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

Dec 25, 2016

Increasing: $\left(- \infty , - 1\right) , \left(1 , \infty\right)$
Decreasing: $\left(- 1 , 1\right)$

#### Explanation:

Expand to avoid having to use the product rule.

$y = {x}^{3} - 3 x$

Use the power rule to differentiate.

$y ' = 3 {x}^{2} - 3$

Find the critical values. These will occur when the derivative equals $0$ or is undefined. There are no undefined points on a quadratic function.

$0 = 3 {x}^{2} - 3$

$0 = 3 \left({x}^{2} - 1\right)$

$0 = {x}^{2} - 1$

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

$x = - 1 \mathmr{and} 1$

We have three separate intervals we have to check here. There is $\left(- \infty , - 1\right) , \left(- 1 , 1\right) \mathmr{and} \left(1 , \infty\right)$.

Pick test points. If when substituted into the derivative the derivative is negative, the function is deceasing over that interval. If the derivative is positive, the function is increasing over that interval.

Test Point 1: $x = - 2$

$y ' = 3 {x}^{2} - 3 \to y ' = 3 {\left(- 2\right)}^{2} - 3 = 3 \left(4\right) - 3 = 9$

$\therefore$The function is increasing over $\left(- \infty , - 1\right)$.

Test Point 2: $x = 0$

$y ' = 3 {x}^{2} - 3 \to y ' = 3 {\left(0\right)}^{2} - 3 = 3 \left(0\right) - 3 = - 3$

$\therefore$The function is decreasing over $\left(- 1 , 1\right)$

Test Point 3: $x = 2$

$y ' = 3 {x}^{2} - 3 \to y ' = 3 {\left(2\right)}^{2} - 3 = 3 \left(4\right) - 3 = 9$

$\therefore$ The function is increasing over $\left(1 , \infty\right)$.

Hopefully this helps!