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Math Help - Differentiability of functions

  1. #1
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    Differentiability of functions

    Determine whether the following two functions are differentiable at the given value
    a) f(x) = xsin(1/x), f(0)=0: at given value 0
    b)
    f(x) = 1-x if x<1
    x(1-x) if x>= 1 at given value 1

    I think a is differentiable. I found the difference quotient and found that as
    h->0 the limit of the quotient -> 0 also.

    But I can't seem to get started on b!
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  2. #2
    Member mabruka's Avatar
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    for b

    find the left and right limits



    \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1}


    <br />
\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}


    See they are equal so it is differenciable on 1
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  3. #3
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    is it really that simple! I've been looking at that for ages!!!
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  4. #4
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    Do you understand that this means you were wrong about (a) being differentiable?

    The derivative at x= 0 would be given by \lim_{h\to 0}\frac{h sin(1/h)- 0}{h}= \lim_{h\to 0} sin(1/h) and that limit does NOT exist.

    It is also true that, while derivatives are not necessarily continuous, they do have the "intermediate value property". That means that if a function is differentiable, the limits of the two derivatives on each side must be equal.

    For (b), the f'= -1 for x< 1 and f'= 1- 2x for x< 1. The limits as x goes to 1 are the same so the function is differentiable at x= 1.
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