# How do use the first derivative test to determine the local extrema f(x) = x / (x^2+1)?

Sep 29, 2015

See the explanation.

#### Explanation:

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

$f ' \left(x\right) = 0 \iff 1 - {x}^{2} = 0 \iff x = 1 \vee x = - 1$

$f ' \left(x\right) > 0$ $\forall x \in \left(- 1 , 1\right)$ function increasing
$f ' \left(x\right) < 0$ $\forall x \in \left(- \infty , - 1\right) \cup \left(1 , \infty\right)$ function decreasing

$f ' \left(x\right)$ changes sign at $x = - 1$ and $f \left(x\right)$ has a minimum value ${f}_{\min} = f \left(- 1\right) = - \frac{1}{2}$

$f ' \left(x\right)$ changes sign at $x = 1$ and $f \left(x\right)$ has a maximum value ${f}_{\max} = f \left(1\right) = \frac{1}{2}$