# How do you find the local extremas for f(x)=xe^x?

Nov 16, 2016

There is a relative minimum at the point $\left(- 1 , - \frac{1}{e}\right)$.

#### Explanation:

We can say that if ${x}_{0}$ is a turning point of the function $f$ then $f ' \left({x}_{0}\right) = 0$.

Therefore, the return points of the function $f$ will be among the solutions of the equation $f ' \left(x\right) = 0$. So we will equate the derivative of the function to zero and then we will search among the solutions those in which the derivative has a change of sign.

If the derivative is positive, we know that the function is increasing, whereas if the derivative is negative, then the function is decreasing.

When the derivative changes from negative to positive, the function has a local minimum, whereas if the change of sign is reversed, that is, from positive to negative, then the function has a local maximum.

In the case of the function $f \left(x\right) = x \cdot {e}^{x}$, we have:

$f ' \left(x\right) = {e}^{x} + x \cdot {e}^{x} = \left(1 + x\right) \cdot {e}^{x}$

Equaling to zero we have:

$\left(1 + x\right) \cdot {e}^{x} = 0 \textcolor{w h i t e}{.} \iff \textcolor{w h i t e}{.} 1 + x = 0 \textcolor{w h i t e}{.} \iff \textcolor{w h i t e}{.} x = - 1$

It is easy to verify that, for values of $x < - 1$, the derivative is negative, $f ' \left(x\right) < 0$, while for values of $x > - 1$, the derivative is positive, $f ' \left(x\right) > 0$. This means that $x = - 1$ is a relative minimun.

The $y$ coordinate is obtained by replacing the value of $x$ in the equation of the function.