Solution 1: separation constant method;

f(x)=(2x)/(3x- 12)=(2x)/[3(x-4)]=(2/3)[x/(x-4)]=(2/3)[(x-4+4)/(x-4)]

=(2/3)[ 1+4/(x-4)]=(2/3)+8/(3x- 12)

When x→4 and f(x) →∞, so f(x) has a vertical asymptote x = 4；; When x →∞, f(x)→2/3, so f(x) has a horizontal gradient.

Nearline y = 3/2; So its range is (-∞, 2/3)∩(2/3,+∞).

Scheme 2: Inverse Function Method:

y=(2x)/(3x- 12)，(3x- 12)y=2x，(3y-2)x= 12y，x= 12y/(3y-2)，

Exchange the inverse function y= 12x/(3x-2) of x and y, and its domain x≠2/3 is the range of positive function: y≠2/3.

2。 Find the range of y = (1/4) x-2 (1-x)-3.

Solution: y = (1/2) (2x)-2 (1/2) x-3, let (1/2) x = u, then y=u? -2u-3=(u- 1)？ -4≦-4, that is,-4 ≦ y.

When u= 1, that is, x=0, y gets the minimum value of -4.

3。 f[x-( 1/x)]=x？ -( 1/x？ ) +9, find f(x)

Solution: let x-( 1/x)=u, and square both sides to get x? -2+( 1/x？ )=u？ , so x? +( 1/x？ )=u？ +2;

So (x+ 1/x)? =x？ +2+( 1/x？ )=u？ +4，x+( 1/x)= √(u？ +4);

So f [x-( 1/x)]=x? -( 1/x？ )+9 =(x+ 1/x)(x- 1/x)+9 = u √( u？ +4)+9

That is, f (u) = u √ (u? +4)+9;

Replace u with x to get f(x)=x√(x? +4)+9; Or f(x)=-x√(x? +4)+9.