If the domain of the function f(x) = x^2 - 6x | vedclass

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(B) Given the function $f(x) = x^2 - 6x + 7$.We can rewrite the function by completing the square:$f(x) = (x^2 - 6x + 9) - 9 + 7$$f(x) = (x - 3)^2 - 2

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$[-2, \infty)$

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(B) Given the function $f(x) = x^2 - 6x + 7$.We can rewrite the function by completing the square:$f(x) = (x^2 - 6x + 9) - 9 + 7$$f(x) = (x - 3)^2 - 2$Since $(x - 3)^2 \ge 0$ for all real $x$,the minimum value of $(x - 3)^2$ is $0$.Therefore,the minimum value of $f(x)$ is $0 - 2 = -2$.As $x 	o \infty$ or $x 	o -\infty$,$f(x) 	o \infty$.Thus,the range of the function is $[-2, \infty)$.

If the domain of the function $f(x) = x^2 - 6x + 7$ is $(-\infty, \infty)$,then the range of the function is

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