1. **State the problem:** We are given two functions: - $f(x) = 3\sqrt{x}$ - $f(x) = -\sqrt{x} + 3$ We need to find the domain and range for each function. 2. **Recall the domain and range rules for square root functions:** - The expression inside the square root must be greater than or equal to zero for the function to be real-valued. So, for $\sqrt{x}$, the domain is $x \geq 0$. - The square root function $\sqrt{x}$ outputs values $\geq 0$. 3. **Analyze the first function $f(x) = 3\sqrt{x}$:** - Domain: Since $\sqrt{x}$ requires $x \geq 0$, the domain is $[0, \infty)$. - Range: Multiply $\sqrt{x}$ by 3, which scales the output but does not change the sign. Since $\sqrt{x} \geq 0$, $3\sqrt{x} \geq 0$. So the range is $[0, \infty)$. 4. **Analyze the second function $f(x) = -\sqrt{x} + 3$:** - Domain: Again, $x \geq 0$ because of the square root. - Range: $\sqrt{x} \geq 0$, so $-\sqrt{x} \leq 0$. Adding 3 shifts the range up by 3. The maximum value is when $x=0$, giving $f(0) = -0 + 3 = 3$. As $x$ increases, $\sqrt{x}$ increases, so $-\sqrt{x}$ decreases, making $f(x)$ decrease without bound but limited by the domain. Since $x \geq 0$, the minimum value approaches $-\infty$ is not possible because $\sqrt{x}$ grows without bound but negative sign makes $f(x)$ approach $-\infty$? Actually, since $x \geq 0$, $\sqrt{x}$ grows to infinity, so $-\sqrt{x}$ goes to $-\infty$, and $f(x)$ goes to $-\infty + 3 = -\infty$. So the range is $(-\infty, 3]$. **Final answers:** - For $f(x) = 3\sqrt{x}$: - Domain: $[0, \infty)$ - Range: $[0, \infty)$ - For $f(x) = -\sqrt{x} + 3$: - Domain: $[0, \infty)$ - Range: $(-\infty, 3]$