Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 29

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 0negative 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: