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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
    -1<=x<=1
    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.
    NEED TO INSERT GRAPH
    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
    4x^3+3x+1=0
    x=0.5,-1
    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
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