find the natural domain and range for the function f(x)=1/( square root of x + 2)

To find the natural domain and range of the function \( f(x) = \frac{1}{\sqrt{x} + 2} \), let's break it down into steps.

### 1. **Domain**:
The domain of a function consists of all the values of \( x \) for which the function is defined.

For \( f(x) = \frac{1}{\sqrt{x} + 2} \), we have two key considerations:

- **Square root condition**: The expression under the square root, \( x \), must be non-negative. So, \( x \geq 0 \).
- **Denominator condition**: The denominator, \( \sqrt{x} + 2 \), must not be zero. Since \( \sqrt{x} \geq 0 \) for all \( x \geq 0 \), \( \sqrt{x} + 2 > 0 \). Therefore, there are no additional restrictions from this condition.

Thus, the natural domain of \( f(x) \) is all \( x \) such that \( x \geq 0 \).

**Domain**: \( [0, \infty) \)

### 2. **Range**:
The range of a function consists of all possible values that \( f(x) \) can take.

To find the range of \( f(x) \):

- As \( x \) increases, \( \sqrt{x} \) increases, causing \( \sqrt{x} + 2 \) to increase. Since \( f(x) = \frac{1}{\sqrt{x} + 2} \), as \( \sqrt{x} + 2 \) increases, \( f(x) \) decreases.
- At \( x = 0 \), \( \sqrt{x} = 0 \), so \( f(0) = \frac{1}{0 + 2} = \frac{1}{2} \).
- As \( x \) approaches infinity, \( \sqrt{x} \) approaches infinity, so \( \sqrt{x} + 2 \) also approaches infinity, making \( f(x) \) approach 0.

Thus, the smallest value of \( f(x) \) is 0 (approached as \( x \) approaches infinity), and the largest value is \( \frac{1}{2} \) (when \( x = 0 \)).

**Range**: \( (0, \frac{1}{2}] \)

### Summary
- **Domain**: \( [0, \infty) \)
- **Range**: \( (0, \frac{1}{2}] \)

Would you like further details or explanations? 

Here are five related questions you might find helpful:
1. How would the domain and range change if the function were \( f(x) = \frac{1}{\sqrt{x - 1} + 2} \)?
2. What happens to the range if the function was \( f(x) = \frac{1}{\sqrt{x} + 1} \) instead?
3. How do you find the domain and range for the function \( g(x) = \frac{1}{\sqrt{x^2 - 4}} \)?
4. Can you determine the domain and range of \( h(x) = \frac{1}{\sqrt{2x + 1} + 3} \)?
5. What if the function were \( f(x) = \frac{1}{\sqrt{x}} \)? How would the domain and range differ?

**Tip**: Always check for conditions that could make the denominator zero or the expression under a square root negative when finding the domain of a function.

Discover how to determine the natural domain and range of the function f(x)=1/(sqrt(x) + 2) step-by-step. Explore the concepts of function domains and ranges, including conditions for square root functions. Suitable for high school students and anyone interested in understanding function properties.

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