- #1

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**Very important....**

Let

**R**denote the set of real numbers. Find all nondecreasing continuous

functions g :

**R**→

**R**such that

g(x+g(y))=g(g(x))+g(y) for all x,y∈

**R**

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- Thread starter vip89
- Start date

- #1

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Let

functions g :

g(x+g(y))=g(g(x))+g(y) for all x,y∈

- #2

HallsofIvy

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And **why** is that "very important"?

- #3

tiny-tim

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Looks **monotonous** to me …

- #4

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Looksmonotonousto me …

I think u see it easy.....

I read this problem in abook,and I liked it

so can you solve it???

- #5

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Looksmonotonousto me …

Andwhyis that "very important"?

I read this problem in abook,and I liked it

so can you solve it???

- #6

tiny-tim

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Hi vip89!

Hint: try differentiating with respect to x, and y, separately.

Hint: try differentiating with respect to x, and y, separately.

- #7

HallsofIvy

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Oh, I see. That hint didn't seem to make much sense until I **did**!

(Funny how often that happens.)

(Funny how often that happens.)

- #8

tiny-tim

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Oh, I see. That hint didn't seem to make much sense until Idid!

(Funny how often that happens.)

Oh my goodness! Does it work?

I didn't actually try it!

- #9

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oh.......I cant solve it.!!

can u send me the answer??I need it quickly and there is no time to solve it because I have final exams

Thankx very much

- #10

tiny-tim

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… I have final exams

Hi vip89!

It's

(and if it doesn't come up … why are you bothered? )

So … differentiating with respect to x … what do you get?

- #11

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I have acomption with my friend ,who find the answer first

- #12

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Why is the competition with your friend important?

- #13

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ok , it is not important

can u solve it ,pls??

can u solve it ,pls??

- #14

HallsofIvy

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Wouldn't that be cheating?

- #15

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ok , it is not important

can u solve it ,pls??

Actually no, I can't. But I do have some advice for you.

In this forum, you'll pretty much never get other people to solve entire problems for you. People like to see some attempt that you tried to solve it first. So far you haven't done that.

Did you try the suggestion from tiny-tim in message #6 ? If not, then try it. If you did try it then explain how far you got with it, or why you were unable to get very far with it, or just what you don't understand.

And lastly, welcome to physicsforums.com

- #16

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ok

I will try

and I want to say that is not cheating because It is not homework

I will try

and I want to say that is not cheating because It is not homework

- #17

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