Draw the graph of `f(x) = xcosx-sinx, x in [-3pi, 3pi]`

Explanation

`f(x) = xcosx-sinx`
Clearly, domain is R.
Also `f(x)` is non-periodic.
`f^(')(x)=-xsinx`
`f^(')(x) =0 rArr x=-3pi, -2pi, -pi, 0, pi, 2pi, 3pi`
`f^(')(0^(-)) = (-)(-)(-) lt 0` and `f^(')(0^(+)) = (-)(+)(+) lt 0`
So `x=0` is the point of inflection (as the derivative does not change sign in the neighbourhood of `x=0`
Sign scheme of `f^(')(x)` is as follows:

Clearly, f is decreasing at `x=0` and has the point of minima at `x=pi, -2pi` and point of maxima at `x=-pi, 2pi`
Since `f(x)` is an odd function, we check the graph for `x int [0, 3pi]`
`f(0)=0, f(pi)=-pi`
Thus, `f(x)` decreases from `0` to `-pi` in the interval `(0, pi)`
Thus, `f(x)` increases from `-pi` to `2pi` in the interval `(pi, 2pi)`
`f(3pi) =-3pi`
Thus, `f(x)` decreases from `2pi` to `-3pi` in the interval `(2pi, 3pi)`
From the above discussion the graph of `y=f(x)` is as shown in the following figure.