Question

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

Answer 1

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

Explanation:

we have
`f'(x)=x^2cos(x)(3cos(x)-2xsin(x))`
`f''(x)=3x+3xcos(2x)-2x^3cos(2x)-6x^2sin(2x)`
with a numerical method i got the values above.