# Unlimited function of set E

1. Oct 30, 2008

### Дьявол

Здравствуйте!

How will I prove that some function f(x) is unlimited of some set E$\subseteq$Df, 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!

2. Oct 30, 2008

### HallsofIvy

Staff Emeritus
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.

3. Oct 30, 2008

### 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?

Last edited: Oct 30, 2008
4. Oct 30, 2008

### HallsofIvy

Staff Emeritus
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.

5. Oct 30, 2008

### 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 $|f(x)| \leq K$.

6. Oct 31, 2008

### Дьявол

Sorry, for posting in same topic. I did not know if I need to open new topic.
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?

7. Oct 31, 2008

### Office_Shredder

Staff Emeritus
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

8. Oct 31, 2008

### HallsofIvy

Staff Emeritus
do you mean you want to prove that f(x) is unbounded?