Step 1: Analyze the function for injectivity.

A function \(f(x)\) is injective (one-to-one) if for every \(x_1, x_2\) in the domain, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Equivalently, if \(x_1 \neq x_2\), then \(f(x_1) \neq f(x_2)\).

Consider \(f(x) = x^2 - 4x + 5\). We can rewrite this as \(f(x) = (x-2)^2 + 1\). This is a parabola with vertex at \((2, 1)\) opening upwards.

For example, \(f(1) = (1-2)^2 + 1 = 2\) and \(f(3) = (3-2)^2 + 1 = 2\). Since \(f(1) = f(3)\) but \(1 \neq 3\), the function is not injective.

Step 2: Analyze the function for surjectivity.

A function \(f(x)\) is surjective (onto) if for every \(y\) in the codomain, there exists an \(x\) in the domain such that \(f(x) = y\). In other words, the range of the function is equal to the codomain.

Since \(f(x) = (x-2)^2 + 1\), the minimum value of \(f(x)\) is 1 (when \(x = 2\)). Therefore, the range of \(f(x)\) is \([1, \infty)\).

The codomain is \(\mathbb{R}\), which is \((-\infty, \infty)\). Since the range \([1, \infty)\) is not equal to the codomain \((-\infty, \infty)\), the function is not surjective.