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

Answer 1

#f(x)# is not EVEN neither ODD

Given #f(x)#:
#f(x)# is EVEN if: #f(-x)=f(x)#
#f(x)# is ODD if:#f(-x)=-f(x)#

In the question:

#f(x)=x^2+sin(x)#
#:. f(-x)=(-x)^2+sin(-x)=x^2-sin(x)!=f(x)# #f(-x)!=-f(x)#
#=> f(x)# is not EVEN neither ODD
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Answer 2

To determine if ( f(x) = x^2 + \sin(x) ) is an even or odd function, we can apply the following tests:

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

For the function ( f(x) = x^2 + \sin(x) ):

  1. To test for evenness, we substitute ( -x ) into the function: [ f(-x) = (-x)^2 + \sin(-x) = x^2 - \sin(x) ]

  2. To test for oddness, we substitute ( -x ) into the function and compare it to ( -f(x) ): [ -f(x) = -(x^2 + \sin(x)) = -x^2 - \sin(x) ]

Comparing ( f(-x) ) with ( f(x) ) and ( -f(x) ), we see that neither ( f(-x) = f(x) ) nor ( f(-x) = -f(x) ) holds true.

Therefore, ( f(x) = x^2 + \sin(x) ) is neither an even nor 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.

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.

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.

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.

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