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.\n$y=-x+sin 2x,-\frac {5pi }{6}leq xleq \frac {5pi }{6}$
Answer
Explanation:
Step 1: Find the first derivative
Given (y = -x+\sin(2x)), using the sum rule ((u + v)^\prime=u^\prime + v^\prime) where (u=-x) and (v = \sin(2x)). The derivative of (u=-x) is (u^\prime=-1), and using the chain rule ((\sin(u))^\prime=\cos(u)\cdot u^\prime) with (u = 2x) (so (u^\prime=2)), the derivative of (v=\sin(2x)) is (v^\prime = 2\cos(2x)). So (y^\prime=-1 + 2\cos(2x)).
Step 2: Find the second derivative
Differentiate (y^\prime=-1 + 2\cos(2x)) with respect to (x). Using the sum rule again, the derivative of (-1) is (0), and for (2\cos(2x)) using the chain rule ((\cos(u))^\prime=-\sin(u)\cdot u^\prime) with (u = 2x) (so (u^\prime=2)). So (y^{\prime\prime}=-4\sin(2x)).
Step 3: Find inflection points
Set (y^{\prime\prime}=0), so (-4\sin(2x)=0). Then (\sin(2x)=0), which gives (2x = k\pi), (x=\frac{k\pi}{2}), (k\in\mathbb{Z}). For (-\frac{5\pi}{6}\leq x\leq\frac{5\pi}{6}), when (k = 0), (x = 0). When (x = 0), (y=-0+\sin(0)=0).
Answer:
A. The point(s) is/are ((0,0))