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 can follow these steps:

### Step 1: Rewrite the function in a more manageable form
We can complete the square for the quadratic expression \( f(x) = x^2 - 4x + 5 \).

\[ f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1 \]

### Step 2: Analyze the completed square
The expression \((x - 2)^2\) represents a square term, which is always non-negative. Therefore, the smallest value of \((x - 2)^2\) occurs when \(x = 2\).

\[ (x - 2)^2 \geq 0 \quad \text{for all } x \]

### Step 3: Find the minimum value of \(f(x)\)
Substituting \(x = 2\) into the function gives:

\[ f(2) = (2 - 2)^2 + 1 = 0 + 1 = 1 \]

This means the minimum value of \(f(x)\) on the interval \([2, \infty)\) is \(1\).

### Step 4: Determine the behavior of \(f(x)\) as \(x\) approaches infinity
As \(x\) increases beyond \(2\), the term \((x - 2)^2\) increases without bound. Therefore, \(f(x)\) also increases without bound.

\[ \lim_{x \to \infty} f(x) = \infty \]

### Step 5: Conclude the range of the function
Since the minimum value of \(f(x)\) is \(1\) and \(f(x)\) approaches infinity as \(x\) increases, the range of \(f\) 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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