To find the domain and range of the function $f(x) = \sqrt{x+16}$, let's break it down:

Domain:

The expression inside the square root, $x+16$, must be non-negative because the square root of a negative number is not defined in the set of real numbers.

$x + 16 \geq 0$

Solving this inequality:

$x \geq -16$

So, the domain of the function is all values of $x$ that are greater than or equal to $-16$. In interval notation, the domain is:

$\text{Domain} = [-16, \infty)$

Range:

For the range, consider the output of the function $f(x) = \sqrt{x+16}$. Since the square root function always gives non-negative values, the smallest value of $\sqrt{x+16}$ occurs when $x = -16$, and that value is:

$f(-16) = \sqrt{-16 + 16} = \sqrt{0} = 0$

As $x$ increases, the value of $\sqrt{x+16}$ increases without bound. Therefore, the range of the function is:

$\text{Range} = [0, \infty)$

Summary:

• Domain: $[-16, \infty)$
• Range: $[0, \infty)$

Would you like more details or have any further questions?

Here are 5 related questions you could explore:

1. What is the domain and range of $\sqrt{x-5}$?
2. How would the domain and range change if the function were $f(x) = \sqrt{-x+16}$?
3. What is the domain and range of $\sqrt{x^2 + 4x + 4}$?
4. How would you find the domain and range of a cubic function like $f(x) = x^3 + 2x^2 + 3x + 1$?
5. How does adding a constant outside the square root, like $f(x) = \sqrt{x+16} + 2$, affect the range?

Tip: Always check for constraints like non-negative values inside square roots and undefined points (like division by zero) when determining the domain of a function.