# If y= (x^2+1)/(x^2-4) , what are the points of inflection, concavity and critical points?

##### 1 Answer
Sep 20, 2017

The critical point is at $\left(0 , - \frac{1}{4}\right)$ and it is a local maximum.
The intervals of convexity are $\left(- \infty , - 2\right) \cup \left(2 , + \infty\right)$
The interval of concavity is $\left(- 2 , 2\right)$

#### Explanation:

Calculating the first and second derivatives and building the variation charts

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

The domain of $y$ is $x \in \left\{\mathbb{R} - \left(- 2 , 2\right)\right\}$

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

$v \left(x\right) = {x}^{2} - 4$, $\implies$, $v ' \left(x\right) = 2 x$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{u ' v - u v '}{{v}^{2}} = \frac{2 x \left({x}^{2} - 4\right) - 2 x \left({x}^{2} + 1\right)}{{x}^{2} - 4} ^ 2$

$= \frac{2 {x}^{3} - 8 x - 2 {x}^{3} - 2 x}{{x}^{2} - 4} ^ 2$

$= \frac{- 10 x}{{x}^{2} - 4} ^ 2$

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

$\implies$, $- 10 x = 0$, $x = 0$

Building the variation chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a 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 a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\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 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}$↗$\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}$$- \frac{1}{4}$$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$↘

Calculating the second derivative to determine the concavity and the points of inflections

$u \left(x\right) = - 10 x$, $\implies$, $u ' \left(x\right) = - 10$

$v \left(x\right) = {\left({x}^{2} - 4\right)}^{2}$, $\implies$, $v ' \left(x\right) = 4 x \left({x}^{2} - 4\right)$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{u ' v - u v '}{{v}^{2}} = \frac{- 10 {\left({x}^{2} - 4\right)}^{2} + 40 {x}^{2} \left({x}^{2} - 4\right)}{{x}^{2} - 4} ^ 4$

$= \frac{\left({x}^{2} - 4\right) \left(- 10 {x}^{2} + 40 + 40 {x}^{2}\right)}{{x}^{2} - 4} ^ 4$

$= \frac{10 \left(3 {x}^{2} + 4\right)}{{x}^{2} - 4} ^ 3$

When $x = 0$, $\implies$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0$, this is a local max

Building 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}$$\left(- \infty , - 2\right)$$\textcolor{w h i t e}{a a a a}$$\left(- 2 , 2\right)$$\textcolor{w h i t e}{a a a a}$$\left(2 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$S i g n \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$$\textcolor{w h i t e}{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}$$\cup$$\textcolor{w h i t e}{a a a a a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

See the graph below

graph{(x^2+1)/(x^2-4) [-7.9, 7.9, -3.95, 3.95]}