What is the range of the function #f(x)=2/(3x-1)#?

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Answer 1 · 2017-03-19T06:30:20

The range of #f(x)# is #R_(f(x))=RR-{0}#

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

Then,

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

#3x=2/y+1=(2+y)/y#

#x=(2+y)/(3y)#

The inverse function of #f(x)# is

#f^-1(x)=(2+x)/(3x)#

The range of #f(x)# is #=# the domain of #f^-1(x)#

As we cannot divide by #0#, #x!=0#

The domain of #f^-1(x)# is #D_(f^-1(x))=RR-{0}#

The range of #f(x)# is #R_(f(x))=RR-{0}#