Question

If a function \( f : [2, \infty) \to \mathbb{R \) is defined by} \[ f(x) = x^2 - 4x + 5 \] then the range of \( f \) is:

• A. \( [2, \infty) \)
• B. \( [5, \infty) \)
• C. \( \mathbb{R} \)
• D. \( [1, \infty) \)

Hint:

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex occurs at \( x = \frac{-b}{2a} \), and the range depends on whether the parabola opens upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)).

Answer 1

The Correct Option is D

Step 1: Analyze the given function.
The function is a quadratic function of the form:
\[ f(x) = x^2 - 4x + 5 \] We know that the graph of a quadratic function is a parabola, and its range depends on the vertex.

Step 2: Find the vertex of the parabola.
The vertex of a quadratic function \( ax^2 + bx + c \) is given by:
\[ x = \frac{-b}{2a} \] For the given function \( f(x) = x^2 - 4x + 5 \), we have \( a = 1 \), \( b = -4 \), and \( c = 5 \). The x-coordinate of the vertex is:
\[ x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2 \]

Step 3: Calculate the value of \( f(x) \) at the vertex.
Substitute \( x = 2 \) into the function:
\[ f(2) = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1 \] Thus, the vertex is at \( (2, 1) \), and since the parabola opens upwards (because \( a > 0 \)), the minimum value of \( f(x) \) is 1.

Step 4: Determine the range of the function.
Since the parabola opens upwards and the minimum value of \( f(x) \) occurs at \( x = 2 \), the range of \( f \) is all values greater than or equal to 1. Therefore, the range of \( f \) is \( [1, \infty) \).

Step 5: Conclusion.
Thus, the range of \( f \) is \( [1, \infty) \), which corresponds to option (D).