https://file.aiquickdraw.com/mathbot/file/1727653455234jmfr95e5.jpg

Let's analyze each function to determine if they are continuous.

### 1. Function (i):
\[
f(x) = 
\begin{cases} 
x + 2 & \text{for } x \leq 1 \\
3 + 5 \ln(x) & \text{for } x > 1 
\end{cases}
\]
- For \(x \leq 1\), \(f(x) = x + 2\) is a linear function, and hence continuous everywhere in this domain.
- For \(x > 1\), \(f(x) = 3 + 5 \ln(x)\) is continuous where \(x > 1\), since the logarithm function is continuous for \(x > 0\).
- To check the continuity at \(x = 1\), we need to confirm that:
    \[
    \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)
    \]
    - For \(x \to 1^-\), \(f(x) = x + 2\), so \(\lim_{x \to 1^-} f(x) = 1 + 2 = 3\).
    - For \(x \to 1^+\), \(f(x) = 3 + 5 \ln(x)\), so \(\lim_{x \to 1^+} f(x) = 3 + 5 \ln(1) = 3\).
    - Thus, \(f(x)\) is continuous at \(x = 1\).

So, function (i) is continuous for all \(x\).

### 2. Function (ii):
\[
f(x) = 
\begin{cases} 
7e^x - 4 & \text{for } x < 0 \\
3 \cos(x) & \text{for } x \geq 0 
\end{cases}
\]
- For \(x < 0\), \(f(x) = 7e^x - 4\) is an exponential function, which is continuous everywhere.
- For \(x \geq 0\), \(f(x) = 3 \cos(x)\) is a cosine function, which is also continuous everywhere.
- To check the continuity at \(x = 0\), we need to confirm that:
    \[
    \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)
    \]
    - For \(x \to 0^-\), \(\lim_{x \to 0^-} f(x) = 7e^0 - 4 = 7 - 4 = 3\).
    - For \(x \to 0^+\), \(f(x) = 3 \cos(x)\), so \(\lim_{x \to 0^+} f(x) = 3 \cos(0) = 3\).
    - Thus, \(f(x)\) is continuous at \(x = 0\).

So, function (ii) is continuous for all \(x\).

### 3. Function (iii):
\[
f(x) = 
\begin{cases} 
\frac{x^2 - 49}{x - 7} & \text{for } x \neq 7 \\
12 & \text{for } x = 7
\end{cases}
\]
- For \(x \neq 7\), \(f(x) = \frac{x^2 - 49}{x - 7}\) simplifies to \(f(x) = x + 7\), which is continuous everywhere for \(x \neq 7\).
- To check the continuity at \(x = 7\), we need to confirm that:
    \[
    \lim_{x \to 7} f(x) = f(7)
    \]
    - \(\lim_{x \to 7} f(x) = \lim_{x \to 7} (x + 7) = 7 + 7 = 14\).
    - But \(f(7) = 12\), so \(f(x)\) is **not continuous** at \(x = 7\).

Thus, function (iii) is not continuous at \(x = 7\).

### Conclusion:
The functions that are continuous are:
- (i) Continuous everywhere
- (ii) Continuous everywhere
- (iii) Not continuous at \(x = 7\)

Thus, the correct answer is **E) (ii) and (iii) only**.

Would you like further details on any of the steps or functions?

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Here are some related questions:
1. What does continuity mean for a function?
2. How do you check the continuity at a specific point?
3. What is the difference between a piecewise and a non-piecewise function regarding continuity?
4. How do exponential and logarithmic functions behave in terms of continuity?
5. Can a function be continuous at all points except one?

**Tip**: Always check the limits from both sides at a point where the definition of the function changes.

Which of the following functions are continuous on (-∞, ∞)? (i) f(x) = {x+2, x ≤ 1; 3 + 5 ln x, x > 1}, (ii) f(x) = {7e^x - 4, x < 0; 3 cos x, x ≥ 0}, (iii) f(x) = {(x^2 - 49)/(x - 7), x ≠ 7; 12, x = 7}

Learn how to determine which piecewise functions are continuous on (-∞, ∞). This problem explores limits, continuity, and how to evaluate piecewise functions at critical points. Suitable for students in Grades 11-12.

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