Key Concepts and Formulas

• Domain of a Square Root Function: For $\sqrt{g(x)}$ to be defined, $g(x) \ge 0$.
• Finding the Range of a Continuous Function: For a continuous function on a closed interval, the range is the interval between its minimum and maximum values. These extrema can occur at endpoints or critical points.
• Differentiation Rules: The power rule and the chain rule are essential for finding the derivative of composite functions.

Step-by-Step Solution

Step 1: Determine the Domain of the Function The function is $f(x)=\sqrt{3-x}+\sqrt{2+x}$. For the function to be defined, both terms under the square roots must be non-negative.

• For $\sqrt{3-x}$: $3-x \ge 0 \implies x \le 3$.
• For $\sqrt{2+x}$: $2+x \ge 0 \implies x \ge -2$. Combining these conditions, the domain of $f(x)$ is $[-2, 3]$. This is a closed interval, so we can apply the Extreme Value Theorem.

Step 2: Find the Derivative of the Function To find critical points, we need to calculate the derivative of $f(x)$. $f'(x) = \frac{d}{dx}(\sqrt{3-x}) + \frac{d}{dx}(\sqrt{2+x})$ Using the chain rule, $\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$: $\frac{d}{dx}(\sqrt{3-x}) = \frac{1}{2\sqrt{3-x}} \cdot (-1) = -\frac{1}{2\sqrt{3-x}}$ $\frac{d}{dx}(\sqrt{2+x}) = \frac{1}{2\sqrt{2+x}} \cdot (1) = \frac{1}{2\sqrt{2+x}}$ Therefore, $f'(x) = -\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{2+x}}$.

Step 3: Find the Critical Points Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. $f'(x)$ is undefined when $3-x=0$ or $2+x=0$, which means $x=3$ and $x=-2$. These are the endpoints of our domain, which we will consider separately. Now, let's find where $f'(x) = 0$: $-\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{2+x}} = 0$ $\frac{1}{2\sqrt{2+x}} = \frac{1}{2\sqrt{3-x}}$ $\sqrt{2+x} = \sqrt{3-x}$ Squaring both sides: $2+x = 3-x$ $2x = 1$ $x = \frac{1}{2}$ This critical point $x = \frac{1}{2}$ lies within our domain $[-2, 3]$.

Step 4: Evaluate the Function at Endpoints and Critical Points We need to evaluate $f(x)$ at $x=-2$, $x=3$, and $x=\frac{1}{2}$.

• At $x = -2$: $f(-2) = \sqrt{3-(-2)} + \sqrt{2+(-2)} = \sqrt{5} + \sqrt{0} = \sqrt{5}$.
• At $x = 3$: $f(3) = \sqrt{3-3} + \sqrt{2+3} = \sqrt{0} + \sqrt{5} = \sqrt{5}$.
• At $x = \frac{1}{2}$: $f\left(\frac{1}{2}\right) = \sqrt{3-\frac{1}{2}} + \sqrt{2+\frac{1}{2}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{5}{2}} = 2\sqrt{\frac{5}{2}} = 2 \cdot \frac{\sqrt{5}}{\sqrt{2}} = \frac{2\sqrt{10}}{2} = \sqrt{10}$.

Step 5: Determine the Minimum and Maximum Values Comparing the values we found: $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$. The minimum value of the function is $\sqrt{5}$. The maximum value of the function is $\sqrt{10}$.

Step 6: State the Range of the Function Since the function is continuous on the closed interval $[-2, 3]$, its range is the interval from its minimum value to its maximum value. Range = $[\min(f(x)), \max(f(x))] = [\sqrt{5}, \sqrt{10}]$.

Common Mistakes & Tips

• Forgetting the Domain: Always determine the domain of the function first. If the domain is not a closed interval, or if critical points fall outside the domain, the approach for finding the range might need adjustments.
• Errors in Differentiation: Be careful with the chain rule and signs when differentiating square root functions. A small error here can lead to incorrect critical points.
• Squaring Both Sides: When solving $\sqrt{a} = \sqrt{b}$, squaring both sides is valid. However, when solving equations like $\sqrt{a} = b$, you must ensure $b \ge 0$ to avoid introducing extraneous solutions. In this case, both sides were positive square roots.

Summary

To find the range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$, we first determined its domain to be $[-2, 3]$. We then found the derivative $f'(x)$ and identified the critical point $x = \frac{1}{2}$ within the domain. By evaluating the function at the endpoints of the domain ($x=-2$ and $x=3$) and at the critical point ($x=\frac{1}{2}$), we found the function values to be $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$ respectively. The minimum value is $\sqrt{5}$ and the maximum value is $\sqrt{10}$. Therefore, the range of the function is $[\sqrt{5}, \sqrt{10}]$.

The final answer is \boxed{[\sqrt{5}, \sqrt{10}]}.