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Thread: Critical points

  1. #1
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    Critical points

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

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

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

    $\displaystyle F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

    $

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

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

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

    $\displaystyle F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

    $

    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 $\displaystyle p-4$ is not $\displaystyle -4$

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

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

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

    $\displaystyle F'(x)= \frac{-6p^2+8p-8}{(p^2+2)^2}

    $

    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 $\displaystyle F(x)=\frac{p-4}{p^2+2}$. With respect to $\displaystyle x$, 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 $\displaystyle F(x)=\frac{x-4}{x^2+2}$, we have:

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

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

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

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

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

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

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

    1) $\displaystyle F'(a)$ is undefined.
    2) $\displaystyle F(a)$ is defined.

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

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