https://file.aiquickdraw.com/mathbot/file/1726870985993slensbbq.jpg

Let's analyze the problem in the image to answer the questions.

### Given Information:

The function \( f(x) \) is defined piecewise as follows:

\[
f(x) = 
\begin{cases} 
x^2 - 3, & -1 \leq x < 0 \\
4x, & 0 \leq x < 1 \\
6, & x = 1 \\
-3x + 7, & 1 < x < 2 \\
3, & x = 2 \\
1, & 2 < x < 3
\end{cases}
\]

We are asked two questions:

1. **Is \( f \) defined at \( x = 2 \)?**

   From the piecewise definition, we can see that for \( x = 2 \), \( f(x) = 3 \). Hence, **\( f \) is defined at \( x = 2 \)**.

2. **Is \( f \) continuous at \( x = 2 \)?**

   To check continuity at \( x = 2 \), we need to verify three things:
   - \( f(2) \) is defined (it is, \( f(2) = 3 \)).
   - The left-hand limit \( \lim_{x \to 2^-} f(x) \).
   - The right-hand limit \( \lim_{x \to 2^+} f(x) \).
   
   From the piecewise function, for \( 1 < x < 2 \), \( f(x) = -3x + 7 \). As \( x \to 2^- \), \( f(x) = -3(2) + 7 = 1 \).
   
   For \( 2 < x < 3 \), \( f(x) = 1 \), so \( \lim_{x \to 2^+} f(x) = 1 \).
   
   Since both the left-hand and right-hand limits are equal to 1, but \( f(2) = 3 \), the function is **not continuous at \( x = 2 \)**.

### Conclusion:
- \( f \) is defined at \( x = 2 \) (answer to part a: Yes).
- \( f \) is not continuous at \( x = 2 \) (answer to part b: No).

Would you like further clarification on any part of this solution?

Here are five related questions:
1. How can you determine the continuity of a function at any given point?
2. What role do limits play in determining continuity?
3. What does it mean for a function to have a jump discontinuity?
4. Can a function be defined but not continuous at a point?
5. How can you determine if a function is continuous on a given interval?

**Tip:** When checking for continuity at a point, always confirm the function's value at the point and evaluate the left-hand and right-hand limits separately.

Is f defined at x = 2? Is f continuous at x = 2?

Analyze the continuity and definition of a piecewise function at x = 2. This problem involves evaluating the limit from both sides of the point and checking the function's value, covering essential calculus concepts. Suitable for college-level students.

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