# What interval(s) is x / (1+x)^2  increasing and decreasing?

Nov 6, 2016

$f \left(x\right)$ is increasing on the interval (-1,1), and decreasing on the interval $\left(- \infty , - 1\right) \cup \left(1 , \infty\right)$

#### Explanation:

Let $f \left(x\right) = \frac{x}{x + 1} ^ 2$
To find where $f \left(x\right)$ is increasing and decreasing, we have to find where f'(x) is positive and negative, respectively.

We do this by finding f'(x), set the numerator and denominator equal to zero, and determine the signs of values in between these intervals.

Find f'(x) using quotient rule, chain rule, and power rule:
$f \left(x\right) = \frac{x}{x + 1} ^ 2$

$f ' \left(x\right) = \frac{\left(x + 1\right) \cancel{{\left(x + 1\right)}^{2}} - 2 x \cancel{\left(x + 1\right)}}{{\left(x + 1\right)}^{3} \cancel{{\left(x + 1\right)}^{4}}}$

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

By setting the numerator equal to zero, we get $x = 1$, and by setting the denominator equal to zero, we get $x = - 1$, meaning there is an asymptote at $x = - 1$.

After testing values between and around x=-1 and x=1, I found that $f ' \left(x\right)$ is positive from (-1,1) and $f ' \left(x\right)$ is negative from $\left(- \infty , - 1\right) \cup \left(1 , \infty\right)$.