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How do you determine if #f(x) =sin(2x)# is an even or odd function?

Answer 1

Caleb Alexander

#f(x)=sin(2x)# is an odd function.

A function #f(x)# is even if #f(-x)=f(x)# and it is odd if #f(-x)=-f(x)#

Here #f(x)=sin(2x)#

and #f(-x)=sin(2xx(-x))-sin(-2x)=-sin(2x)=-f(x)#

Hence, #f(x)=sin(2x)# is an odd function.

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Answer 2

Elijah Allen

To determine if ( f(x) = \sin(2x) ) is an even or odd function, we evaluate ( f(-x) ) and compare it with ( f(x) ).

Even function: If ( f(-x) = f(x) ) for all ( x ) in the domain of ( f ), then ( f(x) ) is an even function.
Odd function: If ( f(-x) = -f(x) ) for all ( x ) in the domain of ( f ), then ( f(x) ) is an odd function.

For ( f(x) = \sin(2x) ):

Evaluate ( f(-x) ): [ f(-x) = \sin(2(-x)) = \sin(-2x) = -\sin(2x) ]

Compare ( f(-x) ) with ( f(x) ): [ -\sin(2x) \neq \sin(2x) ]

Since ( f(-x) ) is not equal to ( f(x) ), the function ( f(x) = \sin(2x) ) is an odd function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.