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

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Answer 1 · 2017-12-11T08:29:43

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

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#.