A function `f:AtoR` is defined by the equation `f(x)=x^(2)-4x+5`
where A=(1,4). What is the range of the function?

Options

A. (2,5)
B. (1,5)
C. [1,5)
D. [1,5]

Explanation

To find the range of the function \( f(x) = x^2 - 4x + 5 \) where \( A = (1, 4) \), we will follow these steps:

### Step 1: Rewrite the Function
We can rewrite the function in a completed square form:
\[ f(x) = x^2 - 4x + 5 = (x - 2)^2 + 1 \]

### Step 2: Analyze the Completed Square
The expression \( (x - 2)^2 \) is always non-negative (i.e., \( (x - 2)^2 \geq 0 \)). Therefore:
\[ f(x) = (x - 2)^2 + 1 \geq 1 \]
This means the minimum value of \( f(x) \) is 1.

### Step 3: Evaluate the Function at the Endpoints of the Interval
Next, we need to evaluate \( f(x) \) at the endpoints of the interval \( A = (1, 4) \).

1. **At \( x = 1 \)**:
\[ f(1) = (1 - 2)^2 + 1 = 1 + 1 = 2 \]

2. **At \( x = 4 \)**:
\[ f(4) = (4 - 2)^2 + 1 = 2^2 + 1 = 4 + 1 = 5 \]

### Step 4: Determine the Range
Since the function is continuous and \( (x - 2)^2 \) takes values from \( 0 \) to \( (4 - 2)^2 = 4 \) as \( x \) varies from \( 1 \) to \( 4 \), the function \( f(x) \) will take values from:
- Minimum value at \( x = 1 \): \( f(1) = 2 \)
- Maximum value at \( x = 4 \): \( f(4) = 5 \)

### Step 5: Finalize the Range
Since the interval \( A \) is open, we do not include the endpoints:
- The minimum value \( 2 \) is included.
- The maximum value \( 5 \) is not included.

Thus, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = [2, 5) \]