How do you find the critical points for $g (x) = 2 x^{3} - 3 x^{2} - 12 x + 5$ $g(x)=2x^3-3x^2-12x+5$ ?

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Answer 1 · 2016-11-22T04:17:19

Find the first derivative and set the equation to equal zero.

$g'(x) = 6x^2-6x-12$
$=6(x^2-x-2)=6(x-2)(x+1)$

$6(x-2)(x+1)=0$

Critical Values:
$x=2, x=-1$

How do you find the critical points for g(x)=2x^3-3x^2-12x+5g(x)=2x3−3x2−12x+5g(x)=2x^3-3x^2-12x+5?