Range of the function f(x) = frac{x^2 + x + 2 | vedclass

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(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.L

Vedclass · Accepted Answer

$(1, 7/3]$

Vedclass · Answer

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.Let $g(x) = x^2 + x + 1$. The minimum value of $g(x)$ occurs at $x = -1/2$,which is $g(-1/2) = (-1/2)^2 - 1/2 + 1 = 1/4 - 1/2 + 1 = 3/4$.Since $x^2 + x + 1$ is a parabola opening upwards,its range is $[3/4, \infty)$.Therefore,the range of $\frac{1}{x^2 + x + 1}$ is $(0, 1 / (3/4)] = (0, 4/3]$.Adding $1$ to this range,we get the range of $f(x)$ as $(1 + 0, 1 + 4/3] = (1, 7/3]$.

Range of the function $f(x) = \frac{x^2 + x + 2}{x^2 + x + 1}; x \in R$ is

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