# For what values of x is f(x)=(2x^2−4x)e^x concave or convex?

Jan 22, 2018

The function is convex for $x \in \left(- \infty , - 1 - \sqrt{3}\right) \cup \left(- 1 + \sqrt{3} , + \infty\right)$ and concave for $x \in \left(- 1 - \sqrt{3} , - 1 + \sqrt{3}\right)$

#### Explanation:

$\text{Reminder}$

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

Calculate the first and second derivatives

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

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

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

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

${x}^{2} + 2 x - 2 = 0$

Solving this quadratic equation for $x$

$x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot \left(- 2\right)}}{2} = \frac{- 2 \pm \sqrt{12}}{2}$

$= \frac{- 2 \pm 2 \sqrt{3}}{2}$

$= - 1 \pm \sqrt{3}$

The roots are

${x}_{1} = - 1 - \sqrt{3}$

${x}_{2} = - 1 + \sqrt{3}$

We can build the variation chart

$\textcolor{w h i t e}{a a a}$$\textcolor{w h i t e}{a a a}$$\text{Interval}$$\textcolor{w h i t e}{a a a}$$\left(- \infty , {x}_{1}\right)$$\textcolor{w h i t e}{a a a}$$\left({x}_{1} , {x}_{2}\right)$$\textcolor{w h i t e}{a a a}$$\left({x}_{2} , + \infty\right)$

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

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

graph{(2x^2-4x)e^x [-17.35, 14.68, -7.95, 8.07]}