To find the number of critical points of the function \( f(x) = \max\{\sin x, \cos x\} \) for \( x \) in the interval \( (-2\pi, 2\pi) \), we will follow these steps:
### Step 1: Identify the points where \( \sin x \) and \( \cos x \) intersect.
To find the critical points, we first need to determine where the two functions \( \sin x \) and \( \cos x \) are equal, as these points will be where the maximum function changes from one function to the other.
Set \( \sin x = \cos x \). This can be rewritten as:
\[
\tan x = 1
\]
The solutions to this equation are:
\[
x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z}
\]
### Step 2: Find the specific solutions in the interval \( (-2\pi, 2\pi) \).
We can find the specific values of \( x \) in the interval \( (-2\pi, 2\pi) \):
- For \( n = -2 \):
\[
x = \frac{\pi}{4} - 2\pi = \frac{\pi}{4} - \frac{8\pi}{4} = -\frac{7\pi}{4}
\]
- For \( n = -1 \):
\[
x = \frac{\pi}{4} - \pi = \frac{\pi}{4} - \frac{4\pi}{4} = -\frac{3\pi}{4}
\]
- For \( n = 0 \):
\[
x = \frac{\pi}{4}
\]
- For \( n = 1 \):
\[
x = \frac{\pi}{4} + \pi = \frac{\pi}{4} + \frac{4\pi}{4} = \frac{5\pi}{4}
\]
- For \( n = 2 \):
\[
x = \frac{\pi}{4} + 2\pi = \frac{\pi}{4} + \frac{8\pi}{4} = \frac{9\pi}{4} \quad \text{(not in the interval)}
\]
Thus, the intersection points in the interval \( (-2\pi, 2\pi) \) are:
\[
x = -\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}
\]
### Step 3: Determine the behavior of \( f(x) \) around the intersection points.
The function \( f(x) \) will switch from \( \sin x \) to \( \cos x \) or vice versa at these intersection points. We will also check the endpoints of the interval \( -2\pi \) and \( 2\pi \) to see if they are critical points.
### Step 4: Evaluate \( f(x) \) at the endpoints.
- At \( x = -2\pi \):
\[
f(-2\pi) = \max\{\sin(-2\pi), \cos(-2\pi)\} = \max\{0, 1\} = 1
\]
- At \( x = 2\pi \):
\[
f(2\pi) = \max\{\sin(2\pi), \cos(2\pi)\} = \max\{0, 1\} = 1
\]
### Step 5: Count the critical points.
From our analysis, we have:
- Four intersection points: \( -\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4} \)
- Two endpoints: \( -2\pi \) and \( 2\pi \)
Thus, the total number of critical points is:
\[
4 + 2 = 6
\]
### Conclusion
The number of critical points of \( f(x) = \max\{\sin x, \cos x\} \) in the interval \( (-2\pi, 2\pi) \) is **6**.
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