# Is f(x)=e^x/(x-e^x) concave or convex at x=0?

Nov 15, 2016

The curve is convex at $x = 0$

#### Explanation:

Let's calculate the derivative

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

Here, $u = {e}^{x}$ ; $\implies$$u ' = {e}^{x}$

$v = x - {e}^{x}$ $\implies$ ; $v ' = 1 - {e}^{x}$

therefore, $f ' \left(x\right) = \frac{{e}^{x} \left(x - {e}^{x}\right) - {e}^{x} \left(1 - {e}^{x}\right)}{x - {e}^{x}} ^ 2$

$= \frac{x {e}^{x} - {e}^{2 x} - {e}^{x} + {e}^{2 x}}{x - {e}^{x}} ^ 2 = \frac{x {e}^{x} - {e}^{x}}{x - {e}^{x}} ^ 2$

$= \frac{{e}^{x} \left(x - 1\right)}{x - {e}^{x}} ^ 2$

$f ' \left(x\right) = 0$ when $x = 1$

Let's do a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a}$$\uparrow$

As the curve is decreasing from $- \infty$ to $1$, the curve is convex at $x = 0$

graph{e^x/(x-e^x) [-5.546, 5.55, -2.773, 2.774]}