Is this an even or odd function

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Homework Help Overview

The discussion revolves around determining whether the function \( f(x) = \sin(4x) + \cos(4x) \) is even, odd, or neither. Participants are exploring the definitions and properties of even and odd functions in the context of trigonometric identities.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are examining the function's behavior under transformations, specifically evaluating \( f(-x) \) and comparing it to \( f(x) \) and \(-f(x)\). There are discussions about the implications of the results and the definitions of even and odd functions.

Discussion Status

There is an ongoing exploration of the function's characteristics, with some participants questioning the original poster's reasoning and suggesting alternative approaches to verify the function's parity. Multiple interpretations of the function's behavior are being considered, and no consensus has been reached.

Contextual Notes

Participants note that many functions do not fit neatly into the categories of even or odd, and there is an emphasis on the importance of thorough evaluation to determine the function's classification.

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1. (sin4x + cos4x)



Homework Equations





The Attempt at a Solution



(sin4x + cos4x)
= (sin4(-x) + cos4(-x))
= -sin4x + cos4x

im thinking it is an even function
 
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For an even function, f(-x) = f(x). An odd function has f(-x) = -f(x). Is your final result equal to either f(x) or -f(x) ? Note that most functions are neither even nor odd.
 
physics=world said:
1. (sin4x + cos4x)



Homework Equations





The Attempt at a Solution



(sin4x + cos4x)
= (sin4(-x) + cos4(-x))
= -sin4x + cos4x
You started off by saying sin(4x) + cos(4x) = sin(-4x) + cos(-4x), which your later work shows isn't true. Your work should start with sin(-4x) + cos(-4x). If you can show this is equal to sin(4x) + cos(4x), then the function is even. If it turns out to be equal to -sin(4x) - cos(4x), then the function is odd. If it results in neither, then the function is neither even nor odd.
physics=world said:
im thinking it is an even function
 
so is it an odd function since it does not look like the original function?
 
If you want to decide whether a function f is odd, you can start by looking at f(0). If that's ≠0, the function definitely isn't odd. (It could still be even though.)
 
physics=world said:
so is it an odd function since it does not look like the original function?

A function is odd if it has a very specific behavior. For every value of x, f(-x) must equal -f(x). Find the expression for f(-x) and find the expression for -f(x). Compare the two expressions. Are they equal for every possible value of x? If they are, your function is odd. If they differ for any single value of x, your function is not odd.
A quick way to compare two expressions is to subtract one from the other. It doesn't matter which one comes first, as if they are the same value, the subtraction will give you 0. If you don't get 0, the expressions are different.
If your function is not odd, that does not mean it is even. To be even it must satisfy a different extremely specific rule. You must compare the expression for f(-x) to the expression for f(x). Are they equal for every possible value of x? If so, the function is even. If they differ for any specific value of x, then your function is not even either.
Most functions are neither even nor odd.
 
Last edited:
physics=world said:
so is it an odd function since it does not look like the original function?
You appear to thinking that any function that is not "even" must be "odd". That is not true. slider142, in the first response to your post, told you that a function is odd if and only if f(-x)= -f(x). slider142 also told you that "most functions are neither even nor odd."
 
When the concepts of "even functions" and "odd functions" are taught, it might be best to tell students that most functions are neither. Just like most numbers are neither even nor odd, for instance 1.5 isn't, nor is pi.
 

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