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!

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 $f(-2)<\lim_\limits{x\to-\infty}f(x) < f(1)$
$[f(-2), f(1)]$
