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Let's break down the problem part by part and solve it step-by-step.

### Part (a): Continuity at \( x = -1 \)

To ensure continuity at \( x = -1 \), we need the following condition from the definition of continuity:
\[
\lim_{x \to -1^-} f(x) = f(-1) = \lim_{x \to -1^+} f(x)
\]

For \( f(x) \) to be continuous at \( x = -1 \), the left-hand limit, the function value at \( x = -1 \), and the right-hand limit must all be equal. 

- **Left-hand limit** \( (x \to -1^-) \): 
  The piece of the function for \( x < -1 \) is \( f(x) = m^2 x - 12 \). So:
  \[
  \lim_{x \to -1^-} f(x) = m^2(-1) - 12 = -m^2 - 12
  \]

- **Value at \( x = -1 \)**: 
  From the hint, \( f(-1) = -m^2 - 12 \).

- **Right-hand limit** \( (x \to -1^+) \): 
  The piece of the function for \( -1 \leq x \leq m \) is \( f(x) = 7mx \). So:
  \[
  \lim_{x \to -1^+} f(x) = 7m(-1) = -7m
  \]

For continuity at \( x = -1 \), these must be equal:
\[
-m^2 - 12 = -7m
\]

Solve for \( m \):
\[
m^2 + 12 = 7m
\]
\[
m^2 - 7m + 12 = 0
\]

Now, use the quadratic formula:
\[
m = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)}
\]
\[
m = \frac{7 \pm \sqrt{49 - 48}}{2}
\]
\[
m = \frac{7 \pm \sqrt{1}}{2}
\]
\[
m = \frac{7 \pm 1}{2}
\]

Thus, \( m = 4 \) or \( m = 3 \).

### Part (b): Differentiability at \( x = -1 \)

For differentiability at \( x = -1 \), the left-hand derivative and right-hand derivative must be equal.

- **Left-hand derivative \( (x < -1) \)**:
  \( f(x) = m^2 x - 12 \), so:
  \[
  f'(x) = m^2
  \]
  Hence, \( f'(-1^-) = m^2 \).

- **Right-hand derivative \( (-1 \leq x \leq m) \)**:
  \( f(x) = 7mx \), so:
  \[
  f'(x) = 7m
  \]
  Hence, \( f'(-1^+) = 7m \).

For differentiability at \( x = -1 \), these must be equal:
\[
m^2 = 7m
\]

Solve for \( m \):
\[
m(m - 7) = 0
\]

So, \( m = 0 \) or \( m = 7 \).

Since \( m \) is given to be positive, we discard \( m = 0 \), leaving \( m = 7 \).

### Part (c): Discontinuity at \( x = m \)

From part (a), we found that \( m = 4 \) and \( m = 3 \) ensure continuity at \( x = -1 \). Now, let’s examine \( x = m \) for one of these values, say \( m = 4 \).

For \( x = 4 \):
- **Left-hand limit** \( (x \to 4^-) \): 
  \( f(x) = 7mx = 7(4)x \), so:
  \[
  \lim_{x \to 4^-} f(x) = 7(4)(4) = 112
  \]
  
- **Right-hand limit** \( (x \to 4^+) \): 
  \( f(x) = \frac{m}{x - 2} \), so:
  \[
  \lim_{x \to 4^+} f(x) = \frac{4}{4 - 2} = 2
  \]

Since the left-hand and right-hand limits are not equal, there is a **jump discontinuity** at \( x = m \).

---

Would you like more details on any part? Here are some further questions you could explore:

1. Why are both continuity and differentiability important for piecewise functions?
2. How does the quadratic formula help in finding critical values of parameters like \( m \)?
3. What types of discontinuities exist, and how do they relate to this problem?
4. How would the solution change if the condition for differentiability were relaxed?
5. Can you graph the function for different values of \( m \) to visualize the behavior at \( x = -1 \) and \( x = m \)?

**Tip:** Always check continuity before differentiability when analyzing piecewise functions. Continuity is a necessary condition for differentiability.

Let m be a positive constant. Consider the piecewise function: f(x) = {m^2x - 12, x < -1; 7mx, -1 ≤ x ≤ m; m / (x - 2), x > m}. Use the definition of continuity to find all values of m, if any, that guarantee that f is continuous at x = -1. Use the definition of the derivative to find all values of m, if any, that guarantee that f is differentiable at x = -1. For one of the values of m that you found in part (a), determine the type of discontinuity that f has at x = m.

Learn how to analyze the continuity and differentiability of a piecewise function, ensuring the function is smooth at x = -1 and determining the type of discontinuity at x = m. This problem applies algebraic techniques like solving quadratic equations and evaluating limits.

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