# How do you find the intervals of increasing and decreasing using the first derivative given y=2xsqrt(9-x^2)?

Jun 20, 2017

Increasing: $x : \left(- \frac{3 \sqrt{2}}{2} , + \frac{3 \sqrt{2}}{2}\right)$
Decreasing: $x : \left[- 3 , - \frac{3 \sqrt{2}}{2}\right) \cup \left(+ \frac{3 \sqrt{2}}{2} , + 3\right]$

#### Explanation:

$y = 2 x \sqrt{9 - {x}^{2}}$

First let's find the domain of $y$

$y \in \mathbb{R}$ where $\left(9 - {x}^{2}\right) \ge 0 \to \left\mid x \right\mid \le 3$

Hence the domain of $y$ is: $- 3 \le x \le + 3$

Next we will find the turning points of $y$ using $y ' = 0$

$y = 2 x {\left(9 - {x}^{2}\right)}^{\frac{1}{2}}$

Applying the product rule and chain rule

$y '$ = $2 x \cdot \frac{1}{2} \cdot {\left(9 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right) + 2 \cdot {\left(9 - {x}^{2}\right)}^{\frac{1}{2}}$

$= 2 {\left(9 - {x}^{2}\right)}^{\frac{1}{2}} - \frac{2 {x}^{2}}{{\left(9 - {x}^{2}\right)}^{\frac{1}{2}}}$

$= \frac{2 \left(9 - {x}^{2}\right) - 2 {x}^{2}}{\sqrt{9 - {x}^{2}}}$

$= \frac{18 - 4 {x}^{2}}{\sqrt{9 - {x}^{2}}}$

For critical $y$:

$\frac{18 - 4 {x}^{2}}{\sqrt{9 - {x}^{2}}} = 0 \to 18 - 4 {x}^{2} = 0$

$2 {x}^{2} = 9 \to x = \pm \sqrt{\frac{9}{2}}$

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

This question asks for the intervals of $x$ of increasing and decreasing $y$ using the $y '$. To avoid using the second derivative, it is now helpful to observe the graph of $y$ below:

graph{2xsqrt(9-x^2) [-20.28, 20.26, -10.13, 10.15]}

Since we know the turning points are where $x = \pm \frac{3 \sqrt{2}}{2}$ and that the domain of $y$ is $x : \left[- 3 , + 3\right]$

We can see that $y$ is increasing for $x : \left(- \frac{3 \sqrt{2}}{2} , + \frac{3 \sqrt{2}}{2}\right)$

and $y$ is decresing for $x : \left[- 3 , - \frac{3 \sqrt{2}}{2}\right) \cup \left(+ \frac{3 \sqrt{2}}{2} , + 3\right]$