identify the inflection points and local maxima and minima of the graphed function. identify the open…

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\nfind the inflection point(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the point(s) is/are (type an ordered pair. use a comma to separate answers as needed.)\nb. there are no inflection points.

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.\nfind the inflection point(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the point(s) is/are (type an ordered pair. use a comma to separate answers as needed.)\nb. there are no inflection points.

Answer

Explanation:

Step1: Find the second - derivative of the function

Given (y = x+\sin(2x)). First, find the first - derivative (y^\prime) using the sum rule ((u + v)^\prime=u^\prime + v^\prime) where (u = x) and (v=\sin(2x)). The derivative of (y=x) is (y^\prime_1 = 1), and using the chain rule ((\sin(u))^\prime=\cos(u)\cdot u^\prime) with (u = 2x) and (u^\prime=2), the derivative of (y = \sin(2x)) is (y^\prime_2=2\cos(2x)). So, (y^\prime=1 + 2\cos(2x)). Then, find the second - derivative (y^{\prime\prime}). The derivative of (y^\prime=1+2\cos(2x)) is (y^{\prime\prime}=-4\sin(2x)) (since the derivative of a constant (1) is (0) and using the chain rule for (2\cos(2x)): ((2\cos(u))^\prime=- 2\sin(u)\cdot u^\prime) with (u = 2x) and (u^\prime = 2)).

Step2: Solve (y^{\prime\prime}=0) for (x) in the interval (-\frac{2\pi}{3}\leq x\leq\frac{2\pi}{3})

Set (y^{\prime\prime}=-4\sin(2x)=0). Then (\sin(2x)=0). Let (t = 2x), so (t = k\pi), (k\in\mathbb{Z}). Substituting back (x=\frac{k\pi}{2}). For (k=-1), (x=-\frac{\pi}{2}) (since (-\frac{\pi}{2}\in[-\frac{2\pi}{3},\frac{2\pi}{3}])); for (k = 0), (x = 0); for (k = 1), (x=\frac{\pi}{2}). When (x=-\frac{\pi}{2}), (y=-\frac{\pi}{2}+\sin(-\pi)=-\frac{\pi}{2}). When (x = 0), (y=0+\sin(0)=0). When (x=\frac{\pi}{2}), (y=\frac{\pi}{2}+\sin(\pi)=\frac{\pi}{2}).

Answer:

((-\frac{\pi}{2},-\frac{\pi}{2}),(0,0),(\frac{\pi}{2},\frac{\pi}{2}))