Find the highest and lowest possible value of function $f(x)={3x^2+6x+6\over x^2+4x+5}$ 
Find the highest and lowest value of the next function without the use of derivatives
$$f(x)={3x^2+6x+6\over x^2+4x+5}$$

Okay so my teacher gave me this problem, and told me to strictly solve it without using derivatives. Obviously with using derivatives it's pretty easy, just find where $f'(x_0)=0$, and check it's second derivative in $x_0$ to see if it's a maximum or minimum.
So I was thinking about this:
Since this function is written as a fraction, the lowest possible value should be when the numerator value is the lowest, while the denominator is the highest possible value.
BUT, again to find the highest and lowest value of both I could use derivatives again, which again isn't the point.
I can see that both the numerator and denominator of this fraction is always positive for every $x$, so maybe there's a hint that I can't see there?
Maybe I could use limits too, but I'm not sure how.
 A: If $\kappa$ belongs to the range of $f$, it follows that
$$3x^2+6x+6 = \kappa(x^2+4x+5)$$
has a real solution, hence the discriminant of
$$ (3-\kappa)x^2+(6-4\kappa)x+(6-5\kappa) $$
i.e. $-4\kappa^2+36\kappa-36$, is non-negative. It follows that the extreme values of $f$ are given by the roots of $-4\kappa^2+36\kappa-36$, i.e. $\color{red}{\frac{3}{2}\left(3\pm\sqrt{5}\right)}$.
I believe this (exploiting discriminants) was the ancient method for the determination of the stationary values of rational functions, before the Method of Fluxions. But relics have their elegance!
A: 
One can cast this issue into an eigenvalue problem. 
I recognize that, evidently, this involves much more computation than the straightforward method given by @Jack D'Aurizio, but see remarks at the bottom.
Moreover, I understand that probably, you, the OP, haven't already seen these notions, but, who knows... 
Let us define $A=\begin{pmatrix}3&3\\3&6\end{pmatrix}$ and $B=\begin{pmatrix}1&2\\2&5\end{pmatrix}$. 
Let $B=CC^T$ be the Cholesky factorization of $B$, with 
$$C= \begin{pmatrix}1&2\\0&1\end{pmatrix}.$$ 
Let $D:=C^{-1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}.$
Then the extreme values are the eigenvalues $\frac32(3\pm \sqrt{5})$ of 
$$M:=D^TAD=\begin{pmatrix} 3&-3\\
    -3&6\end{pmatrix}.$$
Explanation : setting $x=\frac{s}{t}$ in the given fraction transforms it into the quotient of 2 quadratic forms :
$$\frac{3x^2+6x+6}{x^2+4x+5}=\frac{3s^2+6st+6t^2}{s^2+4st+5t^2}=\frac{X^TAX}{X^TBX} \ \  with \ X:=\binom{y}{t}\tag{1}$$
(with matrices $A$ and $B$ defined upwards). 
Then transform (1) into the following Rayleigh quotient : 
$$\frac{(CX)^T(D^TAD)(CX)}{(CX)^T(CX)} = \frac{U^T M U}{U^TU}$$ 
which is known to take all its values in  $[\lambda_{min},\lambda_{max}]$.
Remark : A natural question is : what is the information brought by eigenvectors $U_1=\binom{s}{t}=\binom{-0.8507}{-0.5257}$ and $U_2=\binom{s}{t}=\binom{-0.5257}{0.8507}$. Using 
$X_1=C^{-1}U_1=\binom{0.2008}{-0.5257}$ and $X_2=C^{-1}U_2=\binom{ -2.2270}{0.8507}$, the reader will easily see that they provide information about the abscissas $x=\frac{s'}{t'}=\frac{0.2008}{-0.5257}=-0.3820$ and $x=\frac{-2.2270}{0.8507}=-2.6178$ of points where the minimum and the maximum occur.
Remarks : The advantage of this method lies on the methodological side :


*

*most people working in applied mathematics will appreciate to have a method handling eigenvalues instead of specific numbers for which one has to rediscover properties, like sensibility to variations.

*it can be extended to as many variables as desired, like finding the min and max values of $$\frac{3x^2+5y^2+6z^2-2xz+3yz}{x^2+2y^2+3z^2-2xz}.$$ (for any $x,y,z$).
