What is the domain and range of #f(x) = 1/(x + 3)#?

1 Answer
Mar 14, 2017

The domain of is #=RR-{-3}#
The range is #=RR-{0}#

Explanation:

As you cannot divide by #0#, #x!=-3#

The domain of #f(x)# is #D_f(x)=RR-{-3}#

To find the range, we need the domain of #f^-1(x)#

Let, #y=1/(x+3)#

#x+3=1/y#

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

Therefore,

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

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

So, the range is #=RR-{0}#