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How do you determine if #f(x) = cos^2 x + cos x - 3# is an even or odd function?

Answer 1

Benjamin Achtenberg

#f# is even .

A given fun. #g# is odd #iff AA x,g(-x)=-g(x)#.

A fun. #g# is even #iff AA x, g(-x)=g(x)#.

Now, for our fun. #f, f(x)=(cosx)^2+cosx-3#

#:. f(-x)=(cos(-x))^2+cos(-x)-3#

But, #cos(-x)=cosx#, hence,

#f(-x)=(cosx)^2+cosx-3=f(x)#

Therefore, #f# is even .

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Answer 2

Ariana Alvarez

To determine if ( f(x) = \cos^2(x) + \cos(x) - 3 ) is an even or odd function, we evaluate ( f(-x) ) and compare it with ( f(x) ).

If ( f(x) = f(-x) ) for all ( x ) in the domain, then the function is even.

If ( f(x) = -f(-x) ) for all ( x ) in the domain, then the function is odd.

For ( f(x) = \cos^2(x) + \cos(x) - 3 ):

Evaluate ( f(-x) ): [ f(-x) = \cos^2(-x) + \cos(-x) - 3 ]

Using the even and odd properties of cosine: [ \cos(-x) = \cos(x) ] [ \cos^2(-x) = \cos^2(x) ]

Therefore: [ f(-x) = \cos^2(x) + \cos(x) - 3 ]

Since ( f(-x) = f(x) ), the function is even.

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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.