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(D) Let $y = \frac{x^2+x+1}{x}$.$yx = x^2 + x + 1$$x^2 + (1-y)x + 1 = 0$.Since $f(x)$ is a real-valued function,$x$ must be a real number.Therefore,th

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$(-\infty, -1] \cup [3, \infty)$

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(D) Let $y = \frac{x^2+x+1}{x}$.$yx = x^2 + x + 1$$x^2 + (1-y)x + 1 = 0$.Since $f(x)$ is a real-valued function,$x$ must be a real number.Therefore,the discriminant $D \geq 0$.$(1-y)^2 - 4(1)(1) \geq 0$$1 + y^2 - 2y - 4 \geq 0$$y^2 - 2y - 3 \geq 0$$(y-3)(y+1) \geq 0$.Solving the inequality,we get $y \in (-\infty, -1] \cup [3, \infty)$.Thus,the range of $f(x)$ is $(-\infty, -1] \cup [3, \infty)$.

The range of the real valued function $f(x) = \frac{x^2+x+1}{x}$ is

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