# 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.