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.\n( y = 2 x + sin 4 x , - \frac { pi } { 3 } leq x leq \frac { pi } { 3 } )

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.\n( y = 2 x + sin 4 x , - \frac { pi } { 3 } leq x leq \frac { pi } { 3 } )

Answer

Explanation:

Step1: Find the first derivative

For (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) with respect to (x) 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), then (2+4\cos(4x)=0), (\cos(4x)=-\frac{1}{2}). Since (x\in[-\frac{\pi}{3},\frac{\pi}{3}]), then (4x\in[-\frac{4\pi}{3},\frac{4\pi}{3}]). (4x=-\frac{2\pi}{3}) or (4x=\frac{2\pi}{3}), so (x =-\frac{\pi}{6}) or (x=\frac{\pi}{6}).

Step2: Use the second - derivative test

Find the second derivative. (y^{\prime\prime}=-16\sin(4x)). When (x =-\frac{\pi}{6}), (y^{\prime\prime}=-16\sin(-\frac{2\pi}{3})=-16\times(-\frac{\sqrt{3}}{2}) = 8\sqrt{3}>0). So (x =-\frac{\pi}{6}) is a local minimum. When (x=\frac{\pi}{6}), (y^{\prime\prime}=-16\sin(\frac{2\pi}{3})=-16\times\frac{\sqrt{3}}{2}=-8\sqrt{3}<0). So (x=\frac{\pi}{6}) is a local maximum.

Step3: Find inflection points

Set (y^{\prime\prime}=0), (-16\sin(4x)=0), then (4x = k\pi), (x=\frac{k\pi}{4}), (k\in\mathbb{Z}). For (x\in[-\frac{\pi}{3},\frac{\pi}{3}]), (k = 0) gives (x = 0).

Step4: Determine concavity

Test intervals:

  • For the interval ((-\frac{\pi}{3},0)), let (x=-\frac{\pi}{6}), (y^{\prime\prime}=-16\sin(-\frac{2\pi}{3})>0). The function is concave up on ((-\frac{\pi}{3},0)).
  • For the interval ((0,\frac{\pi}{3})), let (x=\frac{\pi}{6}), (y^{\prime\prime}=-16\sin(\frac{2\pi}{3})<0). The function is concave down on ((0,\frac{\pi}{3})). The function (y = 2x+\sin(4x)) is differentiable on ((-\frac{\pi}{3},\frac{\pi}{3})) (since (y^\prime = 2 + 4\cos(4x)) exists for all (x\in(-\frac{\pi}{3},\frac{\pi}{3}))).

Answer:

  • Local maximum: There is one local maximum value of (y(\frac{\pi}{6})=2\times\frac{\pi}{6}+\sin(\frac{2\pi}{3})=\frac{\pi}{3}+\frac{\sqrt{3}}{2}) at (x = \frac{\pi}{6}).
  • Local minimum: There is one local minimum value of (y(-\frac{\pi}{6})=2\times(-\frac{\pi}{6})+\sin(-\frac{2\pi}{3})=-\frac{\pi}{3}-\frac{\sqrt{3}}{2}) at (x=-\frac{\pi}{6}).
  • Inflection point: The inflection point is at (x = 0) (since (y(0)=0)).
  • Differentiable interval: The function is differentiable on ((-\frac{\pi}{3},\frac{\pi}{3})).
  • Concavity: The function is concave up on ((-\frac{\pi}{3},0)) and concave down on ((0,\frac{\pi}{3})).