Let's analyze the piecewise function again to determine the domain, range, and graph. The function is given by:

[ f(x) = \begin{cases} 3 & , x \in [-7, -3] \ 1 - 2x & , x \in (-2, 2) \ |x - 4| - 6 & , x \in (2, 7] \end{cases} ]

1. Domain

The domain of a function is the set of all possible input values $x$ for which the function is defined:

• The first piece $f(x) = 3$ is defined for $x \in [-7, -3]$.
• The second piece $f(x) = 1 - 2x$ is defined for $x \in (-2, 2)$. Notice this interval is open, which means the endpoints $-2$ and $2$ are not included.
• The third piece $f(x) = |x - 4| - 6$ is defined for $x \in (2, 7]$. This interval is open on the left at $x = 2$ and closed on the right at $x = 7$.

Combining these intervals, the domain of the function is:

[ [-7, -3] \cup (-2, 2) \cup (2, 7] ]

2. Range

The range is the set of all possible output values $y$ the function can produce:

• For the first piece $f(x) = 3$ over $x \in [-7, -3]$, the output is a constant $3$. Thus, this part contributes $\{3\}$ to the range.

• For the second piece $f(x) = 1 - 2x$ over $x \in (-2, 2)$:

  • At $x = -2$ (not included in the domain), $f(x) = 1 - 2(-2) = 5$.
  • At $x = 2$ (not included in the domain), $f(x) = 1 - 2(2) = -3$.

  Since this is a linear function, it decreases from $5$ (just above $x = -2$) to $-3$ (just below $x = 2$). Thus, the range of this part is $(-3, 5)$.

• For the third piece $f(x) = |x - 4| - 6$ over $x \in (2, 7]$:

  • At $x = 2$ (not included), $f(x) = |2 - 4| - 6 = 2 - 6 = -4$.
  • At $x = 4$, $f(x) = |4 - 4| - 6 = -6$ (the minimum value).
  • At $x = 7$, $f(x) = |7 - 4| - 6 = 3 - 6 = -3$.

  The function decreases to $-6$ at $x = 4$ and then increases to $-3$. Thus, this part contributes $(-6, -3]$ to the range.

Combining these, the overall range is:

$[-6, -3] \cup (3, 5)$

3. Graph

To sketch the graph, we plot each piece of the function over its respective interval:

1. For $x \in [-7, -3]$, $f(x) = 3$ is a horizontal line at $y = 3$.

2. For $x \in (-2, 2)$, $f(x) = 1 - 2x$ is a line with a negative slope. It passes through $(0, 1)$, decreases, approaching $(2, -3)$, and increases approaching $(-2, 5)$. The line never actually reaches these values because the interval is open.

3. For $x \in (2, 7]$, $f(x) = |x - 4| - 6$ is a V-shaped function:

   • Decreases from just above $y = -4$ at $x = 2$ to $y = -6$ at $x = 4$.
   • Increases from $y = -6$ at $x = 4$ to $y = -3$ at $x = 7$.

Summary

• Domain: $[-7, -3] \cup (-2, 2) \cup (2, 7]$
• Range: $[-6, -3] \cup \{3\} \cup (3, 5)$

Would you like a more detailed explanation or a graph sketch?

Here are some related questions:

1. How does the domain change if we change the endpoints of the intervals?
2. What is the significance of the absolute value in the third piece of the function?
3. How do discontinuities in the function affect its range?
4. What are the implications of having open versus closed intervals in the domain?
5. Can we determine the function's behavior outside of the given domain?

Tip: When dealing with piecewise functions, carefully analyze each interval and consider both endpoints to avoid missing important details about the function's behavior.