Question

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

Answer 1

Please see below.

Explanation:

Point of inflection of `f(x)=xsinx` is where an increasing slope starts decreasing or vice-versa. At this point second derivative `(d^2f(x))/(dx^2)=0`.

As such using product formula `f(x)=xsinx`,

`(df(x))/(dx)=sinx+xcosx` and

`(d^2f(x))/(dx^2)=cosx+cosx-xsinx=2cosx-xsinx`

Now `2cosx-xsinx=0` i.e. `xsinx=2cosx`

or `x=2cotx`

and solution is given by the points where the function `x-2cotx` cuts `x`-axis.
graph{x-2cotx [-2, 8, -2.5, 2.5]}

Below we give the graph of `f(x)=xsinx` and observe that at these points slope of the curve changes accordingly.
graph{xsinx [-2, 8, -5, 5]}