Question

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

• A. $(1, \infty)$
• B. $(1, \frac{11}{7}]$
• C. $(1, \frac{7}{3}]$
• D. $(1, \frac{7}{5}]$

Answer 1

Answer: (C)

$(1, \frac{7}{3}]$

Answer 2

The function $f(x) = \frac{x^2+x+2}{x^2+x+1}$ can be rewritten as $1 + \frac{1}{x^2+x+1}$.

Let $t = x^2+x+1$. The minimum value of this quadratic expression occurs at $x=-1/2$, which is $t_{min} = (-1/2)^2+(-1/2)+1 = 1/4-1/2+1 = 3/4$. So, $t \in [3/4, \infty)$.

Now, substitute this range into $y = 1 + \frac{1}{t}$.

As $t$ ranges from $3/4$ to $\infty$, $\frac{1}{t}$ ranges from $\frac{1}{3/4} = \frac{4}{3}$ down to $0$ (exclusive). So, $\frac{1}{t} \in (0, \frac{4}{3}]$.

Therefore, $y = 1 + \frac{1}{t} \in (1+0, 1+\frac{4}{3}] = (1, \frac{7}{3}]$.