# How do you find the inflection point of (x+1)/(x^(2)+1)?

Jun 3, 2018

We get the Points
P_1(1;1),P_2(-2-sqrt(3),1/4*(1-sqrt(3))),P_3(-2+sqrt(3),1/4(1+sqrt(3)))

#### Explanation:

By the Quotient rule we get
$f ' \left(x\right) = \frac{- {x}^{2} - 2 x + 1}{{x}^{2} + 1} ^ 2$
$f ' ' \left(x\right) = \frac{2 {x}^{3} + 6 {x}^{2} - 6 x + 2}{{x}^{2} + 1} ^ 3$
so we have to solve
$2 \left({x}^{3} + 3 {x}^{2} - 3 x - 1\right) = 0$
This is
$\left(x - 1\right) \left({x}^{2} + 4 x + 1\right) = 0$
Solutions are
${x}_{1} = 1$
${x}_{2} = - 2 - \sqrt{3}$
${x}_{3} = - 2 + \sqrt{3}$

Jun 3, 2018

${x}_{1} = 1$
${x}_{2} = - 2 - \sqrt{3}$
${x}_{3} = - 2 + \sqrt{3}$

#### Explanation:

For a point of inflection $g$:
$f ' ' \left(g\right) = 0$
$f ' ' ' \left(g\right) \ne 0$

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

Using the Quotient rule:
$f ' \left(x\right) = \frac{1 \cdot \left({x}^{2} + 1\right) - \left(x + 1\right) \left(2 x\right)}{{x}^{2} + 1} ^ 2$
$= \frac{{x}^{2} + 1 - 2 {x}^{2} - 2 x}{{x}^{2} + 1} ^ 2$
$= \frac{- {x}^{2} - 2 x + 1}{{x}^{2} + 1} ^ 2$
$= - \frac{{\left(x + 1\right)}^{2} - 2}{{x}^{2} + 1} ^ 2$

Again, using the Quotient rule:
$f ' ' \left(x\right) = - \frac{2 \cdot \left(x + 1\right) \cdot {\left({x}^{2} + 1\right)}^{2} - \left({\left(x + 1\right)}^{2} - 2\right) \left(2 \cdot 2 x \left({x}^{2} + 1\right)\right)}{{x}^{2} + 1} ^ 4$
$= - \frac{\left(2 x + 2\right) \cdot {\left({x}^{2} + 1\right)}^{2} - 4 x {\left(x + 1\right)}^{2} \left({x}^{2} + 1\right) - 4 x \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 4$
$= - \frac{\left(2 x + 2\right) \cdot \left({x}^{2} + 1\right) - 4 x {\left(x + 1\right)}^{2} - 4 x}{{x}^{2} + 1} ^ 3$

$f ' ' \left(x\right) = 0$
$0 = - \frac{\left(2 x + 2\right) \cdot \left({x}^{2} + 1\right) - 4 x {\left(x + 1\right)}^{2} - 4 x}{{x}^{2} + 1} ^ 3 | \cdot {\left({x}^{2} + 1\right)}^{3}$
$0 = - \left(2 x + 2\right) \cdot \left({x}^{2} + 1\right) - 4 x {\left(x + 1\right)}^{2} - 4 x$
$0 = - 2 {x}^{3} + 2 x - 2 {x}^{2} - 2 + 4 {x}^{3} + 8 {x}^{2} - 4 x - 4 x$
$0 = 2 {x}^{3} + 6 {x}^{2} - 6 x - 2$

${x}_{1} = 1$
${x}_{2} = - 2 - \sqrt{3}$
${x}_{3} = \sqrt{3} - 2$

Again, using the Quotient rule:
$f ' ' ' \left(x\right) = - \frac{6 \left({x}^{4} + 4 {x}^{3} + 4 {x}^{2} - 4 x - 1\right)}{{x}^{2} + 1} ^ 4$
$f ' ' ' \left(1\right) = - \frac{3}{2}$
$f ' ' ' \left(- 2 - \sqrt{3}\right) = \frac{3}{32} \cdot \left(4 \cdot \sqrt{3} - 7\right) \approx - 0.0067$
$f ' ' ' \left(- 2 + \sqrt{3}\right) = - \frac{3}{32} \cdot \left(4 \cdot \sqrt{3} + 7\right) \approx - 1.31$