If `f : [2,oo) to R` be the function defined by `f(x)=x^(2)-4x+5,` then the range of f is

Options

A. R
B. `[1,oo)`
C. `[4,oo)`
D. `[5,oo)`

Explanation

To find the range of the function \( f(x) = x^2 - 4x + 5 \) defined on the interval \( [2, \infty) \), we will follow these steps:

### Step 1: Rewrite the function in a standard form
We start with the function:
\[ f(x) = x^2 - 4x + 5 \]
To analyze this function, we can complete the square.

### Step 2: Complete the square
We can rewrite the quadratic expression:
\[ f(x) = (x^2 - 4x) + 5 \]
Now, we complete the square for the expression \( x^2 - 4x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Thus, we have:
\[ f(x) = (x - 2)^2 - 4 + 5 = (x - 2)^2 + 1 \]

### Step 3: Determine the minimum value of \( f(x) \)
Since \( (x - 2)^2 \) is a perfect square, it is always non-negative. The minimum value occurs when \( (x - 2)^2 = 0 \), which happens at \( x = 2 \):
\[ f(2) = (2 - 2)^2 + 1 = 0 + 1 = 1 \]

### Step 4: Analyze the behavior of \( f(x) \) as \( x \) increases
As \( x \) increases beyond 2, \( (x - 2)^2 \) increases, and thus \( f(x) \) will also increase. Therefore, \( f(x) \) can take any value greater than or equal to 1.

### Step 5: State the range of \( f(x) \)
Since the function \( f(x) \) is defined for \( x \in [2, \infty) \) and the minimum value of \( f(x) \) is 1, the range of \( f(x) \) is:
\[ \text{Range of } f = [1, \infty) \]

### Final Answer:
The range of the function \( f(x) = x^2 - 4x + 5 \) for \( x \in [2, \infty) \) is \( [1, \infty) \).

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