Is the Function f(x) = x + 2 Unlimited on the Set of Real Numbers?

[Дьявол]

Здравствуйте!

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