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 = 2x+\sin(4x)). Using the sum rule ((u + v)^\prime=u^\prime + v^\prime) where (u = 2x) and (v=\sin(4x)). The derivative of (u = 2x) is (u^\prime=2), and using the chain - rule ((\sin(u))^\prime=\cos(u)\cdot u^\prime) for (v=\sin(4x)) (where (u = 4x) and (u^\prime = 4)), we get (v^\prime = 4\cos(4x)). So (y^\prime=2 + 4\cos(4x)). Set (y^\prime=0) for critical points: [ \begin{align*} 2+4\cos(4x)&=0\ \cos(4x)&=-\frac{1}{2} \end{align*} ] Since (x\in[-\frac{\pi}{3},\frac{\pi}{3}]), then (4x\in[-\frac{4\pi}{3},\frac{4\pi}{3}]). (4x=\pm\frac{2\pi}{3}+2k\pi,k\in\mathbb{Z}). For (k = 0), (x=\pm\frac{\pi}{6}).

Step2: Use the second - derivative test

Find the second derivative. Differentiate (y^\prime=2 + 4\cos(4x)) with respect to (x). Using the sum rule and chain - rule again. The derivative of a constant (2) is (0), and for (y_1 = 4\cos(4x)) (where (u = 4x) and (u^\prime = 4)), (y^{\prime\prime}=-16\sin(4x)). Evaluate (y^{\prime\prime}) at (x = \frac{\pi}{6}): (y^{\prime\prime}(\frac{\pi}{6})=-16\sin(4\times\frac{\pi}{6})=-16\sin(\frac{2\pi}{3})=-16\times\frac{\sqrt{3}}{2}=-8\sqrt{3}<0). Evaluate (y) at (x=\frac{\pi}{6}): (y(\frac{\pi}{6})=2\times\frac{\pi}{6}+\sin(4\times\frac{\pi}{6})=\frac{\pi}{3}+\sin(\frac{2\pi}{3})=\frac{\pi}{3}+\frac{\sqrt{3}}{2}).

Step3: Analyze concavity

Set (y^{\prime\prime}=-16\sin(4x)=0). (\sin(4x)=0), then (4x = k\pi,k\in\mathbb{Z}), or (x=\frac{k\pi}{4}). For (x\in(-\frac{\pi}{3},\frac{\pi}{3})), (x = 0). Test intervals:

  • For the interval ((-\frac{\pi}{3},0)), let (x=-\frac{\pi}{6}). Then (y^{\prime\prime}=-16\sin(4\times(-\frac{\pi}{6}))=-16\sin(-\frac{2\pi}{3})=8\sqrt{3}>0), so the function is concave up on ((-\frac{\pi}{3},0)).
  • For the interval ((0,\frac{\pi}{3})), let (x=\frac{\pi}{6}). Then (y^{\prime\prime}=-16\sin(4\times\frac{\pi}{6})=-8\sqrt{3}<0), so the function is concave down on ((0,\frac{\pi}{3})). The inflection point is at (x = 0) (since the concavity changes at (x = 0)), and (y(0)=0).

Answer:

  • Inflection point: ((0,0))
  • Local maximum: There is one local maximum value of (\frac{\pi}{3}+\frac{\sqrt{3}}{2}) at (x=\frac{\pi}{6})
  • The function is differentiable on ((-\frac{\pi}{3},\frac{\pi}{3})), concave up on ((-\frac{\pi}{3},0)) and concave down on ((0,\frac{\pi}{3}))