# How do you determine the intervals for which the function is increasing or decreasing given f(x)=x/(x^2+1)?

Nov 11, 2016

The limits when $f \left(x\right)$ is increasing is $x \in \left[- 1 , 1\right]$
The limits when $f \left(x\right)$ is decreasing is  x in ] -oo,-1 [ uu ] 1, oo[

#### Explanation:

We must calculate the derivative $f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$

$u = x$$\implies$$u ' = 1$
$v = {x}^{2} + 1$$\implies$$v ' = 2 x$

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

$f ' \left(x\right) = 0$ when $x = 1$ and $x = - 1$

We do a sign chart for $f ' \left(x\right)$

$\textcolor{w h i t e}{a a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$
$\textcolor{w h i t e}{a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a a a}$$\downarrow$

The limits when $f \left(x\right)$ is increasing is $x \in \left[- 1 , 1\right]$
The limits when $f \left(x\right)$ is decreasing is  x in ] -oo,-1 [ uu ] 1, oo[
graph{x/(x^2+1) [-3.077, 3.08, -1.538, 1.54]}