Question

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

Answer 1

f(x) is convex at x=0

Explanation:

Let's calculate the first and the second derivative of f(x):

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

`f''(x)=4e^(2x)-((2x-2)(x-1)^2-2(x-1)(x^2-2x))/(x-1)^4`
`=4e^(2x)-(2(x-1)^(cancel3^2)-2cancel((x-1))(x^2-2x))/(x-1)^(cancel4^3)`
`=4e^(2x)-(2(x-1)^2-2(x^2-2x))/(x-1)^3`

Then let's look at the sign of f''(0):

`f''(0)=4e^0-2/-1=4+2=6>0`

Then f(x) is convex at x=0

graph{e^(2x)-x^2/(x-1)-1 [-5.164, 5.17, -2.583, 2.58]}