How do you find the intervals of increasing, decreasing and concavity for $f(x) = 2x^3 + 3x^2 - 432x$ ?

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Answer 1 · 2015-05-23T22:49:11

Values of $f$ increase when $f'(x)>0$ and decrease when $f'(x)<0$ . Also $f$ is concave when $f''(x)<0$ .

So:
$f'(x)=2*3x^2+3*2x-432=6x^2+6x-432=$
$=6(x^2+x-72)=6(x+9)(x-8)$
$f'(x)<0 <=> x in (-9;8)$
$f'(x)>0 <=> x in (-oo;-9) or x in (8;+oo)$

$f''(x)=6*2x+6=12x+6=12(x+1/2)$
$f''(x)<0 <=> x in (-oo;-1/2)$

How do you find the intervals of increasing, decreasing and concavity for f(x) = 2x^3 + 3x^2 - 432x f(x) = 2x^3 + 3x^2 - 432x ?