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.

b. there are no inflection points

find each local maximum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

a. there is one local maximum value of $-\frac{\pi}{6}+\frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{6}$

b. there are two local maxima. in increasing order of x - value, the values are $\square$ at $x = \square$ and $\square$ at $x = \square$

c. there are no local maxima

$y=-x + \sin 2x,-\frac{5\pi}{6}\leq x\leq\frac{5\pi}{6}$

Explanation:

Step1: Find the first derivative

The function is \(y = -x+\sin(2x)\). Using the sum rule and the chain rule, the first derivative \(y^{\prime}=-1 + 2\cos(2x)\).
Set \(y^{\prime}=0\), then \(2\cos(2x)-1 = 0\), so \(\cos(2x)=\frac{1}{2}\).
Since \(x\in[-\frac{5\pi}{6},\frac{5\pi}{6}]\), then \(2x\in[-\frac{5\pi}{3},\frac{5\pi}{3}]\). Solving \(\cos(2x)=\frac{1}{2}\), we get \(2x=-\frac{\pi}{3},\frac{\pi}{3},\frac{5\pi}{3}\) (but \(\frac{5\pi}{3}\) is out of the domain when considering \(x\) in the original domain). So \(x =-\frac{\pi}{6},\frac{\pi}{6}\).
We use the second - derivative test.

Step2: Find the second derivative

Differentiate \(y^{\prime}=-1 + 2\cos(2x)\) with respect to \(x\). Using the chain rule, \(y^{\prime\prime}=-4\sin(2x)\).
When \(x =-\frac{\pi}{6}\), \(y^{\prime\prime}=-4\sin(-\frac{\pi}{3})=2\sqrt{3}>0\), so \(x =-\frac{\pi}{6}\) is a local minimum.
When \(x=\frac{\pi}{6}\), \(y^{\prime\prime}=-4\sin(\frac{\pi}{3})=-2\sqrt{3}<0\).
Substitute \(x = \frac{\pi}{6}\) into \(y=-x+\sin(2x)\), \(y=-\frac{\pi}{6}+\sin(\frac{\pi}{3})=-\frac{\pi}{6}+\frac{\sqrt{3}}{2}\).

Answer:

A. There is one local maximum value of \(-\frac{\pi}{6}+\frac{\sqrt{3}}{2}\) at \(x = \frac{\pi}{6}\)