Answer to Question #256758 in Algebra for Artemis

Question

Find the domain and range of the following rational function. Use any notation. 1) f(x) = 3 / x-1 2) f(x) = 2x / x-4 3) f(x) = x+3 / 5x-5 4) f(x) = 2+x / 2x 5) f(x) = (x^2+4x+3) / x^2-9

Answer 1

$1)$ $When\, (x-1) =0\,,x=1\,f(x)$

Is not defined

Domain is: x<1 or x>1

$If \,y=\frac{3}{x-1}$

Then $\,x=\frac{3+y}{y}$ , y cannot be zero

Range is: f(x)<0 or f(x)>0

2) For x-4=0, x=4, and f(x) cannot be defined

Domain is: x<4 or x>4

Let $y= \frac{2x}{x-4}\>\implies x=\frac{4y}{y-2}$

y-2, cannot be zero, Therefore y cannot be 2

Range is: f(x)<2 or f(x)>2

3) For, 5x-5=0, implying that x=1, f(x) cannot be defined

Domain is: x<1 or x>1

Let $y=\frac{x+3}{5x-5}\; \implies x=\frac{5y+3}{5y-1}$

(5y-1), cannot be zero, Therefore, y, cannot be $\frac{1}{5}$

Range is: $f(x)<\frac{1}{5} \>or\>f(x)>\frac{1}{5}$

4) 2x cannot be zero

Domain is: x<0 or x>0

Let $y=\frac{2+x)}{2x}\> \implies x=\frac{2}{2y-1}\>$

(2y-1), cannot be zero, therefore y cannot be ½

Range is: f(x)<½ or f(x)>½

5) For x²-9=0, x=3 or -3 and f(x) cannot be defined at those points.

Domain is: x<-3 or -3<x<3 or x>3

Let $y=\frac{x^2+4x+3}{x²-9}$

$x=\frac{-4^+_-\sqrt{\smash[b]{16-4(1-y)(3+9y)}}}{2(1-y)}$

(1-y) cannot be zero, therefore y cannot be 1

Discriminant cannot be negative, therefore y cannot be greater than ⅓

Range is: f(x)<⅓ or ⅓<f(x)<1 or f(x)>1