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Math Help - curve sketching clarification

  1. #1
    Junior Member
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    Post curve sketching clarification

    My equation is f(x)=x/x^2-4

    So from that, heres what I have gotten so far...
    -domain: X is a real number that cannot be 2, -2
    - x-intercept is 0
    - y intercept is 0
    -There is a horizontal asymptote at y=0
    -There are vertical asymptotes at x=2 and x=-2

    -To find the critical numbers, I have to take the derivative and I did so using the quotient rule and came up with
    f'(x)= -2x^2/x^2-4

    1) How do I find the critical numbers by setting the function equal to 0? Am i dealing with the numerator? denominator?
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  2. #2
    Member Nacho's Avatar
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    Exist many critical number, for example when the grafh change the concavity or when change from increasing to decreasing

    Which do you look for?
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  3. #3
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    Quote Originally Posted by Nacho View Post
    Exist many critical number, for example when the grafh change the concavity or when change from increasing to decreasing

    Which do you look for?
    because its the first derivative, I'm looking for intervals of increase or decrease.
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  4. #4
    Math Engineering Student
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    then solve for f'(x)<0 and f'(x)>0.
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  5. #5
    Member Nacho's Avatar
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    Quote Originally Posted by linearalgebra View Post
    because its the first derivative, I'm looking for intervals of increase or decrease.

    Ok, first you must divide tha analysis, dependin the interval, because have point where the function is indetermined

    <br />
f'(x) =  - \frac{{2x^2 }}<br />
{{x^2  - 4}}<br />

    If <br />
x \in \left( { - \infty ,2} \right) \Rightarrow f'(x) < 0<br />
. Therefore is decrease

    And follow

    Here I upload the grafh, like that you will have an idea

    curve sketching clarification-math.png
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