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in Applications of Differential Calculus by (48.7k points)
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Find the intervals of concavity and the points of inflection of the following function.

f(θ) = sin 2θ in (θ, π)

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Best answer

f'(θ) = sin 2θ 

f’(θ) = (cos 2θ) (2) = 2 cos 2θ 

f”(θ)= 2[-sin 2θ] (2) = – 4 sin 2θ

f”(θ) = 0 ⇒ – 4 sin 2θ = 0

f(θ) is concave upward in the interval ((π/2),π)

So, at θ = π/2 we get a point of inflection. At θ = π/2, f(θ) = sin 2(π/2) = 0

So, the curve is concave upward in ((π/2),π) and concave downward in (0,(π/2)).

The point of inflection is ((π/2),0).

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