# How do you find the intervals of increasing and decreasing using the first derivative given y=2x^3+3x^2-12x?

Jan 18, 2017

The intervals of increasing are x in ] -oo,-2 [ uu ] -7, oo[
The interval of decreasing is $\left[- 2 , 1\right]$

#### Explanation:

We need

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

We calculate the derivative

$y = 2 {x}^{3} + 3 {x}^{2} - 12 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} + 6 x - 12$

To find the critical points, let $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$6 {x}^{2} + 6 x - 12 = 6 \left({x}^{2} + x - 2\right) = 0$

$6 \left(x - 1\right) \left(x + 2\right) = 0$

Therefore, $x = 1$ and $x = - 2$

We can make a variation table

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\frac{\mathrm{dy}}{\mathrm{dx}}$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a}$↗$\textcolor{w h i t e}{a a}$$20$$\textcolor{w h i t e}{a a}$↘$\textcolor{w h i t e}{a}$$- 7$$\textcolor{w h i t e}{a a a}$↗

The intervals of increasing are x in ] -oo,-2 [ uu ] -7, oo[

The interval of decreasing is $\left[- 2 , 1\right]$