Question

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

Answer 1

The curve is convex at `x=0`

Explanation:

Let's calculate the derivative

`(u/v)'=(u'v-uv')/v^2`

Here, `u=e^x` ; `=>` `u'=e^x`

`v=x-e^x` `=>` ; `v'=1-e^x`

therefore, `f'(x)=(e^x(x-e^x)-e^x(1-e^x))/(x-e^x)^2`

`=(xe^x-e^(2x)-e^x+e^(2x))/(x-e^x)^2=(xe^x-e^x)/(x-e^x)^2`

`=(e^x(x-1))/(x-e^x)^2`

`f'(x)=0` when `x=1`

Let's do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `0` `color(white)(aaaaa)` `1` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `darr` `color(white)(aaaaa)` `darr` `color(white)(aaa)` `uarr`

As the curve is decreasing from `-oo` to `1`, the curve is convex at `x=0`

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