ha ha

[red and white kop!]

can a function be odd and even? :?

 

[Aladdin]

http://www.purplemath.com/modules/fcnnot3.htm.

Surely,

 

[red and white kop!]

that link was not very helpful. i knew all of that, but hey its the intention that counts. anyway, you didnt answer my question very fully.
would the enigmatic 'Surely,' mean 'yes, all functions, would they be polynomial, logistic, logarithmic, exponential or trigonometric, can be odd and even'?
now that would be helpful

 

[Aladdin]

There are many cases that the equation is not a function. Graphically you can tell with the vertical line test.

Logically speaking . . .graph each of the functions you mentiond and see if the y axis is the axis of symmetry or the center is the center of symmetry .

Why is your Title -- ha ha -- Are you embarresed by asking this question ?

 

[Mrspi]

can a function be odd and even? :?

An even function is symmetric about the y-axis. For an even function, f(-x) = f(x)....that is, replacing "x" with "-x" does not change the value of the function.

An odd function is symmetric about the origin. For an odd function, f(-x) = - f(x).....that is, replacing "x" with "-x" gives the OPPOSITE of the original function.

I THINK that the only function which can be considered BOTH odd and even at the same time is the constant function, f(x) = 0.

It is entirely possible, and COMMON, for a function to be NEITHER odd nor even. Consider this function:

f(x) = x[sup:ev0fj0wi]2[/sup:ev0fj0wi] + 2x

If you graph this function, it should be obvious that it is NOT symmetric about either the y-axis or the origin.

And, let's look at it algebraically.

What is f(-x)?

If f(x) = x[sup:ev0fj0wi]2[/sup:ev0fj0wi] + 2x, then
f(-x) = (-x)[sup:ev0fj0wi]2[/sup:ev0fj0wi] + 2(-x)
f(-x) = x[sup:ev0fj0wi]2[/sup:ev0fj0wi] - 2x

Now, clearly this is NOT the same as f(x). And it is not the opposite of f(x), because if f(x) = x[sup:ev0fj0wi]2[/sup:ev0fj0wi] + 2x, then -f(x) = -(x[sup:ev0fj0wi]2[/sup:ev0fj0wi] + 2x) = -x[sup:ev0fj0wi]2[/sup:ev0fj0wi] - 2x

Thus, f(x) in this case is neither odd nor even.

I hope this helps you.

 

[stapel]

that link was not very helpful.

Sorry 'bout that.

To answer your question, yes, it is possible for a function to be both even and odd, the example being (as shown in the previous reply) the constant function f(x) = 0.

As the link showed (and you already knew), an even function obeys f(-x) = f(x), and an odd function obeys f(-x) = -f(x). For the function to be both even and odd, you must then have f(x) = -f(x), so 2f(x) = 0, which means f(x) = 0 for all x. So the given example would appear to be the only such function.

Hope that helps!

 

[red and white kop!]

thanks, but are you sure of this? nobody seems to be 100% certain. no trig functions, no complex numbers.... :?:

 

[Deleted member 4993]

thanks, but are you sure of this? nobody seems to be 100% certain. no trig functions, no complex numbers.... :?:

f(x) can be any function - trig, complex, hyperbolic, etc. - as long as it has a range of 0 ? f(x) = 0.

 

[red and white kop!]

thanks
and aladdin my title is ha ha not because i am embarrassed (???why would that be???) but because i really didnt want to waste precious seconds thinking up an original and attention-grabbing title such as 'calc', 'function', 'graph' or 'I NEED HELP NOWWWWWWW!!!!!!!!!!!! PLEASE!!!!!!'
there

 

[DrMike]

that link was not very helpful. i knew all of that, but hey its the intention that counts. anyway, you didnt answer my question very fully.
would the enigmatic 'Surely,' mean 'yes, all functions, would they be polynomial, logistic, logarithmic, exponential or trigonometric, can be odd and even'?
now that would be helpful

Well....

* an even function has f(x) = f(-x) for all x.

* an odd function has f(x) = -f(-x) for all x.

So far, I've only quoted stuff you'll find in your nots/textbook. Now let's think about the question...

* Can a function be both odd and even ?

Well, let me ask you...

* Is it possible for f(x) to equal both f(-x) and -f(-x) for all x? If so, how?

When you answer that, you'll have the answer to your question...