Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 30

Find the inflection points and discuss the concavity for the function f(x)=x+2cos(x) on the interval [o,2pi]:
Inflection points are found when the second derivative is zero (and has changed sign.)
f'(x)=1-2sin(x)
f''(x)=-2cos(x)
-2cos(x)=0 ==> x=pi/2 and 3pi/2
The second derivative is:
negative for 0---------------------------------------------------------------------------------------
There are an inflection points at x=pi/2 and x=3pi/2. The function is concave down on (0,pi/2), concave up on (pi/2,3pi/2), and concave down on (3pi/2,2pi)
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The graph: