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

Jan 24, 2018

The function is convex for $x \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

$\text{Reminder}$

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

$\left(u v\right) ' = u ' v + u v '$

Calculate the first and second derivatives

$f \left(x\right) = {e}^{x} / \left(5 {x}^{2}\right) + 1$

$f ' \left(x\right) = \frac{{e}^{x} \cdot \left(5 {x}^{2}\right) - {e}^{x} \cdot \left(10 x\right)}{25 {x}^{4}} = \frac{\left(x - 2\right) {e}^{x}}{5 {x}^{3}}$

$\left(x - 2\right) ' = 1$

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

$f ' ' \left(x\right) = \frac{{e}^{x} \left(x - 1\right) \left(5 {x}^{3}\right) - \left(x - 2\right) {e}^{x} \left(15 {x}^{2}\right)}{25 {x}^{6}}$

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

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

The inflection points are when $f ' ' \left(x\right) = 0$

The discriminant of the quadratic equation ${x}^{2} - 4 x + 6 = 0$

is

$\Delta = {b}^{2} - 4 a c = {\left(- 4\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(6\right) = 16 - 24 = - 8$

As the discriminant is negative, the second derivative

$f ' ' \left(x\right) > 0$, $\forall x \in \mathbb{R} - \left\{0\right\}$

The variation chart is as follows

$\textcolor{w h i t e}{a a a a}$$\text{Interval}$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , 0\right)$$\textcolor{w h i t e}{a a a a}$$\left(0 , + \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}$$+$$\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}$$\text{f(x)}$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

graph{e^x/(5x^2)+1 [-9.71, 10.29, -1.96, 8.04]}