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
To find the domain of the function, we need to ensure that the expressions under the square roots are non-negative.

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

2. For \( \sqrt{3-x} \):
\[ 3 - x \geq 0 \implies x \leq 3 \]

Combining these inequalities, we find:
\[ 1 \leq x \leq 3 \]
Thus, the domain of \( f(x) \) is \( [1, 3] \).

### Step 2: Find the Critical Points
Next, we differentiate \( f(x) \) to find the critical points where the function may attain its maximum or minimum values.

The derivative \( f'(x) \) is calculated as follows:
\[ f'(x) = \frac{d}{dx}(\sqrt{x-1}) + 2 \cdot \frac{d}{dx}(\sqrt{3-x}) \]

Using the derivative formula \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \):

1. For \( \sqrt{x-1} \):
\[ \frac{d}{dx}(\sqrt{x-1}) = \frac{1}{2\sqrt{x-1}} \]

2. For \( \sqrt{3-x} \):
\[ \frac{d}{dx}(\sqrt{3-x}) = \frac{-1}{2\sqrt{3-x}} \]

Thus,
\[ f'(x) = \frac{1}{2\sqrt{x-1}} - \frac{1}{2\sqrt{3-x}} \]

### Step 3: Set the Derivative to Zero
Setting the derivative equal to zero to find critical points:
\[ \frac{1}{2\sqrt{x-1}} - \frac{1}{2\sqrt{3-x}} = 0 \]

This simplifies to:
\[ \sqrt{3-x} = \sqrt{x-1} \]

Squaring both sides:
\[ 3 - x = x - 1 \]
\[ 3 + 1 = 2x \implies 4 = 2x \implies x = 2 \]

### Step 4: Evaluate the Function at Critical Points and Endpoints
Now we evaluate \( f(x) \) at the critical point \( x = 2 \) and the endpoints of the domain \( x = 1 \) and \( x = 3 \).

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

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

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

### Step 5: Determine the Range
Now we compare the values:
- \( f(1) = 2\sqrt{2} \approx 2.828 \)
- \( f(2) = 3 \)
- \( f(3) = \sqrt{2} \approx 1.414 \)

The minimum value is \( \sqrt{2} \) and the maximum value is \( 3 \).

Thus, the range of the function \( f(x) \) is:
\[ [\sqrt{2}, 3] \]