分式函數的定義域和值域

1.
y = (2x + 2) / (5x - 3)
Domain of function=R\{3/5}
Make x the subject: x=(3y+2)/(5y-2).
Range of function=R\{2/5}

2.
y = (2x² - 3x + 3) / (3x² + x - 1)= (2x² - 3x + 3) / [3(x-(-1+√13)/6)(x-(-1-√13)/6)]
Domain of function=R\{(-1+√13)/6,(-1-√13)/6}
y = (2x² - 3x + 3) / (3x² + x - 1)⇒(3y-2)x²+(y+3)x+(y+3)=0
As x is real, Δ=(y+3)²-4(3y-2)(y+3)≥0.
y²+2y-3≤0⇒(y+3)(y-1)≤0⇒-3≤y≤1.
Range of function={y∈R:-3≤y≤1}

3.
y = (x + 1) / (2x³ - x² - 8x + 4) = (x + 1) / [(2x-1)(x-2)(x+2)]
Domain of function=R\{1/2, 2, -2}
When x=-1, y=0. ∴0∈range of function.
For any y∈R\{0}, rewrite the function into 2x³ - x²-(8+1/y)x+(4-1/y)=0....(*)
(*) is a cubic equation with real coefficients, there must be a real solution.
Thus y∈range of function.
In conclusion, range of function=R.

2012-08-21 11:37:11 補充：
多謝指正。現改正值域如下：

2.(值域部份)
y = (2x² - 3x + 3) / (3x² + x - 1)⇒(3y-2)x²+(y+3)x-(y+3)=0
As x is real, Δ=(y+3)²+4(3y-2)(y+3)≥0.
13y²+34y-15≥0⇒(13y-5)(y+3)≥0⇒y≤-3 or y≥5/13.
Range of function={y∈R：y≤-3 or y≥5/13}