Question

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

Answer 1

Convex

Explanation:

The second derivative is determinative of concavity. Start by finding the first derivative.

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

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

`f'(x) = 1 - (xe^x - 2)/x^3`

Now differentiate this.

`f''(x) = 0 - ((e^x + xe^x)x^3 - (xe^x - 2)(3x^2))/(x^3)^2`

`f''(x) = - (x^3e^x + x^4e^x - 3x^3e^x + 6x^2)/x^6`

`f''(x) = -(x^4e^x - 2x^3e^x + 6x^2)/x^6`

We now evaluate the point within the second derivative.

`f''(-1) = -((-1)^4e^-1 - 2(-1)^3e^-1 + 6(-1)^2)/(-1)^6`

`f''(-1) = -(1/e -2(-1)/e + 6)/1`

`f''(-1) =- (1/e + 2/e + 6)/1`

This is obviously negative, therefore, the function is convex at `x = -1`. If the second derivative had had a positive value at `x = -1`, then we would have dubbed it concave at that point. Here is a graphical confirmation:

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

Hopefully this helps!