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

1 Answer
George C.
Mar 15, 2016

The domain is #(-oo, 2) uu (2, oo)#, i.e. #RR "\" { 2 }#

The range is #(-oo, 1) uu (1, oo)#, i.e. #RR "\" { 1 }#

Explanation:

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

When #x = 2#, the denominator of #f(x)# is #0#, but the numerator is non-zero. So #f(x)# is undefined and has a vertical asymptote at #x = 2#.

If we let #y = f(x) = 1+1/(x-2)#, then:

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

So:

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

So:

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

So:

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

This is well defined for all #y in RR# except #y=1#, where the inverse function #f^(-1)(y)# has a vertical asymptote - so the original function has a horizontal asymptote #y=1#.

Since #y=1# is not in the domain of the inverse function, it is not in the range of the original function.

To summarise:

The domain of #(x-1)/(x-2)# is #RR "\" { 2 } = (-oo, 2) uu (2, oo)#

The range of #(x-1)/(x-2)# is #RR "\" { 1 } = (-oo, 1) uu (1, oo)#

graph{(x-1)/(x-2) [-8.13, 11.87, -3.88, 6.12]}

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