even and decreasing

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

Is it possible that function is even as well as decreasing in its domain?

Shadow · **Posted:** Thu, 10 May 2012 19:25:58 UTC

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be $(0,\infty)$ and let the function be -x^2

.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

No. If the function is decreasing for x > 0, it must be increasing for x < 0 and vice versa.

y = -x^2 is a parabola pointing downward, so it is decreasing for X > 0 and increasing for x < 0, with the max at x = 0.

alstat · **Posted:** Fri, 11 May 2012 00:06:38 UTC

Shadow wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be $(0,\infty)$ and let the function be -x^2

.

Since the definition of an even function is that f(-x) = f(x)

, can you have an even function whose domain is on only one side of zero?

Shadow · **Posted:** Fri, 11 May 2012 00:22:28 UTC

mathematic wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

No. If the function is decreasing for x > 0, it must be increasing for x < 0 and vice versa.

y = -x^2 is a parabola pointing downward, so it is decreasing for X > 0 and increasing for x < 0, with the max at x = 0.

My function is only defined on x>0

you said "on its domain", and in its domain it is indeed decreasing. In fact, I can take any decreasing function on x>0

and with that domain restriction we can say it satisfies the condition vacuously.

If you demand that the function be defined everywhere, or on a neighborhood of 0, then the answer should be clear to you as a "no" by the MVT (unless you allow for weakly decreasing, in which case 0 works), but as-stated my example works just fine.

Shadow · **Posted:** Fri, 11 May 2012 00:23:55 UTC

alstat wrote:

Shadow wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be $(0,\infty)$ and let the function be -x^2

.

Since the definition of an even function is that f(-x) = f(x)

, can you have an even function whose domain is on only one side of zero?

The definition is $\forall x\in U$ such that $-x\in U$ f(-x)=f(x)

, it just so happens that my set of

such that

is also in the domain is empty, so since we have a universal quantifier applied to an empty set, it is vacuously true that my function is even.

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

Shadow wrote:

alstat wrote:

Shadow wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be $(0,\infty)$ and let the function be -x^2

.

Since the definition of an even function is that f(-x) = f(x)

, can you have an even function whose domain is on only one side of zero?

The definition is $\forall x\in U$ such that $-x\in U$ f(-x)=f(x)

, it just so happens that my set of

such that

is also in the domain is empty, so since we have a universal quantifier applied to an empty set, it is vacuously true that my function is even.

Good evening sir
sorry sir but one of my great sir has given this explanation (which confused me more )as he disagree with your explanation .His explanation is following
Any of the common definitions for even functions
A function such that f(x) = f(-x)

for all

in the domain of

; or

A function having its graph symmetrical with respect to the

axis;

in fact implies that the domain

of an even function

is itself symmetrical with respect to the origin, i.e. D = -D

. This is for example how the notion of even function is introduced in Romanian textbooks: a function $f\colon D \to \mathbb{R}$ where $D\subseteq \mathbb{R}$ with D=-D

, and

for all $x\in D$ .
Thus speaking about an even function defined on $(0,\infty)$ is disallowed, and not vacuously true (only because

does not exist).

Shadow · **Posted:** Sat, 12 May 2012 18:22:01 UTC

mun wrote:

Shadow wrote:

alstat wrote:

Shadow wrote:

mun wrote:

Is it possible that function is even as well as decreasing in its domain?

Of course, let the domain be $(0,\infty)$ and let the function be -x^2

.

Since the definition of an even function is that f(-x) = f(x)

, can you have an even function whose domain is on only one side of zero?

The definition is $\forall x\in U$ such that $-x\in U$ f(-x)=f(x)

, it just so happens that my set of

such that

is also in the domain is empty, so since we have a universal quantifier applied to an empty set, it is vacuously true that my function is even.

Good evening sir
sorry sir but one of my great sir has given this explanation (which confused me more )as he disagree with your explanation .His explanation is following
Any of the common definitions for even functions
A function such that f(x) = f(-x)

for all

in the domain of

; or

A function having its graph symmetrical with respect to the

axis;

in fact implies that the domain

of an even function

is itself symmetrical with respect to the origin, i.e. D = -D

. This is for example how the notion of even function is introduced in Romanian textbooks: a function $f\colon D \to \mathbb{R}$ where $D\subseteq \mathbb{R}$ with D=-D

, and

for all $x\in D$ .
Thus speaking about an even function defined on $(0,\infty)$ is disallowed, and not vacuously true (only because

does not exist).

I've never seen that assumption made, before. If that is your definition of an even function, then you may ignore what I've said, but I don't see any reason why one should assume the domain is symmetric and I never have. It's easy to *extend* even functions to a maximal, symmetric domain, but there's no reason do do that ahead of time IMO.

In that case you need to see if weakly decreasing is possible, if so, then the identically zero function works, otherwise the answer is no for the reason explained above (I'm assuming the function is C^1