Let f:[2, infty) rightarrow R be the function | 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}

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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$$f(x) = (x - 2)^{2} + 1$Since the domain is $x \in [2, \infty)$,we have $x - 2 \geq 0$.Therefore,$(x - 2)^{2} \geq 0$.Adding $1$ to both sides,we get $(x - 2)^{2} + 1 \geq 1$.Thus,$f(x) \geq 1$.The range of the function $f$ is $[1, \infty)$.

Let $f:[2, \infty) \rightarrow R$ be the function defined by $f(x)=x^{2}-4x+5$. Then the range of $f$ is:

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