Question 1022005

In order for it to be positive before and negative after {{{x=2}}}, then it must look like this,
{{{f(x)=a(x-2)^3}}}
or else you will have an additional set of directional changes.
So now solve for the constant.
After x=2 the cubic will be positive so the constant needs to be negative.
So,
{{{f(x)=-a(x-2)^3}}}
{{{f(0)=-a(0-2)^3=4}}}
{{{a=4/8}}}
{{{a=1/2}}}
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{{{f(x)=-(x-2)^3/2}}}
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*[illustration fq10.JPG].