https://file.aiquickdraw.com/mathbot/file/1726548624908shxoibcv.jpg

Let's analyze the function and check its continuity at \( a = 15 \).

### Function Definition:
\[
f(x) = \begin{cases}
\frac{x^2 - 225}{x - 15} & \text{if } x \neq 15 \\
1 & \text{if } x = 15
\end{cases}
\]
We need to determine whether this function is continuous at \( x = 15 \).

### Continuity Conditions:
For a function to be continuous at \( x = a \), the following three conditions must hold:
1. **\( f(a) \) is defined.**
2. **\( \lim_{x \to a} f(x) \) exists.**
3. **\( \lim_{x \to a} f(x) = f(a) \).**

Let's check each condition.

### Step 1: Is \( f(15) \) defined?
Yes, \( f(15) = 1 \). So, this condition is satisfied.

### Step 2: Does \( \lim_{x \to 15} f(x) \) exist?
For \( x \neq 15 \), \( f(x) = \frac{x^2 - 225}{x - 15} \). This can be simplified as follows:
\[
\frac{x^2 - 225}{x - 15} = \frac{(x - 15)(x + 15)}{x - 15}.
\]
For \( x \neq 15 \), we can cancel the \( (x - 15) \) terms:
\[
f(x) = x + 15 \quad \text{for} \quad x \neq 15.
\]
Thus, \( \lim_{x \to 15} f(x) = 15 + 15 = 30 \). So, the limit exists and equals 30.

### Step 3: Is \( \lim_{x \to 15} f(x) = f(15) \)?
No, \( f(15) = 1 \), but \( \lim_{x \to 15} f(x) = 30 \). So, this condition is not satisfied.

### Conclusion:
The function is **not continuous** at \( x = 15 \) because while \( f(15) \) is defined and the limit exists, the limit does not equal the value of the function at \( x = 15 \).

The correct answer is **B**:
- The function is not continuous at \( a = 15 \) because although \( f(15) = 1 \) is defined and \( \lim_{x \to 15} f(x) = 30 \) exists, \( f(15) \neq \lim_{x \to 15} f(x) \).

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Do you want more details or have any questions?

Here are 5 related questions:
1. What happens if the limit of a function at a point equals the function's value at that point?
2. How do you handle piecewise functions when checking continuity?
3. Can a function be discontinuous if the limit exists at a given point?
4. How would the function be modified to make it continuous at \( x = 15 \)?
5. What role does factoring play in simplifying limits?

**Tip**: When simplifying rational expressions to find limits, always check if cancellation of terms changes the domain of the function!

Determine whether the following function is continuous at a = 15: f(x) = (x^2 - 225) / (x - 15) if x ≠ 15; f(x) = 1 if x = 15. Use the continuity checklist to justify your answer.

Learn how to determine the continuity of a piecewise function at x = 15. This problem involves the concepts of limits, factoring, and checking the value of the function at specific points. Suitable for Grade 12 or first-year college students studying calculus.

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