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

Dec 11, 2016

The intervals of decreasing are x in ] -oo,-sqrt3 ] uu [sqrt3, oo[

The intervals of increasing are $x \in \left[- \sqrt{3} , 0 \left[\cup\right] 0 , \sqrt{3}\right]$

#### Explanation:

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{0\right\}$

We calculate the derivative to find the intervals of increasing and decreasing.

Here, we have a fraction of 2 functions

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$ and $\left({x}^{n}\right) ' = n {x}^{n - 1}$

$u = 3 {x}^{2} - 3$, $\implies$ $u ' = 6 x$

$v = {x}^{3}$, $\implies$, $v ' = 3 {x}^{2}$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{4} - 9 {x}^{4} + 9 {x}^{2}}{{x}^{6}}$

$= \frac{9 {x}^{2} - 3 {x}^{4}}{{x}^{6}} = \frac{3 {x}^{2} \left(3 - {x}^{2}\right)}{{x}^{6}}$

$= \frac{3 \left(3 - {x}^{2}\right)}{x} ^ 4$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, when $3 - {x}^{2} = 0$

$x = - \sqrt{3}$ and $x = \sqrt{3}$

Now, we can make our sign chart

$\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}$$- \sqrt{3}$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$\sqrt{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a a}$∣∣$\textcolor{w h i t e}{a a}$$\uparrow$$\textcolor{w h i t e}{a a a}$$\downarrow$

So,
The intervals of decreasing are x in ] -oo,-sqrt3 ] uu [sqrt3, oo[

The intervals of increasing are $x \in \left[- \sqrt{3} , 0 \left[\cup\right] 0 , \sqrt{3}\right]$

graph{(3x^2-3)/(x^3) [-10, 10, -5, 5]}