Question

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

Answer 1

The function `f(x)` is convex at `x=-1`

Explanation:

The derivative of a quotient is

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

We calcule the first derivative

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

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

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

Now, we calculate the second derivative,

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

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

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

At `x=-1`

`f''(-1)=e^(-1-1)-(2)/(-1-1)^3=0.135+0.25=0.385`

As `f''(-1)>0`, we conclude that `f(x)` is convex at `x=-1`

graph{e^(x-1)-x^2/(x-1)-1 [-10, 10, -5, 5]}