Even, Odd,or Neither Odd or Even Function

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The function h(x) = 2x - x² is neither even nor odd because h(-x) does not equal h(x) or -h(x). An even function satisfies f(-x) = f(x), while an odd function satisfies f(-x) = -f(x). The discussion highlights the importance of testing values to understand the function's behavior, as initial assumptions may lead to confusion. The concept of even and odd functions can also be explored through their algebraic components. Understanding these definitions and testing with specific values can clarify why h(x) does not fit either category.
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Dear People,
I'm sorry to put up such an easy question. But would someone please explain a little further.

h(x)=2x-x2
h(-x)=2(x)-(-x)2 = -2x-x2

Since h(-x) doesn't equal h(x) and h(-x) doens't equal -h(x), we conclude that H is neither even nor odd.

I would have put that this fuction is odd on the test. Why is it neither? The answer is above I was wondering if someone could explain in a different way that i may be able to undestand.
 
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Albert2008 said:
I would have put that this fuction is odd on the test. Why is it neither? The answer is above I was wondering if someone could explain in a different way that i may be able to undestand.
Why would you have said that it was odd? What is the definition of an odd function?
 
An odd function is a function (behaving like an odd power of x) for which f(-x)=-f(x).
An even function is a function (behaving like an even power of x) for which f(-x)=f(x).

Any function can be resolved into its even and odd components:

f(x) = f_{odd}(x) + f_{even}(x)
where
f_{even}(x) = \frac{f(x)+f(-x)}{2}
and
f_{odd}(x) = \frac{f(x)-f(-x)}{2}

Note for example that the even and odd components of the exponential function are the hyperbolic trig functions.
e^x = \cosh(x)+\sinh(x)

The use of the "even" and "odd" terms comes from the even and odd components of a polynomial which will be respectively the sum of the even degree terms and the sum of the odd degree terms.

Similarly with "infinite polynomials" i.e. the power series expansion of analytic functions.

FWIW
Another way to look at this is to define oddness and evenness in terms of commutativity or anti-commutativity with the negation function under composition:

f is even iff \{f,N\}_\circ \equiv f\circ N + N\circ f = 0

f is odd iff [f,N]_\circ \equiv f\circ N - N\circ f = 0

where N(x) = -x
 
Dear Dr. Baugh,
Thank you so much for taking time to answer my question.
 
If you don't see why algebraically why it's not odd, try to numerically. Just keep plugging numbers into h(x) a d h(-x) and see if h(-x)=-h(x). Be warned, your teacher probably picked this function because for what would be most people's first two choices of numbers to plug in, it would <i>seem</i> like h(x) is odd.

Then when you find an x that contradicts h(-x)=-h(x), you'll probably see <i>why</i> it didn't hold.
 
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