# How do you find the critical points for xe^x?

Jan 13, 2017

$x = - 1$ is a local minimum for $x {e}^{x}$

#### Explanation:

Critical points for a function are identified equating its first derivative to zero.

As:

$\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right) = \left(x + 1\right) {e}^{x}$

its critical points are the roots of the equation:

$\left(x + 1\right) {e}^{x} = 0$

or, as ${e}^{x} > 0$ for every $x$:

$x + 1 = 0$

that is:

$x = - 1$

We can easily see that:

$\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right) < 0$ for $x < - 1$

$\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right) > 0$ for $x > - 1$

which means that $x = - 1$ is a local minimum.