What is the domain and range of y =1/(2x-4)?

2 Answers
Narad T.
May 2, 2017

The domain of y is =RR-{2}
The range of y, =RR-{0}

Explanation:

As you cannot divide by 0,

2x-4!=0

x!=2

Therefore, the domain of y is D_y=RR-{2}

To determine the range, we calculate y^-1

y=1/(2x-4)

(2x-4)=1/y

2x=1/y+4=(1+4y)/y

x=(1+4y)/(2y)

So,

y^-1=(1+4x)/(2x)

The domain of y^-1 is D_(y^-1)=RR-{0}

This is the range of y, R_y=RR-{0}
graph{1/(2x-4) [-11.25, 11.25, -5.625, 5.625]}

Jim G.
May 2, 2017

"domain " x inRR,x!=2

"range " y inRR,y!=0

Explanation:

The denominator of y cannot be zero as this would make y color(blue)"undefined".Equating the denominator to zero and solving gives the value that x cannot be.

"solve " 2x-4=0rArrx=2larrcolor(red)" excluded value"

"domain " x inRR,x!=2

"to find excluded value/s in the range"

"Rearrange the function making x the subject"

rArry(2x-4)=1

rArr2xy-4y=1

rArr2xy=1+4y

rArrx=(1+4y)/(2y)

"the denominator cannot be zero"

"solve " 2y=0rArry=0larrcolor(red)" excluded value"

"range " y inRR,y!=0
graph{1/(2x-4) [-10, 10, -5, 5]}

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