# What are the points of inflection of f(x)=x^3cos^2x  on the interval x in [0,2pi]?

May 25, 2018

I have found:
$x = 0$
$x \setminus \approx 0.63584893101835810541 , 1.34300713055201783266 , 2.8772796137343975171 , 4.2206536081793615994 , 5.7678994523455943589$

#### Explanation:

we have
$f ' \left(x\right) = {x}^{2} \cos \left(x\right) \left(3 \cos \left(x\right) - 2 x \sin \left(x\right)\right)$
$f ' ' \left(x\right) = 3 x + 3 x \cos \left(2 x\right) - 2 {x}^{3} \cos \left(2 x\right) - 6 {x}^{2} \sin \left(2 x\right)$
with a numerical method i got the values above.