Question

Let f : [2, ∞) → R be the function defined by f(x) = x² - 4x + 5 , then the range of f is

• A. (-∞, ∞)
• B. [1, ∞)
• C. (1, ∞)
• D. [5, ∞)

Answer 1

The Correct Option is B

Step 1: Complete the square to rewrite the quadratic function. \[ f(x) = x^2 - 4x + 5 \] Complete the square: \[ f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1 \] 

Step 2: Analyze the function's behavior. 

- The function \( f(x) = (x - 2)^2 + 1 \) is a parabola that opens upwards (since the coefficient of \( (x - 2)^2 \) is positive). - The minimum value occurs at \( x = 2 \), since the vertex of the parabola is at \( x = 2 \). At \( x = 2 \): \[ f(2) = (2 - 2)^2 + 1 = 0 + 1 = 1 \] Step 3: Determine the range. 

Since the parabola opens upwards and the minimum value at \( x = 2 \) is 1, the range of \( f(x) \) is \( [1, \infty) \). 

The range of \( f(x) \) is [1, ∞), so the correct option is (B).

Answer 2

Let \(f: [2, \infty) \rightarrow R\) be the function defined by \(f(x) = x^2 - 4x + 5\). We need to find the range of f.

First, complete the square for the quadratic function:

\(f(x) = x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1\)

Since the domain is \([2, \infty)\), let's find the minimum value of \(f(x)\) in this interval. The vertex of the parabola \(y = (x-2)^2 + 1\) is at x = 2. Since the coefficient of \(x^2\) is positive, the parabola opens upwards, so x = 2 corresponds to a minimum.

When \(x = 2\), \(f(2) = (2 - 2)^2 + 1 = 0^2 + 1 = 1\).

As x increases from 2 to infinity, \((x-2)^2\) also increases from 0 to infinity. Therefore, \(f(x) = (x-2)^2 + 1\) increases from 1 to infinity.

Thus, the range of f is \([1, \infty)\).

Therefore, the correct option is (B) \([1, \infty)\).