The range of the function ` f(x) = sqrt(4 - x^2) + sqrt(x^2 - 1)` is

Options

A. [`sqrt(3) , sqrt(7)`]
B. `[sqrt(3) , sqrt(5)]`
C. `[(sqrt(2) , sqrt(3)]`
D. `[(sqrt(3) , sqrt(6))]`

Explanation

To find the range of the function \( f(x) = \sqrt{4 - x^2} + \sqrt{x^2 - 1} \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function contains two square root terms, so we need to ensure that the expressions inside the square roots are non-negative.

1. For the first term \( \sqrt{4 - x^2} \):
\[ 4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2 \]

2. For the second term \( \sqrt{x^2 - 1} \):
\[ x^2 - 1 \geq 0 \implies x^2 \geq 1 \implies x \leq -1 \text{ or } x \geq 1 \]

Combining these two conditions, we find the domain of \( f(x) \):
\[ x \in [-2, -1] \cup [1, 2] \]

### Step 2: Analyze the Function on the Domain
Next, we will evaluate the function at the endpoints and critical points within the domain.

1. Evaluate \( f(-2) \):
\[ f(-2) = \sqrt{4 - (-2)^2} + \sqrt{(-2)^2 - 1} = \sqrt{4 - 4} + \sqrt{4 - 1} = 0 + \sqrt{3} = \sqrt{3} \]

2. Evaluate \( f(-1) \):
\[ f(-1) = \sqrt{4 - (-1)^2} + \sqrt{(-1)^2 - 1} = \sqrt{4 - 1} + \sqrt{1 - 1} = \sqrt{3} + 0 = \sqrt{3} \]

3. Evaluate \( f(1) \):
\[ f(1) = \sqrt{4 - 1^2} + \sqrt{1^2 - 1} = \sqrt{4 - 1} + \sqrt{1 - 1} = \sqrt{3} + 0 = \sqrt{3} \]

4. Evaluate \( f(2) \):
\[ f(2) = \sqrt{4 - 2^2} + \sqrt{2^2 - 1} = \sqrt{4 - 4} + \sqrt{4 - 1} = 0 + \sqrt{3} = \sqrt{3} \]

### Step 3: Find Maximum Value
To find the maximum value of \( f(x) \), we can differentiate \( f(x) \) and set the derivative to zero.

1. Differentiate \( f(x) \):
\[ f'(x) = \frac{-x}{\sqrt{4 - x^2}} + \frac{x}{\sqrt{x^2 - 1}} \]

Setting \( f'(x) = 0 \):
\[ \frac{-x}{\sqrt{4 - x^2}} + \frac{x}{\sqrt{x^2 - 1}} = 0 \]
This implies:
\[ \frac{x}{\sqrt{x^2 - 1}} = \frac{x}{\sqrt{4 - x^2}} \]
Squaring both sides and simplifying leads to:
\[ x^2 - 1 = 4 - x^2 \implies 2x^2 = 5 \implies x = \pm \sqrt{\frac{5}{2}} \]

### Step 4: Evaluate at Critical Points
Now we need to check if \( \sqrt{\frac{5}{2}} \) lies within the domain:
\[ 1 < \sqrt{\frac{5}{2}} < 2 \]
Thus, we can evaluate \( f\left(\sqrt{\frac{5}{2}}\right) \):
\[ f\left(\sqrt{\frac{5}{2}}\right) = \sqrt{4 - \frac{5}{2}} + \sqrt{\frac{5}{2} - 1} = \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} = 2\sqrt{\frac{3}{2}} = \sqrt{6} \]

### Step 5: Conclusion
The minimum value of \( f(x) \) is \( \sqrt{3} \) and the maximum value is \( \sqrt{6} \). Therefore, the range of the function is:
\[ \text{Range of } f(x) = [\sqrt{3}, \sqrt{6}] \]

### Final Answer
The range of the function \( f(x) = \sqrt{4 - x^2} + \sqrt{x^2 - 1} \) is \( [\sqrt{3}, \sqrt{6}] \).