Question

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

Answer 1

Neither. The function is undefined at `x=0` since `ln0` is undefined.

Explanation:

However, you could take the limit as x approaches 0 of the concavity of the function. First, let's take the second derivative of the function.

`d/dx f(x) = -1/(x-1)^2-2x(1/x)-2lnx`

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

`therefore (d^2y)/(dx^2) = 2/(x-1)^3 - 2/x`

Now we need to take the limit of this second derivative as x approaches 0. Since ln(x) is not defined for negatives, we only need to worry about when x approaches 0 from the positive direction.

`lim_(x->0)(2/(x-1)^3 - 2/x)`

`= 2/(0-1)^3 - lim_(x->0)2/x`

`= -2 - oo`

`= -oo`

So we can say that the function is concave down as x approaches 0.