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Thread: calculus

  1. #1
    Junior Member
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    Angry calculus

    let f be a function defined by f(x)=
    2x- x^2 for x is less than or equal to 1
    x^2 +kx +p for x>1

    for what values of k and p will f be continuous and differentiable at x=1

    i got p=6 and k=-2 but i don't think that is correct, also shouldn't there be more than one answer? please help!!
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  2. #2
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    Quote Originally Posted by andyaddition View Post
    let f be a function defined by f(x)=
    2x- x^2 for x is less than or equal to 1
    x^2 +kx +p for x>1

    for what values of k and p will f be continuous and differentiable at x=1

    i got p=6 and k=-2 but i don't think that is correct, also shouldn't there be more than one answer? please help!!
    You've got:

    $\displaystyle f(x)=-x^2+2x~and~f'(x)=-2x+2$
    and
    $\displaystyle g(x)=x^2+kx+p$

    You know f(1)=1 that means the endpoint of f is P(1,1) and the slope of the graph at P is zero.

    $\displaystyle \lim_{x\rightarrow1}(x^2+kx+p)=1~\implies~k = -p$

    $\displaystyle \lim_{x\rightarrow1}(2x+k)=0~\implies~k = -2$

    Therefore p = 2

    The equation of g: $\displaystyle g(x) = x^2-2x+2$
    Attached Thumbnails Attached Thumbnails calculus-continuous.png  
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  3. #3
    MHF Contributor
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    Hi

    The continuity at x=1 is expressed by
    $\displaystyle \lim_{x \rightarrow 1^+} f(x) = f(1)$

    $\displaystyle \lim_{x \rightarrow 1^+} (x^2 + kx + p) = 1$

    $\displaystyle 1 + k + p = 1$

    $\displaystyle k + p = 0$

    The differentiability at x=1 is expressed by
    $\displaystyle \lim_{x \rightarrow 1^+} f'(x) = f'(1)$

    $\displaystyle \lim_{x \rightarrow 1^+} (2x + k) = 0$

    $\displaystyle 2 + k = 0$

    Then $\displaystyle k = -2$ and $\displaystyle p = 2$
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