If `f(x)=sqrt(4-x^2)+sqrt(x^2-1)` , then the maximum value of `(f(x))^2` is ____________

Explanation

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

### Step 1: Define the function
We start with the function:
\[ f(x) = \sqrt{4 - x^2} + \sqrt{x^2 - 1} \]

### Step 2: Determine the domain of \( f(x) \)
The expression \( \sqrt{4 - x^2} \) is defined when \( 4 - x^2 \geq 0 \) which gives:
\[ x^2 \leq 4 \implies -2 \leq x \leq 2 \]
The expression \( \sqrt{x^2 - 1} \) is defined when \( x^2 - 1 \geq 0 \) which gives:
\[ x^2 \geq 1 \implies x \leq -1 \text{ or } x \geq 1 \]
Combining these two conditions, the domain of \( f(x) \) is:
\[ [-2, -1] \cup [1, 2] \]

### Step 3: Analyze \( f(x) \) on the intervals
We will analyze \( f(x) \) on the intervals \( [-2, -1] \) and \( [1, 2] \).

#### Interval 1: \( x \in [-2, -1] \)
- At \( x = -2 \):
\[ f(-2) = \sqrt{4 - (-2)^2} + \sqrt{(-2)^2 - 1} = \sqrt{4 - 4} + \sqrt{4 - 1} = 0 + \sqrt{3} = \sqrt{3} \]
- At \( x = -1 \):
\[ f(-1) = \sqrt{4 - (-1)^2} + \sqrt{(-1)^2 - 1} = \sqrt{4 - 1} + \sqrt{1 - 1} = \sqrt{3} + 0 = \sqrt{3} \]

#### Interval 2: \( x \in [1, 2] \)
- At \( x = 1 \):
\[ f(1) = \sqrt{4 - 1^2} + \sqrt{1^2 - 1} = \sqrt{4 - 1} + \sqrt{1 - 1} = \sqrt{3} + 0 = \sqrt{3} \]
- At \( x = 2 \):
\[ f(2) = \sqrt{4 - 2^2} + \sqrt{2^2 - 1} = \sqrt{4 - 4} + \sqrt{4 - 1} = 0 + \sqrt{3} = \sqrt{3} \]

### Step 4: Find maximum value of \( f(x) \)
From the calculations above, we see that \( f(x) \) takes the value \( \sqrt{3} \) at all endpoints of the intervals. Therefore, the maximum value of \( f(x) \) is:
\[ \max f(x) = \sqrt{3} \]

### Step 5: Calculate \( (f(x))^2 \)
Now, we find the maximum value of \( (f(x))^2 \):
\[ (f(x))^2 = (\sqrt{3})^2 = 3 \]

### Final Answer
Thus, the maximum value of \( (f(x))^2 \) is:
\[ \boxed{3} \]