The range of the function `f(x)=sqrt(x-1)+2sqrt(3-x)` is

Options

A. `[sqrt(2),2sqrt(2)]`
B. `[sqrt(2),sqrt(10)]`
C. `[2sqrt(2),sqrt(10)]`
D. [1,3]

Answer

To find the range of the function \( f(x) = \sqrt{x - 1} + 2\sqrt{3 - x} \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function \( f(x) \) consists of two square root terms. For these square roots to be defined, we must have:
1. \( x - 1 \geq 0 \) which implies \( x \geq 1 \)
2. \( 3 - x \geq 0 \) which implies \( x \leq 3 \)

Thus, the domain of \( f(x) \) is the intersection of these two conditions:
\[ x \in [1, 3] \]

### Step 2: Evaluate the Function at the Endpoints of the Domain
Next, we will evaluate \( f(x) \) at the endpoints of the domain to find the minimum and maximum values.

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

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

### Step 3: Check for Maximum Value in the Interval
To find the maximum value of \( f(x) \) within the interval \( [1, 3] \), we can also evaluate \( f(x) \) at a point within the interval, such as \( x = 2 \).

- **At \( x = 2 \)**:
\[ f(2) = \sqrt{2 - 1} + 2\sqrt{3 - 2} = \sqrt{1} + 2\sqrt{1} = 1 + 2 = 3 \]

### Step 4: Determine the Range
Now we have the following values:
- \( f(1) = 2\sqrt{2} \)
- \( f(2) = 3 \)
- \( f(3) = \sqrt{2} \)

Calculating the approximate values:
- \( \sqrt{2} \approx 1.414 \)
- \( 2\sqrt{2} \approx 2.828 \)
- \( 3 \) is clearly the maximum value.

Thus, the minimum value of \( f(x) \) is \( \sqrt{2} \) and the maximum value is \( 3 \).

### Conclusion
The range of the function \( f(x) \) is:
\[ \text{Range} = [\sqrt{2}, 3] \]