# range question

Joined Dec 29, 2004
83
Hi
please I need some help to dtermine the range of the following function:

f(x,y)=(e^x - e^y)/ (e^x +e^y)

the domain is all the real for x and y.

I know that the answer for the range is [-1,1] but I do not know how they found it.

Thank you

Joined Jan 19, 2004
220
one easy way to find this out would be to find maxima -minima using derivatives
since its a a function of two independent variables
the following conditions are to be satisfied:-
∂ f/ ∂ x = 0
∂ f/ ∂ y = 0
this will give you the pair of critical points(x,y) where optima may exist

let r= ∂ ^2 f/ ∂x^2 ; s= ∂ ^2 f/ ∂y^2; t= ∂ ^2 f/ ∂x ∂ y;

substituting each of the above critical points into the second order derivatives we check for the following conditions:
1. rt-s*s <0 no maxima or minima
2. rt-s*s > 0 => maxima if r<0; minima r>0
3. rt-s*s = 0 => test is inconclusive and an alternate means must be established

try attemptin a solution using this fact

Joined Dec 29, 2004
83
Originally posted by haditya@Oct 7 2005, 05:00 AM
one easy way to find this out would be to find maxima -minima using derivatives
since its a a function of two independent variables
the following conditions are to be satisfied:-
∂ f/ ∂ x = 0
∂ f/ ∂ y = 0
this will give you the pair of critical points(x,y) where optima may exist

let r= ∂ ^2 f/ ∂x^2 ; s= ∂ ^2 f/ ∂y^2; t= ∂ ^2 f/ ∂x ∂ y;

substituting each of the above critical points into the second order derivatives we check for the following conditions:
1. rt-s*s <0 no maxima or minima
2. rt-s*s > 0 => maxima if r<0; minima r>0
3. rt-s*s = 0 => test is inconclusive and an alternate means must be established

try attemptin a solution using this fact
[post=10846]Quoted post[/post]​
I didi it it!
Thanks a lot

#### barrygale2003

Joined Oct 11, 2005
2
Here is a simpler method:

If x can be any number (positive or negative) then e^x varies between "0" (it can't actually reach zero, but can get as close as you like) and + infinity. Same for e^y of course.

Let u=e^x and v=e^y. Then the fraction is (u-v)/(u+v), which is easier to think about.

To get a feel for the fraction, look at some extreme cases :
When u>>v (v is negligible by comparison with u) then the fraction becomes u/u which is +1.
When u<<v (u is negligible by comparison with v) the fraction becomes -v/v which is -1.
When u=v the fraction becomes 0/2u which is 0 (provided u is not zero).

This suggests that the fraction varies between +1 and -1, but isn't conclusive - there might be some weird behaviour in between. We need a logical proof.

Because u and v are both positive their sum (u+v) is always going to be bigger than their difference (u-v). So the fraction will always be "bottom heavy", ie less than 1 (ignoring the sign). It could only reach 1 if one of u or v is exactly zero, but that can't happen (e^x never reaches zero no matter how large and negative x is).

What if both u and v were zero? Then we could have a problem to decide about the value of 0/0. Well, neither can reach zero, so there's no problem.

So the fraction varies between (but never actually equals) +1 and -1.

Originally posted by haditya@Oct 7 2005, 11:00 AM
one easy way to find this out would be to find maxima -minima using derivatives
since its a a function of two independent variables
the following conditions are to be satisfied:-
∂ f/ ∂ x = 0
∂ f/ ∂ y = 0
this will give you the pair of critical points(x,y) where optima may exist

let r= ∂ ^2 f/ ∂x^2 ; s= ∂ ^2 f/ ∂y^2; t= ∂ ^2 f/ ∂x ∂ y;

substituting each of the above critical points into the second order derivatives we check for the following conditions:
1. rt-s*s <0 no maxima or minima
2. rt-s*s > 0 => maxima if r<0; minima r>0
3. rt-s*s = 0 => test is inconclusive and an alternate means must be established

try attemptin a solution using this fact
[post=10846]Quoted post[/post]​