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)...

We are asked
to find the points of inflection for the graph of `f(x)=6sin(x)+sin(2x)`

Points of inflection occur when the concavity of the graph changes. (Graphs can be
concave up, concave down, or in the case of straight lines have no concavity.) Points of
inflection will be found at the points where the second derivative of the function changes sign.
Typically we solve for the zeros of the second derivative and check to make sure that the sign
of the second derivative changes.

`f'(x)=6cos(x)+2cos(2x)` (using the chain
rule for the second term.)

`f''(x)=-6sin(x)-4sin(2x)`

Using the identity `sin(2x)=2sin(x)cos(x)` we get:

`f''(x)=-6sin(x)-8sin(x)cos(x)`

We set the second derivative equal
to zero and solve:

`-6sin(x)-8sin(x)cos(x)=0`

`-2sin(x)(3+4cos(x))=0`

`-2sin(x)=0 ==> x=k * pi, k in
NN`

`3+4cos(x)=0 ==> x=cos^(-1)(-3/4)=tan^(-1)(-sqrt(7)/3)`

Exact answers: `x=k*pi, x=2*pi*k +- 2tan^(-1)(sqrt(7))` for k an integer.

Approximate answers on `[0,2*pi]` :

0, 2.419, 3.142, 3.864,
6.283

The graph:

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