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Math Help - Sketching A Curve

  1. #1
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    Sketching A Curve

    I have to find all the information (intercepts, asymptotes, increasing/decreasing, concavity, min/max) on the function f(x) = 2x/(x^2+1)

    From this I can tell that the x and y intercepts are both 0, the horizontal asymptote is 0, and there is no vertical asymptote. However, there is a slope asymptote since the highest power of x is different in the numerator than in the denominator. I'm not sure how to find this though.

    Then I move on to the derivative (found from the quotient rule) f'(x) = [-2(x+1)(x-1)]/[(x^2+1)(x^2+1)]

    From this I can tell that the critical points are +1 and -1 and there is no vertical tangent line. I can also see that the function is falling when x<-1, rising when -1<x<1, and falling when 1<x This means that there is a minimum at x=-1 and a maximum at x=1

    Then I move on to the second derivative, which I'm having trouble finding. I think if I could get help finding it (probably through the quotient rule) then I would be able to identify the critical points and find the concavity on the different intervals in the graph.

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Adrian View Post
    I have to find all the information (intercepts, asymptotes, increasing/decreasing, concavity, min/max) on the function f(x) = 2x/(x^2+1)

    From this I can tell that the x and y intercepts are both 0, the horizontal asymptote is 0, and there is no vertical asymptote. However, there is a slope asymptote since the highest power of x is different in the numerator than in the denominator. I'm not sure how to find this though.

    Then I move on to the derivative (found from the quotient rule) f'(x) = [-2(x+1)(x-1)]/[(x^2+1)(x^2+1)]

    From this I can tell that the critical points are +1 and -1 and there is no vertical tangent line. I can also see that the function is falling when x<-1, rising when -1<x<1, and falling when 1<x This means that there is a minimum at x=-1 and a maximum at x=1

    Then I move on to the second derivative, which I'm having trouble finding. I think if I could get help finding it (probably through the quotient rule) then I would be able to identify the critical points and find the concavity on the different intervals in the graph.

    Thanks
    we have:

    f'(x) = \frac {2 - 2x^2}{ \left( x^2 + 1 \right)^2}

    By the quotient rule:

    f''(x) = \frac {-4x \left(x^2 + 1 \right)^2 - 2 \left( 1 - x^2 \right) \cdot 2 \left( x^2 + 1 \right) \cdot 2x}{\left( x^2 + 1 \right)^4}

    = \frac {-4x \left( x^2 + 1 \right)^2 - 8x \left( 1 - x^2 \right) \left( x^2 + 1 \right) }{\left( x^2 + 1 \right)^4}

    = \frac {-4x \left( x^2 + 1 \right) - 8x \left( 1 - x^2 \right)}{\left( x^2 + 1 \right)^3}

    and you can simplify further if you want
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