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

1 Answer
Aug 16, 2015

Answer:

Domain: #(-oo, 2) uu (2, + oo)#
Range: #(-oo, 0) uu (0, + oo)#

Explanation:

Your function is defined for any value of # in RR# except the one that can make the denominator equal to zero.

#x-2 = 0 implies x = 2#

This means that #x = 2# will be excluded from the domain of the function, which will thus be #RR - {2}#, or #(-oo, 2) uu (2, + oo)#.

The range of the function will be affected by the fact that the only way a fraction can be equal to zero is if the numerator is equal to zero.

In your case, the numerator is constant, euqal to #1# regardless of the value of #x#, which implies that the function can never be equal to zero

#f(x) != 0", "(AA)x in RR-{2}#

The range of the function will thus be #RR - {0}#, or #(-oo, 0) uu (0, + oo)#.

graph{1/(x-2) [-10, 10, -5, 5]}