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

Question

Lily Anderson · Accepted Answer

odd function

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]}

Anna Allen · Answer

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

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 #

Sadie Alexander · Answer

undefined To determine if a function $ f(x) = x^3 + x $ is even or odd, we can analyze its symmetry properties.

1. Even Function:
A function is even if it satisfies the condition $ f(x) = f(-x) $ for all values of $ x $ in its domain.

Let's test this condition for $ f(x) = x^3 + x $:
$$ f(x) = x^3 + x $$
$$ f(-x) = (-x)^3 + (-x) = -x^3 - x $$

Since $ f(x) $ is not equal to $ f(-x) $ for all values of $ x $, $ f(x) = x^3 + x $ is not an even function.

2. Odd Function:
A function is odd if it satisfies the condition $ f(x) = -f(-x) $ for all values of $ x $ in its domain.

Let's test this condition for $ f(x) = x^3 + x $:
$$ f(x) = x^3 + x $$
$$ -f(-x) = -(-x^3 - x) = x^3 + x $$

Since $ f(x) $ is equal to $ -f(-x) $ for all values of $ x $ in its domain, $ f(x) = x^3 + x $ is an odd function.