# Problem regarding relations and functions .

1. Apr 12, 2012

### sankalpmittal

Problem regarding "relations and functions".

There are 3 very mini problems so I thought it would be rather fine to adjust them in a single thread.

1. The problem statement, all variables and given/known data

(i) Find the range of the function :

f(x) = 2-3x , x$\in$R , x>0
Note : R is universal set containing real numbers in all the three problems.

(ii) Let f={(x,x2/(1+x2)):x$\in$R} be a function from R to R , i.e f:R→R . Determine the range of f.

(iii) Let f be the subset of Z x Z defined by f = {(ab,a+b): a,b$\in$Z}. Is f a function from Z to Z ? Justify your answer.

2. Relevant equations

No equations. Only the concepts regarding the relations and functions.

3. The attempt at a solution

(i) Since x>0 so f(x) can be any .. Hmm. If we take f(1) , f(2) etc then we get f(x)<0. But if we take f(0.1) etc. then we will get f(x) >0 and if we take f(0.6666666......) then we get f(x) = 0. So according to me the answer will be just a subset of R. Am I missing something ?

(ii) f:R→R
so range means all second elements of the function f in the form of {(x,y) , (a,b) , ....}
where range = { y,b,....}

For every value of x such that x belongs to R we get x2/(1+x2) , right ?
But here elements of range cannot be negative. So Range = { 0 , +∞}

(iii)Since for every element ab there is one and only one unique value of a+b and so f is a function.

Thanks in advance ! Where am I going wrong ?

2. Apr 12, 2012

### mtayab1994

Re: Problem regarding "relations and functions".

Why would you say a subset of R for the first one ?

3. Apr 12, 2012

### mtayab1994

Re: Problem regarding "relations and functions".

For the second one you won't get 0 in your range because no matter how large or how small x gets f(x) will never equal zero.

4. Apr 12, 2012

### tiny-tim

hi sankalpmittal!
yes, but what subset?

write it something like [a,b)
i don't understand the question

either f is wrongly written, or f is from R to R2

if the latter, isn't the range is a curve in R2 ?
no, a subset {(x,y)} of A x A is a function f from A to A if f(x) = y defines a function

5. Apr 12, 2012

### sankalpmittal

Re: Problem regarding "relations and functions".

My reference textbook says that answer to my first question will be :
range = {-∞,2}
But this is not correct ! If we take x=0.1 or x=0.2 etc. we will get a positive value for f(x) , i.e f(x) >0. Here we cannot precisely tell the range so I said that answer will be any subset of R.

f(x) = x2/(1+x2)

Well take x=0
f(0) = 0/1 = 0

@ tiny-tim

And yes f is a function from set R to R written like f:R->R
In other words we say f is a function on set R as it is itself a subset of Cartesian product of R x R and for every element of domain in it , it has a unique one and only one element of range that means element of domain is not repeating twice or more.

I understood (iii) question. Answer is no because there can be many values of y for single value of x , as tiny-tim said.
Eg. a = 1/2 and b = 2
so ab = 1
and a+b = 5/2

a = 1 and b=1
so ab=1
but a+b = 2
so ab is repeating twice. (I'm taking an example).

Last edited: Apr 12, 2012
6. Apr 12, 2012

### tiny-tim

i don't understand that at all

anyway, x = 0.9 gives f(x) = -0.7
ok, then "it has a unique one and only one element of range that means element of domain is not repeating twice or more" is what you have to prove (and you haven't yet)

7. Apr 12, 2012

### sankalpmittal

Re: Problem regarding "relations and functions".

Again see my post #5. I have edited a lot there.

Please leave question (iii) as I have understood it.

And first let me solve question (i).

In question (i) take values of x such that 0<x<0.7 and I am sure you will get f(x)>0 which of course is not complying with textbook answer.

If I take x >=0.7 then I get f(x) < 0 and if I take x=0.66666666666...... then I get f(x) =0

So how can I precisely find the range ?

range = { -∞ , 0 } + { 2-3x , if 0<x<0.7}

Or let me write :
range = { -∞ , x>=0.7}
range = {0 , x=0.66666666666......}
range={ 2-3x , 0<x<0.7}

8. Apr 12, 2012

### HallsofIvy

Staff Emeritus
Re: Problem regarding "relations and functions".

I have no clue why you are choosing "0.7". Surely by the time you are working on problems like this you should be able to recognize that the graph of "y= 2- 3x" is a straight line. When x= 0, what is y? What does the graph look like for x> 0?

So what? The problem does NOT ask what happens around f(x)= 0, it asks what happens around x= 0.

\
No, it isn't. You are confusing values of x and values of f(x).

These two don't even mean the same thing. The first is a set of values of x, the second values of f(x). You are again confusing the two.

Last edited: Apr 21, 2012
9. Apr 12, 2012

### sankalpmittal

Re: Problem regarding "relations and functions".

Here was my first question :
We are defining the function like this

f = {(x,2-3x): x$\in$R , x>0}

where f(x) = y = 2-3x
Yes , the line is a straight line and if x=0 then y=2 and here x>0 graph with respect to f(x) will be a straight line.

But my point is that by trial and error I can figure that when I take 0<x<0.7 I get f(x)>0 but according to my textbook range of function f = { -∞ , 2 } and f(x) = 2-3x for 0<x<0.7 is not coming in the range which is given in my textbook.

f(x) = y = 2-3x for every value of x which is a real number and x>0.

10. Apr 12, 2012

### Mentallic

Re: Problem regarding "relations and functions".

Do you know how to graph y=2-3x ? Now you just need to cut out the function where x<0, what are you left with?

No. First of all, f(x)>0 for 0<x<2/3. How do we know this? Because we have f(x)=2-3x so if f(x)>0 then we are looking for all the points such that 2-3x>0. A little algebra:

2>3x

x<2/3

And since 2/3 = 0.666... that means that for x=0.7, f(x)<0. Just plug x=0.7 in yourself to see.

Because it said for all x>0, not just 0<x<2/3. What about for x>2/3?

11. Apr 20, 2012

### sankalpmittal

Re: Problem regarding "relations and functions".

Have been offline for a couple of says. Am I too late here ?

Yes , I know how to graph the function and it'll be a straight line.

Question says that (emphasizing bold parts :)

Find the range of the function :

f(x) = 2-3x , x∈R , x>0
Note : R is universal set containing real numbers

Here I have to cut out x<0 and x=0 so that I'm left with x>0 only.

Range is the set containing second elements or "images" of set "f from two non empty sets R and its subset" defined like {(x,f(x))} where range = y = f(x).

f:R→R

Ok when I take 2/3>x>0 , I get f(x)>0. When I take x=2/3 , I get f(x)=0 . When I take x>2/3 , I get f(x)<0. So how can I precisely get the range of this function ?

Yes range =? One element we can write as -∞. Other we can write 0. But what about when 2/3>x>0 ?

That's what I wrote....
x>2/3 , I get f(x)<0 .

x>2/3 , I get f(x)<0 .
f(x)>0 for 0<x<2/3
When I take x=2/3 , I get f(x)=0

That covers the range of x>0.

12. Apr 20, 2012

### Mentallic

Re: Problem regarding "relations and functions".

Right, but I'm troubled as to why you can't figure out what the range is.
What's f(0)?
It's a straight line so where is it headed as x gets large?

Graphing the function should help.

13. Apr 21, 2012

### HallsofIvy

Staff Emeritus
Re: Problem regarding "relations and functions".

NO. Yor text book does NOT say the range is {-∞, 2}. It says the range is (-∞, 2) and all of those values you calculated are in that range.

14. Apr 21, 2012

### sankalpmittal

Re: Problem regarding "relations and functions".

Oh yeah !

Range = (-∞,2) denotes open interval ! And this covers all aspects which I got :

That's the answer which was in my textbook.

Yes !!

Because range = {-∞, 2} is correct but makes no sense here because -∞ cannot be regarded as element. Am I right ?

I can also right the range like this :
range = r => r$\in$ (-∞, 2)

Ok now , second question :

(ii) Let f={(x,x2/(1+x2)):x∈R} be a function from R to R , i.e f:R→R . Determine the range of f.

So I guess ,
Range = [0,1) ?

Yes !

In my textbook ,
Range = {x: 0<=x<1}

Which is what I got !

What's difference between {-∞, 2} and (-∞, 2) ?

15. Apr 21, 2012

### Mentallic

Re: Problem regarding "relations and functions".

The first has the two values -∞ and 2 in the set of solutions, the second is a range between those two values, so it has an infinite number of solutions.