# How do you find the inflection point of the function f(x)=xe^(-2x)?

Jul 12, 2018

Below

#### Explanation:

You can find the inflexion point of a function by first finding the second derivative and letting the equation equal to 0.

$f \left(x\right) = x {e}^{- 2 x}$
$f ' \left(x\right) = x \times - 2 {e}^{- 2 x} + {e}^{- 2 x}$
$f ' \left(x\right) = {e}^{- 2 x} \left(1 - 2 x\right)$
$f ' ' \left(x\right) = {e}^{- 2 x} \times \left(- 2\right) + \left(1 - 2 x\right) \left(- 2 {e}^{- 2 x}\right)$
$f ' ' \left(x\right) = - 2 {e}^{- 2 x} - 2 {e}^{- 2 x} + 4 x {e}^{- 2 x}$
$f ' ' \left(x\right) = 4 x {e}^{- 2 x} - 4 {e}^{- 2 x}$

To find the inflexion point, $f ' ' \left(x\right) = 0$

$4 x {e}^{- 2 x} - 4 {e}^{- 2 x} = 0$
$4 {e}^{- 2 x} \left(x - 1\right) = 0$
$4 {e}^{- 2 x} = 0$ or $x - 1 = 0$

$4 {e}^{- 2 x} = 0$ has no solution because ${e}^{- 2 x} > 0$

$x - 1 = 0$
$x = 1$

Test $x = 1$ (YOU MUST TEST YOUR POINT FOR CONCAVITY)

When $x = 0$,
$f ' ' \left(0\right) = - 4$
When $x = 1$,
$f ' ' \left(1\right) = 0$
When $x = 2$,
$f ' ' \left(2\right) = 0.07326 \ldots$

Therefore, there is a change in concavity and so there is a point of inflexion at $\left(1 , {e}^{- 2}\right)$