Hello!

We are given the function `f ( x ) = 7 sin x + sin ( 2 x )` on `[ 0 , 2 pi ] .`

This is an elementary function defined everywhere, and as such it is continuous and infinitely differential everywhere.

To find the inflection points, determine the first and the second derivatives of `f ( x ) :`

`f ' ( x ) = 7 cos x + 2 cos ( 2 x ),`

`f '' ( x ) = - 7 sin x - 4 sin ( 2 x ) .`

We need to determine where the second derivative changes sign. To do this, factor it using the formula `sin ( 2 x ) = 2 sin x cos x :`

`f '' ( x ) = - sin x ( 7 + 8 cos x ) .`

Now we see that `f ''` has some roots on `[ 0, 2 pi ]` and changes sign at each of them because they all have multiplicity `1 .` Specifically, `f '' ( x ) = 0` at

`x_1 = 0, x_2 = pi, x_3 = 2 pi, x_4 = arccos ( 7 / 8 ), x_5 = 2 pi - arccos ( 7 / 8 ) .`

`x_1` and `x_3` are the endpoints of the given interval, so we ignore them.
The remaining roots are the points of inflection, approximately
**2.636**, **3.142**, **3.647**.

Note that for the same function considered on a wider interval, `0` and `2pi` are also the points of inflection.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.