# How do you find the critical numbers for f(x)= (x-1) e^x to determine the maximum and minimum?

Feb 28, 2017

The only critical point is for $x = 0$ and the function has a local minimum in that point.

#### Explanation:

Critical points are found by equating the first derivative to zero:

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

Solving:

$x {e}^{x} = 0$

the only critical point is $x = 0$

Now consider that:

$f ' \left(x\right) = x {e}^{x} < 0$ for $x < 0$

$f ' \left(x\right) = x {e}^{x} > 0$ for $x > 0$

We have then that $f \left(x\right)$ is decreasing in $\left(- \infty , 0\right)$ and increasing in $\left(0 , + \infty\right)$, therefore $x = 0$ is a local minimum.

graph{(x-1)e^x [-10, 10, -5, 5]}