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Math Help - Critical point? Help me please

  1. #1
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    Critical point? Help me please

    Hello i have a test tomorrow and i can't figure this out.

    I need to find the critical points, local extrema, intervals of decrease and increase, intervals of concavity, inflection point, and the vertical and horizontal asymptotes.

    for this equation

    f(x)= x/((x^2)-9)

    Any help at all would be great, thanks
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  2. #2
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    Quote Originally Posted by bdavidson View Post
    Hello i have a test tomorrow and i can't figure this out.

    I need to find the critical points, local extrema, intervals of decrease and increase, intervals of concavity, inflection point, and the vertical and horizontal asymptotes.

    for this equation

    f(x)= x/((x^2)-9)

    Any help at all would be great, thanks
    Do you know how to find the derivative?

    Do you know the relationship between the derivative and "critical points, local extrema, intervals of decrease and increase, intervals of concavity, inflection point"?

    Horizontal aymptote is y = 0 (why?). Vertical asymptotes found by solving x^2 - 9 = 0 (why?)
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  3. #3
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    Quote Originally Posted by bdavidson View Post
    Hello i have a test tomorrow and i can't figure this out.

    I need to find the critical points, local extrema, intervals of decrease and increase, intervals of concavity, inflection point, and the vertical and horizontal asymptotes.

    for this equation

    f(x)= x/((x^2)-9)

    Any help at all would be great, thanks

    1.Take the derivative(use the Quotient Rule)
    2.Find the critical points by setting f'(x)=0, and solve for x
    3.Use the little table thing, plug in the points to find the interval of dec&inc
    4.Take the second derivative
    5.Same thing as #3, but to find the concavity
    6.Inflection point, I believe you get al lthe points from the second derivative, and find it by plugging it into the original function. (x, f(x))
    7.Guy above me said it
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  4. #4
    Junior Member toraj58's Avatar
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    for the vertical asymptote solve this equation:

    x^2-9=0;

    for horizontal asymptote the solution is like this:

    \lim_ {x\to\infty}f(x)
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