 2x , x </= 1
    1/2x^(1/2) , x > 1

    f'(1) = 2
    lim f'(x) = 1/2
    x~> 1+
    lim f'(x) = 2
    x~> 1-
    since f'(1) =/= lim f'(x)
    x~> 1
    the derivative of the function is not continuous at x = 1 which means the two functions meets at x = 1 with different slopes so the function is not differentiable at x = 1.

    however i'm not sure about this next problem:
    f(x) = x^2 cos(1/x), x=/= 0
    0 , x = 0

    f(0) = 0
    lim f(x) = 0 (by the squeeze theorem)
    x~> 0
    since f(0) = lim f(x)
    x~>0
    the function is continuous at x = 0. then to check differentiability i took the derivative:
    f'(x) = sin(1/x) + 2x cos(1/x), x =/= 0 (i used the product and chain rules)
    0 , x = 0

    f'(0) = 0
    lim f'(x) = DNE (since sin(1/x) oscillates between 1 and -1)
    x~>0

    therefore i concluded that since f'(0) =/= lim f'(x)
    x~>0
    the function was not differentiable at x = 0 since the derivative wasn't continuous at x = 0 which meant that the 2 functions met with different slopes at x = 0. but on many websites with this same problem, it says that it's differentiable at ALL x. how is this possible? i don't understand how the logic from the first problem i did couldn't be applied to this problem. help will be very much appreciated!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    The derivative of \cos{\frac{1}{x}}=(-\sin\frac{1}{x})(-\frac{1}{x^2})=\frac{1}{x^2}\sin\frac{1}{x}

    So by the product rule we have

    x^2*\frac{1}{x^2}\sin\frac{1}{x}+\cos\frac{1}{x}*2  x\Longleftrightarrow\sin\frac{1}{x}+2x\cos\frac{1}  {x}

    I'm sorry, I was focused on the wrong thing.

    The function is defined everywhere, is it not? So as x\to0 you are right in saying that f(x) osillates \mid{f(x)}\mid\leq{1}, but the fuction has been defined to be 0 when x=0, therefore the function ossilates from the left and the right, but when x reaches 0 f(x)=0, and therefore, \lim_{x\to{0}}f(x)=0. Further more, \lim_{x\to0}f(x)=f(0).

    Does that help at all?
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  3. #3
    MHF Contributor

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    The derivative at 0 is \lim_{h\rightarrow 0}\frac{f(h)- f(0)}{h} which is \lim_{h\rightarrow 0} \frac{h^2 cos(1/h)}{h} = \lim_{h\rightarrow 0} h cos(1/h)= 0

    It is true that, for this function, \lim_{x\rightarrow 0} f'(x) does not exist so this derivative is not continuous at 0, but the derivative does exist at x= 0.
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