Question

Is `f(x)=e^x-x^2/(x-1)` concave or convex at `x=0`?

Answer 1

The function is convex at `x=0`

Explanation:

The derivative of a quotient is

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

The function is

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

The first derivative is

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

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

The second derivative is

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

`=e^x-((2x-2)(x-1)-(2x^2-4x))/(x-1)^3`

`=e^x-(2x^2-4x+2-2x^2+4x)/(x-1)^3`

`=e^x-2/(x-1)^3`

When `x=0`

`f''(0)=e^0-2/(-1)^3=1-2=3`

As `f''(0) >0`, the function is convex at `x=0`

graph{e^x-x^2/(x-1) [-14.24, 14.23, -7.12, 7.12]}