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Re: Enrichment
Posted:
Aug 23, 2012 8:49 AM
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On Aug 20, 2012, at 5:29 PM, Dave L. Renfro <renfr1dl@cmich.edu> wrote:
> Mike's function, f(x) = (x-1)(x-2) / (x+1)(x+2),
> is a quadratic divided by a quadratic.
This is how I used to attack these problems back in the algebra days. I guess you would call it "street fighting" math, a term coined in the following MIT article...
http://mitpress.mit.edu/catalog/item/default.asp?ttype=2&tid=12156
My object in this approach is to find the value of x where the function is at a min/max. I do this by mangling the original function, but in ways that do not alter the min/max point (the value of x). We can add or remove constants, scale by constants and, if you are brave, take the reciprocal of the function. The ultimate goal is to end up with a function that has an easily seen min/max. Of course, I assume that the problem has an answer and that the function acts nicely around it.
A simple example is the general quadratic...
ax^2 + bx + c
The "c" term is constant and we can drop that (doesn?t affect x at max/min).
ax^2 + bx
Divide by "a"
x^2 + (b/a)x
Add a constant to complete the square (this won?t affect the maximum).
x^2 + (b/a)x + (b/2a)^2
Factor...
(x +b/2a)^2
This is now just a square and its max/min occurs when x + b/2a = 0
x = -b/2a
So I thought I would apply this to Mike?s problem...
(x-1)(x-2)/(x+1)(x+2)
Expand...
(x^2 - 3x + 2) / (x^2 +3x + 2)
Change the numerator...
((x^2 + 3x + 2) - 6x) / (x^2 + 3x + 2)
This simplifies to...
1 - 6x/(x^2 + 3x + 2)
We can get rid of the constant (1) and the multiplicative constant (-6)...
x / (x^2 + 3x + 2)
Take the reciprocal of this (remember my disclaimer)...
(x^2 + 3x + 2) / x
This becomes...
x + 3 + 2/x
Get rid of the constant (3) and we end up with...
(x^2 + 2) / x
We can add a constant (-2sqrt(2)) which will complete the square in the numerator...
(x^2 + 2) / x - 2sqrt(2) = (x^2 - 2*sqrt(2)x + 2) / x
= (x - sqrt(2))^2 / x
= [(x - sqrt(2)) / sqrt(x) ]^2
This is a square and its min/max occurs when the inner contents is zero which would be when (x - sqrt(2)) is zero.
Thus x = sqrt(2)
Bob Hansen
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