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

Oct 21, 2015

${y}_{\max} = 2$

#### Explanation:

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

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

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

$y ' = \frac{- 2 x}{{x}^{2} + 1} ^ 2$

$y ' = 0 \iff - 2 x = 0 \iff x = 0$

$\forall x > 0 \implies y ' < 0$

$\forall x < 0 \implies y ' > 0$

The first derivative changes sign at $x = 0$ so it's the extrema point:
${y}_{\max} = y \left(0\right) = \frac{2}{1} = 2$