# How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for f(x)=(x^2+1)/(x^2-4)?

Mar 18, 2017

There are no points of inflection.
The discontinuities are when $x = - 2$ and $x = 2$
$f \left(x\right)$ is concave up for x in ]-oo,-2[ uu]2.+oo[
$f \left(x\right)$ is concave down for x in ]-2,2[

#### Explanation:

As we cannot divide by $0$, $x \ne - 2$ and $x \ne 2$

The domain of $f \left(x\right)$ is ${d}_{f} \left(x\right) = \mathbb{R} - \left\{- 2 , 2\right\}$

Therefore, the discontinuities are when $x = - 2$ and $x = 2$

The derivative of a quotient is

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

We calculate the first derivative of $f \left(x\right)$

$u = {x}^{2} + 1$, $\implies$, $u ' = 2 x$

$v = {x}^{2} - 4$, $\implies$, $v ' = 2 x$

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

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

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

We can calculate the second derivative of $f \left(x\right)$

$u = - 10 x$, $\implies$, $u ' = - 10$

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

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

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

$= \frac{30 {x}^{2} + 40}{{x}^{2} - 4} ^ 3$

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

The points of inflexion are when, $f ' ' \left(x\right) = 0$

$f ' ' \left(x\right) \ne 0$, there are no points of inflection.

Let build the chart for concavities

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l s$$\textcolor{w h i t e}{a a a a}$]-oo,-2[$\textcolor{w h i t e}{a a a a}$]-2,2[$\textcolor{w h i t e}{a a a a}$]2,+oo[

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

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

$\textcolor{w h i t e}{a a a a}$$S i g n f ' ' \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{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 a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

Therefore,

$f \left(x\right)$ is concave up for x in ]-oo,-2[ uu]2.+oo[

$f \left(x\right)$ is concave down for x in ]-2,2[