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

Jul 11, 2017

There is a local maximum at $\left(1 , {e}^{- 1}\right)$

There is a non-stationary point of inflection at $\left(2 , 2 {e}^{- 2}\right)$.

#### Explanation:

We have:

$y = x {e}^{- x}$
graph{xe^(-x) [-5, 10, -5, 5]}

Firstly, let us look for critical points, that is coordinates where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$:

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

For a critical point:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \left(1 - x\right) {e}^{- x} = 0$

Which has a single solution $x = 1$, as ${e}^{a} > 0 \forall a \in \mathbb{R}$, and

$x = 1 \implies y = {e}^{- 1}$

Thus there is a single critical point at $\left(1 , {e}^{- 1}\right)$. To determine the nature of the critical point we must examine the second derivative:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - {e}^{- x} - \left\{\left(1 - x\right) {e}^{- x}\right\}$
$\text{ } = - \left(2 - x\right) {e}^{- x}$
$\text{ } = \left(x - 2\right) {e}^{- x}$

When $x = 1 \implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(1 - 2\right) {e}^{- x} < 0$

As the second derivative is negative at the critical point then we can conclude the critical point $\left(1 , {e}^{- 1}\right)$ is a maximum.

Secondly, we look for inflection points, which are coordinates where the second derivative vanishes:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0 \implies \left(x - 2\right) {e}^{- x} = 0$

Which has a single solution $x = 2$, for which we have:

$y = 2 {e}^{- 2}$

We already know that $x = 2$ does not correspond to a critical point and thus we have a non-stationary point of inflection at $\left(2 , 2 {e}^{- 2}\right)$.