What is the range of f given by F(x)= 1/(x-1)(x+1), x>1?

[matt_crouch]

Homework Statement

Find the range of f

Homework Equations

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

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

 

[tiny-tim]

Hi matt_crouch!

Find the range of f
…
F(x)= 1/(x-1)(x+1) , x>1

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?

 

[Mark44]

Homework Statement

Find the range of f

Homework Equations

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

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)).

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

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.

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

 

[matt_crouch]

How can y decrease to infinity?

Because its a 1/x relationship isn't the graph asymptotic?
Isnt the range a part of the graph where a horizontal line cuts the graph at two points?

 

[tiny-tim]

Isnt the range a part of the graph where a horizontal line cuts the graph at two points?

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?

 

[Gauged]

Because its a 1/x relationship isn't the graph asymptotic?
Isnt the range a part of the graph where a horizontal line cuts the graph at two points?

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.

 

[matt_crouch]

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

 

[tiny-tim]

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

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

 

[matt_crouch]

ahh i think i got it.. is the range

0<y<1/3

?

 

[tiny-tim]

ahh i think i got it.. is the range

0<y<1/3

?

how did you get that?

hmm … I've just looked again at your original post …

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

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.

 

[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

 

[Gauged]

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

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

 

[matt_crouch]

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.

As x>1 then isn't the largest value that f(x) can be a 1/3 ? so isn't the range 0 < y < 1/3 ?

 

[tiny-tim]

As x>1 then isn't the largest value that f(x) can be a 1/3 ? so isn't the range 0 < y < 1/3 ?

No … as you said …

F(x)= 1/(x-1)(x+1) , x>1
…
I started by substituting x=2 and x=3 this gives me 1/3 and 1/8

… f(2) = 1/3.

f(1+) = … ?

 

[matt_crouch]

1/0 which is undefined?

 

[tiny-tim]

1/0 which is undefined?

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.

 

[matt_crouch]

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

 

[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.
You were asked for x>1.
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.

 

[matt_crouch]

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