# How do you find points of inflection and determine the intervals of concavity given y=xe^(1/x)?

Dec 30, 2017

The function $y = x {e}^{\frac{1}{x}}$ is convex for $x > 0$ and concave for $x < 0$. It has no inflection points. It has discontinuity at $x = 0$

#### Explanation:

What does it mean?
To determine concavity and points of inflection we need to find second derivative.
If $y$ is continuous at some point and:
$y ' ' > 0 \iff$function is convex (or concave up)
$y ' ' < 0 \iff$function is concave (or concave down)
$y ' ' = 0$ and it changes sign$\iff$inflection point.
We will discover various features of this function and sketch the curve as we go.
If you want to see how I got any of the following limit let me know in the comments.

Before deriving
$y = x {e}^{\frac{1}{x}}$
Domain: $x \in \mathbb{R} \text{\} \left\{0\right\}$ or $x \ne 0$ - no y intercept
$y \ne 0$ for $x \ne 0$ - no x intercept

Finding limits:
${\lim}_{x \to {0}^{-}} x {e}^{\frac{1}{x}} = 0$
${\lim}_{x \to {0}^{+}} x {e}^{\frac{1}{x}} = + \infty$
${\lim}_{x \to {0}^{-}} y \ne {\lim}_{x \to {0}^{+}} y$ so the limit at 0 doesn't exist and we have discontinuity. Assuming $x \ne 0$ from now on.
${\lim}_{x \to \pm \infty} x {e}^{\frac{1}{x}} = \pm \infty$
Because these limits are infinite, there could be up to 2 diagonal asymptotes $y = a x + b$. We find them by formulas
$a = {\lim}_{x \to \pm \infty} \frac{y}{x} = {\lim}_{x \to \pm \infty} {e}^{\frac{1}{x}} = 1$
and
$b = {\lim}_{x \to \pm \infty} y - a x = {\lim}_{x \to \pm \infty} x {e}^{\frac{1}{x}} - x = 1$.
It appears that both asymptotes exist and are the same.

First derivative
$y = x {e}^{\frac{1}{x}}$
$y ' = {e}^{\frac{1}{x}} + x {e}^{\frac{1}{x}} \cdot \frac{- 1}{x} ^ 2$ (product rule and chain rule)
$y ' = \left(1 - \frac{1}{x}\right) {e}^{\frac{1}{x}}$
Setting it to 0 to find extrema
$\left(1 - \frac{1}{x}\right) {e}^{\frac{1}{x}} = 0$
$\left(1 - \frac{1}{x}\right) = 0$
$1 = \frac{1}{x}$
$x = 1$ - candidate for extremum.
It must be a minimum, because here $y '$ changes sign from negative to positive.

Most limits of first derivative are already given from general function behavior.
${\lim}_{x \to {0}^{+}} \frac{\mathrm{dy}}{\mathrm{dx}} = - \infty$
${\lim}_{x \to \pm \infty} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
One left to calculate
${\lim}_{x \to {0}^{-}} \frac{x - 1}{x} {e}^{\frac{1}{x}} = 0$

Second derivative
$y ' = \left(1 - \frac{1}{x}\right) {e}^{\frac{1}{x}}$
$y ' ' = \frac{1}{x} ^ 2 {e}^{\frac{1}{x}} + \left(1 - \frac{1}{x}\right) {e}^{\frac{1}{x}} \cdot \frac{- 1}{x} ^ 2$ (product rule and chain rule)
$y ' ' = \frac{1}{x} ^ 2 {e}^{\frac{1}{x}} \left(1 - \left(1 - \frac{1}{x}\right)\right)$
$y ' ' = {e}^{\frac{1}{x}} / {x}^{3} \ne 0$ (thus no inflection points)
Looking at sign we see that $y ' '$ has the same sign as $x$, so your function $y = x {e}^{\frac{1}{x}}$ is convex for $x > 0$ and concave for $x < 0$.

Graph of $x {e}^{\frac{1}{x}}$ together with its diagonal asymptote:
graph{(y-xe^(1/x))(y-x-1)sqrt(|x|-0.03)=0 [-10, 10, -5, 5]}