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

\[
f(x) =
\begin{cases}
1 & \text{if } -6 \leq x \leq -1 \\
x + 2 & \text{if } -1 < x \leq 5
\end{cases}
\]

### (a) Domain of \( f(x) \)
The domain of the function combines both intervals:
- From the first piece, \( -6 \leq x \leq -1 \).
- From the second piece, \( -1 < x \leq 5 \).

Therefore, the overall domain is \( -6 \leq x \leq 5 \), which matches option **A** in your selection.

### (b) Evaluation of \( f(-5), f(0), f(3) \)

1. \( f(-5) \): Since \( -6 \leq -5 \leq -1 \), we use the first piece of the function \( f(x) = 1 \). Thus, \( f(-5) = 1 \).
   
2. \( f(0) \): Since \( -1 < 0 \leq 5 \), we use the second piece of the function \( f(x) = x + 2 \). So, \( f(0) = 0 + 2 = 2 \).
   
3. \( f(3) \): Since \( -1 < 3 \leq 5 \), we again use the second piece \( f(x) = x + 2 \). Thus, \( f(3) = 3 + 2 = 5 \).

These match the correct answers you selected.

### (c) Graph of \( f(x) \)

To graph the function, we consider both pieces:
- From \( -6 \leq x \leq -1 \), the function is constant at \( f(x) = 1 \). This is a horizontal line from \( x = -6 \) to \( x = -1 \).
- From \( -1 < x \leq 5 \), the function is \( f(x) = x + 2 \), which is a line with slope 1 and intercept 2.

The correct graph should show:
- A horizontal line at \( y = 1 \) for \( -6 \leq x \leq -1 \).
- A diagonal line starting at \( f(-1) = 1 \) (but excluding \( x = -1 \)) and continuing upward to \( f(5) = 7 \).

The graph you selected, **D**, is correct.

### (d) Continuity of \( f(x) \)

The function is continuous if there are no jumps or breaks in its graph. At \( x = -1 \), the function's value for the first piece is \( f(-1) = 1 \), and for the second piece, \( \lim_{x \to -1^+} f(x) = -1 + 2 = 1 \). Since the function values agree at the transition point, the function is continuous on its entire domain.

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Would you like more details on any part? Here are five related questions you can explore:

1. How would the graph change if the second piece was \( f(x) = x - 2 \) instead of \( x + 2 \)?
2. What is the range of the function \( f(x) \)?
3. How can we determine if a piecewise function is differentiable at a transition point?
4. What would happen to the continuity if \( f(x) \) had been defined differently at \( x = -1 \)?
5. How would the domain change if the function was defined as \( f(x) = 1 \) for \( -6 \leq x < -1 \)?

**Tip:** When graphing piecewise functions, always check for points of discontinuity at the boundaries of the pieces to ensure the function is continuous.

Determine the domain of f(x), evaluate f(-5), f(0), and f(3), graph f(x), and assess the continuity of f(x) for the given piecewise function.

Explore how to find the domain, evaluate specific values, graph, and assess the continuity of the piecewise function f(x). This problem covers key concepts in evaluating and graphing piecewise functions and verifying continuity at transition points. Suitable for high school students.

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