# How do you find the intervals of increasing and decreasing using the first derivative given y=(x^5-4x^3+3x)/(x^2-1)?

Aug 14, 2017

The intervals of increasing are $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$
The interval of decreasing is $x \in \left(- 1 , 1\right)$

#### Explanation:

We need

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

Our function is

$y = \frac{{x}^{5} - 4 {x}^{3} + 3 x}{{x}^{2} - 1}$

$u \left(x\right) = {x}^{5} - 4 {x}^{3} + 3 x$, $\implies$, $u ' \left(x\right) = 5 {x}^{4} - 12 {x}^{2} + 3$

$v \left(x\right) = {x}^{2} - 1$, $\implies$, $v ' \left(x\right) = 2 x$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(5 {x}^{4} - 12 {x}^{2} + 3\right) \left({x}^{2} - 1\right) - \left({x}^{5} - 4 {x}^{3} + 3 x\right) \left(2 x\right)}{{x}^{2} - 1} ^ 2$

$= \frac{\left(5 {x}^{4} - 12 {x}^{2} + 3\right) \left({x}^{2} - 1\right) - x \left({x}^{2} - 1\right) \left({x}^{2} - 3\right) \left(2 x\right)}{{x}^{2} - 1} ^ 2$

$= \frac{\left({x}^{2} - 1\right) \left(5 {x}^{4} - 12 {x}^{2} + 3 - 2 {x}^{4} + 6 {x}^{2}\right)}{{x}^{2} - 1} ^ 2$

$= \frac{3 {x}^{4} - 6 {x}^{2} + 3}{{x}^{2} - 1}$

$= \frac{3 \left({x}^{4} - 2 {x}^{2} + 1\right)}{{x}^{2} - 1}$

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

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

The critical points are when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

We can build the variation chart

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , - 1\right)$$\textcolor{w h i t e}{a a a a}$$\left(- 1 , 1\right)$$\textcolor{w h i t e}{a a a a}$$\left(1 , + \infty\right)$

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

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

The intervals of increasing are $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

The interval of decreasing is $x \in \left(- 1 , 1\right)$

We could have obtained the same result by simplifying $y$

$y = \frac{{x}^{5} - - 4 {x}^{3} + 3 x}{{x}^{2} - 1} = \frac{x \left({x}^{2} - 3\right) \left({x}^{2} - 1\right)}{{x}^{2} - 1} = {x}^{3} - 3 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 3 = 3 \left({x}^{2} - 1\right) = 3 \left(x + 1\right) \left(x - 1\right)$

graph{(x^5-4x^3+3x)/(x^2-1) [-10, 10, -5, 5]}