What are the critical points of f(x) = x / (x^2 + 4)?

Nov 24, 2016

We have a min at $\left(- 2 , - \frac{1}{4}\right)$ and a max at $\left(2 , \frac{1}{4}\right)$

We have points of inflexions at $\left(0 , 0\right)$ ; $\left(2 \sqrt{3} , \frac{\sqrt{3}}{8}\right)$ and $\left(- 2 \sqrt{3} , - \frac{\sqrt{3}}{8}\right)$

Explanation:

We need to differentiate a quotient

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

Here
$u = x$, $\implies$, $u ' = 1$

$v = {x}^{2} + 4$,, $\implies$,$v ' = 2 x$

Therefore,

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

$= \frac{{x}^{2} + 4 - 2 {x}^{2}}{{x}^{2} + 4} ^ 2 = \frac{4 - {x}^{2}}{{x}^{2} + 4} ^ 2$

$= \frac{\left(2 + x\right) \left(2 - x\right)}{{x}^{2} + 4} ^ 2$

The critical values are when $f ' \left(x\right) = 0$

$\left(2 + x\right) \left(2 - x\right) = 0$

$x = - 2$ and $x = - 2$

Let's calculate $f ' ' \left(x\right)$

$u = 4 - {x}^{2}$, $\implies$, $u ' = - 2 x$

$v = {\left({x}^{2} + 4\right)}^{2}$, $\implies$, $v ' \left(x\right) = 4 x \left({x}^{2} + 4\right)$

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

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

$= \frac{- 2 {x}^{3} - 8 x - 16 x + 4 {x}^{3}}{{x}^{2} + 4} ^ 3$

$= \frac{2 {x}^{3} - 24 x}{{x}^{2} + 4} ^ 3$

$= \frac{2 x \left({x}^{2} - 12\right)}{{x}^{2} + 4} ^ 3$

When $f ' ' \left(x\right) = 0$, we have inflexion points

$2 x \left({x}^{2} - 12\right) = 0$ when $x = 0$ ; $x = \sqrt{12} = 2 \sqrt{3}$ and $x = - \sqrt{12} = - 2 \sqrt{3}$

When $x = - 2$ , $f ' ' \left(x\right) > 0$, so we have a min

When $x = 2$ , $f ' ' \left(x\right) < 0$, so we have a max

graph{x/(x^2+4) [-3.895, 3.9, -1.948, 1.947]}