`f(x) = 2sin(x) + sin(2x), [0, 2pi]` Find the points of inflection and discuss the concavity of the graph of the function.

Given f(x)=2sin(x)+sin(2x) on the interval [0,2pi], find the points of inflection and discuss the concavity:

Find the second derivative:

f'(x)=2cos(x)+2cos(2x)

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

Inflection points occur when the second derivative is zero (and changes sign.)

-2sinx-4sin(2x)=0

-8sin(x)cos(x)-2sin(x)=0    Using the identity sin(2x)=2sin(x)cos(x)

-2sin(x)[4cos(x)+1]=0

-2sin(x)=0 ==> x=0,pi,2pi on the interval

4cos(x)+1=0 ==> cos(x)=-1/4 ==> x=1.823,4.460

The second derivative is:

positive to the immediate left of 0
negative on (0,1.823)
positive on (1.823,pi)
negative on (pi,4.460)
positive on (4.460,2pi)
negative to the immediate right of 2pi

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There are inflection points at 0,1.823,pi,4.460,2pi

The function is concave down on (0,1.823), up on (1.823,pi), down on (pi,4.460), and up on (4.460,2pi)

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The graph: