How to find the range of the given function? 
Find the range of $$f(x)=\dfrac{x^2+14x+9}{x^2+2x+3}$$ where $x\in \mathbb R$ 

I thought of finding derivative but this will get too complicated so i am completely blank.
Thanks in advance!
 A: I think you'd have to find the derivative for this problem. But don't worry, finding the derivative is not as cumbersome as you think. Just use the quotient rule and you'll get something reasonable.
You should get $-\frac{12(x^2+x-2)}{(x^2+2x+3)^2}$ and setting that equal to 0 and solving for $x$ should be straightforward.
A: Its not that hard to differentiate.
$\frac {d}{dx} \dfrac {x^2 + 14x + 9}{x^2 + 2x + 3} = 0\\
\dfrac {(x^2 + 2x + 3)(2x + 14) - (2x + 2)(x^2 + 14x + 9)}{(x^2 + 2x + 3)^2} = 0$
Since the denominator is clearly not $0.$  We can multiply through by the denominator.
$2x^3 + 4x^2 + 6x + 14x^2 + 28x + 42 - 2x^3 - 2x^2 - 28x^2 - 28x -18x - 18 = 0\\-12 x^2 -12x +24=0\\
-12 (x+2)(x-1) = 0$
And that gives you your local minimum and maximum.
Next show that these are absolute max and min, too.
$ f(-2) <\lim_\limits{x\to\infty} f(x) <f(1)$ and $f(-2)<\lim_\limits{x\to-\infty}f(x) < f(1)$
$[f(-2), f(1)]$
A: Just some general notes on finding the range for a rational function:
First of all a rational function is continuous where it's defined so if $y_1$ and $y_2$ is in the range then so is any value inbetween $y_1$ and $y_2$ (the intermediate value theorem). This is a key fact needed to determine the range.
First we check that the denominator is not zero. If the denominator has a zero at $x=x_*$ then $\lim_{x\to x_*}|f(x)| = \infty$ so the range will extend to $\infty$ or $-\infty$ (or both). You can determine this by computing the one-sided limits $\lim_{x\to x_*^{\pm}} f(x)$. For your function we have $x^2+2x+3 = (x+1)^2+2 \geq 2$ so this is not the case here.
Next we determine the asymptotic values by computing $\lim_{x\to\pm\infty} f(x)$. Again if one or both of these limits are $\pm\infty$ then the range will exend out to these values. For your case these limits are $1$ and togeather with the previous information we can conclude that the function is bounded.
In after any of the steps above if the range has beens shown to be $\mathbb{R}$ then we can stop. But if this is not the case then there is one more step needed. We need to determine the local max/min values of the function. This is done by solving $f'(x) = 0$ and computing $f(x)$ using the values found. For your function we find the extremal points $x=-2$ and $x=1$ and $f(-2) = -5$ and $f(1) = 4$. The point $x=-2$ is a global minimum and $x=1$ is a global maximum.
Combinding all the information found by going through the steps above you should be able to determine the range quite easily. For your function it's not hard to see that the range is $[-5,4]$. For this last step it's often quite useful to draw a sketch based on the information previously found. For example I drew up this rough sketch:

A: Since $\lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(x) = 1$, the maximum value of $f$'s range is the maximum value of $f$, and similarly for the minimum value (unless of course, it were asymptotic -- think exponentials).
Do you know how to find the extrema of functions?
A: A simpler (pre-calculus) solution. Let we consider the equation $f(x)=k$:
$$\frac{x^2+14x+9}{x^2+2x+3}=k\tag{1}$$
that is equivalent to:
$$ (1-k) x^2 +(14-2k) x + (9-3k) = 0 \tag{2} $$
whose discriminant equals:
$$ \Delta = (14-2k)^2-4(1-k)(9-3k) = 8(k+5)(k-4).\tag{3}$$
Such discriminant vanishes at the boundary of $\text{range}(f)$, hence:

$$\color{red}{ \text{range}(f) = [-5,4]}.\tag{4}$$

The endpoints are attained at $x=-2$ and $x=1$ by solving $(1)$ for $k=-5$ and $k=4$.
A: $$(y-1)x^2+(2y-14)x+(3y-9)=0$$
$$\Delta=(2y-14)^2-4(y-1)(3y-9)=4(-2y^2-2y+40)$$
then
$$x=\frac{-(2y-14)\pm  2\sqrt{-2y^2-2y+40} }{2(y-1)}=\frac{7-y\pm  \sqrt{-2y^2-2y+40} }{y-1}$$
$$R_f=[-5,4]$$
