As x varies over all real numbers, the range of the function $f(x)=\frac{x^2-3x+4}{x^2+3x+4}$ is (1) $[\frac{1}{7},7]$, (2) $[-\frac{1}{7},7]$, (3) $[-7,7]$ (4) $(-\infty,\frac{1}{7})\bigcup(7,\infty)$.
Trial:$\begin{align} \frac{x^2-3x+4}{x^2+3x+4} &=\frac{(x-\frac{3}{2})^2+\frac{7}{4}}{(x+\frac{3}{2})^2+\frac{7}{4}} >0 \end{align}$ So,(1) will be the right answer. But how I show that $\frac{1}{7} \leq f(x) \leq 7$. Please help.
Note: Originally it was stated that
$f(x)=\frac{x^2-3x+1}{x^2+3x+1},$
and this is what the initial answers were based on.