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Math Help - Question on Differentiation

  1. #1
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    Question on Differentiation

    I am stuck at this one question, mainly because I can't understand the question. Any help will be appreciated!!

    Qn

    In the following question, c=3
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  2. #2
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    Because it is talking about the derivative, this Q(h) is the "difference quotient", \frac{f(0+h)- f(0)}{h}.

    Because f, here, is defined by different formulas for x< 0 and x> 0, you must use the limits "from the right" and "from the left".

    As h approaches 0 "from the left, 0+ h= h< 0 so f(0+h)= f(h)= h+ h^2. f(0)= sin(3(0))= 0 so, for h< 0, \frac{f(0+h)- f(0)}{h}= \frac{h+ h^2}{h}.

    \lim_{h\to 0^-} Q(h)= \lim_{h\to 0}\frac{h+ h^2}{h}.

    As h approaches 0 "from the right", 0+ h= h> 0 so f(0+h)= f(h)= sin(3h). for h> 0, \frac{f(0+h)- f(0)}{h}= \frac{sin(3h)}{h}.

    \lim_{h\to 0^+}Q(h)= \lim_{h\to 0}\frac{sin(3h)}{h}.

    To find that limit, you will need to remember a limit formula for \lim_{x\to 0}\frac{sin(x)}{x} and use the fact that \frac{sin(3h)}{h}= 3\frac{sin(3h)}{(3h)}.

    The function, f, will be differentiable at x= 0 if and only if those two one sided limits are the same.
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  3. #3
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    Thanks for your help!

    Wow, the maths syntax is hard to use.

    Anyway, the answer is 3 since lim h->0(sinh/h)=1

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