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We are given a piecewise function \( g(x) \) defined as follows:

\[
g(x) =
\begin{cases} 
-\frac{1}{2}x + 5 & \text{if } x < 6 \\
\sqrt{x - 6} & \text{if } x \geq 6 
\end{cases}
\]

We need to evaluate this function at \( g(-6), g(0), g(6), \) and \( g(12) \).

### Step-by-step Calculation:

1. **For \( g(-6) \):**
   Since \( -6 < 6 \), we use the first part of the piecewise function:
   \[
   g(-6) = -\frac{1}{2}(-6) + 5 = 3 + 5 = 8
   \]

2. **For \( g(0) \):**
   Since \( 0 < 6 \), we again use the first part of the piecewise function:
   \[
   g(0) = -\frac{1}{2}(0) + 5 = 0 + 5 = 5
   \]

3. **For \( g(6) \):**
   Since \( 6 \geq 6 \), we use the second part of the piecewise function:
   \[
   g(6) = \sqrt{6 - 6} = \sqrt{0} = 0
   \]

4. **For \( g(12) \):**
   Since \( 12 \geq 6 \), we use the second part of the piecewise function:
   \[
   g(12) = \sqrt{12 - 6} = \sqrt{6}
   \]
   Approximating \( \sqrt{6} \), we get:
   \[
   g(12) \approx 2.449
   \]

### Final Answers:
- \( g(-6) = 8 \)
- \( g(0) = 5 \)
- \( g(6) = 0 \)
- \( g(12) \approx 2.449 \)

Would you like more details or clarifications on any of the steps?

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Here are 5 related questions for further practice:
1. What is the domain of the function \( g(x) \)?
2. Evaluate \( g(10) \) using the same function.
3. How does the function \( g(x) \) behave as \( x \to -\infty \)?
4. Find the inverse of \( g(x) \) for \( x \geq 6 \).
5. What is the range of the function \( g(x) \)?

**Tip:** When dealing with piecewise functions, always check which part of the function applies based on the value of \( x \).

Let g be the function defined by:
g(x) = 
- (1/2)x + 5 if x < 6
sqrt(x - 6) if x >= 6
Find g(-6), g(0), g(6), and g(12).

Learn how to evaluate a piecewise function with both linear and square root components at specific points, including g(-6), g(0), g(6), and g(12). This solution includes step-by-step calculations for each case.

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