How do you determine if #h(x)= (2x)/(x^3 - x)# is an even or odd function?

Question

1348 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T19:42:47

Simplify and analyse #h(x)# to find that it is an even function.

#h(x) = (2x)/(x^3-x) = 2/(x^2-1)#

with exclusion #x != 0#

Since #(-x)^2 = x^2#, we find #h(-x) = h(x)# for all #x# in the domain of #h(x)#

So #h(x)# is an even function.

#color(white)()#
Another quick method of finding that this is an even function is to look at the numerator and denominator polynomials.

They both consist solely of terms with odd degree. So #h(x)# is a quotient of two odd functions, so is an even function.

To see that the quotient of any two odd functions is an even function, suppose that #f(x)# and #g(x)# are both odd functions.

By definition #f(-x) = -f(x)# and #g(-x) = -g(-x)# for all #x#.

So we find:

#f(-x)/g(-x) = (-f(x))/(-g(x)) = f(x)/g(x)# for all #x#