What is the domain of # f(x)= 1/(x^2-4x)#?

1 Answer
YoshiMan007
Oct 15, 2017

All real numbers except #x=0# and #x=4#

Explanation:

The domain of a function is simply the set of all #x#-values that will output real #y#-values. In this equation, not all #x#-values will work as we cannot divide by #0#. Thus, we need to find when the denominator will be #0#.

#x^2-4x=0#

#x*(x-4)=0#

Using the Zero Property of Multiplication, if #x=0# or #x-4=0#, then #x^2-4x=0# will be #0#.

Thus, #x=0# and #x=4# should not be part of the domain as they would result in a non-existent #y#-value.

This means the domain is all real numbers except #x=0# and #x=4#.

In set notation, this can be written as #x in RR " such that " x!=0 and x!=4#

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