Question

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

Answer 1

`f'(x)=cos(x)-xsin(x)`
`f"(x)=-2sin(x)-xcos(x)`
set `f"(x)=0` for points of inflection, x=0, 2.29

Explanation:

On your calculator let equation 1 be `y=2sin(x)`
let equation 2 be `y=-xcos(x)`
Find the points of intersection in `[0, 2pi]`