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

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<x<pi/2
positive for pi/2<x<3pi/2
negative for 3pi/2<x<2pi

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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: