Is #f(x)=1/(x-1)-2xlnx# concave or convex at #x=0#?
1 Answer
Jun 11, 2017
Neither. The function is undefined at
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.
