How do you determine if f(x) = x^3 + xf(x)=x3+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]}