# Find the points of inflection of the graph of the function. (Round your answers to three decimal places. If an answer does not exist, enter DNE.) f(x) = 7 sin x + sin 2x, [0, 2π] 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,...

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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.

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