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

Explanation

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

### Step 1: Determine the domain of \( f(x) \)

The function \( f(x) \) involves square roots, which means the expressions inside the square roots must be non-negative.

1. For \( \sqrt{x - 1} \) to be defined, we need:
\[ x - 1 \geq 0 \implies x \geq 1 \]

2. For \( \sqrt{5 - x} \) to be defined, we need:
\[ 5 - x \geq 0 \implies x \leq 5 \]

Combining these inequalities, we find the domain of \( f(x) \):
\[ 1 \leq x \leq 5 \]

### Step 2: Rewrite the function in terms of \( y \)

Let \( y = f(x) = \sqrt{x - 1} + \sqrt{5 - x} \).

### Step 3: Square both sides to eliminate the square roots

Squaring both sides gives:
\[ y^2 = (\sqrt{x - 1} + \sqrt{5 - x})^2 \]
Expanding the right side:
\[ y^2 = (x - 1) + (5 - x) + 2\sqrt{(x - 1)(5 - x)} \]
This simplifies to:
\[ y^2 = 4 + 2\sqrt{(x - 1)(5 - x)} \]

### Step 4: Isolate the square root

Rearranging gives:
\[ y^2 - 4 = 2\sqrt{(x - 1)(5 - x)} \]
Dividing by 2:
\[ \frac{y^2 - 4}{2} = \sqrt{(x - 1)(5 - x)} \]

### Step 5: Square again to eliminate the square root

Squaring both sides again:
\[ \left(\frac{y^2 - 4}{2}\right)^2 = (x - 1)(5 - x) \]
This expands to:
\[ \frac{(y^2 - 4)^2}{4} = -x^2 + 6x - 5 \]

### Step 6: Find the maximum and minimum values of \( f(x) \)

To find the maximum and minimum values of \( f(x) \), we can evaluate \( f(x) \) at the endpoints of the domain:

1. At \( x = 1 \):
\[ f(1) = \sqrt{1 - 1} + \sqrt{5 - 1} = 0 + 2 = 2 \]

2. At \( x = 5 \):
\[ f(5) = \sqrt{5 - 1} + \sqrt{5 - 5} = 2 + 0 = 2 \]

3. At \( x = 3 \) (the midpoint):
\[ f(3) = \sqrt{3 - 1} + \sqrt{5 - 3} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

### Step 7: Conclusion about the range

From the evaluations:
- The minimum value of \( f(x) \) is \( 2 \).
- The maximum value of \( f(x) \) is \( 2\sqrt{2} \).

Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = [2, 2\sqrt{2}] \]