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Math Help - Differentiable FUnctions

  1. #1
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    Differentiable FUnctions

    a) Prove from definition that if
    f, g : R -->R are differentiable functions, then the product function fg is also differentiable.
    (b) Prove that if
    h : R --->R is a differentiable function that satisties
    |h(x)-h(y)|< or = |x-y|^1.5
    for all x; y elements of R, then h(x)=0 for all x.
    (c) Give an example of a function
    k : R --->R such that k is differentiable but not twice differentiable.

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  2. #2
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    For functions on the real numbers, the definition of "differentiable" at a point is that the derivative exist at that point. So to prove that "if f and g are differentiable at x= a, then fg is differentiable there", the best thing to do is to follow the derivation of the product rule.
    The derivative of fg at x= a is \displaystyle\lim_{h\to 0}\frac{f(a+h)(g(a+h)- f(a)g(a)}{h}= \lim_{h\to 0}\frac{f(a+h)g(a+h)- f(a)g(a+h)+ f(a)g(a+h)- f(a)g(a)}{h} and now separate into two fractions. Use the fact that f and g, separately are differentiable. Also, you will need to use the fact that, since f is differentiable, f is continuous.

    For (b), use the mean value theorem: for any x and y, there exist c between x and y, such that h(x)- h(y)= h'(c)(x- y). Take the absolute value of each side and use the fact that |h(x)- h(y)|\le |x- y|^{1.5}.

    For (c), start with a function that is continuous but not differentiable (there is a well known example) and integrate it!
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    For functions on the real numbers, the definition of "differentiable" at a point is that the derivative exist at that point. So to prove that "if f and g are differentiable at x= a, then fg is differentiable there", the best thing to do is to follow the derivation of the product rule.
    The derivative of fg at x= a is \displaystyle\lim_{h\to 0}\frac{f(a+h)(g(a+h)- f(a)g(a)}{h}= \lim_{h\to 0}\frac{f(a+h)g(a+h)- f(a)g(a+h)+ f(a)g(a+h)- f(a)g(a)}{h} and now separate into two fractions. Use the fact that f and g, separately are differentiable. Also, you will need to use the fact that, since f is differentiable, f is continuous.

    For (b), use the mean value theorem: for any x and y, there exist c between x and y, such that h(x)- h(y)= h'(c)(x- y). Take the absolute value of each side and use the fact that |h(x)- h(y)|\le |x- y|^{1.5}.

    For (c), start with a function that is continuous but not differentiable (there is a well known example) and integrate it!
    For (b) i obtain |h'(c)(x- y)|=|h(x)- h(y)|\le |x- y|^{1.5}.
    therefore |h'(c)(x- y)|\le |x- y|^{1.5}. Is there any clue how I can get h(x)=0 for all x?
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  4. #4
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    Dividing both sides by |x- y|, |h'(c)|\le |x- y|^{.5} since that is a positive power, if we take x and y very close together, the right side can be as close as we please to 0. That implies that h'(c) is 0 for all c- that is, that h(x) is a constant. Now show that that constant must be 0.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Dividing both sides by |x- y|, |h'(c)|\le |x- y|^{.5} since that is a positive power, if we take x and y very close together, the right side can be as close as we please to 0. That implies that h'(c) is 0 for all c- that is, that h(x) is a constant. Now show that that constant must be 0.
    So in order to show that the constant must be 0, do we show that h(c) passes through the origin at (0,0)?
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