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I see that you have uploaded an image of a math problem in Indonesian. The problem asks to solve a function division based on two given functions, \( f(x) = \sqrt{x} - 3 \) and \( g(x) = 2x - 8 \), followed by four sub-questions (a to d) that deal with the composite functions \( h(x) = \frac{f(x)}{g(x)} \) and \( p(x) = \frac{f(x)}{g(x)} \), as well as evaluating these functions at specific values of \( x \).

### Let's solve this step by step:

Given:
- \( f(x) = \sqrt{x} - 3 \)
- \( g(x) = 2x - 8 \)

#### a. Find \( h(x) = \frac{f(x)}{g(x)} \) and the domain (daerah asal) of \( h(x) \)
- \( h(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x} - 3}{2x - 8} \)

The domain of \( h(x) \) is determined by two conditions:
1. \( f(x) = \sqrt{x} - 3 \) implies that \( x \geq 0 \) (since \( \sqrt{x} \) is defined for non-negative values).
2. The denominator \( g(x) = 2x - 8 \) should not be zero, i.e., \( 2x - 8 \neq 0 \Rightarrow x \neq 4 \).

Thus, the domain of \( h(x) \) is \( x \geq 0 \) but \( x \neq 4 \). Therefore, the domain is \( x \in [0, \infty) \setminus \{4\} \).

#### b. Find \( p(x) = \frac{f(x)}{g(x)} \) and the domain (daerah asal) of \( p(x) \)
Since both functions \( h(x) \) and \( p(x) \) are described as the same, this step is essentially the same as part (a):
- \( p(x) = \frac{\sqrt{x} - 3}{2x - 8} \)
- The domain of \( p(x) \) is also \( x \in [0, \infty) \setminus \{4\} \).

#### c. Evaluate \( h(4), h(6) \), and \( h(8) \)

To evaluate \( h(x) \) at specific points, plug in the values of \( x \) into the function \( h(x) = \frac{\sqrt{x} - 3}{2x - 8} \).

- \( h(4) = \frac{\sqrt{4} - 3}{2(4) - 8} = \frac{2 - 3}{8 - 8} \) is undefined because the denominator becomes zero. Hence, \( h(4) \) is not defined.
- \( h(6) = \frac{\sqrt{6} - 3}{2(6) - 8} = \frac{\sqrt{6} - 3}{12 - 8} = \frac{\sqrt{6} - 3}{4} \).
- \( h(8) = \frac{\sqrt{8} - 3}{2(8) - 8} = \frac{\sqrt{8} - 3}{16 - 8} = \frac{\sqrt{8} - 3}{8} \).

#### d. Evaluate \( p(4), p(6) \), and \( p(8) \)
As previously mentioned, \( p(x) = h(x) \), so:

- \( p(4) \) is also undefined (as \( h(4) \) is).
- \( p(6) = h(6) = \frac{\sqrt{6} - 3}{4} \).
- \( p(8) = h(8) = \frac{\sqrt{8} - 3}{8} \).

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Would you like more details on any step, or should we go deeper into any particular part?

### 5 Related Questions:
1. What would happen if the denominator was a different linear function?
2. Can you find the derivative of \( h(x) \) using quotient rule?
3. What are the conditions for \( p(x) \) to be continuous over the domain?
4. How do you simplify \( \sqrt{8} \) or other square root expressions in function evaluations?
5. How can we find the asymptotes for \( h(x) \)?

**Tip:** Always check the denominator for zeroes when determining the domain of rational functions.

Selesaikan pembagian fungsi: f(x) = √x - 3 dan g(x) = 2x - 8. Tentukan: a) h(x) = f(x) / g(x) dan daerah asal h(x), b) p(x) = f(x) / g(x) dan daerah asal p(x), c) h(4), h(6), h(8), d) p(4), p(6), p(8)

Learn how to solve function division problems with square root functions and linear denominators. This problem includes calculating composite functions and determining their domains, along with evaluating specific points. Suitable for Grades 10-12 students.

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