1. ## functions

R is a set of real numbers. A function f from R into R is given by f(x)=x/(x^2 +1).
Show that if a and b are real numbers with a>b>=1, then f(b) -f(a)=((a-b)(ab-1))/((1+a^2)(1+b^2))

well, I felt kind of confused here, as I thought I just have to do the sum, but then what's the point of providing "a>b>=1"?

And, deduce that f(b)>f(a)

same here, I thought that was pretty obvious. How should I deduce it???

2. ## Re: functions

What do you mean by "the sum"? Mathematics and vagueness do not mix very well.

Have you proven yet that:

$f(b) - f(a) = \dfrac{(a - b)(ab - 1)}{(1 + a^2)(1 + b^2)}$?

The point being, if we are to show it is positive, then both numerator and denominator have to have the same sign. Now the denominator is the product of two sums of squares. Squares are always positive, so the sum of squares is always positive, so the product in the denominator is positive. Therefore the whole fraction is positive if (and only if) the numerator is positive. How would you show that:

$a - b > 0$ and $ab - 1 > 0$?

3. ## Re: functions

The proviso is because dividing by small fractions makes quantities bigger.

4. ## Re: functions

Hi

On the matter of signs - each factor could potentially be + or -

eg (+)(+)/(+)(+) , (+)(-)/(+)(+) ...

As the two in the denominator can never be -ve as already stated, the numerator possibilities are (+)(+) , (+)(-) , (-)(+) , (-)(-)

The first and last will be +ve so this is where the a > b >= 1 condition is used.

Hope this helps

Regards