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

Question

Serenity Jones · Accepted Answer

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#. Thus, we discover: #f(-x)/g(-x) = (-f(x))/(-g(x)) = f(x)/g(x)# for all #x#

Elias Ackerman · Answer

undefined To determine if the function $ h(x) = \frac{2x}{x^3 - x} $ is even or odd, we can use the properties of even and odd functions.

1. Even functions satisfy the property $ f(-x) = f(x) $ for all $ x $ in the function's domain.
2. Odd functions satisfy the property $ f(-x) = -f(x) $ for all $ x $ in the function's domain.

For the function $ h(x) = \frac{2x}{x^3 - x} $:

1. Substitute $ -x $ for $ x $ in the function: $ h(-x) = \frac{2(-x)}{(-x)^3 - (-x)} $
2. Simplify the expression: $ h(-x) = \frac{-2x}{-x^3 + x} $
3. Compare $ h(-x) $ with $ h(x) $ to determine if it satisfies the properties of even or odd functions.

Since $ h(-x) = \frac{-2x}{-x^3 + x} $ and $ h(x) = \frac{2x}{x^3 - x} $, we can see that $ h(-x) $ is equal to $ -h(x) $, which means the function satisfies the property of an odd function.

Therefore, $ h(x) = \frac{2x}{x^3 - x} $ is an odd function.