# For what values of x is f(x)= e^x/(5x^2 +1 concave or convex?

Jun 23, 2018

The function is convex in the intervals $x \in \left(- \infty , - 0.175\right) \cup \left(0.355 , + \infty\right)$ and concave in the interval $x \in \left(- 0.175 , 0.355\right)$

#### Explanation:

The function is

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

Calculate the first and second derivatives

The first derivative is the derivative of a quotient

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

$u = {e}^{x}$, $\implies$, $u ' = {e}^{x}$

$v = 5 {x}^{2} + 1$, $\implies$, $v ' = 10 x$

$f ' \left(x\right) = \frac{{e}^{x} \left(5 {x}^{2} + 1\right) - {e}^{x} \left(10 x\right)}{5 {x}^{2} + 1} ^ 2$

$= \frac{{e}^{x} \left(5 {x}^{2} - 10 x + 1\right)}{5 {x}^{2} + 1} ^ 2$

The second derivative is the derivative of a quotient

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

$u = {e}^{x} \left(5 {x}^{2} - 10 x + 1\right)$, $\implies$, $u ' = {e}^{x} \left(5 {x}^{2} - 10 x + 1\right) + 10 {e}^{x} \left(x - 1\right)$

$v = {\left(5 {x}^{2} + 1\right)}^{2}$, $\implies$, $v ' = 2 \cdot 5 x \cdot \left(5 {x}^{2} + 1\right) = 10 x \left(5 {x}^{2} + 1\right)$

Therefore,

$f ' ' \left(x\right) = \frac{\left({e}^{x} \left(5 {x}^{2} - 10 x + 1\right) + 10 {e}^{x} \left(x - 1\right)\right) \cdot {\left(5 {x}^{2} + 1\right)}^{2} - \left({e}^{x} \left(5 {x}^{2} - 10 x + 1\right)\right) \left(10 x \left(5 {x}^{2} + 1\right)\right)}{5 {x}^{2} + 1} ^ 4$

$= \frac{{e}^{x} \left(\left(5 {x}^{2} - 10 x + 1 + 10 x - 10\right) \left(5 {x}^{2} + 1\right) - 10 x \left(5 {x}^{2} - 10 x + 1\right)\right)}{5 {x}^{2} + 1} ^ 3$

$= \frac{{e}^{x} \left(25 {x}^{4} - 100 {x}^{3} + 160 {x}^{2} - 20 x - 9\right)}{5 {x}^{2} + 1} ^ 3$

The points of inflections are when

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

$\implies$, $\frac{{e}^{x} \left(25 {x}^{4} - 100 {x}^{3} + 160 {x}^{2} - 20 x - 9\right)}{5 {x}^{2} + 1} ^ 3 = 0$

$\implies$, $25 {x}^{4} - 100 {x}^{3} + 160 {x}^{2} - 20 x - 9 = 0$

Solving $25 {x}^{4} - 100 {x}^{3} + 160 {x}^{2} - 20 x - 9 = 0$ graphically

graph{25x^4-100x^3+160x^2-20x-9 [-3.465, 3.464, -1.73, 1.734]}

The points of inflection are $\left(- 0.175 , 0.712\right)$ and $\left(0.355 , 0.86\right)$

Let's build a variation chart to determine the concavities

$\textcolor{w h i t e}{a a}$$\text{Interval}$$\textcolor{w h i t e}{a a}$$\left(- \infty , - 0.175\right)$$\textcolor{w h i t e}{a}$$\left(- 0.175 , 0.355\right)$$\textcolor{w h i t e}{a}$$\left(0.355 , + \infty\right)$

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

$\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}$$\cup$$\textcolor{w h i t e}{a a a 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 a}$$\cup$

The function is convex in the intervals $x \in \left(- \infty , - 0.175\right) \cup \left(0.355 , + \infty\right)$ and concave in the interval $x \in \left(- 0.175 , 0.355\right)$

graph{e^x/(5x^2+1) [-3.465, 3.464, -1.73, 1.734]}