Question

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.
find the inflection point(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the point(s) is/are
(type an ordered pair. use a comma to separate answers as needed.)
b. there are no inflection points.

Explanation:

Step1: Find the first derivative

Differentiate \(y = 2x+\sin(4x)\) using the sum rule \((u + v)^\prime=u^\prime + v^\prime\) and the chain rule \((\sin(u))^\prime=\cos(u)\cdot u^\prime\).
\(y^\prime=\frac{d}{dx}(2x)+\frac{d}{dx}(\sin(4x))=2 + 4\cos(4x)\)

Step2: Find the second derivative

Differentiate \(y^\prime = 2+4\cos(4x)\) again.
\(y^{\prime\prime}=\frac{d}{dx}(2)+\frac{d}{dx}(4\cos(4x))=0-16\sin(4x)=- 16\sin(4x)\)

Step3: Solve for inflection points

Set \(y^{\prime\prime}=0\), so \(-16\sin(4x)=0\), which implies \(\sin(4x) = 0\).
\(4x = k\pi\), \(x=\frac{k\pi}{4}\), \(k\in\mathbb{Z}\).
For \(x\in[-\frac{\pi}{3},\frac{\pi}{3}]\), when \(k = 0\), \(x = 0\); when \(k=- 1\), \(x=-\frac{\pi}{4}\); when \(k = 1\), \(x=\frac{\pi}{4}\)
Substitute \(x = 0\) into \(y = 2x+\sin(4x)\), \(y(0)=2\times0+\sin(0)=0\)
Substitute \(x=-\frac{\pi}{4}\) into \(y = 2x+\sin(4x)\), \(y(-\frac{\pi}{4})=2\times(-\frac{\pi}{4})+\sin(-\pi)=-\frac{\pi}{2}\)
Substitute \(x=\frac{\pi}{4}\) into \(y = 2x+\sin(4x)\), \(y(\frac{\pi}{4})=2\times\frac{\pi}{4}+\sin(\pi)=\frac{\pi}{2}\)

Answer:

A. The point(s) is/are \((-\frac{\pi}{4},-\frac{\pi}{2}),(0,0),(\frac{\pi}{4},\frac{\pi}{2})\)