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The problem asks which of the following functions are continuous on \( (-\infty, \infty) \).

Let's analyze each piecewise-defined function one by one for continuity.

### (i) \( f(x) = \begin{cases} 
x + 2 & \text{if } x \leq 1 \\
3 + 5 \ln(x) & \text{if } x > 1 
\end{cases} \)

1. **Continuity of individual pieces:**
   - \( x + 2 \) is a linear function and continuous for all \( x \leq 1 \).
   - \( 3 + 5 \ln(x) \) involves the natural logarithm, which is continuous for \( x > 0 \). However, the natural logarithm is undefined for \( x \leq 0 \), meaning this function is defined only for \( x > 1 \).

2. **Checking at the boundary point \( x = 1 \):**
   - The left-hand limit as \( x \to 1^- \) for \( x + 2 \) is \( 1 + 2 = 3 \).
   - The right-hand limit as \( x \to 1^+ \) for \( 3 + 5 \ln(x) \) is \( 3 + 5 \ln(1) = 3 \).
   - The function value at \( x = 1 \) from the left is 3 and from the right is also 3.

Thus, \( f(x) \) is continuous for all \( x \in (-\infty, \infty) \).

### (ii) \( f(x) = \begin{cases} 
7e^x - 4 & \text{if } x < 0 \\
3 \cos(x) & \text{if } x \geq 0 
\end{cases} \)

1. **Continuity of individual pieces:**
   - \( 7e^x - 4 \) is continuous for all \( x \in (-\infty, 0) \).
   - \( 3 \cos(x) \) is continuous for all \( x \in [0, \infty) \).

2. **Checking at the boundary point \( x = 0 \):**
   - The left-hand limit as \( x \to 0^- \) for \( 7e^x - 4 \) is \( 7e^0 - 4 = 7 - 4 = 3 \).
   - The right-hand limit as \( x \to 0^+ \) for \( 3 \cos(x) \) is \( 3 \cos(0) = 3 \cdot 1 = 3 \).
   - The function value at \( x = 0 \) is 3.

Thus, \( f(x) \) is continuous for all \( x \in (-\infty, \infty) \).

### (iii) \( f(x) = \begin{cases} 
\frac{x^2 - 49}{x - 7} & \text{if } x \neq 7 \\
12 & \text{if } x = 7 
\end{cases} \)

1. **Simplifying the expression:**
   \[
   \frac{x^2 - 49}{x - 7} = \frac{(x - 7)(x + 7)}{x - 7} = x + 7 \quad \text{for} \, x \neq 7.
   \]
   Therefore, \( f(x) = x + 7 \) for \( x \neq 7 \), which is a linear function and continuous for all \( x \neq 7 \).

2. **Checking at \( x = 7 \):**
   - The limit as \( x \to 7 \) for \( x + 7 \) is \( 7 + 7 = 14 \).
   - The function value at \( x = 7 \) is given as 12.

Since the limit as \( x \to 7 \) (which is 14) does not equal the function value \( f(7) = 12 \), the function has a discontinuity at \( x = 7 \).

Thus, \( f(x) \) is **not** continuous on \( (-\infty, \infty) \).

### Conclusion:
- (i) is continuous on \( (-\infty, \infty) \).
- (ii) is continuous on \( (-\infty, \infty) \).
- (iii) is **not** continuous on \( (-\infty, \infty) \).

Would you like more details on any step, or further explanation? Here are some related questions:

1. What does it mean for a function to be continuous at a point?
2. How do we compute limits for piecewise functions at boundary points?
3. What are common types of discontinuities?
4. Why does the natural logarithm cause restrictions on domain?
5. How can algebraic simplifications help in identifying continuity?

**Tip:** Always check the boundary points of piecewise functions for continuity by ensuring the limit from both sides matches the function value.

Which of the following functions are continuous on (-∞, ∞)?

This page explains how to determine the continuity of piecewise functions involving logarithms, exponentials, and trigonometric functions, with a detailed analysis of boundary points and algebraic simplification. Suitable for undergraduate calculus students.

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