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How do you determine if #y=1-sin(x)# is an even or odd function?

Answer 1

Theodore Amidon

Function is neither odd nor even.

If #f(-x)=f(x)# then the function is even,

but if #f(-x)=-f(x)# then the function is odd.

As #f(x)=1-sinx#

#f(-x)=1-sin(-x)=1-(-sinx)=1+sinx#

As #-f(x)=-1+sinx#,

hence neither #f(-x)=f(x)# nor #f(-x)=-f(x)#

Hence the given function is neither odd nor even.

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Answer 2

Theodore Amidon

To determine if ( y = 1 - \sin(x) ) is an even or odd function, we can analyze its symmetry properties:

Even functions satisfy the condition ( f(-x) = f(x) ) for all ( x ) in their domain.
Odd functions satisfy the condition ( f(-x) = -f(x) ) for all ( x ) in their domain.

For ( y = 1 - \sin(x) ):

Substitute ( -x ) into the function: ( y(-x) = 1 - \sin(-x) = 1 + \sin(x) ).
Compare this with the original function: ( y(x) = 1 - \sin(x) ).

Since ( y(-x) ) does not equal ( y(x) ) and also does not equal the negative of ( y(x) ), the function ( y = 1 - \sin(x) ) does not satisfy the conditions for being even or odd. Therefore, it 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.