# How do you find the intervals of increasing and decreasing given y=-x^4+3x^2-3?

Apr 1, 2017

The intervals of increasing are $x \in \left(- \infty , - 1.225\right) \cup \left(0 , 1.225\right)$.
The intervals of decreasing are $x \in \left(- 1.225 , 0\right) \cup \left(1.225 , + \infty\right)$

#### Explanation:

We calculate the first derivative and build a chart.

$y = - {x}^{4} + 3 {x}^{2} - 3$

$y ' = - 4 {x}^{3} + 6 x$

We have, $y ' = 0$ when

$- 4 {x}^{3} + 6 x = 0$

$x \left(- 4 {x}^{2} + 6\right) = 0$

$x \left(6 - 4 {x}^{2}\right) = 0$

$x = 0$,

and $x = \pm \sqrt{\frac{3}{2}} = \pm 1.225$

The chart is :

$\textcolor{w h i t e}{a}$$I$$\textcolor{w h i t e}{a}$$\left(- \infty , - 1.225\right)$$\textcolor{w h i t e}{a}$$\left(- 1.225 , 0\right)$$\textcolor{w h i t e}{a}$$\left(0 , 1.225\right)$$\textcolor{w h i t e}{a}$$\left(1.225 , + \infty\right)$

$\textcolor{w h i t e}{a}$$y '$$\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 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 a a a a}$$-$

$\textcolor{w h i t e}{a}$$y$$\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 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 a a a a}$↘