# What are the points of inflection, if any, of f(x)=2x(2x-3)^3 ?

Nov 14, 2016

$\left(\frac{3}{2} , 0\right)$ is the only point of inflection

#### Explanation:

$f \left(x\right) = 2 x {\left(2 x - 3\right)}^{3}$
graph{2x(2x-3)^3 [-2, 5, -11, 11]}

Using the product rule and chain rule we have;
$f ' \left(x\right) = \left(2 x\right) 3 {\left(2 x - 3\right)}^{2} \left(2\right) + \left(2\right) {\left(2 x - 3\right)}^{3}$
$\therefore f ' \left(x\right) = 12 x {\left(2 x - 3\right)}^{2} + 2 {\left(2 x - 3\right)}^{3}$
$\therefore f ' \left(x\right) = {\left(2 x - 3\right)}^{2} \left(12 x + 2 \left(2 x - 3\right)\right)$
 :. f'(x) = (2x-3)^2(12x + 4x - 6))
$\therefore f ' \left(x\right) = {\left(2 x - 3\right)}^{2} \left(16 x - 6\right)$
$\therefore f ' \left(x\right) = 2 {\left(2 x - 3\right)}^{2} \left(8 x - 3\right)$ .... 

At critical points (min/max/poi) then $f ' \left(x\right) = 0$
$f ' \left(x\right) = 0 \to 2 {\left(2 x - 3\right)}^{2} \left(8 x - 3\right) = 0$
$\therefore x = \frac{3}{8} , \frac{3}{2}$

To determine the nature of these points we need to examine the second derivative to see if f'(x) is increasing (min), decreasing (max) or 0 (inflection):

Differentiating  wrt $x$ we get;
$f ' ' \left(x\right) = \left(2 {\left(2 x - 3\right)}^{2}\right) \left(8\right) + \left(2 \left(2\right) \left(2 x - 3\right) \left(2\right)\right) \left(8 x - 3\right)$
$\therefore f ' ' \left(x\right) = 16 {\left(2 x - 3\right)}^{2} + 8 \left(2 x - 3\right) \left(8 x - 3\right)$

When $x = \frac{3}{8} \implies f ' ' \left(\frac{3}{8}\right) = 16 {\left(\frac{6}{8} - 3\right)}^{2} > 0$ (ie min)
When $x = \frac{3}{2} \implies f ' ' \left(\frac{3}{2}\right) = 0$ (ie poi)

When $x = \frac{3}{2} \implies f \left(\frac{3}{2}\right) = 0$

Hence $\left(\frac{3}{2} , 0\right)$ is the only point of inflection