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From: W^3 on 20 May 2010 22:55
In article
<6b8aaa1a-fe95-4a74-988d-ea3ffe80772f(a)v18g2000vbc.googlegroups.com>,
Peter <pwolynes(a)yahoo.com> wrote:

> Hi,
>
>
> I am working with a rational function where both the numerator and
> denominator are quadratic polynomials,
> P(x) = q(x)/r(x)
> where
> q(x) = x^2 + b*x + c, and q(x) > 0 for all x in R
> r(x) = 1 + x^2
>
> I can show treat the problem as a simple unconstrained optimization
> problem and find that the stationary points of P are the roots to a
> quadratic eqn, one root of which is the max and the other is the min.
>
> I am wondering however if there is a simple way one can argue the
> existence of a max and min based on the general properties of the
> denominator and numerator given.

Write P(x) = 1 + [bx + (c-1)]/(x^2+1) = 1 + s(x). Note that s(x) -> 0
as x -> +/- oo.

The easy case is when b = 0. If b = 0 and c > 1, then s(x) is positive
everywhere, even, and strictly decreasing on [0, oo). It follows that
P has an absolute max of c at 0, and no minimum. Similarly, if b = 0
and c < 1, then P has an absolute min of c at 0, and no maximum. If b
= 0 and c = 1, then P is identically 1.

For nonzero b, s(x) will be positive for some values of x, negative
for others. So P < 1 some of the time and P > 1 some of the time, and
as already mentioned by Arturo, this implies P attains both an
absolute minimum (necessarily < 1) and and an absolute maximum
(necessarily > 1) on R.
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