How do you determine if #y = (x^4 + 1) / (x^3 - 2x)# is an even or odd function?

Question

Serenity Johnson · Accepted Answer

#y = (x^4+1)/(x^3-2x)# is an odd function.

An even function is one for which #f(-x) = f(x)# for all #x# in its domain. An odd function is one for which #f(-x) = -f(x)# for all #x# in its domain. Let #f(x) = (x^4+1)/(x^3-2x)# Then: #f(-x) = ((-x)^4+1)/((-x)^3-2(-x))# #=(x^4+1)/(-x^3+2x)# #=-(x^4+1)/(x^3-2x)# #=-f(x)# So #y = (x^4+1)/(x^3-2x)# is an odd function.

Serenity Johnson · Answer

Odd

You can calculate f(-x) and see if : 1)f(-x)=f(x), in this case y is even 2)f(-x)=-f(x), in this one it is odd so #f(-x)=((-x)^4+1)/((-x)^3-2(-x))# #=(x^4+1)/(-x^3+2x)# #=-(x^4+1)/(x^3-2x)# Therefore f(-x)=-f(x) and y is odd

James Allen · Answer

undefined To determine if $ y = \frac{x^4 + 1}{x^3 - 2x} $ is an even or odd function, you need to check its symmetry properties.

1. Even function: If $ f(x) = f(-x) $ for all $ x $ in the domain, then the function is even.
2. Odd function: If $ f(x) = -f(-x) $ for all $ x $ in the domain, then the function is odd.

For $ y = \frac{x^4 + 1}{x^3 - 2x} $:

- Substitute $ -x $ for $ x $ in the equation and simplify.
- If you get the original function back (without any changes), the function is even.
- If you get the negative of the original function, the function is odd.

Performing this substitution:

$ f(-x) = \frac{(-x)^4 + 1}{(-x)^3 - 2(-x)} $

Simplify the numerator and denominator.

$ f(-x) = \frac{x^4 + 1}{-x^3 + 2x} $

This is not equal to the original function, nor is it the negative of the original function. Therefore, $ y = \frac{x^4 + 1}{x^3 - 2x} $ is neither even nor odd.