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.