https://file.aiquickdraw.com/mathbot/file/17262133682496u6jk0lx.jpg

The image contains a mathematical problem regarding the continuity of a piecewise function \( f(x) \), defined as:

\[
f(x) = \begin{cases} 
\frac{x^2 - 9}{x - 3}, & \text{for } x \neq 3 \\
9, & \text{for } x = 3 
\end{cases}
\]
The problem asks whether this function is continuous or not.

### Step-by-step solution:

1. **Simplify the function for \( x \neq 3 \):**

   The expression \( \frac{x^2 - 9}{x - 3} \) can be simplified using the difference of squares formula:
   \[
   x^2 - 9 = (x - 3)(x + 3)
   \]
   So,
   \[
   \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \quad \text{for } x \neq 3.
   \]
   Thus, for \( x \neq 3 \), \( f(x) = x + 3 \).

2. **Check the limit as \( x \to 3 \):**

   Since for \( x \neq 3 \), \( f(x) = x + 3 \), we can find the limit of \( f(x) \) as \( x \) approaches 3:
   \[
   \lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6.
   \]

3. **Check the value of \( f(3) \):**

   From the definition of the function, \( f(3) = 9 \).

4. **Determine continuity:**

   A function is continuous at \( x = 3 \) if:
   \[
   \lim_{x \to 3} f(x) = f(3).
   \]
   In this case, the limit as \( x \to 3 \) is 6, but \( f(3) = 9 \). Since \( \lim_{x \to 3} f(x) \neq f(3) \), the function is **not continuous** at \( x = 3 \).

### Conclusion:
The function is not continuous at \( x = 3 \), so the correct answer is **False**.

Let me know if you need further clarification, or if you have more questions!

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Here are five related questions for practice:

1. What does it mean for a function to be continuous at a point?
2. How do you check if a piecewise function is continuous at the boundary point?
3. Can a function be discontinuous at one point but continuous everywhere else?
4. What is the difference between a removable discontinuity and a jump discontinuity?
5. How can limits be used to determine continuity at a point?

**Tip:** When simplifying rational functions, always check if the factor you are canceling makes the denominator zero, as it indicates a point of potential discontinuity.

Determine if the following piecewise function is continuous: 
 f(x) = { (x^2 - 9)/(x - 3), x ≠ 3; 
 9, x = 3 }.

Solve a piecewise function problem involving limits and continuity at a specific point. Learn how to check continuity using limit definitions and algebraic simplification. Suitable for Grades 11-12 and introductory calculus courses.

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