Function f:(2, infty) rightarrow R defined by | vedclass

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Question

(C) 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

Vedclass · Accepted Answer

$(1, \infty)$

Vedclass · Answer

(C) 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 $x \in (2, \infty)$,we have $x > 2$.Subtracting $2$ from both sides,we get $x - 2 > 0$.Squaring both sides,we get $(x - 2)^2 > 0$.Adding $1$ to both sides,we get $(x - 2)^2 + 1 > 1$.Therefore,$f(x) > 1$.Thus,the range of the function $f$ is $(1, \infty)$.

Function $f:(2, \infty) \rightarrow R$ defined by $f(x) = x^2 - 4x + 5$. The range of $f$ is $=$ . . . . . . .

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