# Use the first and second derivatives of the function y = x^3-3x^2-9x+15 to find the intervals of increase and decrease. Help!?

Jun 3, 2018

The function is decreasing for $x \in \left[- 1 , 3\right]$, and is decreasing for all x outside this interval, $x \notin \left[- 1 , 3\right]$.

#### Explanation:

You may benefit from looking at some videos where the derivatives are explained, for instance

It is always a good idea to start with drawing a graph of your 3rd degree function (I've used Geogebra):

The graph shows that for $x \in \left[- 1 , 3\right]$ the function is decreasing, and it is decreasing for all x outside this interval, $x \notin \left[- 1 , 3\right]$.

Let's use the 1st and 2nd derivatives to find this, as requested in the task:
$y = {x}^{3} - 3 {x}^{2} - 9 x + 15$
$y ' = 3 {x}^{2} - 6 x - 9 = 3 \left(x + 1\right) \left(x - 3\right)$
$y ' ' = 6 x - 6 = 6 \left(x - 1\right)$

$y '$ gives the slope in each point of the curve, and $y ' '$ tels if it's increasing or decreasing in the point. Therefore, for what values of x is $y ' = 0$?

The first derivative$y ' = 3 \left(x + 1\right) \left(x - 3\right)$
We see that $y ' = 0$ when $x = - 1$ and $x = 3$, so the slope changes direction in these points.

Next: the 2nd derivative $y ' ' = 6 \left(x - 1\right)$.
$x = - 1$: $y ' ' = 6 \left(x - 1\right) = 6 \left(- 2\right) = - 12 < 0$
A negative value tells us that the slope is changing from an increase to decrease

$x = 3$: $y ' ' = 6 \left(x - 1\right) = 6 \cdot 2 = 12 > 0$
A positive value tells us that the slope is changing from a decrease to increase.

This is actually in agreement with what we could see from the graph.