How do you know if #f(x) =-x^3 + 6x# is an even or odd function?

Question

Lily Anderson · Accepted Answer

Odd.

The parity of a function can be found through finding #f(-x)#: Here, #f(-x)=-(-x)^3+6(-x)# #f(-x)=-(-(x^3))-6x# #f(-x)=x^3-6x# #f(-x)=-(-x^3+6x)# #f(-x)=-f(x)# Thus, the function is odd. Notice how the function is comprised totally of odd-powered variables. Odd functions have the special property of being "origin symmetrical", meaning it's a reflection of itself over the #x# and #y# axes. graph{-x^3+6x [-16.02, 16.02, -8.01, 8.01]}

Liam Brennan · Answer

undefined To determine if a function $ f(x) $ is even or odd, we examine its symmetry properties:

1. Even functions satisfy $ f(-x) = f(x) $ for all $ x $ in their domain.
2. Odd functions satisfy $ f(-x) = -f(x) $ for all $ x $ in their domain.

For $ f(x) = -x^3 + 6x $, we evaluate $ f(-x) $ and $ -f(x) $:

1. $ f(-x) = -(-x)^3 + 6(-x) = -(-x^3) - 6x = -(-x^3) - 6x = -x^3 - 6x $
2. $ -f(x) = -(-x^3 + 6x) = x^3 - 6x $

Comparing $ f(-x) $ and $ -f(x) $:

- Since $ f(-x) = -x^3 - 6x $ and $ -f(x) = x^3 - 6x $, the function does not satisfy the condition for even functions ($ f(-x) \neq f(x) $).
- Since $ f(-x) = -x^3 - 6x $ and $ -f(x) = x^3 - 6x $, the function does not satisfy the condition for odd functions ($ f(-x) \neq -f(x) $).

Therefore, $ f(x) = -x^3 + 6x $ is neither an even nor an odd function.