Question 1029546
{{{d(x)=3-x^2+x^3-(1+x^2+x^3)}}}


{{{d(x)=3-x^2+x^3-1-x^2-x^3}}}


{{{d(x)=2-2x^2}}}, a parabola with vertex as a maximum, something which is wanted.  This maximum will be in the exact middle of the two zeros of d(x).


{{{2-2x^2=0}}}
{{{1-x^2=0}}}
{{{1=x^2}}}
{{{x=0+- 1}}}
The vertex x-coordinate is therefore at {{{x=0}}}.


Question is find the maximum difference between the two, or, evaluate d(x) for {{{x=0}}}.


{{{d(0)=2-x^2}}}
{{{d(0)=2-0^2}}}
{{{highlight(d(0)=2)}}}