1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Calculate M = max|f(x)|>1/4 for f(x) = x^3+ ax^2+ bx + c

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = x^3+ ax^2+ bx + c with a, b, c real. Show that
    M = max|f(x)|>1/4
    and find all cases where equality occurs

    2. Relevant equations

    3. The attempt at a solution

    Let h(x)=x^3, then let g(x)=mx, we need to find m so that g(x) can get as close to h(x) with an even minimum distance between them, between [-1,1].
    By trail and error we get m=3/4 and are left with
    f(x)=x^3-3/4 x where a=0,b=-3/4 and c=0.
    So 1,-1,x_1 and x_2 are the critical points. So when is g(x)-h(x)=1/4?
    3/4 x-x^3=1/4
    x^3-3/4 x+1/4=0
    So the only solution are when x=0.5 or -1 and we have
    f(0.5)=0.125-0.385=-0.26 so |f(0.5)|=0.26>1/4
    f(-1)=-1+3/4=-0.25 so |f(-1)|=0.25=1/4
    So when x=0.5 or -1,M≥1/4.
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted