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

2 Answers
Mar 27, 2016

It is an odd function since #f(-x)=-f(x) AA x in RR#

Explanation:

A function #f(x)# is even if #f(-x)=f(x)AAx in RR#, and it is odd if #f(-x)=-f(x)AAx#.

So in this case hence it is odd.

For example, select #x=2#.
Then #f(-x)=f(-2)=(-2)^3-2=-10=-f(x)=2^2+2=10#.

Note also that all odd functions which are piecewise continuous and differentiable and having period #2L# may be represented as an infinite Fourier power series containing only sine terms, ie of the form
#sum_(n=1)^oo b_n sin((npix)/L)#, where #b_n=1/Lint_0^L f(x)sin((npix)/L)dx #

Mar 27, 2016

odd function

Explanation:

To determine if a function is even/odd the following applies.

• If f(x) = f( -x) then f(x) is even , # AAx #

Even functions have symmetry about the t-axis.

• If f(-x) = - f(x) then f(x) is odd , #AAx#

Odd functions have symmetry about the origin.

Test for even :

f( -x) = #(-x)^3 + (-x) = - x^3 - x ≠ f(x)# , hence not even

Test for odd :

#- f(x) = - (x^3 + x) = - x^3 - x =f(-x) #, hence function is odd.

Here is the graph of f(x). Note symmetry about O.
graph{x^3+x [-10, 10, -5, 5]}