Find the range of `f(x)sqrt(x-1)+sqrt(5-1)`

Explanation

To find the range of the function \( f(x) = \sqrt{x - 1} + \sqrt{5 - x} \), we will follow these steps:

### Step 1: Define the function
Let \( y = f(x) = \sqrt{x - 1} + \sqrt{5 - x} \).

### Step 2: Square both sides
Squaring both sides gives:
\[ y^2 = (\sqrt{x - 1} + \sqrt{5 - x})^2 \]
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \), we have:
\[ y^2 = (x - 1) + (5 - x) + 2\sqrt{(x - 1)(5 - x)} \]
This simplifies to:
\[ y^2 = 4 + 2\sqrt{(x - 1)(5 - x)} \]

### Step 3: Isolate the square root
Rearranging gives:
\[ y^2 - 4 = 2\sqrt{(x - 1)(5 - x)} \]
Dividing both sides by 2:
\[ \frac{y^2 - 4}{2} = \sqrt{(x - 1)(5 - x)} \]

### Step 4: Square both sides again
Squaring both sides again results in:
\[ \left(\frac{y^2 - 4}{2}\right)^2 = (x - 1)(5 - x) \]
Expanding the left-hand side:
\[ \frac{(y^2 - 4)^2}{4} = (x - 1)(5 - x) \]
The right-hand side can be expanded as:
\[ (x - 1)(5 - x) = 5x - x^2 - 5 + x = -x^2 + 6x - 5 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ -x^2 + 6x - 5 = \frac{(y^2 - 4)^2}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ -4x^2 + 24x - 20 = (y^2 - 4)^2 \]

### Step 6: Completing the square
To complete the square for the quadratic in \( x \):
\[ -4(x^2 - 6x + 5) = (y^2 - 4)^2 \]
Completing the square inside the parentheses:
\[ -4((x - 3)^2 - 4) = (y^2 - 4)^2 \]
This simplifies to:
\[ -4(x - 3)^2 + 16 = (y^2 - 4)^2 \]

### Step 7: Finding the minimum and maximum values
For \( y \) to have real values, the right-hand side must be non-negative. Thus:
\[ -4(x - 3)^2 + 16 \geq 0 \]
This implies:
\[ (x - 3)^2 \leq 4 \]
So:
\[ -2 \leq x - 3 \leq 2 \implies 1 \leq x \leq 5 \]

### Step 8: Evaluating the function at the endpoints
- At \( x = 1 \):
\[ f(1) = \sqrt{1 - 1} + \sqrt{5 - 1} = 0 + 2 = 2 \]
- At \( x = 5 \):
\[ f(5) = \sqrt{5 - 1} + \sqrt{5 - 5} = 2 + 0 = 2 \]
- At \( x = 3 \):
\[ f(3) = \sqrt{3 - 1} + \sqrt{5 - 3} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

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