A function is injective if for every \(x_1, x_2\) in the domain, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Let's check if this holds for \(f(x) = x^2 - 4x + 5\). Suppose \(f(x_1) = f(x_2)\). Then, $$x_1^2 - 4x_1 + 5 = x_2^2 - 4x_2 + 5$$ $$x_1^2 - x_2^2 - 4x_1 + 4x_2 = 0$$ $$(x_1 - x_2)(x_1 + x_2) - 4(x_1 - x_2) = 0$$ $$(x_1 - x_2)(x_1 + x_2 - 4) = 0$$ This implies either \(x_1 = x_2\) or \(x_1 + x_2 = 4\). Since \(x_1 + x_2 = 4\) does not necessarily mean \(x_1 = x_2\), the function is not injective. For example, \(f(1) = 1 - 4 + 5 = 2\) and \(f(3) = 9 - 12 + 5 = 2\), so \(f(1) = f(3)\) but \(1 \neq 3\).
A function is surjective 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 the codomain is \(\mathbb{R}\), we need to check if the range of \(f(x)\) is \(\mathbb{R}\). We can rewrite \(f(x)\) by completing the square: $$f(x) = x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$$ Since \((x - 2)^2 \geq 0\) for all \(x \in \mathbb{R}\), we have \(f(x) = (x - 2)^2 + 1 \geq 1\). Therefore, the range of \(f(x)\) is \([1, \infty)\). Since the range is not equal to the codomain \(\mathbb{R}\), the function is not surjective.
The function is neither injective nor surjective.
Final Answer: neither injective nor surjective.<\/span>
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