given f(x)=square root of x and g(x) = square root of x^2-16 find (f/g)(x) and its domain (express it in interval notation)

Given the functions \( f(x) = \sqrt{x} \) and \( g(x) = \sqrt{x^2 - 16} \), we need to find \( \left( \frac{f}{g} \right)(x) \) and determine its domain.

### Step 1: Define \( \left( \frac{f}{g} \right)(x) \)
The function \( \left( \frac{f}{g} \right)(x) \) is defined as:
\[
\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{\sqrt{x^2 - 16}}
\]
This expression is valid as long as both the numerator and denominator are defined and the denominator is not zero.

### Step 2: Determine the domain
#### Domain restrictions:
1. **For \( f(x) = \sqrt{x} \):**
   - The square root function \( \sqrt{x} \) requires \( x \geq 0 \), because the square root of a negative number is not defined in the real number system. 
   - So, \( x \geq 0 \).

2. **For \( g(x) = \sqrt{x^2 - 16} \):**
   - The square root function \( \sqrt{x^2 - 16} \) requires \( x^2 - 16 \geq 0 \), because we cannot take the square root of a negative number.
   - Solving \( x^2 - 16 \geq 0 \):
     \[
     x^2 \geq 16
     \]
     \[
     x \geq 4 \quad \text{or} \quad x \leq -4
     \]
     So, \( x \in (-\infty, -4] \cup [4, \infty) \).

3. **Denominator cannot be zero:**
   - \( g(x) = \sqrt{x^2 - 16} \) must not be zero, so we require \( x^2 - 16 \neq 0 \), or \( x^2 \neq 16 \), which means \( x \neq \pm 4 \).

#### Combine restrictions:
- From \( f(x) \), we know \( x \geq 0 \).
- From \( g(x) \), we have \( x \in (-\infty, -4] \cup [4, \infty) \), but combining this with the restriction \( x \geq 0 \) gives \( x \in [4, \infty) \).
- Finally, we must exclude \( x = 4 \), because \( g(4) = 0 \) would make the denominator zero.

### Domain of \( \left( \frac{f}{g} \right)(x) \):
The domain is \( (4, \infty) \).

### Final Answer:
\[
\left( \frac{f}{g} \right)(x) = \frac{\sqrt{x}}{\sqrt{x^2 - 16}}, \quad \text{with domain } (4, \infty)
\]

Would you like more details or have any questions?

Here are five related questions:
1. What happens if the function \( f(x) \) is replaced by a cubic root function?
2. How would the domain change if \( g(x) = \sqrt{x^2 - 25} \)?
3. Can we simplify \( \left( \frac{f}{g} \right)(x) \) any further?
4. What is the behavior of \( \left( \frac{f}{g} \right)(x) \) as \( x \to 4^+ \)?
5. How does the function behave as \( x \to \infty \)?

**Tip:** Always check both the numerator and denominator for restrictions when finding the domain of a quotient function.

Given f(x) = square root of x and g(x) = square root of x^2 - 16, find (f/g)(x) and its domain (express it in interval notation).

Solve the problem of finding the quotient of the square root functions f(x) = √x and g(x) = √(x² - 16). This solution covers how to define the function (f/g)(x) and determine its domain, using domain restrictions and square root properties. Suitable for students in Grades 9-12.

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