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Math Help - Continuous functions

  1. #1
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    Continuous functions

    If possible, choose k so that the following function is continuous on any interval.

    f(x)= (4x^(3)-8x^(2))/(x-2) for x not= 2

    k for x=2

    alright, i need help on how to start this one, any suggestions would be very helpful, thank you
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  2. #2
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    \frac{{4x^3  - 8x^2 }}{{x - 2}} = 4x^2 ,\quad x \ne 2
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    Quote Originally Posted by Plato View Post
    \frac{{4x^3  - 8x^2 }}{{x - 2}} = 4x^2 ,\quad x \ne 2

    great, so what exactly do you do to find that it = 4x^2?
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  4. #4
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    Question

    You cannot find some value k such that f(2)=k is continuous with the rest of the function.
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    Quote Originally Posted by mathlete View Post
    great, so what exactly do you do to find that it = 4x^2?
    Do you know what it means for the function to be continuous at x=2?
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  6. #6
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    Quote Originally Posted by Plato View Post
    Do you know what it means for the function to be continuous at x=2?
    oh, ok i got it...duh, thanks alot
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  7. #7
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    Quote Originally Posted by mathlete View Post
    If possible, choose k so that the following function is continuous on any interval.

    f(x)= (4x^(3)-8x^(2))/(x-2) for x not= 2

    k for x=2

    alright, i need help on how to start this one, any suggestions would be very helpful, thank you

    \frac{4x^3-8x^2}{x-2} =4x^2 \frac{x-2}{x-2}

    so when x \ne 2 :


    \frac{4x^3-8x^2}{x-2} =4x^2

    Hence:

    \lim_{x \to 2} \frac{4x^3-8x^2}{x-2} =4\times 2^2=16

    So if you make k=16 then f will be continuous.

    RonL
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