# Finding the Range?

1. Dec 19, 2008

### matt_crouch

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

Find the range of f

2. Relevant equations

F(x)= 1/(x-1)(x+1) , x>1

3. The attempt at a solution

I started by substituting x=2 and x=3 this gives me 1/3 and 1/8 respectively so i get that as x increases y decreases to infinity but how do i represent that as

A<y<B

its just the notation and stuff i am struggling on.
cheers

2. Dec 19, 2008

### tiny-tim

Hi matt_crouch!
Be systematic …

wirte it out as an equation in x …

you should get a quadratic equation …

for what values of f does it have real solutions?

3. Dec 19, 2008

### Staff: Mentor

Do you mean this as $$F(x) = \frac{1}{(x -1)(x + 1)}$$? If so, you should have put another pair of parentheses around the two factors involving x, like so: F(x) = 1/((x - 1)(x + 1)).
How can y decrease to infinity?

You're given that x > 1, so both factors in the denominator (I'm assuming that's what you meant) are going to be positive. The numerator is 1, which is positive. The closer x gets to 1, the larger F(x) will be. As x gets large, F(x) will get closer to zero, but will remain positive.

4. Dec 19, 2008

### matt_crouch

Because its a 1/x relationship isnt the graph asymptotic?
Isnt the range a part of the graph where a horizontal line cuts the graph at two points?

5. Dec 19, 2008

### tiny-tim

No, the range is those parts of the vertical axis where a horizontal line cuts the graph at at least one point.

Anyway, what quadratic equation did you get?

6. Dec 19, 2008

### Gauged

Yes, the graph of 1/x is asymptotic by definition. However, when a horizontal line (that is not part of the original graph) intersects the graph at more than one point, then the function in not one-to-one.

7. Dec 20, 2008

### matt_crouch

Im not quite sure how you can write it out as an quatratic. should i find the inverse of f(x)

8. Dec 20, 2008

### tiny-tim

x2 - 1 - 1/f = 0 … for what values of f are there solutions?

9. Dec 20, 2008

### matt_crouch

ahh i think i got it.. is the range

0<y<1/3

?

10. Dec 20, 2008

### tiny-tim

how did you get that?

hmm … i've just looked again at your original post …
which didn't seem to me to make any sense originally, but now it's dawned on me that you probably meant "zero" not "infinity" …

in which case the answer to your original question "how do i represent that as A<y<B" is that if you know that y decreases to zero (but never quite makes it), then you write that simply as y > 0.

11. Dec 20, 2008

### matt_crouch

ye i was getting confused because as the graph was asymtopic i was thinking that it would go to infinity but the domain would go on for infinity whereas the range which is the "output" would never quite reach it.

cheers though :D

12. Dec 21, 2008

### Gauged

The "easy" way to find the range is to find the domain of a functions inverse.

13. Dec 22, 2008

### matt_crouch

As x>1 then isnt the largest value that f(x) can be a 1/3 ? so isnt the range 0 < y < 1/3 ?

14. Dec 22, 2008

### tiny-tim

No … as you said …
… f(2) = 1/3.

f(1+) = … ?

15. Dec 22, 2008

### matt_crouch

1/0 which is undefined?

16. Dec 22, 2008

### tiny-tim

Sort-of …

f(1+) = ∞ (and f(1-) = -∞) …

but we're not really supposed to regard ∞ as a number

so instead of 0 < f < ∞, we just say 0 < f.

17. Dec 22, 2008

### matt_crouch

ahhhh is seeeee aaawesome..
massive help cheers Tiny Tim.

18. Dec 22, 2008

### epenguin

The substance (I don't know what jawbreaking terminology you are required to use but once you get the substance you can adapt it to that )

Either do or don't write it as 1/(x2-1) according to what seems to you easiest.
So where do you start? At x=1 What is f(1)?
Where do you end? At x=∞. What is f(∞)?
f is continuous between these two values of x. So f has to include all values between f(1) and f(∞)?

For x between those 1 and ∞, for f to go out of the range between f(1) and
f(∞) f would have to have an extremum in the range of x. First does it look like it?

Then you could formally demonstrate it hasn't.

Actually I think the simplest argument is f would have to be increasing with x somewhere in the range. You don't need calculus to show that it can't.

Last edited: Dec 22, 2008
19. Dec 22, 2008

### matt_crouch

Ye cheers epenguin and tiny tim u have really helped me out =]