How do you find the domain and range of # f(x)=2/(3x-1)#?

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Answer 1 · 2018-05-26T08:39:06

The domain is #x in (-oo, 1/3) uu(1/3,+oo)#. The range is #y in (-oo,0) uu(0,+oo)#

The denominator must be #!=0#

Therefore,

#3x-1!=0#

#=>#, #x!=1/3#

The domain is #x in (-oo, 1/3) uu(1/3,+oo)#

To find the range, proceed as follows :

Let #y=2/(3x-1)#

#<=>#, #y(3x-1)=2#

#<=>#, #3yx-y=2#

#<=>#, #3yx=2+y#

#<=>#, #x=(2+y)/(3y)#

The denominator must be #!=0#

#3y!=0#

#=>#, #y!=0#

The range is #y in (-oo,0) uu(0,+oo)#

graph{2/(3x-1) [-12.66, 12.65, -6.33, 6.33]}