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Concave up and down of f (solving inequality)

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    f(x)=(2x^3+2x^2-5x-2) / 2(x^(2)-1)

    f''(x)=(-12x^5-24x^3+36x)/(4x^8-16x^6+24x^4-16x^2+4)

    Find the intervals where f is concave up.

    2. The attempt at a solution (I am having trouble interpreting the results at the end or if I've made a mistake somewhere):

    Attempt at finding intervals of concave up:
    f''(x)>0 where f(x) is concave up.
    Therefore,
    (-12x^5-24x^3+36x)/(4x^8-16x^6+24x^4-16x^2+4)>0

    -12x^5-24x^3+36x > 0 (multiplied denominator up)

    x(-12x^4-24x^2+36) > 0 (factored out x)

    So, I've stated that there are 2 cases where this is true, when the left side is + , + or
    -, -.

    So, for case 1 (+, +):
    x>0 and -12x^4-24x^2+36>0
    let x^2=u
    -12u^2-24u+36>0
    -12(u^2+2u-3)>0
    u^2+2u-3<0
    (u+3)(u-1)<0. At this point, I have again stated that there are 2 cases for where this is true, when the left side is +, + or -, - (am I supposed to do this?)

    So, for case 1 of this factored expression in case 1 of the entire inequality:
    u+3>0 and u-1>0
    u>-3 and u>1
    x^2>-3 (ignore due to non-real solutions) and x^2>1
    x>+-sqrt(1)
    Therefore, x>1 and x<-1.

    And for case 2 of this factored expression in case 1 of the entire inequality:
    u+3<0 and u-1<0
    u<-3 and u<1
    x^2<-3 (ignore again) and x^2<1
    x<+-sqrt(1)
    Therefore, x<1 and x>-1

    So overall from case 1 of the entire inequality (+,+), I have 2 sets of solutions.
    1) x>0, x>1 and x<-1
    2) x>0, x<1 and x>-1

    I don't want to type out the entire of case 2, however the general jist of what I have done is similar to the case 1 working out and from case 2 of the entire inequality (-, -) I have also have 2 sets of solutions:
    1) x<0, x<-1 and x>1
    2) x<0, x>-1 and x<1

    At this point, I'm having trouble determining the overall intervals of where concave up occurs. I have checked the actual solution, and it is concave up when x<-1 and 0<x<1.

    Any help would be greatly appreciated.
     
    Last edited: May 11, 2012
  2. jcsd
  3. May 11, 2012 #2
    Do you remember using sign charts in pre-calc to solve rational inequalities? That's the way to go here.
     
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