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Math Help - Continuity problem

  1. #1
    Senior Member Paze's Avatar
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    Continuity problem

    Find values of the constants a and b for which the following function is continuous but not differentiable.

    f(x)=\left\{\begin{array}{cc}ax+b,&\mbox{ if }x>0;\\sin2x, & \mbox{ if } x\leq 0\end{array}\right.

    In other words, the graph of the function should have a sharp corner at the pont (0,f(0)).

    I'm not even sure what they are asking. This is part of a limit course and I am very familiar with limits.
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  2. #2
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    Re: Continuity problem

    Quote Originally Posted by Paze View Post
    Find values of the constants a and b for which the following function is continuous but not differentiable.

    f(x)=\left\{\begin{array}{cc}ax+b,&\mbox{ if }x>0;\\sin2x, & \mbox{ if } x\leq 0\end{array}\right.

    In other words, the graph of the function should have a sharp corner at the pont (0,f(0)).

    I'm not even sure what they are asking. This is part of a limit course and I am very familiar with limits.
    1. Replace x by 0 in the appropriate part of the function. You'll get f(0) = 0

    2. Therefore b must be zero.

    3. The function has the slope f'(0) = 2

    4. Thus for a \ne 2 the function is not differentiable at x = 0.

    5. I've attached a sketch of the situation.
    Attached Thumbnails Attached Thumbnails Continuity problem-nichtdiffbar.png  
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  3. #3
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    Re: Continuity problem

    You will need to know at least something about derivatives to solve this problem (otherwise how can you be sure that it's not differentiable). The derivative of \sin{2x} at 0 is 2, and the derivative of ax+b is a, so since they want it to be not differentiable, a\ne{2}.

    For it to be continuous, the limit from the right (using ax+b) must be equal to the limit from the left (using \sin{2x}). You should be able to calculate those two limits and set them equal to get an equation for b.

    - Hollywood

    P.S. earboth beat me to the answer.
    Last edited by hollywood; January 30th 2013 at 11:25 PM.
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  4. #4
    Senior Member Paze's Avatar
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    Re: Continuity problem

    I understand derivatives, I should probably have mentioned that.

    However I don't understand why the derivative of sin2x at 0 is 2. The derivative of sinx is cosx so I assume the derivative of sin2x is cos2x...The derivative of sin(2*0) is 0 and the derivative of cos(2*0) is 1...How do you get 2?

    EDIT: Something tells me I am supposed to be using chain/product rule. If so then I have to vent some frustration at these courses all over the net. They ask you to solve problems which you have not learned every single time when teaching derivatives.

    EDIT2: Solved with chain rule. Jesus Christ, I like how these morons shoe-horn chain rule in the problems before properly teaching derivatives! Thanks sent to you both!
    Last edited by Paze; January 30th 2013 at 11:50 PM.
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    Re: Continuity problem

    So it sounds like you understand why a \ne 2. Did you figure out the equation for b?

    - Hollywood
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  6. #6
    Senior Member Paze's Avatar
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    Re: Continuity problem

    Quote Originally Posted by hollywood View Post
    So it sounds like you understand why a \ne 2. Did you figure out the equation for b?

    - Hollywood
    It seems to me that the function is continuous but not differentiable for any a except 2? In other words I'm getting y=(any a except 2)*x+0
    Am I understanding this wrong?
    Last edited by Paze; January 31st 2013 at 05:01 PM.
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  7. #7
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    Re: Continuity problem

    That is correct - you need b=0 to make it continuous and a\ne{2} to make it not differentiable.

    - Hollywood
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  8. #8
    Senior Member Paze's Avatar
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    Re: Continuity problem

    Cool! Thanks for the help to both of you!
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