# What are the points of inflection of f(x)=2xe^-x + x^2e^x ?

May 1, 2017

$f \left(x\right)$ has one point of inflection, which is at $x \approx .161$

#### Explanation:

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

To find points of inflection, find where $f ' ' \left(x\right)$ changes sign.

Use the product rule and chain rule:
$f ' \left(x\right) = \left(2\right) \left({e}^{- x}\right) + \left(2 x\right) \left(- {e}^{- x}\right) + \left(2 x\right) \left({e}^{x}\right) + \left({x}^{2}\right) \left({e}^{x}\right)$

Rewrite/simplify:
$f ' \left(x\right) = 2 {e}^{- x} - 2 x {e}^{- x} + 2 x {e}^{x} + {x}^{2} {e}^{x}$

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

More simplification...
$f ' ' \left(x\right) = - 4 {e}^{- x} + 2 x {e}^{- x} + 2 {e}^{x} + 4 x {e}^{x} + {x}^{2} {e}^{x}$

Set $f ' ' \left(x\right)$ equal to zero using graphing calculator:
$0 = f ' ' \left(x\right)$

$x \approx .16059649$