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The range of the function $f(x) = \large\frac{x^2+x+2}{x^2+x+1}$, $x \in (-\infty, \infty)$ is:

(A) $(0, \frac{4}{3}$$]$ $\quad$ (B) $(1, \frac{7}{3}$$]$$\quad$ (C) $(1, \frac{4}{5}$$] \quad$ (D) $(1, \infty)$

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We can rewrite $\;f(x) = \large\frac{x^2+x+2}{x^2+x+1}$$ = \large\frac{x^2+x+1}{x^2+x+1}$$ + \large\frac{1}{x^2+x+1}$$ = 1 + \large\frac{1}{x^2+x+1}$$ = 1 + g(x)$
The idea is to determine the range for $g(x)$ and then shift that to the left and determine the range for $f(x)$.
Since the numerator of $g(x) \;=\;$ constant, we can study the denominator as $x$ varies from $-\infty$ to $\infty$.
Denominator $ = x^2+x+1 = (x+\large\frac{1}{2})^2$$ + \large\frac{3}{4} $$\neq 0$
It is always positive and it's min value $=\large\frac{3}{4}$, and there is no maximum.
As $x$ ranges from $\large\frac{-1}{2}$ to $\infty$, the denominator increases from $\large\frac{3}{4}$ to $\infty$
$\Rightarrow$ Range for $g(x)$ is $(0, \large\frac{4}{3})$.
Now, since $f(x) = 1 + g(x)$, we can get the range for $f(x)$ by shifting this to the right, i.e, $(1, \large\frac{7}{3})$
answered Jan 6, 2014 by sreemathi.v
edited Mar 7 by sharmaaparna1

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