Is f(x)=1/(x-1)-1/(x+1)^2 increasing or decreasing at x=0?

1 Answer
Dec 11, 2017

Since f'(0)>0, f(x) is increasing at x=0.

Explanation:

We need to find the derivative. First I like to rewrite f(x) for these types of functions:

f(x)=(x-1)^(-1)-(x+1)^(-2)

now use the chain rule:

f'(x)=-1(x-1)^(-2)+2(x+1)^-3

Now substitute x=0:

f'(0)=-1(0-1)^(-2)+2(0+1)^-3=-1+2=1

Since f'(0)>0, f(x) is increasing at x=0.