Step-by-step Solution:

Step 1: Rewrite the Function We start by completing the square for the quadratic function: \[ f(x) = x^2 - 4x + 5 \] \[ = x^2 - 4x + 4 + 1 \] \[ = (x - 2)^2 + 1 \] Step 2: Analyze the Completed Square The expression \( (x - 2)^2 \) is always non-negative, i.e., \( (x - 2)^2 \geq 0 \). Therefore: \[ f(x) = (x - 2)^2 + 1 \geq 1 \] Step 3: Determine the Range Since \( (x - 2)^2 \) can take any value from 0 to \( \infty \), the minimum value of \( f(x) \) is 1, and it can increase without bound. Thus, the range of \( f(x) \) is: \[ [1, \infty) \] Step 4: Conclusion The range of the function \( f(x) = x^2 - 4x + 5 \) is: \[ \boxed{[1, \infty)} \] Correct Answer: \(\boxed{B}\)