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.\nseparate answers as needed.)\n○ b. there are no inflection points\nfind each local maximum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\n○ a. there is one local maximum value of at\nx=\n○ b. there are two local maxima. in increasing order of x - value, the values are at\nx = and at x=\n○ c. there are no local maxima

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.\nseparate answers as needed.)\n○ b. there are no inflection points\nfind each local maximum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\n○ a. there is one local maximum value of at\nx=\n○ b. there are two local maxima. in increasing order of x - value, the values are at\nx = and at x=\n○ c. there are no local maxima

Answer

Explanation:

Step1: Find the first derivative

The function is (y = x+\sin(2x)). Using the sum rule ((u + v)^\prime=u^\prime + v^\prime) and the chain rule ((\sin(u))^\prime=\cos(u)\cdot u^\prime) (where (u = 2x) and (u^\prime=2)), we get (y^\prime=1 + 2\cos(2x)).

Step2: Find the critical points

Set (y^\prime = 0), so (1+2\cos(2x)=0). Then (\cos(2x)=-\frac{1}{2}). Since (x\in[-\frac{2\pi}{3},\frac{2\pi}{3}]), then (2x\in[-\frac{4\pi}{3},\frac{4\pi}{3}]). The solutions of (\cos(2x)=-\frac{1}{2}) are (2x=\pm\frac{2\pi}{3}+ 2k\pi,k\in\mathbb{Z}). For (k = 0), (2x=\frac{2\pi}{3}) gives (x=\frac{\pi}{3}) and (2x=-\frac{2\pi}{3}) gives (x =-\frac{\pi}{3}).

Step3: Use the second - derivative test

Find the second derivative (y^{\prime\prime}=-4\sin(2x)).

  • When (x =-\frac{\pi}{3}), (y^{\prime\prime}=-4\sin(-\frac{2\pi}{3})=-4\times(-\frac{\sqrt{3}}{2}) = 2\sqrt{3}>0), so (x =-\frac{\pi}{3}) is a local minimum.
  • When (x=\frac{\pi}{3}), (y^{\prime\prime}=-4\sin(\frac{2\pi}{3})=-4\times\frac{\sqrt{3}}{2}=-2\sqrt{3}<0). Substitute (x = \frac{\pi}{3}) into the original function (y=x+\sin(2x)), (y=\frac{\pi}{3}+\sin(\frac{2\pi}{3})=\frac{\pi}{3}+\frac{\sqrt{3}}{2}).

Answer:

There is one local maximum value of (\frac{\pi}{3}+\frac{\sqrt{3}}{2}) at (x=\frac{\pi}{3}).