# How do you find the maximum, minimum and inflection points and concavity for the function y = xe^x?

Dec 18, 2016
• $x = - 1$ is a local minimum and the absolute minimum of the function $y = x {e}^{x}$ because $y '$ changes from negative to positive at $x = - 1$.
• $x = - 2$ is an inflection point of the function $y = x {e}^{x}$ because $y ' '$ changes from negative to positive at $x = - 2$.
• $y = x {e}^{x}$ is concave down (convex) on $x \in \left(- \infty , - 2\right)$, and concave up on $x \in \left(- 2 , \infty\right)$

#### Explanation:

$y = \left(x\right) \left({e}^{x}\right)$

To find maxima and minima, find where $y '$ is equal to zero:
$y ' = \left(x\right) \left({e}^{x}\right) + \left(1\right) \left({e}^{x}\right)$
$y ' = \left({e}^{x}\right) \left(x + 1\right)$
$0 = \left({e}^{x}\right) \left(x + 1\right)$
$0 = x + 1$
$x = - 1$
To check whether $x = - 1$ is a max or a min, use the first derivative test (plug in values less than and greater than -1):
$y ' \left(- 100\right) = \text{negative}$
$y ' \left(100\right) = \text{positive}$
$x = - 1$ is a local minima and the absolute minimum of the function $y = x {e}^{x}$ because $y '$ changes from negative to positive at $x = - 1$.

Do the second derivative test to find inflection points and concavity:
$y ' = \left({e}^{x}\right) \left(x + 1\right)$
$y ' ' = \left({e}^{x}\right) \left(1\right) + \left({e}^{x}\right) \left(x + 1\right)$
$y ' ' = \left({e}^{x}\right) \left(x + 2\right)$
$0 = \left({e}^{x}\right) \left(x + 2\right)$
$0 = x + 2$
$x = - 2$
$y ' ' \left(- 100\right) = \text{negative}$
$y ' ' \left(100\right) = \text{positive}$
$x = - 2$ is an inflection point of the function $y = x {e}^{x}$ because $y ' '$ changes from negative to positive at $x = - 2$.
$y = x {e}^{x}$ is concave down (convex) on $x \in \left(- \infty , - 2\right)$, and concave up on $x \in \left(- 2 , \infty\right)$