Question

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

Answer 1

`x_1=0` or `x_2=pi` or `x_3=2*pi`

Explanation:

We have
`f(x)=x+sin(x)`
Then
`f'(x)=1+cos(x)`
`f''(x)=-sin(x)`
Solving `sin(x)=0` for `x`in the given interval we get
`x=0`
`x=pi`
`x=2pi`