The range of the function `f(x) = sqrt(2-x)+sqrt( 1+x)`

Options

A. `[sqrt3, sqrt6]`
B. `[0, sqrt6]`
C. `(sqrt3, sqrt6)`
D. none of these

Explanation

To find the range of the function \( f(x) = \sqrt{2 - x} + \sqrt{1 + x} \), we will follow these steps:

### Step 1: Determine the Domain
The function involves square roots, which means the expressions inside the square roots must be non-negative.

1. For \( \sqrt{2 - x} \):
\[ 2 - x \geq 0 \implies x \leq 2 \]

2. For \( \sqrt{1 + x} \):
\[ 1 + x \geq 0 \implies x \geq -1 \]

Thus, the domain of \( f(x) \) is:
\[ [-1, 2] \]

### Step 2: Find the Minimum and Maximum Values of \( f(x) \)
Next, we will evaluate \( f(x) \) at the endpoints of the domain and check for any critical points within the interval.

1. **Evaluate at the endpoints**:
- At \( x = -1 \):
\[ f(-1) = \sqrt{2 - (-1)} + \sqrt{1 + (-1)} = \sqrt{3} + 0 = \sqrt{3} \]
- At \( x = 2 \):
\[ f(2) = \sqrt{2 - 2} + \sqrt{1 + 2} = 0 + \sqrt{3} = \sqrt{3} \]

2. **Check for critical points**:
To find critical points, we can differentiate \( f(x) \) and set the derivative to zero.
\[ f'(x) = \frac{-1}{2\sqrt{2 - x}} + \frac{1}{2\sqrt{1 + x}} \]
Setting \( f'(x) = 0 \):
\[ \frac{-1}{2\sqrt{2 - x}} + \frac{1}{2\sqrt{1 + x}} = 0 \]
\[ \frac{1}{\sqrt{1 + x}} = \frac{1}{\sqrt{2 - x}} \]
Squaring both sides:
\[ 1 + x = 2 - x \]
\[ 2x = 1 \implies x = \frac{1}{2} \]

3. **Evaluate \( f(x) \) at the critical point**:
- At \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \sqrt{2 - \frac{1}{2}} + \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} = 2\sqrt{\frac{3}{2}} = \sqrt{6} \]

### Step 3: Determine the Range
Now we have the values of \( f(x) \) at the endpoints and the critical point:
- \( f(-1) = \sqrt{3} \)
- \( f(2) = \sqrt{3} \)
- \( f\left(\frac{1}{2}\right) = \sqrt{6} \)

Since \( f(x) \) is continuous on the interval \([-1, 2]\), the range of \( f(x) \) is from the minimum value \( \sqrt{3} \) to the maximum value \( \sqrt{6} \).

Thus, the range of the function \( f(x) \) is:
\[ [\sqrt{3}, \sqrt{6}] \]

### Final Answer
The range of the function \( f(x) = \sqrt{2 - x} + \sqrt{1 + x} \) is:
\[ [\sqrt{3}, \sqrt{6}] \]