# Let f(x) = x / (x^2+1), how do you use the first derivative test to determine which critical numbers, if any give relative extrema?

Oct 29, 2016

The critical numbers are when $x = - 1$ and $x = 1$
there is a minimum at $\left(- 1 , - \frac{1}{2}\right)$
and a maximum at $\left(1 , \frac{1}{2}\right)$

#### Explanation:

The derivative of a quotient is $\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

$f \left(x\right) = \frac{x}{{x}^{2} + 1}$
so $f ' \left(x\right) = \frac{1 \left({x}^{2} + 1\right) - \left(x \cdot 2 x\right)}{{x}^{2} + 1} ^ 2 = \frac{{x}^{2} + 1 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2$
$= \frac{1 - {x}^{2}}{{x}^{2} + 1} ^ 2 = \frac{\left(1 + x\right) \left(1 - x\right)}{{x}^{2} + 1} ^ 2$
The denominator is always $> 0$
so
$f ' \left(x\right) = 0$ when $x = - 1$ and when $x = 1$
Let's do a sign chart

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

so, there is a minimum at $x = - 1$ and a maximum at $x = 1$
Also, the limit of $f \left(x\right)$ as $x \to \pm \infty$ is
limit $f \left(x\right) = {0}^{-}$
$x \to - \infty$

limit $f \left(x\right) = {0}^{+}$
$x \to + \infty$

Also, $f \left(- x\right) = - f \left(x\right)$, there is a symmetry about the Origin

graph{x/(x^2+1) [-2.222, 2.222, -1.111, 1.11]}