Question 87302
Remember, a function is even when this equality is true:


{{{f(x)=f(-x)}}}


and a function is odd when this equality is true:


{{{-f(x)=f(-x)}}}


Lets see if the function is even:


{{{y=x^3+x}}} Start with the given equation

Here's {{{f(x)}}}:

{{{f(x)=x^3+x}}} This is the given equation

Here's {{{f(-x)}}}:

{{{f(-x)=(-x)^3+(-x)}}} Replace x with {{{-x}}}


{{{f(-x)=-x^3-x}}} Simplify

Since {{{x^3+x}}} does not equal {{{-x^3-x}}}, this means the equation {{{f(x)=f(-x)}}} is <b>not</b> true.


Since {{{f(x)}}} does <b>not</b> equal {{{f(-x)}}}, the equation {{{y=x^3+x}}} is <b>not</b> an even function.




Now lets see if the function is odd:


{{{y=x^3+x}}} Start with the given equation

Here's {{{f(-x)}}}:

{{{f(-x)=-x^3-x}}} Remember we solved for this previously

Here's {{{-f(x)}}}:

{{{-f(x)=-(x^3+x)}}} Negate the whole function


{{{f(-x)=-x^3-x}}} Distribute the negative and simplify

Since {{{-x^3-x}}} equals {{{-x^3-x}}}, this means the equation {{{-f(x)=f(-x)}}} is true.


Since {{{-f(x)}}} equals {{{f(-x)}}}, the equation {{{y=x^3+x}}} is an odd function. This means the equation has symmetry with respect to the origin.


When we graph the equation {{{y=x^3+x}}}, we can see if the equation has any symmetry:

{{{ graph( 500, 500, -10, 10, -10, 10, x^3+x) }}}

and we can clearly see that the equation is symmetrical with respect to the origin

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