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

May 23, 2015

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

So:
$f ' \left(x\right) = 2 \cdot 3 {x}^{2} + 3 \cdot 2 x - 432 = 6 {x}^{2} + 6 x - 432 =$
$= 6 \left({x}^{2} + x - 72\right) = 6 \left(x + 9\right) \left(x - 8\right)$
f'(x)<0 <=> x in (-9;8)
f'(x)>0 <=> x in (-oo;-9) or x in (8;+oo)

$f ' ' \left(x\right) = 6 \cdot 2 x + 6 = 12 x + 6 = 12 \left(x + \frac{1}{2}\right)$
f''(x)<0 <=> x in (-oo;-1/2)