How do you determine whether #f(x)=x^5-x# is even, odd, or neither?

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Answer 1 · 2017-01-06T07:27:22

The function is odd

#f(x)=x^5-x#

We calculate

#f(-x)=-x^5+x#

#-f(x)=-(x^5-x)=-x^5+x#

Therefore,

#f(-x)=-f(x)#

The function is odd, that is, symmetric wrt the origin #O#

A function is even if #f(-x)=f(x)#

graph{x^5-x [-4.385, 4.385, -2.19, 2.194]}