# How do you find the inflection points of the graph of the function:  f(x) = (6x)/(x^2 + 16)?

##### 1 Answer
Aug 11, 2015

$x = 0 , f \left(0\right) = 0 \text{ and } x = \pm 4 \sqrt{3} , f \left(\pm 4 \sqrt{3}\right) = \pm \frac{3 \sqrt{3}}{8}$

#### Explanation:

A point of inflection can be found when the second derivative of f(x) is equal to zero i.e. $f ' ' \left(x\right) = 0$

Using quotient rule:
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{6 \left({x}^{2} + 16\right) - 12 {x}^{2}}{{x}^{2} + 16} ^ 2 = - \frac{6 \left({x}^{2} - 16\right)}{{x}^{2} + 16} ^ 2$
$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = - \frac{12 x \left({x}^{2} + 16\right) - 24 x \left({x}^{2} - 16\right)}{{x}^{2} + 16} ^ 3 = \frac{12 x \left({x}^{2} - 48\right)}{{x}^{2} + 16} ^ 3 = 0$

$x = 0 , f \left(0\right) = 0 \text{ and } x = \pm 4 \sqrt{3} , f \left(\pm 4 \sqrt{3}\right) = \pm \frac{3 \sqrt{3}}{8}$