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## 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

## Homework Equations

## 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.