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Homework Help: Curve Sketching for f(x)=6x^2/(x^2-2x-15)

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Please state the following components of this rational function?
    f(x)=6x^2/(x^2-2x-15)

    -intervals of increase/decrease
    -local maximum and minimum values
    -intervals of concave up/concave down
    -points of inflection


    2. Relevant equations
    N/A
    3. The attempt at a solution
    There is a horizontal asymptote at y=6.
    There are vertical asymptotes at x=-3 and x=5
    The first derivative is -12x(x+15)/(x^2-2x-15)^2
    The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3
    According to the first derivative, the critical numbers would be -15, and 0.
    After the first derivative test, I found out that a min. point occurs when x=-15 and a max. point occurs when x=0. However, I got (-15,5.625) and (0,0). How can the min. point have a larger y-value than the max. point?
    Also, when I graphed this function on GraphCalc, it seems that there does not seem to be a minimum point at x=-15. Why is that?
    When I graphed the function, there seems to be 3 branches; in the left, middle, and right. The left branch seems to go through the horizontal asymptote (y=6) but then it slowly approaches it from below y=6.
    Please graph f(x)=6x^2/(x^2-2x-15) at http://math.ucalgary.ca/undergraduate/webwork/graphing-calculator [Broken] to see what I am describing. I cannot justify why there would be a minimum at x=-15.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 2, 2010 #2

    Mark44

    Staff: Mentor

    What did you get for the derivative? The link you posted doesn't seem to have anything to do with this problem.

    A local minimum point can have a larger y value than a local maximum point. Such points have smaller y values than all other points in some interval.
     
  4. Mar 2, 2010 #3

    Please take a look the edited information that I typed above in my first post.
    I would appreciate it if you can help me out.
     
  5. Mar 2, 2010 #4

    Mark44

    Staff: Mentor

    Everything is fine to here. I didn't check the 2nd derivative, though.
    These are the correct critical numbers. As noted in my other post, a local minimum point can have a larger y value than another local maximum point.
    Change your scale and it will show up better. Going to the left from the y axis, the graph drops down to its local min point, and then gradually moves up the the hor. asymptote.
     
    Last edited by a moderator: May 4, 2017
  6. Mar 2, 2010 #5
    Will I need to use the values of -15, -3, 0, and 5 for the chart to find intervals of increase and decrease?

    Also, how do I find the intervals of concave up and concave down and the points of inflection?
    The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3. For what values of x do I test for in the 2nd derivative chart? Are -3 and 5 two numbers that I must include in this chart, since either one of them makes the second derivative undefined? Please help.
     
  7. Mar 2, 2010 #6

    Mark44

    Staff: Mentor

    Yes.
    Concave up - where y'' > 0. Concave down - where y'' < 0.
    Your intervals of concave up/concave down can't include x = -3 or x = 5, since the function is not defined there.
     
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