Domain Range Square Root A7366C

1. **State the problem:** Find the domain and range of the function $$f(x) = \sqrt{x + 8}$$. 2. **Domain:** The expression inside the square root must be non-negative because the square root of a negative number is not a real number. 3. Set the inside of the square root $$x + 8 \geq 0$$. 4. Solve for $$x$$: $$x + 8 \geq 0$$ $$x \geq -8$$ 5. Therefore, the domain in set notation is: $$\{x \mid x \geq -8\}$$ 6. **Range:** The square root function outputs values greater than or equal to zero because square roots are always non-negative. 7. Since $$f(x) = \sqrt{x + 8}$$, the smallest value is when $$x = -8$$, which gives $$f(-8) = 0$$. 8. As $$x$$ increases, $$f(x)$$ increases without bound. 9. Therefore, the range in set notation is: $$\{y \mid y \geq 0\}$$ **Final answers:** - Domain: $$\{x \mid x \geq -8\}$$ - Range: $$\{y \mid y \geq 0\}$$