Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function.
\[
y=2 \sqrt{3} x+4 \cos x, 0 \leq x \leq 2 \pi
\]
Answer
\(\boxed{\text{Final Answer:}}\) The local and absolute extreme points are at \(x = \frac{\pi}{3}\) with \(y = 2 + 2\sqrt{3}\frac{\pi}{3}\), \(x = \frac{2\pi}{3}\) with \(y = -2 + 4\sqrt{3}\frac{\pi}{3}\), \(x = 0\) with \(y = 4\), and \(x = 2\pi\) with \(y = 4 + 4\sqrt{3}\pi\). The inflection points are at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
Step 1 ：First, we need to find the derivative of the function \(y = 2\sqrt{3}x + 4\cos(x)\). The derivative is \(y' = -4\sin(x) + 2\sqrt{3}\).
Step 2 ：Setting the derivative equal to zero gives us the critical points. The critical points are \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\).
Step 3 ：Evaluating the function at these critical points and at the endpoints of the interval \([0, 2\pi]\) gives us the values \((\frac{\pi}{3}, 2 + 2\sqrt{3}\frac{\pi}{3})\), \((\frac{2\pi}{3}, -2 + 4\sqrt{3}\frac{\pi}{3})\), \((0, 4)\), and \((2\pi, 4 + 4\sqrt{3}\pi)\).
Step 4 ：Next, we need to find the second derivative of the function. The second derivative is \(y'' = -4\cos(x)\).
Step 5 ：Setting the second derivative equal to zero gives us the potential inflection points. The potential inflection points are \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
Step 6 ：Checking the sign of the second derivative on either side of these points confirms that the concavity changes, indicating that these are indeed inflection points.
Step 7 ：\(\boxed{\text{Final Answer:}}\) The local and absolute extreme points are at \(x = \frac{\pi}{3}\) with \(y = 2 + 2\sqrt{3}\frac{\pi}{3}\), \(x = \frac{2\pi}{3}\) with \(y = -2 + 4\sqrt{3}\frac{\pi}{3}\), \(x = 0\) with \(y = 4\), and \(x = 2\pi\) with \(y = 4 + 4\sqrt{3}\pi\). The inflection points are at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
