How do you find the critical points for $y = x^3 + 12x^2 + 6x + 8$ ?

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Answer 1 · 2015-11-04T19:34:30

Since $lim_(x->+-oo) f(x)=+-oo$ there is no global maximum or minimum.

It has a local maximum at $x=-4-sqrt14$ which is

$f(-4-sqrt14)=28(4+sqrt14)$

and it has local minimum at $x=-4+sqrt14$ which is

$f(-4+sqrt14)=-28(sqrt14-4)$

How do you find the critical points for y = x^3 + 12x^2 + 6x + 8 y = x^3 + 12x^2 + 6x + 8 ?