The given function is:

$f(x) = \sqrt{0.5x + 2} - 1$

Domain:

For the square root function, the expression inside the square root must be non-negative, meaning:

$0.5x + 2 \geq 0$

Solve for $x$:

$0.5x \geq -2 \quad \Rightarrow \quad x \geq -4$

Thus, the domain is:

$x \in [-4, \infty)$

Range:

The square root function, $\sqrt{0.5x + 2}$, is always non-negative, meaning $\sqrt{0.5x + 2} \geq 0$. Subtracting 1 from this gives:

$\sqrt{0.5x + 2} - 1 \geq -1$

Therefore, the minimum value of the function is $-1$. As $x$ increases, the function grows without bound. Thus, the range is:

$f(x) \in [-1, \infty)$

Correct Answer:

• Domain: $(-4, \infty)$
• Range: $[-1, \infty)$

This corresponds to the last option.

Would you like further details or have any questions?

Here are some related questions:

1. How do you find the domain of a square root function in general?
2. What are the properties of the square root function and its transformations?
3. How does the addition or subtraction outside a function affect its range?
4. Why does $f(x) = \sqrt{0.5x + 2}$ require non-negative values inside the square root?
5. What happens to the range of $f(x)$ if the function was $f(x) = -\sqrt{0.5x + 2} - 1$?

Tip: Always check for domain restrictions when dealing with square root or logarithmic functions!