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Math Help - Continuity question

  1. #1
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    Continuity question

    The function f is defined by

    f(x) = ax^3 + bx^2 for x < 1
    f(x) = x^2 + x for x \geq 1

    Both f and its derivative are continuous for all values of x. Find the values of the constants a and b.

    I find the derivative for both parts as

    f'(x) = 3ax^2 + 2bx for x < 1
    f'(x) = 2x + 1 for x \geq 1

    I know how to find a and b if they are continuous for one particular value for x - the function and the derivative are solved by means of simultaneous equations.

    I tried to do this method for this question, but found that I achieved varying answers - I tried x=1, x=2 and this resulted in different values for a and b, so I'm guessing that this question requires something a bit different? How would you attempt to solve this?

    Thanks for your help
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  2. #2
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    From the given we get this system:
    a+b=2~\&~3a+2b=3.
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  3. #3
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    in case some explanation is needed; A function is continuous if its limit from above is the same as its limit from below.

    First, apply this condition to the derivative:

     \lim_{x  \to  1^{-}} f'(x) = \lim_{x  \to  1^{+}} f'(x)

     \lim_{x  \to  1^{-}} 3ax^{2} +2bx = \lim_{x  \to  1^{+}} 2x+1

      3a +2b =  2+1

      3a +2b =  3


    now, do the same on the main function
     \lim_{x  \to  1^{-}} ax^{3}+bx^{2} = \lim_{x  \to  1^{+}} x^{2} + x

    a + b = 2
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  4. #4
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    Quote Originally Posted by Plato View Post
    From the given we get this system:
    a+b=2~\&~3a+2b=3.
    And I solved this to get a = -1, b = 3.

    But the question said that the function is continuous for all values of x - so shouldn't a and b be the same whatever value of x is used?
    I've obviously misunderstood the idea of continuity, but I would like to know how my above statement can be proved wrong if it is
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  5. #5
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    Quote Originally Posted by db5vry View Post
    And I solved this to get a = -1, b = 3.

    But the question said that the function is continuous for all values of x - so shouldn't a and b be the same whatever value of x is used?
    I've obviously misunderstood the idea of continuity, but I would like to know how my above statement can be proved wrong if it is

    If you look at x values other than 1, the function is continuous for any values of a and b. The only possible point of discontinuity is at x=1 (where the functional form changes) so we only have to inspect it there.
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  6. #6
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    Quote Originally Posted by SpringFan25 View Post
    If you look at x values other than 1, the function is continuous for any values of a and b. The only possible point of discontinuity is at x=1 (where the functional form changes) so we only have to inspect it there.
    It makes a lot more sense now seeing that! Thanks for your and everyone's time and help
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