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Math Help - Continuity problem

  1. #1
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    Red face Continuity problem

    Hi everyone, this is a problem from a previous exam. I manage to solve half of it, but am struggling with the second half.

    f(x) = x2 + x - 1 for x ≥ 0
    f(x) = a cos(x) + b sin(x) for x < 0

    Find a and b so that f(x) is continuous and differentiable for all x = 0

    Can someone help solve this?

    Thank you all!
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  2. #2
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    Re: Continuity problem

    Quote Originally Posted by Nora314 View Post
    f(x) = x2 + x - 1 for x ≥ 0
    f(x) = a cos(x) + b sin(x) for x < 0

    Find a and b so that f(x) is continuous and differentiable for all x = 0
    Well, I found the required a and b for some x = 0, but I am not sure about all of them.

    Equate the values of the two functions at 0, as well as the values of their derivatives at 0. You'll get two equations in a and b.
    Thanks from Nora314
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  3. #3
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    Re: Continuity problem

    Hi, thanks for the reply!

    I wrote it a bit weirdly, but the functions I wrote are actually supposed to be one function. Would what you said still work then? I found that a = 1, by investigating the limit as x approaches 0.
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  4. #4
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    Re: Continuity problem

    Quote Originally Posted by Nora314 View Post
    the functions I wrote are actually supposed to be one function. Would what you said still work then?
    Yes.

    Quote Originally Posted by Nora314 View Post
    I found that a = 1, by investigating the limit as x approaches 0.
    x + x - 1 = -1 when x = 0 and a cos(x) + b sin(x) = a when x = 0. Therefore a = -1. Now find the left and right derivatives at 0 and equate them.
    Thanks from Nora314
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  5. #5
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    Re: Continuity problem

    Thanks for the help, I did that. Let me see, derivative on the right is 2x + 1 and on the left -sin x + b cosx. When I put them equal to each other and x = 0, I got b = 1.

    In the answer key it says the opposite values, it says that a = 1 and b = -1. I manged to get that answer too by using another method. Perhaps it does not matter if the values of a and b are switched around ? (lol, I always think that).

    Ok, I solved it both ways now do you know why the answers are switched around when we use this method?

    And btw, I really like your signature ^^
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  6. #6
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    Re: Continuity problem

    Quote Originally Posted by Nora314 View Post
    Let me see, derivative on the right is 2x + 1 and on the left -sin x + b cosx.
    Since a = -1, the left derivative is sin(x) + b cos(x), but since sin(0) = 0, the sign of sin(x) does not matter.

    Quote Originally Posted by Nora314 View Post
    In the answer key it says the opposite values, it says that a = 1 and b = -1.
    a = -1 and b = 1 gives the red graph, while a = 1 and b = -1 gives the blue one.



    Perhaps the textbook authors thought that the left function is a sin(x) + b cos(x).

    Quote Originally Posted by Nora314 View Post
    And btw, I really like your signature ^^
    Thanks!
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  7. #7
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    Re: Continuity problem

    Thank you so so much for your help ^^ It makes a lot more sense now. It is really nice that you made a graph and everything!
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