If f:[2, infty) rightarrow R is defined by f( | vedclass

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(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.We can rewrite the function by completing the square:$f(x) = (x^2 - 4

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$[1, \infty)$

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(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.We can rewrite the function by completing the square:$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.Since the domain is $x \in [2, \infty)$, the minimum value of $(x - 2)^2$ is $0$ (at $x = 2$).As $x \rightarrow \infty$, $(x - 2)^2 \rightarrow \infty$.Therefore, the range of $f(x)$ is $[0 + 1, \infty) = [1, \infty)$.

If $f:[2, \infty) \rightarrow R$ is defined by $f(x)=x^2-4x+5$, then the range of $f$ is

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