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To find the points of discontinuity of the given piecewise function:

\[
f(x) = \begin{cases} 
3x - 1 & \text{for } x \leq -2 \\
x^2 - 3 & \text{for } -2 < x < 1 \\
5 \cos(x - 1) - 7 & \text{for } x \geq 1 
\end{cases}
\]

We need to check for discontinuities at the boundaries between the pieces of the function, which occur at \(x = -2\) and \(x = 1\). Specifically, we'll check if the limits from the left and right at these points are equal to the value of the function at these points, as this will determine whether the function is continuous there.

### 1. Check at \(x = -2\):
- Left-hand limit as \(x \to -2^-\) (from the first piece, \(3x - 1\)):
  \[
  \lim_{x \to -2^-} (3x - 1) = 3(-2) - 1 = -6 - 1 = -7
  \]
- Right-hand limit as \(x \to -2^+\) (from the second piece, \(x^2 - 3\)):
  \[
  \lim_{x \to -2^+} (x^2 - 3) = (-2)^2 - 3 = 4 - 3 = 1
  \]
  
Since \(-7 \neq 1\), there is a discontinuity at \(x = -2\).

### 2. Check at \(x = 1\):
- Left-hand limit as \(x \to 1^-\) (from the second piece, \(x^2 - 3\)):
  \[
  \lim_{x \to 1^-} (x^2 - 3) = (1)^2 - 3 = 1 - 3 = -2
  \]
- Right-hand limit as \(x \to 1^+\) (from the third piece, \(5 \cos(x - 1) - 7\)):
  \[
  \lim_{x \to 1^+} \left( 5 \cos(x - 1) - 7 \right) = 5 \cos(1 - 1) - 7 = 5 \cos(0) - 7 = 5(1) - 7 = -2
  \]

Both the left-hand and right-hand limits at \(x = 1\) are equal, and since the function is defined for \(x \geq 1\) with \(f(1) = -2\), the function is continuous at \(x = 1\).

### Conclusion:
There is a **discontinuity at \(x = -2\)** because the left-hand and right-hand limits do not match. The function is **continuous at \(x = 1\)**.

Would you like further clarification or a deeper exploration of the continuity concept? Here are some related questions you can explore:

1. What are the different types of discontinuities (e.g., jump, infinite, removable)?
2. How do we formally define a piecewise function?
3. How would you graph this piecewise function, and where would the discontinuities appear?
4. Can a function have both removable and non-removable discontinuities?
5. What is the difference between limits and actual values in determining continuity?

**Tip:** When checking for discontinuities in piecewise functions, always compare the left-hand and right-hand limits at the boundary points, and don't forget to check if the function is actually defined at those points!

Find all points of discontinuity, if any: f(x) = {3x - 1, x <= -2; x^2 - 3, -2 < x < 1; 5 cos(x - 1) - 7, x >= 1}.

Find all points of discontinuity for the piecewise function f(x) = {3x - 1 for x <= -2, x^2 - 3 for -2 < x < 1, and 5 cos(x - 1) - 7 for x >= 1}. Explore the process of checking limits from the left and right to determine discontinuity at x = -2 and x = 1.

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