# How do you find the intervals of increasing and decreasing given y=x^3-11x^2+39x-47?

Mar 13, 2017

The intervals of increasing are ]-oo,3[ and ]13/3,+oo[
The interval of decreasing is ]3,13/3[

#### Explanation:

We calculate the first derivative

$y = {x}^{3} - 11 {x}^{2} + 39 x - 47$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 22 x + 39$

We need $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$3 {x}^{2} - 22 x + 39 = \left(3 x - 13\right) \left(x - 3\right) = 0$

The critical points are

$x = \frac{13}{3}$ and $x = 3$

Now, we construct the chart

$\textcolor{w h i t e}{a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a}$$|$$\textcolor{w h i t e}{a a a}$]-oo,3[$\textcolor{w h i t e}{a a a}$$|$$\textcolor{w h i t e}{a a a}$]3,13/3[$\textcolor{w h i t e}{a a a}$$|$]13/3,+oo[

$\textcolor{w h i t e}{a a}$$x - 3$$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{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}$$+$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a}$$3 x - 13$$\textcolor{w h i t e}{a a a}$$|$$\textcolor{w h i t e}{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}$$-$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$+$

$\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}$$|$$\textcolor{w h i t e}{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}$$-$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a 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}$$|$$\textcolor{w h i t e}{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}$↘$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$↗

The intervals of increasing are ]-oo,3[ and ]13/3,+oo[

The interval of decreasing is ]3,13/3[