`f(x) = sin(x/2), [0, 4pi]` Find the points of inflection and discuss the concavity of the graph of the function.

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Chapter 3, 3.4 - Problem 25 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This function is defined everywhere and is infinitely differentiable.

At the points of inflection of f the second derivative must be zero:

f'(x)=(1/2)*cos(x/2),

f''(x)=-(1/4)*sin(x/2).

It is =0 at x/2=k*pi, x=2k*pi where k is any integer. Because we consider f to be defined on [0, 4pi] there is only one such point inside (0, 2pi), it is x0=2pi.

f'' is negative at the left of x0 and positive at the right. So f is concave downward at (0, 2pi) and concave upward at (2pi, 4pi). And 2pi is the point of inflection.

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