Question 983091
First find the second derivative.
First derivative,
{{{dg/dx=4x^3-12x-1}}}
Second derivative,
{{{(d2g)/(dx2)=12x^2-12=12((x-0)^2-1)}}}
So the second derivative, in vertex form, has a vertex of (0,-1).
Since the multiplier is positive {{{12>0}}}, it opens upwards so the value at the vertex is the minimum.
({{{0}}},{{{-12}}})
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The maximum occurs at the endpoints specifically at {{{x=2}}} since it is further away from the axis of symmetry {{{x=0}}}.
({{{0}}},{{{36}}})