Is the function #f(x)=x^2/(x-1)# even, odd or neither?

2 Answers
Sep 1, 2015

Answer:

It is neither odd or even.

Explanation:

First of all if you want to discuss if a function is odd or even, the domain of the function must be symetrical to 0 (in other words if #x in D# then #-x in D#).

Function given in this task does not satisfy this condition, because #-1 in D# but #1 cancel(in) D#.

Sep 1, 2015

Answer:

#f# is neither even nor odd.

Explanation:

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

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

#f(-x) != f(x)# (that is #f(-x)# is not always the same as #f(x)#), so #f# is not even

#f(-x) != -f(x)# (that is #f(-x)# is not always the same as #-f(x)#, so #f# is not odd.

#f# is neither even nor odd.