Range of a function $f(x)$ 
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.
 A: Let $\dfrac{x^2-3x+1}{x^2+3x+1} = y$. We then have $$(y-1)x^2 + 3(y+1)x + (y-1) = 0$$
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$9(y+1)^2 - 4 (y-1)^2 \geq 0 \implies 5y^2 +26y+5 \geq 0 \implies (5y+1)(y+5) \geq 0$$
This gives us $y \geq -\dfrac15$ and $y \leq -5$. Hence, the right answer is
$$\left(-\infty,-5\right] \cup \left[ -\dfrac15, \infty \right)$$
EDIT
If the constant term is $4$ i.e. if $\dfrac{x^2-3x+4}{x^2+3x+4} = y$. We then have $$(y-1)x^2 + 3(y+1)x + 4(y-1) = 0$$
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$9(y+1)^2 - 16 (y-1)^2 \geq 0 \implies -7y^2 +50y-7 \geq 0 \implies 7y^2 - 50y +7 \leq 0 \implies (7y-1)(y-7) \leq 0$$
This gives us $y \in \left[\dfrac17,7 \right]$. Hence, the right answer is
$$\left[\dfrac17,7\right]$$
A: Answer to modified question: Note that the line $y=1$ is an asymptote to the curve $y=f(x)$.
Calculate $f'(x)$.  It turns out to be $\frac{6x^2-24}{(x^2+3x+4)^2}$. So $f(x)$ is increasing from $-\infty$ to $-2$, then decreasing until $2$, then increasing. 
For negative $x$, the curve is above the asymptote, and for positive $x$ it is below the asymptote. 
The maximum value (at $x=-2$) is $7$, and the minimum value (at $x=2$) is $\frac{1}{7}$. So the range is $[1/7,7]$. 
