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Math Help - Critical points

  1. #1
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    Critical points

    F(x) = \frac{p-4}{p^2+2}

    f(x) = p-4
    f'(x) = -4
    g(x) = p^2 + 2
    g'(x) = 2p

    F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}

    F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}<br /> <br />

    I understand that a critical number is where F'(x) is either c or DNE.

    But as far as I can tell, neither the bottom nor the top can be equal to 0, which I think neglects both cases.
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  2. #2
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    Quote Originally Posted by CFem View Post
    F(x) = \frac{p-4}{p^2+2}

    f(x) = p-4
    f'(x) = -4
    g(x) = p^2 + 2
    g'(x) = 2p

    F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}

    F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}<br /> <br />

    I understand that a critical number is where F'(x) is either c or DNE.

    But as far as I can tell, neither the bottom nor the top can be equal to 0, which I think neglects both cases.
    the derivative of p-4 is not -4

    you also did the quotient rule incorrectly.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by CFem View Post
    F(x) = \frac{p-4}{p^2+2}

    f(x) = p-4
    f'(x) = -4
    g(x) = p^2 + 2
    g'(x) = 2p

    F'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}

    F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}<br /> <br />

    I understand that a critical number is where F'(x) is either c or DNE.

    But as far as I can tell, neither the bottom nor the top can be equal to 0, which I think neglects both cases.
    For one thing, you have F(x)=\frac{p-4}{p^2+2}. With respect to x, this is a constant function (and therefore has no critical points). I'm not sure if that's what you meant, but with the case F(x)=\frac{x-4}{x^2+2}, we have:

    f(x)=x-4
    f'(x)=1
    g(x)=x^2+2
    g'(x)=2x

    Therefore, F'(x)=\frac{x^2+2-2x(x-4)}{(x^2+2)^2}=\frac{-x^2+8x+2}{(x^2+2)^2}

    Using the quadratic formula on the numerator, we get that F has two critical points (with horizontal tangent lines) at:

    x=\frac{-8\pm\sqrt{64+8}}{-2}=4\pm3\sqrt{2}

    F'(x) is defined everywhere, so there are no vertical tangents.

    ----------------------

    On a related note, it only makes sense to define a vertical tangent of F at x=a when:

    1) F'(a) is undefined.
    2) F(a) is defined.

    Otherwise, you'll just get the equation for the asymptote.

    Take for example, F(x)=\sqrt{x} and investigate the behavior of the tangent line at x=0.
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