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Re: Enrichment
Posted:
Aug 20, 2012 5:29 PM
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Peter Duveen wrote (in part):
http://mathforum.org/kb/message.jspa?messageID=7870219
> In the current presentation I am working on, quotients
> drop out before the limits are taken. So there is no need
> to explain the limit of the quotient of two quantities,
> both of which are approaching zero at the same time.
You can find "secant --> tangent" arguments in most of
the older analytic geometry texts (specifically, U.S. texts
roughly dating from 1910 to 1950), where it was considered
as not employing calculus methods, even though limits were
involved (but only for polynomial functions that can be
dealt with algebraically).
Even less calculus needs to be brought onto the scene
for sufficiently simple functions. For example, no limits
are needed to determine the maximum or minimum of quadratic
functions, as this can be done by completing the square
and using observations such as: The values of -2(x - 3)^2 + 5
are always less than or equal to 5.
I wrote about a slightly more involved method that works
for more complicated functions nearly 4 years ago in
an AP-calculus post (archived at Math Forum). Below, between
the horizontal ##-lines, is an excerpt from that post.
The entire post is at this URL:
http://mathforum.org/kb/message.jspa?messageID=6528441
One advantage of looking at something like what I give
below is that it certainly qualifies (in today's high
school environment) as enrichment material, and it also
serves to greatly strengthen algebraic skills in a context
where you are actually applying those skills.
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Mike's function, f(x) = (x-1)(x-2) / (x+1)(x+2),
is a quadratic divided by a quadratic. This is a
type of function I've seen discussed in a number
of papers in old volumes of elementary journals
and in old algebra texts. An example of such a
paper is Darboux's paper below, but I've seen many
others as well. I've also seen quadratics divided
by quadratics on a lot of French admissions/scholarship
exams that one can find reprinted in "Journal de
Mathematics Elementaries", "Mathesis Recueil
Mathematique", etc. These papers, texts, and
exam appearances are mostly from the 1880s to
the early 20th century, by the way.
Jean Gaston Darboux, "Discussion de la fraction
(ax^2 + bx + c)/(a'x^2 + b'x + c')", Nouvelles
Annales de Mathematiques (2) 8 (1869), 81-86.
http://books.google.com/books?id=jRgAAAAAMAAJ&pg=PA81
I essentially said this back on 27 August 2008,
along with a challenge:
- ----------------------------------------------
http://mathforum.org/kb/message.jspa?messageID=6352389
Those interested in a challenge might want to consider
how you can use non-calculus concepts (quadratic formula
and max/min of parabolas topics) to investigate where
the graph of y = (ax^2 + bx + c)/(dx^2 + ex + f) has
a local maximum value and/or a local minimum value.
This used to be a standard problem given in "contests
for admission" to French universities, starting around
1880 or so through at least (I think) the 1920s. The topic
can also be found in most of the more advanced algebra
texts from this period, many dozens of which have now
been digitized by google.
- ----------------------------------------------
No one replied that I know of, so I'll demonstrate
the non-calculus method using Mike's function.
I'll go through the calculus method first.
[calculus solution omitted in math-teach re-posting]
For the "precalculus method", we note that (for
sufficiently nice functions) we can detect the
presence of a strict local extrema by investigating
the intersection of a variable horizontal line with
the graph as the horizontal line is moved vertically.
For example, if we get no intersection points with
the line y = c when c < 4, exactly one intersection
point with the line y = 4, and exactly two intersection
points with the line y = c when c > 4, then y = 4 is
a local minimum value. The same is true if, instead
of "each c < 4" and "each c > 4", this holds for all
values of c in some left neighborhood of 4 and all
values of c in some right neighborhood of 4.
We begin by rewriting the equation in implicit form
as a quadratic expression in the variable x:
x^2y + 3xy + 2y = x^2 - 3x + 2
(y-1)x^2 + 3(y+1)x + 2(x-1) = 0
We can see from this that, for each value of y,
there exist at most 2 values of x such that the
pair (x,y) belongs to the graph. Since specifying a
value of y corresponds to looking at the intersection
of the graph with a certain horizontal line, it
follows from the previous paragraph that we want
to determine for which values of y there is no
solution for x, for which value(s) of y there
is exactly one solution for x, and for which
values of y there are two solutions for x.
But this is exactly what the discriminate of the
quadratic tells us!
If b^2 - 4ac < 0, there are no (real) solutions.
If b^2 - 4ac = 0, there is exactly one solution.
If b^2 - 4ac > 0, there are two solutions.
In our case,
b^2 - 4ac = 9(y+1)^2 - 8(y-1)^2
= 9(y^2 + 2y + 1) - 8(y^2 - 2y + 1)
= y^2 + 34y + 1
This is equal to zero when
y = [-34 +/- sqrt(34*34 - 4*1*1)] / 2
y = -17 +/- sqrt(1152)/2
y = -17 +/- 12*sqrt(2),
the same values of y we got earlier.
Also, since b^2 - 4ac = y^2 + 34y + 1 is
positive for y < -17 - 12*sqrt(2) (I know this
immediately from knowing what the graph of an
upward opening parabola looks like relative to
the horizontal axis), zero for y = -17 - 12*sqrt(2),
and negative for all values of y in a certain right
neighborhood of -17 - 12*sqrt(2), I know that
y = -17 - 12*sqrt(2) corresponds to a local maximum
value.
To find the x-coordinate for this local maximum,
we substitute y into the original equation and
solve for x.
(y-1)x^2 + 3(y+1)x + 2(x-1) = 0
The task is much easier than you might think,
since we already know that the square root part
of the quadratic formula is zero.
Thus, using the quadratic formula, we have
x = [-3(y+1) +/- sqrt(0)] / 2(y-1)
x = (-3/2) * (y+1)/(y-1)
= (-3/2) * (-16 - 12*sqrt(2)) / (-18 - 12*sqrt(2))
= (-3/2) * (8 + 6*sqrt(2)) / (9 + 6*sqrt(2))
{now rationalize the denominator}
= (-3/2) * [(8 + 6*sqrt(2))*(9 - 6*sqrt(2))] / (81 - 72)
= (-3/2) * [72 + (-48 + 54)*sqrt(2) - 72] / 9
= (-3/2) * [6*sqrt(2)] / 9
= (-3/2) * (2/3) * sqrt(2)
= -sqrt(2),
which is the value of x we found earlier, using
calculus to find where the maximum occurs.
In the same way, we can show that y = -17 + 12*sqrt(2)
is a local minimum value and show that it corresponds
to exactly one value of x, namely x = +sqrt(2).
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Dave L. Renfro
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