# For what values of x is f(x)=4/(x^2-1 concave or convex?

May 14, 2018

#### Explanation:

The function is

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

The domain of $f \left(x\right)$ is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , + \infty\right)$

Calulate the first derivative with the quotient rule

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

$u = 4$, $\implies$, $u ' = 0$

$v = {x}^{2} - 1$, $\implies$, $v ' = 2 x$

Therefore,

$f ' \left(x\right) = \frac{0 \cdot \left({x}^{2} - 1\right) - 4 \cdot 2 x}{{x}^{2} - 1} ^ 2 = - \frac{8 x}{{x}^{2} - 1} ^ 2$

$f ' \left(x\right) = 0$, $\implies$, $x = 0$

There is a critical point at $\left(0 , - 4\right)$

Calulate the second derivative with the quotient rule

$u = - 8 x$, $\implies$, $u ' = - 8$

$v = {\left({x}^{2} - 1\right)}^{2}$, $\implies$, $v ' = 4 x \left({x}^{2} - 1\right)$

$f ' ' \left(x\right) = \frac{- 8 {\left({x}^{2} - 1\right)}^{2} + 32 {x}^{2} \left({x}^{2} - 1\right)}{{x}^{2} - 1} ^ 4$

$= \frac{- 8 {x}^{2} + 8 + 32 {x}^{2}}{{x}^{2} - 1} ^ 3$

$= \frac{24 {x}^{2} + 8}{{x}^{2} - 1} ^ 3$

Therefore,

$f ' ' \left(x\right) \ne 0$, $\forall x \in \text{domain}$

Build a variation chart to determine the concavities

$\textcolor{w h i t e}{a a a a}$$\text{Interval}$$\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}$$\text{Sign f''(x)}$$\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}$$-$$\textcolor{w h i t e}{a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\text{ f(x)}$$\textcolor{w h i t e}{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$

Finally,

$f \left(x\right)$ is convex for $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

$f \left(x\right)$ is concave for $x \in \left(- 1 , 1\right)$