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

## Homework Statement

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

## 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.