If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an alternative what I did is
$y = \frac{x^2 + 2x + 4}{x^2 - 2x + 4}$
Then$ \frac{1+y}{1-y} =\frac{x^2+4}{2x} $
Therefore$ \frac{1+y}{1-y} = {x/2+2/x}$
If we go further putting rhs range to lhs the answer gets altered . Is this method wrong if so why if not where did the process go wrong.

HINT
I would recommend you to notice that
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{(x^{2} + 2x + 4) - 4x}{x^{2} + 2x + 4}\\\\
& = 1 - \frac{4x}{x^{2} + 2x + 4}
\end{align*}
Consequently, the problem reduces to study the critical points of the last expression.
In order to do so, let us take its derivative:
\begin{align*}
f(x) = 1 - \frac{4x}{x^{2} + 2x + 4} & \Rightarrow f'(x) = -\frac{4(x^{2} + 2x + 4) - 4x(2x + 2)}{(x^{2} + 2x + 4)^{2}}\\\\
& \Rightarrow f'(x) = \frac{4x^{2} - 16}{(x^{2} + 2x + 4)^{2}}
\end{align*}
Based on its expression, we conclude the critical points are given by $x = -2$ and $x = 2$.
Can you take it from here?
EDIT
Let us approach it by parts. Firstly, let us set that $y = x + \frac{4}{x}$, where $x > 0$.
Then one concludes that $y\geq 4$ based on AM-GM inequality, for example.
Hence we obtain the lower bound according to:
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{x - 2 + \frac{4}{x}}{x + 2 + \frac{4}{x}} = \frac{y - 2}{y + 2} = 1 - \frac{4}{y + 2} \geq 1 - \frac{2}{3} = \frac{1}{3}
\end{align*}
Now we can suppose that $x < 0$. If we set the same change of variable $y = x + \frac{4}{x}$, one gets $y\leq -4$.
Consequently, we obtain the upper bound as follows:
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{x - 2 + \frac{4}{x}}{x + 2 + \frac{4}{x}} = \frac{y - 2}{y + 2} = 1 - \frac{4}{y + 2} \leq 1 + 2 = 3
\end{align*}
and we are done.
Hopefully this helps!
