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

Dec 11, 2017

Since $f ' \left(0\right) > 0$, $f \left(x\right)$ is increasing at $x = 0$.

#### Explanation:

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

$f \left(x\right) = {\left(x - 1\right)}^{- 1} - {\left(x + 1\right)}^{- 2}$

now use the chain rule:

$f ' \left(x\right) = - 1 {\left(x - 1\right)}^{- 2} + 2 {\left(x + 1\right)}^{-} 3$

Now substitute $x = 0$:

$f ' \left(0\right) = - 1 {\left(0 - 1\right)}^{- 2} + 2 {\left(0 + 1\right)}^{-} 3 = - 1 + 2 = 1$

Since $f ' \left(0\right) > 0$, $f \left(x\right)$ is increasing at $x = 0$.