1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding extrema when derivative has no rational roots.

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data
    How may I find stationary points, increase/decrease intervals, concavity for f(x)=(x^3-2x^2+x-2)/(x^2-1)?


    2. Relevant equations



    3. The attempt at a solution
    I am familiar with how it should be done, except that here I get f'(x)=x^4-4x^2+8x-1 for the numerator of the derivative and am unable to figure out how to find extrema. I'd appreciate some advice.
     
  2. jcsd
  3. Jan 2, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Are you expected to find the exact locations of the stationary points, or merely whereabouts they lie? E.g. you can easily show there's one between 0 and 1.
     
  4. Jan 2, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Can you see why there exists (at least) one positive root, and one negative root? Of course, there are two other roots, which are either both real or are complex conjugates of one another.

    There are formulas for solving 4th degree polynomials, but the results are so complicated as to be almost useless. For example, Maple gives a positive root as

    -1/6*6^(1/2)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+1/6*((48*(82+9*83^(1/2))^(1/3)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)*(82+9*83^(1/2))^(2/3)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+72*6^(1/2)*(82+9*83^(1/2))^(1/3))/(82+9*83^(1/2))^(1/3)/((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2))^(1/2)

    and a negative root as

    -1/6*6^(1/2)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-1/6*((48*(82+9*83^(1/2))^(1/3)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)*(82+9*83^(1/2))^(2/3)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+72*6^(1/2)*(82+9*83^(1/2))^(1/3))/(82+9*83^(1/2))^(1/3)/((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2))^(1/2)

    Usually in such problems we just use numerical methods; that is why such methods were invented.
     
    Last edited: Jan 2, 2013
  5. Jan 3, 2013 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For the purposes of taking the 1st & 2nd derivatives, it may be useful to rewrite you function as:

    [itex]\displaystyle f(x)=\frac{x^3-2x^2+x-2}{x^2-1}[/itex]
    [itex]\displaystyle =
    x-2+\frac{2(x-2)}{x^2-1}[/itex]

    [itex]\displaystyle =x-2+\frac{3}{x+1}-\frac{1}{x-1}[/itex]​
     
  6. Jan 3, 2013 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Very nice!!

    To the OP: this is a technique that always comes in very handy. If you want to see how SammyS did this, then you should research partial fraction decomposition. See http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Examples
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook