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

2 Answers
May 15, 2017

The domain is restricted as the denominator cannot be zero.

i.e. if x + 1 = 0, then x = -1

Hence, the domain can take all real values except x = -1

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

i.e. #f(x) = (x + 1)/(x + 1) - 3/(x + 1)#

or, #f(x) = 1 - 3/(x + 1)#

as x gets very small or very big, 3/(x + 1) tends towards zero

i.e. f(x) approaches a limit of 1

Hence, the range of f(x) takes all real values except 3

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May 15, 2017

Answer:

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

Explanation:

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 #f^-1(x)#

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

Interchange #y# and #x#

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

We rewrite #y# as function of #x#

#x(y+1)=y-2#

#xy+x=y-2#

#y(1-x)=x+2#

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

So,

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

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

This is the range of #f(x)#, #R_f(x)=RR-{1}#