Question 146668
Is the function {{{g(x)=2x^3- 3x}}} ??? Get back to me if I'm wrong






Remember, if {{{f(x)=f(-x)}}} then the function is an even function. If {{{f(-x)=-f(x)}}} then the function is an odd function.




First, let's see if {{{f(x)=2x^3-3x}}} is an even function.



{{{f(x)=2x^3-3x}}} Start with the given function.



{{{f(-x)=2(-x)^3-3(-x)}}} Replace each x with -x.



{{{f(-x)=-2x^3+3x}}} Simplify. Note: only the terms with an <b>odd</b> exponent will change in sign.


So this shows us that {{{2x^3-3x<>-2x^3+3x}}} which means that {{{f(x)<>f(-x)}}}

Since {{{f(x)<>f(-x)}}}, this shows us that {{{f(x)=2x^3-3x}}} is <b>not</b> an even function.



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Now, let's see if {{{f(x)=2x^3-3x}}} is an odd function.


{{{f(x)=2x^3-3x}}} Start with the given function.



{{{-f(x)=-(2x^3-3x)}}} Negate the entire function by placing a negative outside the function.



{{{-f(x)=-2x^3+3x}}} Distribute and simplify.



So this shows us that {{{-2x^3+3x=-2x^3+3x}}} which means that {{{f(-x)=-f(x)}}}

Since {{{f(-x)=-f(x)}}}, this shows us that {{{f(x)=2x^3-3x}}} is an odd function.



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Answer:

So the function {{{f(x)=2x^3-3x}}} is an odd function.