Answer to Question #256758 in Algebra for Artemis

Question #256758
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
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Expert's answer
2021-10-28T14:21:19-0400

1)1) When(x1)=0,x=1f(x)When\, (x-1) =0\,,x=1\,f(x)

Is not defined


Domain is: x<1 or x>1


Ify=3x1If \,y=\frac{3}{x-1}


Then x=3+yy\,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=2xx4     x=4yy2y= \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=x+35x5      x=5y+35y1y=\frac{x+3}{5x-5}\; \implies x=\frac{5y+3}{5y-1}


(5y-1), cannot be zero, Therefore, y, cannot be 15\frac{1}{5}


Range is: f(x)<15orf(x)>15f(x)<\frac{1}{5} \>or\>f(x)>\frac{1}{5}


4) 2x cannot be zero


Domain is: x<0 or x>0


Let y=2+x)2x     x=22y1y=\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=x2+4x+3x²9y=\frac{x^2+4x+3}{x²-9}


x=4+164(1y)(3+9y)2(1y)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








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