unlimited function of set E

[Дьявол]

Здравствуйте!

How will I prove that some function f(x) is unlimited of some set E\subseteqDf, where Df is the domain of the function, or the values of x, which can be used in the function.
For example:

How will I prove that this function f(x)=x+2 is unlimited for E=R (set of real numbers)?

Спасибо за помощь!

[HallsofIvy]

By "unlimited" I presume you mean what I would call unbounded "unbounded".
Pretty much any time you are asked to prove "f is ****", where **** is some word, you prove it by showing that f satisfies the definition of ****.

A function is unbounded if it is both unbounded above and unbounded below.

A function is unbounded above if, given any number Y, there exist x such that f(x)> Y.
A function is unbounded below if, given any number Y, there exist x such that f(x)< Y.


Given any Y, is there an x such that x+ 2> Y?

Given any Y, is there an x such that x+ 2< Y?

For that simple function, it is just a matter of solving the inequalities.

[Дьявол]

I think you're right, but also in my book there is |f(x)|>Y, and that means f(x)>Y and -f(x)>Y, or f(x)<-Y.

So x+2>Y and
x+2<-Y

x>Y-2

Let's say x=Y-1, so:

|f(x)|=|x+2|=|Y-1+2|=|Y+1|=Y+1>Y, for every Y \geq 0.

x<-2-Y

x=-3-Y

|f(x)=|x+2|=|-3-Y+2|=|-Y-1|=-Y-1<-Y, for every Y \geq 0

So |f(x)|>K , for K \in \mathbb{R}.

Did I solved it correctly?

[HallsofIvy]

I think you're right, but also in my book there is |f(x)|>Y, and that means f(x)>Y and -f(x)>Y, or f(x)<-Y.
Well, yes, but those are equivalent definitions. If you can find a number X so that f(x)> X and a number Z so that f(x)< Z, take Y to be the larger of |X| and |Z|.

So x+2>Y and
x+2<-Y

x>Y-2

Let's say x=Y-1, so:

|f(x)|=|x+2|=|Y-1+2|=|Y+1|=Y+1>Y, for every Y \geq 0.

x<-2-Y

x=-3-Y

|f(x)=|x+2|=|-3-Y+2|=|-Y-1|=-Y-1<-Y, for every Y \leq 0

So |f(x)|>K , for K \in \mathbb{R}.

Did I solved it correctly?[/QUOTE]

Yes.

[Дьявол]

Just to correct, one mistake of mine. It is Y \geq 0, down on the second function written with LateX.

Also I think that in my book, the autors use |f(x)|>K (for unbounded functions), because |f(x)|<K is bounded of it (the opposite one, vice versa).

But also your statement is also good. f(x)>K and f(x)<K. Also in some cases maybe it is better to use |f(x)| \leq K .

[Дьявол]

Sorry, for posting in same topic. I did not know if I need to open new topic.
I want to ask you about this problem.
The problem is to prove that f(x)=x+sinx ; E=R

So, |f(x)|>Y

x+sinx>Y

x=Y+2

Y+2+sin(Y+2)>Y

sin(Y+2)>-2

Which is correct.

So |f(x)|=|x+sinx|=|Y+2+sin(Y+2)|=Y+2+sin(Y+2)>Y

Is this good?

[Office_Shredder]

While the order in which you listed things makes it slightly confusing to see what you are doing, it looks like the right idea is there

[HallsofIvy]

Sorry, for posting in same topic. I did not know if I need to open new topic.
I want to ask you about this problem.
The problem is to prove that f(x)=x+sinx ; E=R
do you mean you want to prove that f(x) is unbounded?

So, |f(x)|>Y

x+sinx>Y

x=Y+2

Y+2+sin(Y+2)>Y

sin(Y+2)>-2

Which is correct.
Which is certainly NOT correct! "x+ sin x> Y" does NOT imply x= Y+2! I think what you meant to say was that -1\le sin x\le 1 for all x so [itex]x- 1\le x+ sin x\le x+ 1.
Now, given in Y, if x> Y+1, x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y.


So |f(x)|=|x+sinx|=|Y+2+sin(Y+2)|=Y+2+sin(Y+2)>Y

Is this good?

[Дьявол]

If -1\leq sinx \leq 1, then -1+x\leq sinx+x\leq 1+x.

So if x+sinx > x-1 >Y; so x-1>Y ; x>Y+1

x=Y+2

|f(x)|=|x+sinx|=Y+2+sin(Y+2)>Y;

I think we got the same thing. Why you think it is not correct?

[Office_Shredder]

Which is certainly NOT correct! "x+ sin x> Y" does NOT imply x= Y+2! I think what you meant to say was that -1\le sin x\le 1 for all x so [itex]x- 1\le x+ sin x\le x+ 1.
Now, given in Y, if x> Y+1, x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y.

I think you got it backwards, reading it I interpreted it as intending to say x=Y+2 implies x+sinx>Y (hence proving the function is unbounded)

[HallsofIvy]

Either way, there is no need to connect the argument of the sine with Y at all.

[Дьявол]

Ok, we got: x+sinx > x-1 >Y; so x-1>Y ; x>Y+1

But where does this x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y comes from?

Thanks in advance.

[HallsofIvy]

Ok, we got: x+sinx > x-1 >Y; so x-1>Y ; x>Y+1

But where does this x+ sin x> Y+1- 1= Y if x< Y+1, x+ sin x< Y+1- 1= Y comes from?

Thanks in advance.
That was a typo on my part. What I meant to say was "If x< Y- 1, then x+ sin x< Y-1+ sin x and, since sin x\le 1, x+ sin x\le Y-1+ 1= Y so that Y is not a lower bound for x+ sin x.