# What are the points of inflection, if any, of f(x) =12x^5+15x^4–240x^3+6?

May 24, 2017

The points of inflection are at:
$x = 0 , \frac{- 3 \pm \sqrt{393}}{8}$

#### Explanation:

The points of inflection are given by when $f ' ' \left(x\right) = 0$.

$f \left(x\right) = 12 {x}^{5} + 15 {x}^{4} - 240 {x}^{3} + 6$
$f ' \left(x\right) = 60 {x}^{4} + 60 {x}^{3} - 720 {x}^{2}$
$f ' ' \left(x\right) = 240 {x}^{3} + 180 {x}^{2} - 1440 x$

$x = 60 x \left(4 {x}^{2} + 3 x - 24\right)$
(1)
$60 x = 0$
$x = 0$

(2) and (3)
$4 {x}^{2} + 3 x - 24 = 0$
$x = \frac{- 3 \pm \sqrt{393}}{8}$