How do you derive this easy to find the max/min points (There aren't actually any stationary points)
$$ \dfrac {-24 x^2 -88 x -18} {16 x^2 +64 x +16} $$
I know how to use the quotient rule, but I think this has to be simplified at first, else it would be an extremely long approach.
Many thanks!
Rewrite it as $-\frac32+\frac{4x+3}{8x^2+32x+8}$, then use the quotient rule.
For the derivative of $\frac{p(x)}{q(x)}$ to be zero, you need
$$\frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}=0\\p'(x)q(x)=p(x)q'(x)$$
so you don't need to write down the denominator.
Unless there is a common factor between the numerator and denominator, which there isn't here, there isn't much you can do. Just plug ahead with the quotient rule.
By Quotient Rule
$$ \dfrac{p(x)}{q(x)} = \dfrac{p^{'}(x)}{q^{'}(x)} $$
It is so convenient to put numerator, denominator into four compartments raster with = sign in between.
Remove common factors from numerator and denominator at the first opportunity to reduce needless calculation with bigger numbers later.
$$ \dfrac{-12 x^2 -44 x -9}{(x^2 + 4 x + 4)} = \dfrac{-24 x -44 }{2 x + 4}= \dfrac{-12 x -22 }{ x + 2 } $$ (was in typo error)
Now easier to cross multiply, collect terms and consolidate into a quadratic equation.
EDIT:
after correction:
$$ \dfrac{12 x^2 + 44 x + 9}{(x^2 + 4 x + 1)} = \dfrac{24 x + 44 }{2 x + 4}= \dfrac{12 x +22 }{ x + 2 }$$
Cross multiplying,
$$ 12 x^3 + 68 x^2 + 97 x + 18 = 12 x^3 + 70 x^2 + 100 x +22, $$
Simplifies to quadratic equation $ 2 x^2 + 3 x + 4 = 0 $
Has no real roots as the discriminant = $ 9 - 4 \cdot 2 \cdot 4 < 0 $
Zeroes are $ @ \pm \infty $ with horizontal asymptotes and with no extrema.
Apologies for using this space for comment replies.
@Michael:There is no confusion between any two things. Did the OP talk about difficulties in evaluating an indeterminate quotient at any given $x=x_1$ point or is he asking about finding extremum value of a quotient at an unevaluated or unknown point $x=x_1 $ yet to be determined? The former calls for L'Hospital's Rule and the latter calls for an application of the Quotient Rule.
If at a known point $ x=x_1, F(x_1) = \dfrac {f(x)}{g(x)} $ takes the form $ 0/0, or, \infty/\infty $, then L’Hospital’s Rule can be applied to evaluate the indeterminate function of x with one , two or more separate differentiations of the numerator and denominator depending on the degree of indeterminacy :
$$ \dfrac{f(x)}{g(x)} = \dfrac {f’(x)}{g’(x)} = \dfrac {f’’(x)}{g’’(x)} etc. $$
On the other hand to find extremum point ($x=x_1)$ and corresponding ($ y=y_1 $) without reference to c or involving c, we can use Quotient Rule $ F(x) = \dfrac {f(x)}{g(x)}= const. = c, then \; each = \dfrac {f’(x)}{g’(x)} @ x=x_1.$ OP wants this only, I replied to this specifically.
There is similarity about the operation or procedure (viz., proceeding with individual differentiation of numerator and denominator) in either situation. However, the objectives of the two operations are entirely different.
As an often quoted example:
$ \dfrac{\sin x}{x}$ is determinate at all x except at $x=0$. It becomes determined ( as unity) using L'Hospital's Rule.
Next, to find all extremum values $x_1$ of this function just as OP has asked, one solves the equation:
$$ \dfrac{\sin (x_1)} {x_1} = \dfrac {\cos( x_1)}{1}, or , tan(x_1) = x_1. $$
Hope it resolved the confusion.