# How do you determine whether the function f(x) = xe^-x is concave up or concave down and its intervals?

Jul 18, 2015

To determine concavity, analyze the sign of $f ' ' \left(x\right)$.

#### Explanation:

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

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

$= {e}^{-} x - x {e}^{-} x$

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

So,
$f ' ' \left(x\right) = \left[- {e}^{-} x \left(- 1\right)\right] \left(x - 1\right) + \left(- {e}^{-} x\right) \left(1\right)$

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

$= {e}^{-} x \left(x - 2\right)$

Now, $f ' ' \left(x\right) = {e}^{-} x \left(x - 2\right)$ is continuous on its domain, $\left(- \infty , \infty\right)$, so the only way it can change sign is by passing through zero. (The only partition numbers are the zeros of $f ' ' \left(x\right)$)

$f ' ' \left(x\right) = 0$ if and only if either ${e}^{-} x = 0$ or $x - 2 = 0$

$e$ to any (real) power is positive, so the only way for $f ' '$ to be $0$ is for $x$ to be $2$.

We partition the number line:

$\left(- \infty , 2\right)$ and $\left(2 , \infty\right)$

On the interval $\left(- \infty , 2\right)$, we have $f ' ' \left(x\right) < 0$ so $f$ is concave down.

On $\left(2 , \infty\right)$, we get $f ' ' \left(x\right) > 0$, so $f$ is concave up.

Inflection point

The point $\left(2 , f \left(2\right)\right) = \left(2 , \frac{2}{e} ^ 2\right)$ is the only inflection point for the graph of this function.