Sign charts and the first derivative test
Dave L. Renfro
to do with sign charts in a non-usenet group devoted
to AP-calculus. The present post is a collection of 8
of those posts that I want to assemble in one place.
Besides the advantage of having them in one place when
I want to refer them to a poster (posters in that group,
posters in sci.math, or elsewhere), or to refer them to
someone asking me a question by private e-mail, each of
the Math Forum archive methods currently has drawbacks
for these posts. Their standard font format doesn't
preserve empty characters put at the beginning of a new
line, which is essential to "see" the sign charts, and
their plaintext format (which does preserve the empty
characters put at the beginning of a new line), for some
reason, doesn't preserve empty lines (used to separate
one paragraph from another, or to separate the sign chart
from the text that appears before and after the sign chart).
The posts are listed in chronological order.
TABLE OF CONTENTS
1. FIRST DERIVATIVE TEST FOR f(x) = (x-2)*(x-3)^2
2. FIRST DERIVATIVE TEST FOR f(x) = [ (ln x)^2 ] / x
WITHOUT USING A CALCULATOR
3. FIRST DERIVATIVE TEST FOR f(x) = 2x*e^(2x) WITHOUT
USING A CALCULATOR
4. WHICH POINTS SHOULD BE USED FOR SIGN CHART DIVISION POINTS?
5. A SIGN CHART FOR sin(x^2) WITHOUT USING A CALCULATOR
6. TERRACE POINTS IN THE FIRST DERIVATIVE TEST
7. FIRST DERIVATIVE TEST FOR f(x) = (x^2 - 15)*e^(-x)
8. FIRST DERIVATIVE TEST FOR f(x) = ax - (a+1)*ln(x+1)
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1. FIRST DERIVATIVE TEST FOR f(x) = (x-2)*(x-3)^2
Posted: Apr 29, 2000 7:58 PM
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Jeanne Benecke <Mygir...@aol.com>
[Tue, 25 Apr 2000 20:12:50 -0400 (EDT)]
wrote
> I do not know if this question has already been addressed
> and if it has, my apologies. I was doing the 1993 AB exam
> with some students and we came upon question # 15 which read:
>
> If f(x)=(x-2)(x-3)^2, which of the following are a relative
> maximium of f?
> (A) -3 (B)? (C) ? (D) 7/3 (E) ?
>
> I do not recall the choices for B,C or E. The correct
> answer was 7/3. This test was a non calculator test so
> I had the students find f ' (x) =0 algebraically and test
> each interval. This was quite a bit of work that did give
> the correct answer, but took way more than the 3 minutes
> that I tell students to spend per question. I can't
> help but shake the feeling that I missed a short cut.
> Any suggestions? Thanks in advance and good luck to all
> on the next 2 weeks of insanity.
> Jeanne Benecke
> Tappan Zee High School
> Orangeburg, NY 10965
There is a certain precalculus topic that used to be taught
before the widespread use of graphing calculators that might
be what was intended. The presence of the non-repeated linear
factor x-2 causes the graph to cross through the x-axis at (2,0)
(with a positive angle), whereas the presence of (x-3)^2 causes
the graph to be tangent to the x-axis at (3,0) (in such a way
that a neighborhood of the curve at (3,0) lies on the same
side of the x-axis). Since the dominant term of (x-2)(x-3)^2 is
x^3, we know that the graph rises from the 3'rd quadrant into
the 1'st quadrant, with possible local detour(s) near the origin
due to the effect of the other terms when |x| isn't large. So,
starting off with large negative x's, the graph is in the 3'rd
quadrant, then passes through the x-axis at x=2, rises up
some -- how much, we don't know nor care -- and then falls back
down to the x-axis at x=3, after which the graph rises up
again for good.
Therefore, if none of the choices for (B), (C), (E) were
values between x=2 and x=3, choice (D) wins by default.
On the other hand, the calculus solution doesn't seem very
involved to me, but maybe I've done too many of these
kinds of problems to be an impartial judge anymore.
f'(x) = 1*(x-3)^2 + 2*(x-2)(x-3) = (x-3)*[x-3 + 2(x-2)]
= (x-3)*(3x-7)
To make a sign chart for f'(x), think about what the graph of
y = (x-3)(3x-7) looks like -- it's a concave up parabola that
passes through the x-axis at x=3 and x=3/7. This is all you
need, and no calculation is necessary if some basic properties
of quadratic graphs are known. Hence, f'(x) is positive to the
left of x=3, negative between 3 and 3/7, and positive to the
right of 3/7. Now apply the first derivative test.
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2. FIRST DERIVATIVE TEST FOR f(x) = [ (ln x)^2 ] / x
WITHOUT USING A CALCULATOR
Posted: May 8, 2000 6:34 PM
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Stephen Jenkins <SJen...@Montgomery.k12.ky.us>
[ap-calculus Mon, 8 May 2000 15:07:50 -0400]
wrote
> I have a question concerning the non-calculator part of the
> exam. Are students required to know approximations of ln's.
> for example, from Venture Publishing exam 1 there is a question
> concerning rel. max of f(x)= ((ln x)^2)/x. the student has to
> plug numbers in the intervals into f '(x) which is
> ((ln x)(2 - ln x))/x^2
>
> are they suppose to be able to find 2- ln 2 on their own to
> ind inc/dec intervals?
>
> There is probably something I'm missing here.
It doesn't appear to me that you need to know anything besides
ln(x) = a ==> x = e^a and the fact that ln(x) is an increasing
function. [Students should know what the graph of ln(x) looks
like--in particular that ln(x) is increasing--but, failing this,
it's automatic from the fact that (ln x)' = 1/x is positive.]
The domain consists of the open interval (0, infinity), and
so we only need to focus on these values. The denominator x^2
is positive for these values, and therefore can be ignored when
determining the sign of f'(x). The numerator is zero if and only
if x = 1 or x = e^2. Finally, ln(x) is an increasing function
and 2 - ln(x) is a decreasing function. [The latter follows
automatically from ln(x) being increasing, but you can also
just observe that (2 - ln x)' = -1/x is negative for the values
of x under consideration.] This is enough to construct a sign
chart for the derivative.
[Use fixed font size for what follows.]
__________________________________
| | |
x 0 1 e^2
ln(x) | ---- | ++++ | ++++
| | |
2 - ln(x) | ++++ | ++++ | ----
| | |
f'(x) | ---- | ++++ | ----
EXPLANATIONS:
Since ln(x) = 0 for x=1 and ln(x) is increasing, we know
that ln(x) is negative for 0 < x < 1 and positive for x > 1.
Since 2 - ln(x) = 0 for x = e^2 and 2 - ln(x) is decreasing,
we know that 2 - ln(x) is positive for 0 < x < e^2 and
negative for x > e^2.
Finally, although a decimal approximation for e^2 is not
needed, you would have to know that e^2 is larger than 1 to
get the correct ordering on this sign chart. However, e^2 > 1
is a consequence of the fact that e^x is increasing and 2 > 0.
[But students SHOULD know without thinking that e^2 > 1.]
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3. FIRST DERIVATIVE TEST FOR f(x) = 2x*e^(2x) WITHOUT
USING A CALCULATOR
Posted: May 24, 2000 9:44 PM
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Keith Ballantine <keith_ba...@eu.odedodea.edu>
[math-teach 23 May 00 10:25:08 -0400 (EDT)]
wrote (in part)
> This fascination with novel problems is something I disagree
> with, and, to me, almost every problem on the AP test seems
> to be a novel problem. In the front of the course description
> they list topics such as "An intuitive understanding of
> continuity" or "The Mean Value Theorem and its geometric
> consequences", and so on. Then a teacher prepares a course
> outline, gets a book, and teaches about these topics and has
> students solve a variety of problems on these topics.
>
> In an ordinary and rational world, it seems to me, a test
> should be designed where a typical student CAN do the problems
> if she or he has learned the topic, and CANNOT do them if
> the topic was not learned.
>
> For example, suppose we want to know if a student has learned
> how to figure out a relative minimum and an absolute minimum
> of a function. In the text they are taught to see if the first
> derivative is zero, and that the second derivative is positive,
> and to check the endpoints of the domain.
>
> So a nice polynomial of some sort will accomplish this nicely,
> since the students who have learned the concept as explained
> in the above paragraph will almost all be able to find the
> derivatives and check the end points or the value of the
> function at + or - infinity.
>
> But a recent AP test had the question
>
> f(x) = 2xe^(2x),
>
> find the absolute minimum value of f. Justify
> that your answer is an absolute minimum.
>
> The problem with this is, although you may discover a
> mathematician, material and can do maximum/minimum problems
> that are new to them (not ones where they've memorized the
> answers) can't do this problem because, even though they know
> what they are supposed to do, they can't figure out how to
> do it.
>
> I advocate the stream of thought in the "ed" world and to some
> extent math that teachers should "test on what is taught", and
> I believe that the ultimate results will be better than testing
> beyond what is taught because more students will stay with us
> and continue to study math.
>
> The hypothesis here is that math can be learned rather than
> being an ability inherited to be discovered by teachers, and
> if the tests follow what is learned rather than proceed it
> I think we'd end up with a lot more people who are good at
> doing math.
Are you sure you copied the problem correctly? The derivative
of what you gave is 2*e^(2x) + (4x)*e^(2x) = [e^(2x)]*(2 + 4x),
which equals zero <==> x = -1/2 (since e^(2x) is never zero).
Moreover, a sign chart for f'(x) is virtually trivial:
--------------------------------------------------
|
-1/2
e^(2x) +++++++++ | +++++++++++
|
2+4x --------- | +++++++++++
|
y' --------- | +++++++++++
The function is decreasing for x < -1/2, increasing for x > 1/2,
and so the absolute minimum must be at x = -1/2.
Most polynomials would be much harder to solve because
of factorization difficulties just to get the critical points.
My students would probably think I made a mistake if I put this
on a major test, although I might put it on a 10 minute
quiz. [They're used to seeing stuff like f(x) = (3x)^(2x),
f(x) = (ln x)^(ln x), f(x) = (ax)*ln(ax) for a > 0, etc. Of
course, I also work them to death on these types before a test.]
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4. WHICH POINTS SHOULD BE USED FOR SIGN CHART DIVISION POINTS?
Posted: May 8, 2001 8:54 AM
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Lynn Fisher <lfi...@woodstockhs.k12.vt.us>
[ap-calculus Sun, 6 May 2001 18:00:20 -0400]
wrote
> Thanks for your attempts, but no one is understanding
> my question!!
>
> My question is not about how to solve a particular problem.
> It is about terminology.
>
> I'm beginning to satisfy myself that there isn't any good
> answer. But, just in case, let me try again:
>
> I want language that can be used to describe the values for
> which an expression may change sign. These are, both you and
> I know, the values for which the expression is zero, and the
> values for which it is undefined. "Critical values" won't do,
> apparently, because this is reserved for values where the
> derivative of the expression is zero.
>
> Anyone want to try again? Thanks, Lynn F. Woodstock Vermont
Don't forget to include values where the function is not
continuous. But, to address your question, I don't know
of a term off hand. I personally think the categories
involved -- zeros, domain, and continuity -- are sufficiently
distinct and individually important that students are better
served by not lumping them together. Also, two out of three
of these categories are not really *values* of the function.
In class I call the x-coordinates of these points the "division
points" of our sign chart, and label them just below the number
line. Above these points I label '0', 'VA', 'hole', etc.
according to what I know is happening at these points. Above
the number line and between these division points I fill
in +++++++++ or ----------, as appropriate.
As an interesting side issue, the derivative of any function
that is (finitely) differentiable throughout an open interval
satisfies the intermediate value property on that interval.
So theoretically speaking, when doing a sign chart for f'
or f'', we only have to consider the zeros and non-existence
categories. For the significance of this, there exist functions
differentiable everywhere (better still, functions having a
bounded derivative) that are discontinuous at each rational
number. Such functions can even be discontinuous on certain
uncountable dense sets as well. In class, of course, the
procedure I follow when labeling sign charts for derivatives
is exactly the same that I do for an arbitrary function.
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5. A SIGN CHART FOR sin(x^2) WITHOUT USING A CALCULATOR
Posted: Jun 5, 2002 12:31 PM
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Jim Mumaugh <teei...@telocity.com>
Mon, 03 Jun 2002 17:51:14 -0800
http://mathforum.org/epigone/ap-calc/chahwholphen
wrote (in part)
> "If the function g is defined by g(x) = int sin t^2 dt with
> lower limit zero and upper limit x on the closed interval
> - 1 <= x <= 3, then g has a local minimum at x = "
>
> a) 0 b) 1.084 c) 1.772 D) 2.171
> e) 2.507 this one is the listed answer.
>
> I took the derivative of g to get g (prime) = sin x^2 .
>
> I set it = zero and concluded that the only zero in the domain
> was x = 0. g(prime)(pi/6) = .2707 > 0 so by sign analysis I
> believe I have a min at x = 0. (I also conclude that I have
> maximums at x = -1 and at x = 3).
>
> g(0) = 0. g(2.507) = .4304 g(prime)(2.507) = .0019 ! Clearly
> not a critical value!
>
> What am I missing? I even double checked that I was using the
> proper key this time.
Solving sin(x^2) = 0 --->
(undo SINE) x^2 = n*Pi, n = 0, 1, -1, 2, -2, ...
(undo square) x = (+/-)*sqrt(n*Pi), n = 0, 1, 2, 3, ...
Using the fact that the signs of SINE are +++ --- +++ --- ...
as we pass through quadrants I, II, III, IV, ... , and using
the fact that sin(x^2) is an even function, we immediately get
the sign chart
[Use fixed font size for what follows.]
__________________________________________________________________
| | | | | |
x -sqrt(2*Pi) -sqrt(Pi) 0 sqrt(Pi) sqrt(2*Pi) sqrt(3*Pi)
| | | | | |
++++++++ | -------- | ++++++ | +++++ | --------- | ++++++++ |
| | | | | |
incr. | decr. | incr. | incr. | decr. | incr. |
| | | | | |
max min neither max min
Since we want - 1 <= x <= 3, only x = 0, sqrt(Pi), and sqrt(2*Pi)
are available. [Clearly, -sqrt(Pi) < -1. Also, since 3*Pi > 9,
we know that sqrt(3*Pi) > sqrt(9) = 3.]
Hence, x = sqrt(2*Pi) is the desired relative minimum.
Looking at the answers, the obvious choice is (e), since
sqrt(2*Pi) > sqrt(6) and sqrt(6) is clearly larger than any
of the other values given. [Moreover, (2.5)^2 is 6.25, which
is in very close agreement with 2*Pi (approximately 6.28).]
Interestingly, I never had to reach for my calculator as I was
working this out.
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6. TERRACE POINTS IN THE FIRST DERIVATIVE TEST
Posted: Jul 4, 2002 8:51 AM
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Dave Slomer <slom...@ucmail.uc.edu>
[ap-calculus Tue, 2 Jul 2002 12:55:57 -0400]
http://mathforum.org/epigone/ap-calc/tranthayswel
wrote (in part):
> Anybody heard of "terrace points?" Ostebee and Zorn call an
> inflection point at which f ' = 0 a "terrace point," which,
> they say, "isn't official math jargon. But maybe it should be
> since it meets a need." (Think about terrace farming.)
A colleague of mine from a few years ago uses the term
"terrace point" at a point x=a where the first derivative is
zero and the first derivative doesn't change sign as you pass
through x=a, and I liked the idea so much that I've been using
it ever since. I didn't know this term was in Ostebee/Zorn,
however. Among other things this notion allows you to label
all four of the possibilities that can show up on a first
derivative sign chart where the derivative is zero --->
++++0++++ ter
++++0---- max
----0++++ min
----0---- ter
Incidentally, this use of "terrace point" is not quite the
same as what Dave Slomer defined. For example, I'm not
assuming second order differentiability. But, more importantly,
the definition I gave has a more immediate intuitive connection
with the behavior of the graph. ++++0++++ says the graph goes up,
levels out instantaneously, then goes back up. [Yes, I know that
in this situation the function is strictly increasing *on an
interval* containing the point where the first derivative
is zero.]
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7. FIRST DERIVATIVE TEST FOR f(x) = (x^2 - 15)*e^(-x)
Posted: Apr 6, 2006 10:13 AM
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Heather Gioia wrote:
http://mathforum.org/kb/thread.jspa?threadID=1360096
> What are the values of x for which the function
> f defined by f(x) = (x^2 -15)e^(-x) is increasing?
>
> It says the correct answer is -3 < x < 5, but I am
> not clear as to how the book got the answer....no
> explaination is given.
The derivative exists everywhere and is given by
(2x - x^2 + 15)*e^(-x).
Since e^(-x) is always positive (in particular,
it's never zero), the critical points will be
where 2x - x^2 + 15 = 0:
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = 5, -3
A sign chart is now easy to construct. We can
ignore e^(-x), since it's always positive. Also,
the graph of y = 2x - x^2 + 15 is a parabola
that opens downward (because the coefficient
of x^2 is negative) and has x-intercepts 5 and -3,
so 2x - x^2 + 15 is positive between 5 and -3.
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8. FIRST DERIVATIVE TEST FOR f(x) = ax - (a+1)*ln(x+1)
Posted: Aug 4, 2006 2:51 PM
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Qibo Jing wrote:
http://mathforum.org/kb/thread.jspa?messageID=4987696
> f(x) = ax - (a+1)ln(x+1), a>=0. Find the regions
> where the function f(x) is monotonic.
f'(x) = a - (a+1)/(x+1), so f'(x) = 0 gives x = 1/a.
Since f is undefined for x <= -1, a sign chart
for f' when 'a' is not equal to 0 or -1 is
(VA is vertical asymptote; use fixed font size):
***** VA neg 0 pos
-------|---------|--------
-1 1/a
Note this assumes -1 is not equal to 1/a, which is
where the restriction 'a' not equal to -1 comes from.
By considering the behavior of f' as x --> -1
on the right, it's clear that f' is negative
for -1 < x < 1/a. By considering the behavior
of f' as x --> oo, it's clear that f' is
positive for x > 1/a.
Therefore, if 'a' is not 0 or -1, f is decreasing
on the interval (-1, 1/a] and f is increasing on
the interval [1/a, oo).
If a = 0, we have f(x) = -ln(x+1), and hence f is
decreasing on the interval (-1, oo).
If a = -1, we have f(x) = -x - (0)*ln(x+1), which
equals -x for x > -1 and is undefined for x <= -1.
Hence, f is decreasing on the interval (-1, oo).
Some people might take f to have domain all real
numbers if a = 0, but my view is that the domain
of the expression ax - (a+1)ln(x+1) is all real
numbers greater than -1 because the instructions
for evaluating this expression, regardless of the
value of 'a', call for me to multiply the value of
(a+1) by the value of ln(x+1). In any event, in
potentially ambiguous situations like this, I think
it's incumbent on the person defining the function
to be explicit about the domain. After all, the
domain *IS* part of what's involved when defining
a function. (If the quibble is only about the
domain, then the argument isn't about mathematics.)
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Dave L. Renfro