# How do you find points of inflection and determine the intervals of concavity given y=3/(x^2+4)?

Apr 3, 2018

Below

#### Explanation:

$y = \frac{3}{{x}^{2} + 4}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{6 x}{{x}^{2} + 4} ^ 2$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 6 \times \frac{4 - 3 {x}^{2}}{{x}^{2} + 4} ^ 3$

For stationary points, $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{6 x}{{x}^{2} + 4} ^ 2 = 0$

ie $- 6 x = 0$
$x = 0$

Test $x = 0$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{3}{8} < 0$

Therefore, it is a maximum and concave down at $x = 0$ $\left(0 , \frac{3}{4}\right)$

For point of inflexion, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$- 6 \times \frac{4 - 3 {x}^{2}}{{x}^{2} + 4} ^ 3 = 0$

$4 - 3 {x}^{2} = 0$
$\frac{4}{3} = {x}^{2}$
$x = \pm \frac{2}{\sqrt{3}}$

Test when $x = - \frac{2}{\sqrt{3}}$

$x = - 1.5$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{1056}{15625}$

$x = - \frac{2}{\sqrt{3}}$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$x = - 1$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{6}{125}$

Therefore, there is a change in concavity so there is a point of inflexion at $x = - \frac{2}{\sqrt{3}}$ $\left(- \frac{2}{\sqrt{3}} , \frac{9}{16}\right)$

Test when $x = \frac{2}{\sqrt{3}}$

$x = 1$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{6}{125}$

$x = \frac{2}{\sqrt{3}}$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$x = 1.5$ $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{1056}{15625}$

Therefore there is a change in concavity so there is a point of inflexion at $x = \frac{2}{\sqrt{3}}$ $\left(\frac{2}{\sqrt{3}} , \frac{9}{16}\right)$