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

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Answer 1 · 2017-02-14T07:21:39

The domain is #=RR-{-1}#
The range is #=RR-{0}#

As you cannot divide by #0#, #x!=-1#

The domain of #f(x)# is #D_f(x)=RR-{-1}#

To find the range, we need to calculate #f^-1(x)#

Let #y=1/(x+1)#

#x+1=1/y#

#x=1/y-1#

#x=(1-y)/y#

Therefore,

#f^-1(x)=(1-x)/x#

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

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

The range is #R_y=RR-{0}#