Discuss the continuity of:
f(x)=⎧⎪
⎪⎨⎪
⎪⎩cos(x)x<−π|cos(x)|−π≤x≤π2cos2(x)x>π2⎫⎪
⎪⎬⎪
⎪⎭

To discuss the continuity of the piecewise function

\[
f(x) = 
\begin{cases} 
\cos(x) & \text{if } x < -\pi \\
|\cos(x)| & \text{if } -\pi \leq x \leq \pi \\
2\cos^2(x) & \text{if } x > \pi 
\end{cases}
\]

we need to examine its continuity across its domain, particularly at the points where the pieces of the function change, namely \(x = -\pi\) and \(x = \pi\).

### 1. Continuity at \(x = -\pi\)
- **Left-hand limit:**  
  Consider \( \lim_{x \to -\pi^-} f(x) \). Since \( f(x) = \cos(x) \) for \(x < -\pi\), we calculate:
  \[
  \lim_{x \to -\pi^-} f(x) = \cos(-\pi) = -1
  \]

- **Right-hand limit:**  
  Consider \( \lim_{x \to -\pi^+} f(x) \). Since \( f(x) = |\cos(x)| \) for \( -\pi \leq x \leq \pi \), we calculate:
  \[
  \lim_{x \to -\pi^+} f(x) = |\cos(-\pi)| = 1
  \]

- **Value of the function at \(x = -\pi\):**  
  Since \( f(x) = |\cos(x)| \) for \( -\pi \leq x \leq \pi \), we have:
  \[
  f(-\pi) = |\cos(-\pi)| = 1
  \]

Since the left-hand limit (\(-1\)) and right-hand limit (\(1\)) at \(x = -\pi\) are not equal, \(f(x)\) is **not continuous** at \(x = -\pi\).

### 2. Continuity at \(x = \pi\)
- **Left-hand limit:**  
  Consider \( \lim_{x \to \pi^-} f(x) \). Since \( f(x) = |\cos(x)| \) for \( -\pi \leq x \leq \pi \), we calculate:
  \[
  \lim_{x \to \pi^-} f(x) = |\cos(\pi)| = 1
  \]

- **Right-hand limit:**  
  Consider \( \lim_{x \to \pi^+} f(x) \). Since \( f(x) = 2\cos^2(x) \) for \( x > \pi \), we calculate:
  \[
  \lim_{x \to \pi^+} f(x) = 2\cos^2(\pi) = 2(-1)^2 = 2
  \]

- **Value of the function at \(x = \pi\):**  
  Since \( f(x) = |\cos(x)| \) for \( -\pi \leq x \leq \pi \), we have:
  \[
  f(\pi) = |\cos(\pi)| = 1
  \]

Since the left-hand limit (\(1\)) and right-hand limit (\(2\)) at \(x = \pi\) are not equal, \(f(x)\) is **not continuous** at \(x = \pi\).

### Conclusion
The function \( f(x) \) is **not continuous** at \(x = -\pi\) and \(x = \pi\), due to mismatched left-hand and right-hand limits at these points. The function is continuous elsewhere.

Would you like further details or have any questions? Here are some related questions:

1. How do we define continuity for a piecewise function?
2. What are the implications of discontinuity in a function?
3. Can a function be continuous everywhere but not differentiable at some points?
4. How would the function \( f(x) \) change if we adjusted the piecewise conditions?
5. What are some examples of real-world phenomena modeled by discontinuous functions?

**Tip:** When checking for continuity at a point in a piecewise function, always compare the left-hand limit, right-hand limit, and the function's value at that point.

Explore the continuity of the piecewise function f(x) = {...} across its domain, analyzing continuity at specific points and comparing left-hand and right-hand limits. Understand the implications of discontinuity and continuity conditions in piecewise-defined functions.

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