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The problem you're working on asks to find the domain and range of the function:

\[
f(x) = \sqrt{x} + 16
\]

### Finding the Domain:
- The function includes a square root, \(\sqrt{x}\). The expression under the square root, \(x\), must be non-negative (i.e., \(x \geq 0\)) because the square root of a negative number is not defined in the set of real numbers.
- Thus, the domain of \(f(x)\) is:
  \[
  x \in [0, \infty)
  \]

### Finding the Range:
- The smallest value inside the square root is when \(x = 0\), which gives \(\sqrt{0} = 0\). Therefore, the smallest value of \(f(x)\) is:
  \[
  f(0) = 0 + 16 = 16
  \]
- As \(x\) increases, \(\sqrt{x}\) also increases, without any upper bound. Therefore, the range of \(f(x)\) starts at 16 and extends to infinity.

- Thus, the range of \(f(x)\) is:
  \[
  y \in [16, \infty)
  \]

### Final Answer:
- **Domain:** \([0, \infty)\)
- **Range:** \([16, \infty)\)

Let me know if you want more details or further explanations!

Here are 5 related questions to deepen your understanding:
1. What is the domain of \( f(x) = \sqrt{x - 5} + 3 \)?
2. How does the domain change if we consider \( f(x) = \frac{1}{\sqrt{x}} \)?
3. What is the range of \( f(x) = \sqrt{x} + 3 \)?
4. How would the range of \( f(x) = \sqrt{x} - 5 \) differ?
5. How would the domain change if \( f(x) = \sqrt{4 - x} \)?

**Tip:** Always check for restrictions in the domain by looking at square roots, fractions, and logarithms, as they impose specific conditions on the inputs.

Find the domain and range of the function f(x) = √x + 16.

Discover how to find the domain and range of the square root function f(x) = √x + 16. This problem involves understanding restrictions on the domain and interpreting the range based on square root properties. Suitable for high school students in Grades 9-11.

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