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.