Finding set of possible values of a given function
newb...@gmail.com
I'm self teaching through a math's book and at the end of a chapter on
inequalities. They're currently being employed to find the possible
values of functions. I can do the exercises but I don't really
understand fully WHY it works!
The last question was asking about (x + 1) / (2x^2 + x + 1).
I converted that into a quadratic equation in 'x':
2y^2 + (y - 1)x + (y - 1) = 0
And the next step was to look for "real roots" of this quadratic;
i.e. that satisfy the inequality b^2 - 4ac >= 0.
It's at this stage where my understanding of the meaning behind the
process breaks down. I'm having trouble understanding the correlation
of real roots of the quadratic with the range of the function.
I thought the roots just showed you where the function intersected the
x-axis. I've seen graphs that don't intersect the x-axis at all (i.e.
graphs of f(x) = x^2 + 1). That doesn't have real roots, but it has an
output range!
I've been working with this for over a week now and even went over the
whole inequality chapter again but I'm still not "getting it"!. Hoping
for some great explanations.
Regards,
Pete
newb...@gmail.com
> I converted that into a quadratic equation in 'x':
>
> 2y^2 + (y - 1)x + (y - 1) = 0
Ray Vickson
Let F(x) = (x + 1) / (2x^2 + x + 1). Note first that the denominator D
(x) = 2x^2 + x + 1 is always > 0 for all x (because it is > 0 at x = 0
and it has no real roots). The denominator is > 0, while the numerator
N(x) = x + 1 changes sign at x = -1. Therefore, F(x) takes on both
positive and negative values. Note also that for large |x|, the
denominator is much, much larger than the numerator, so F(x) behaves
like 1/(2x). In fact, F(x) = x(1 + 1/x)/[2x^2 (1 + 1/(2x) + 1/
(2x^2) )] = (1/(2x)) *[factor that -> 1 as |x| goes to infinity]. In
other words, F(x) --> 0 as |x| --> infinity. Basically, this fact,
together with the fact that the numerator never vanishes, implies that
|F(x)| remains bounded, so F(x) itself remains between some lower
bound L < 0 and some upper bound U > 0. The achievable values (the
range of F) will be the interval L <= y <= U.
We can find L and U by your method, as follows. To find U, you want to
know the smallest value of y so that F(x) <= y always holds. At v = U,
the line y = v should just touch the graph y = F(x) at one point, so
we can look for a root of F(x) = v that is "single". Re-write the
equation F(x) = v as (x+1) = v*(2x^2 + x + 1), or (2v)x^2 + (v-1)x +
(v-1) = 0, whose roots are x = {(1-v) +- sqrt[(v-1)^2 - 4(2v)(v-1)]}/
(4v). Since we have a "single" root at the touching point, we need the
square root part to vanish, so we need (v-1)^2 -(8v)(v-1) = 0, giving
v = 1 or v-1 - 8v = 0, hence v = -1/7. The value v = 1 cannot be the
minimum because we know already that F(x) can have negative values;
therefore, the root v = 1 is the maximum (giving x = 0 in the root-
formula for x). The other root v = -1/7 actually gives the minimum,
because the argument for finding L is the same: the line y = v (v < 0)
must touch the graph y = F(x) at just one point when v = L. Thus, L =
-1/7, occurring at x = -2 (by plugging v = -1/7 in the root formula
for x).
The above method for finding L and U is somewhat ad hoc. A more
generally-applicable method is to use calculus, but I don't know if
you would be in a position to use such techniques. Anyway, if you know
some calculus, you can try to maximize and minimize F(x) by setting
the derivative to zero. The derivative of F is F'(x) = -2x*(x+2)/
(2x^2+x+1)^2, which vanishes at x = 0 and x = -2. We could perform a
second-derivative test to check which point is the maximum and which
the minimum, but we can reason it out more directly. F is < 0 for x <
-1 and is > 0 for x > -1. F has a maximum or minimum at x = 0, is > 0
there, while going to zero as x goes to + infinity. That means that x
= 0 cannot be the minimum (do you see why?); it must be the maximum.
Thus, U = F(0) = 1. It now follows that x = -2 must be the minimum
(actually, the minimizer) of F, so L = F(-2) = -1/7.
R.G. Vickson
>
> Regards,
>
> Pete
newb...@gmail.com
Just a quick note to say thank you. I've been working through your e-
mail for about an hour - especially paragraph 2 about finding L and U.
It's not fully making sense to me, but I think it might be easier
tomorrow as it's late here now.
One thing you do is set the function equal to v. This is really
interesting. I've always seen the function set equal to zero and have
only ever really thought about finding root that are x-intercepts. I
didn't know you could set it equal to another value to find the roots
there. It sounds so fundamental a thing to want to do though.
The next chapter introduces differentiation so at the moment I'm still
pre-calculus. I really want to understand inequalities and quadratics
as much as possible before moving on though.
Will let you know how I get on when I can visit your e-mail again.
Regards,
Pete
rgvi...@gmail.com
> Ray,
>
> Just a quick note to say thank you. I've been working through your e-
> mail for about an hour - especially paragraph 2 about finding L and U.
Think about it graphically. Near the maximum of F(x), the graph y = F
(x) looks like a hilltop. A horizontal line that is too high will miss
the hilltop altogether; this is the case when v is too large, so that
all x give F(x) < v. As we keep lowering v, we get to the scenario
where the line of height v just grazes the hilltop. Now look what
happens if we lower the line just a bit more. The line will now pass
through the hill, entering on one side and leaving on the other. This
would be the case where the equation F(x) = v has two roots (one at
the entry point and the other at the exit point---just like the entry
and exit wounds so commonly mentioned in forensic television dramas).
At the grazing point we want the two roots to coalesce into one. That
would be the case where the quadratic equation has just one root, due
to the vanishing of the +-sqrt() part. The same thing applies to the
minimum, except there we are looking at a valley bottom instead of a
hilltop.
Anyway, the best way to understand what is happening is to make a
simple sketch, drawing a hypothetical graph y = F(x) near the peak
value and looking at horizontal lines y = v for various v near the
maximum.
> It's not fully making sense to me,
As I said: draw a picture.
R.G. Vickson
newb...@gmail.com
> Think about it graphically. Near the maximum of F(x), the graph y = F
> (x) looks like a hilltop. A horizontal line that is too high will miss
> the hilltop altogether; this is the case when v is too large, so that
> all x give F(x) < v. As we keep lowering v, we get to the scenario
> where the line of height v just grazes the hilltop. Now look what
> happens if we lower the line just a bit more. The line will now pass
> through the hill, entering on one side and leaving on the other. This
> would be the case where the equation F(x) = v has two roots (one at
> the entry point and the other at the exit point---just like the entry
> and exit wounds so commonly mentioned in forensic television dramas).
> At the grazing point we want the two roots to coalesce into one. That
> would be the case where the quadratic equation has just one root, due
> to the vanishing of the +-sqrt() part. The same thing applies to the
> minimum, except there we are looking at a valley bottom instead of a
> hilltop.
>
> Anyway, the best way to understand what is happening is to make a
> simple sketch, drawing a hypothetical graph y = F(x) near the peak
> value and looking at horizontal lines y = v for various v near the
> maximum.
That all makes sense and think I already understood most of it.
I've gone over your paragraph 2 from your previous e-mail again and
that's sunk in well now. A good night's sleep does help!
There's something still not clicking into place though. Maybe I can
give another example which doesn't intercept the x-axis to show what I
mean:
y = (1 + x^2) / x.
Changing it into a quadratic in 'x':
x^2 - yx + 1 = 0.
I know that values that satisfy the discriminant >=0 identify the
range of the output.
b^2 - 4ac >=0
--> y^2 - 4 >=0
--> roots are -2 and 2 and the inequality is satisfied when y <= -2
and y >= 2.
But I don't know why. Previously the book's instructed me to inspect
the discriminant of a quadratic and > 0 means 2 x-intercepts, = 0
means one x-intercept (like y = x^2 just touches the x-axis), and < 0
means a non-real/complex x-intercepts (which I've not really learnt
about yet).
So how does applying discriminant >= 0 work when looking at possible
values of y = (1 + x^2) / x?
Thanks for your help so far!
Regards,
Pete
Dave L. Renfro
> So how does applying discriminant >= 0 work when looking
> at possible values of y = (1 + x^2) / x?
Rather than write up a reply specific to what you're asking
(because I only have a few minutes to spare this morning for
"sci.math playing"), below is a copy of a post I wrote in
another group a couple of months ago that may be useful.
**********************************************************
**********************************************************
ap-calculus -- Suggest 'interesting' rational functions?
(Dave L. Renfro; 10 December 2008)
http://mathforum.org/kb/message.jspa?messageID=6528441
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.
Taking the derivative of y = (x^2 - 3x + 2)/(x^2 + 3x + 2)
gives a rational expression whose numerator is
(x^2 + 3x + 2)(2x - 3) - (x^2 - 3x + 2)(2x + 3)
= 6x^2 - 12,
which is zero for x = +/- sqrt(2).
Because 6x^2 - 12 is positive to the left and right
of these values and negative between them (see [1]),
the first derivative test tells us:
-sqrt(2) corresponds to a local maximum
+sqrt(2) corresponds to a local minimum
[1] The graph of 6x^2 - 12 is a parabola that
opens upward. We don't need to worry about the
denominator of the expression for y' -- the
denominator is the square of something, and
hence the sign of y' is the same as the sign
of the numerator expression.
It will be useful later to find the y-coordinates
of the turning points:
y = (2 -/+ 3*sqrt(2) + 2) / (2 +/- 3*sqrt(2) + 2)
y = (4 -/+ 3*sqrt(2)) / (4 +/- 3*sqrt(2))
Rationalizing the denominator (to make our later
comparison easier), we get
y = (4 -/+ 3*sqrt(2))^2 / (4^2 - 9*2)
y = (16 -/+ 24*sqrt(2) + 18) / (-2)
y = -17 +/- 12*sqrt(2)
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 discriminant 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).
**********************************************************
**********************************************************
Dave L. Renfro
Ray Vickson
y = f(x) = x + 1/x. For x > 0 this has a graph that looks like a
valley with a sharp (infinitely high) cliff on one side (as x --> 0
from the right) and a gentle (but infinitely high) slope on the other
(as x --> goes to infinity on the right). For x < 0 the graph is
reversed and upside-down. For y >= 2 you are in the x > 0 region of
the graph, while for y <= -2 you are in the x < 0 region. You can see
a sketch of what y = f(x) looks like (for x > 0) in Figure 2, page 4
of
http://www.cargalmathbooks.com/The%20EOQ%20Formula.pdf , or in the
"total cost" graph in
http://people.brunel.ac.uk/~mastjjb/jeb/or/invent.html . Then, take
its mirror image of the graph and flip it over to see what it looks
like for x < 0.
>
> Changing it into a quadratic in 'x':
>
> x^2 - yx + 1 = 0.
>
> I know that values that satisfy the discriminant >=0 identify the
> range of the output.
>
> b^2 - 4ac >=0
> --> y^2 - 4 >=0
> --> roots are -2 and 2 and the inequality is satisfied when y <= -2
> and y >= 2.
>
> But I don't know why.
Draw a picture. Often it is much easier to understand a concept by
looking first at a graph and only then worrying about the algebra.
> Previously the book's instructed me to inspect
> the discriminant of a quadratic and > 0 means 2 x-intercepts, = 0
> means one x-intercept (like y = x^2 just touches the x-axis), and < 0
> means a non-real/complex x-intercepts (which I've not really learnt
> about yet).
>
> So how does applying discriminant >= 0 work when looking at possible
> values of y = (1 + x^2) / x?
The graph has two separate, unconnected "branches". The x > 0 branch
has y >= 2, while the x < 0 branch has y <= -2. Again, just draw a
picture.
R.G. Vickson
>
quasi
>y = (1 + x^2) / x.
>
>Changing it into a quadratic in 'x':
>
>x^2 - yx + 1 = 0.
>
>I know that values that satisfy the discriminant >=0 identify the
>range of the output.
Let me try to clarify some of the "why" ...
Let f be the function
f(x) = (1 + x^2)/x
Then the graph of f is the graph of the equation
y = (1 + x^2)/x
Thus, the range of f is the set of all y-values for which there is at
least one x-value satisfying the above equation.
Let's apply this concept.
Suppose I asked you if the number 3 is in the range of f. That's
equivalent to asking whether the equation
3 = (1 + x^2)/x
has a real root. Cross multiplying (but noting the restriction that x
can't be zero), you can convert the above equation into the quadratic
equation
x^2 - 3x + 1 = 0
and then, using the discriminant test on this new equation, you can
see immediately, without actually solving for x, that the equation
_does_ have a real root, in fact 2 of them. Hence the value 3 _is_ in
the range of f. Agreed?
Note that the real roots of this new equation do not correspond to
x-intercepts for the original equation, since, relative to the
original equation, the y-value is 3, not 0. In fact, as you noted, the
original equation doesn't have any x-intercepts.
This seems to be one of the past points of confusion for you. Finding
real roots of an equation f(x) = 0 is equivalent to finding
x-intercepts on the graph of y = f(x). But if you take a _different_
equation, say f(x) = 3, the real solutions of this equation are not
x-intercepts for y = f(x). They _can_ be interpreted as x-intercepts
for the new equation y = g(x), where g(x) = f(x) - 3. In general, with
2 different functions, f(x) and g(x), there's no reason to expect the
x-intercepts of one to be the same as the x-intercepts of the other.
Going back to the question about the range f(x), we could just as
easily answer the question as to whether y = 4 is in the range. In
exactly the same way, we could answer it for y = 5/2, or y = 1, or y =
-9 or any other specific value of y. In each case we would get a
quadratic equation (but a different one for each y), and then testing
the discriminant we would know whether that y-value is in the range.
But rather than testing the y-values one at a time, instead, why don't
we leave y unknown? Then, we can do the same thing as before, but this
time the quadratic equation in x has coefficients expressed in terms
of the unknown y. As before, we can apply the discriminant test to get
an inequality condition, expressed in terms of y, which determines
whether or not there will be a real x for the given y. Thus, a y-value
is in the range of f if and only if the inequality is satisfied. In
other words, the set of y-values which satisfy that inequality _is_
the range we seek. Do you follow the reasoning?
quasi
newb...@gmail.com
> On Sun, 22 Feb 2009 01:29:46 -0800 (PST), newbar...@gmail.com wrote:
> >y = (1 + x^2) / x.
>
> >Changing it into a quadratic in 'x':
>
> >x^2 - yx + 1 = 0.
>
> >I know that values that satisfy the discriminant >=0 identify the
> >range of the output.
>
> Let me try to clarify some of the "why" ...
>
>
> But rather than testing the y-values one at a time, instead, why don't
> we leave y unknown? Then, we can do the same thing as before, but this
> time the quadratic equation in x has coefficients expressed in terms
> of the unknown y. As before, we can apply the discriminant test to get
> an inequality condition, expressed in terms of y, which determines
> whether or not there will be a real x for the given y. Thus, a y-value
> is in the range of f if and only if the inequality is satisfied. In
> other words, the set of y-values which satisfy that inequality _is_
> the range we seek. Do you follow the reasoning?
Yes, I understand the reasoning behind it now! I will go away now to
do some more of the exercises with this in mind.
I'll come back (probably tomorrow now) and read Dave Renfro's and Ray
Vickson's replies.
Very grateful for everyone's help. Once this is licked, then it's onto
calculus which I'm really looking forward to but I'm sure will throw
me into confusion again :)
Regards,
Pete
Michael Press
<e713c94f-0036-4392...@n10g2000vbl.googlegroups.com>,
newb...@gmail.com wrote:
Your progress in calculus will be smoothed by acquaintance
with the kinds of questions you are addressing. Consider
the graph of a function y = f(x).
What is the graph of
f(x) + 1
f(x) + a
f(x) - a
-f(x)
a*f(x)
f(x - a)
f(x + a)
f(a*x)
f(x/a)
f((x - a)/b)
c*f((a * x + b)
?
--
Michael Press
newb...@gmail.com
> Your progress in calculus will be smoothed by acquaintance
> with the kinds of questions you are addressing. Consider
> the graph of a function y = f(x).
>
> What is the graph of
I'll try to answer the ones I think I know now and will research the
others. If any of my answers/terminology are suspect, please let me
know.
> f(x) + 1
f(x) translated by one positive unit on the positive y-axis i.e. one
unit "higher".
> f(x) + a
f(x) moved up by a units.
> f(x) - a
f(x) moved down by a units.
> -f(x)
f(x) reflected in the x-axis.
> a*f(x)
Not sure. My instinct tells me a tighter, narrower curve. A similar
shape as f(x) but more exaggerated on the y-axis.
> f(x - a)
Same as f(x) but translated a units right.
> f(x + a)
Same as f(x) but translated a units left.
> f(a*x)
Not sure. This is going to be some kind of left/right translation.
I'll come back about this and the remaining three.
> f(x/a)
> f((x - a)/b)
> c*f((a * x + b)
>
> ?
>
Thanks,
Pete