Range of a rational algebraic function How do we find the range of an algebraic rational function of the form $(a_1x^2+b_1x+c_1)/(a_2x^2+b_2x+c_2)$ for real values of x?
For instance, take the function f(x) = $(16x^2-12x+9)/(16x^2+12x+9)$ .
Now, one thing which I was able to get was that since both the polynomials have their discriminant $D < 0$, so they must always be positive and since both the polynomials are always positive, the expression must be always greater than $0$. However, I don't know how to go from here. How would I go about doing this?
 A: Let $ y = \frac{ 16x^2 - 12 x + 9 } { 16x^2 + 12x + 9 } $.
Since the denominator is non zero, we can multiply by the denominator and shift terms to get
$$ x^2 ( 16y - 16) + x ( 12y + 12 ) + (9y- 9 ) = 0 $$
This is a quadratic equation with real roots, which means that the discriminant is non-negative. Hence,
$$ 0 \leq (12y+12)^2 - (16y-16)(9y-9) = -144 (3-y)(1-3y), $$
so the range is contained within $ [ \frac{1}{3} , 3 ] $.
Note that for this to truly be a quadratic, we require $ 16y - 16 \neq 0$, or that $ y \neq 1$. In this case, we're solving the linear equation $24x = 0 $, which has a solution $x = 0$, so $y=1$ is in the range. Henceforth, let's ignore $ y = 1$ as an edge case.
Conversely, for any $ y \in [ \frac{1}{3} , 3 ] $, we can solve the corresponding quadratic equation since the discriminant is non-negative, to obtain a real $x$. Hence, that $y$ value is in the range.
Hence, the range is exactly $[ \frac{1}{3} , 3 ] $.

Note: This works for rational functions where the denominator is a quadratic with no real roots.
It doesn't easily extend to higher degree polynomials.
A: So you want to know a solution involving calculus, here it is-
We have, $f(x)=\displaystyle\frac{16x^2-12x+9}{16x^2+12x+9}$
Clearly, $\displaystyle\lim_{x \to \infty}f(x)=lim_{x \to -\infty}f(x)=1$ $(1)$
Now $\displaystyle f(x)=1-\frac{24x}{16x^2+12x+9}$
Let denominator be equal to $u$
$\displaystyle\frac{d}{dx}f(x)=-\frac{(16x^2+12x+9)24-24x(32x+12)}{u^2}$
$\displaystyle\frac{d}{dx}f(x)=0$ implies
$16x^2+12x+9-32x^2-12x=0$
$16x^2-9=0$ therefore $f(x)$ has it extremums at $x=\pm \frac{3}{4}$
Now $\displaystyle f(x)=1-\frac{24x}{(4x+3)^2-12x}$
Therefore value of $f(x)$ at extremums $f(0.75)=\frac{1}{3}$ and  $f(-0.75)=3$
Therefore by $(1)$ we can say that these are global maximum and minimum values of $f(x)$ and hence range of $f(x)$ is $\displaystyle \big[\frac{1}{3},3\big]$
