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

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Chapter 3, 3.4 - Problem 29 - Calculus of a Single Variable (10th Edition, Ron Larson).
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embizze | High School Teacher | (Level 1) Educator Emeritus

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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:



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


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


-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


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)


The graph:



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