Question 427186
{{{y = (3x^2-6x)(3x^2+6x) = 9x^4 - 36x^2}}}
==> {{{dy/dx = 36x^3 - 72x}}}
Let this equal 0:
 {{{36x^3 - 72x = 36x(x^2 - 2) = 0}}}==> x = 0, {{{sqrt(2)}}},{{{ -sqrt(2)}}}.
Also, {{{d^2y/dx^2 = 108x^2 - 72 = 36(3x^2 - 2)}}}
When x = 0, {{{d^2y/dx^2 < 0}}} ==> local max there, and so local max value is y = 0.
When x = {{{sqrt(2)}}} or {{{-sqrt(2)}}}, {{{d^2y/dx^2 > 0}}}, ==> local mins there, and so local (absolute) min value is y = -36.
(There is no absolute max value.)