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Sketching curves: intercepts, asymptotes, critical points [answer check]

  1. Sep 19, 2008 #1
    Sketch the following function, showing all work needed to sketch each curve.

    y = [tex]\frac{1}{3 + x^2}[/tex]

    The question is asking for all the work done to find x and y intercepts, vertical, horizontal and slant asymptotes; critical points and points of inflection, i have completed the question but i'm not sure i did it correctly, any guidance is appreciated. My answer is below:

    x - intercept: there are no x - intercepts.
    y - intercept: sub x = 0 into the function and you get y = 1/3

    there are no vertical asymptotes

    horizontal asymptote is y = 0 found by dividing every term in the function by x^2

    critical points:
    to find the critical points find the first derivative

    y1 = -2x [tex]^{-3}[/tex]

    = [tex]\frac{-2}{x^-3}[/tex]

    y1 can never equal zero therefore there are no max or min.

    to find the point of infliction find the second derivative

    y11 = 6x [tex]^{-4}[/tex]
    = [tex]\frac{6}{x^-4}[/tex]

    y11 can never equal zero
    for x [tex]\geq[/tex] 0, y11 is negative and for x [tex]\leq[/tex] 0, y11 is positive

    I have sketched the graph but not added it here i just would like to check if the work i did to get the sketch of the function is correct!
    Thanks A LOT!
    I really appreciate it!! :D
  2. jcsd
  3. Sep 19, 2008 #2


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    Gold Member

    I dont understand how you found the derivative of y. Check that again, because if you graph the function, you see that there is indeed a maximum at x=0.

    EDIT: Same with your second derivative. How exactly are you differentiating these?
  4. Sep 19, 2008 #3
    That is exactly what confused me!
    I found the derivatives as follows:

    y = [tex]\frac{1}{3+x^2}[/tex]
    = 3 + x[tex]^{-2}[/tex]

    first derivative
    y1 = -2x[tex]^{-3}[/tex]
    = [tex]\frac{-2}{x^3}[/tex]

    second derivative
    y11 = -2x[tex]^{-3}[/tex]
    = (-2)(-3)x [tex]^{-4}[/tex]
    = 6x [tex]^{-4}[/tex]
    = [tex]\frac{6}{x^4}[/tex]

    for both of these y cannot equal zero...therefore there are no critical points or points of infliction. however, i too see that there is a max at about (0, 1/3)!!
    any idea of where i am going wrong!
  5. Sep 19, 2008 #4


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    Staff Emeritus
    Science Advisor

    No. [tex]y= (3+ x^2)^{-1}[/tex] You will need to use the chain rule to find the derivative.

  6. Sep 19, 2008 #5


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    Homework Helper

    Your line

    y = \frac 1 {3+x^2} = 3+x^{-2}

    is the source of your error.

    Look at what happens for [tex] x = 1 [/tex] with your approach
    \frac 1 {3 + 1^1} = 3 + 1^{-1} = 4

    which is obviously false. Try writing the formula for [tex] y [/tex] with the entire denominator having a negative exponent.
  7. Sep 20, 2008 #6
    YAY! thanks a lot! i think i got it!
    for the critical points
    the first derivative would be

    y1 = [tex]\frac{-2x}{(3+x^2)^2}[/tex]

    which is equal to zero when x = 0 forming the point (0, 1/3) which i found to be a max.

    for the second derivative and the point of inflection
    y11 = [tex]\frac{6x^2 - 6}{(3 + x^2)^3}[/tex]

    y11 is equal to zero when x = 1 and x = -1. I found the points of inflection to be (1, 1/4) and (-1, 1/4).

    I'm pretty sure this is right i'd just like to make sure!
    thank you!
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