Solution of Term Exam 2
The results. 82 students took the exam; the average
grade was 45.3 and the standard deviation was about 25.
Required thought and response. These results are
disappointing. What went wrong? You are required to think about
it and send me your thoughts, by email, via CCNET or using the
feedback form. Did something go wrong with the way you studied? In your
opinion, was the exam unfair? Did I make serious mistakes in teaching
the material? Did we sink into a routine and forgot to see the bigger
picture? Anything else?
The goal of this exercise is to improve things. Be constructive! Don't
just swallow or throw dirt, that won't help anyone. An indication
that went wrong is fine, but it's better if it comes along with
``and could have fixed it''.
As always, anonymous messages are fine (though signed messages are
better). I guess this means that I cannot verify that you all do this
exercise. Yet it remains morally required, for the benefit of
everybody.
The due date for this task is next Friday, December 8, at 5PM. I
may or may not prepare a synopsis of your responses (with all
identifying details removed) for distribution as a handout early in the
next semester.
Problem 1. Let and be continuous functions defined
on all of
.
- Prove that if
for some
, then there is a
number such that
whenever
.
- Prove that if two continuous functions are equal over the
rationals then they are always equal. That is, if for every
then for all
.
Solution. (Graded by Derek Krepski)
- Let
. Then is continuous and
. Set
. Then
, so using the continuity of
find so that
whenever
. Now
if
then
, so
and in particular . but this means that when
we have
and so
.
- Assume for every
and let be an
arbitrary real number. Assume
. Using part 1 of this
question find such that
whenever
. Between any two different real numbers there is a rational
number, so find
so that
. But then
and so
, but as
. This
is a contradiction, so it can't be that
. Thus for every
we have that .
Problem 2. Let be a continuous function defined on
all of
, and assume that is rational for every
. Prove that is a constant function.
Solution. (Graded by Derek Krepski) Assume by contradiction that
for some real numbers . Between any two real numbers
there is an irrational number, so let be some irrational number between
and . By the intermediate value theorem there is some
with . But then is irrational, contradicting the
assumption that is rational for every
. Thus no such pair
exists and must be a constant function.
Problem 3. We say that a function is locally
bounded on some interval if for every there is an
so that is bounded on
.
Let be a locally bounded function on the interval and let
is bounded on and .
- Justify the definition of : How do we know that exists?
- Prove that .
- Prove that .
- Prove that .
- Can you summarize these results with one catchy phrase?
Solution. (Graded by Shay Fuchs)
- is certainly bounded on the set , so and is
non-empty. Also, all elements of are between and , hence is
bounded from above by . So by P13 has a least upper bound -- in
other words, exists.
- As is locally bounded and as , there is some
for which is bounded on
. But then is
bounded (with the same or smaller bound) on
,
and so
and hence
.
- Assume by contradiction that . As is locally bounded and
as , there is some
for which
and is bounded on
(say by ). As there is some
with
and then by the definition of , is
bounded by some constant on . It follows that is
bounded by
on
and
so
, contradicting the fact that . So
is false and hence . But as is an upper bound of and
is a least upper bound of , we also have that and hence
.
- As is locally bounded and as , there is some
for which is bounded on
. Like before,
is also bounded on some with
, and hence on
. So by the definition of we have that
.
- ``On a closed interval, a locally bounded function is bounded''.
Problem 4.
- Define `` is differentiable at ''.
- Prove that if is differentiable at then it is also continuous
at .
Solution. (Graded by Brian Pigott)
- is said to be differentiable at if is defined in a
neighborhood of and
exists (and
if this limit exists, it is called ).
- If is differentiable at then (using the theorems about limits
of sums and products),
So by the definition of continuity, is continuous at .
Problem 5. Draw an approximate graph of the function
making sure to clearly indicate
(along with clear justifications) the domain of definition of , its
-intercepts and its -intercepts (if any), the behaviour of at
and near points at which is undefined (if any),
intervals on which is increasing/decreasing, its local
minima/maxima (if any) and intervals on which is convex/concave.
Solution. (Graded by Brian Pigott)
- is defined whenever the denominator is non-zero. That is,
whenever
.
- The -intercepts are when . The only solution to that is
when , and that point is also the -intercept.
-
and so
. Also,
as , the numerator of goes to , so its behaviour is
determined by the behaviour of its denominator, which is positive for
, negative for and zero for . Hence
,
,
and
.
-
. The denominator here is always
positive, so when and so is increasing on
and on and when and so is decreasing on
and on
.
- Comparing the intervals of decrease/increase, we find that has a
local max at , and then .
-
. Here the numerator is always
positive so the sign is determined by the denominator. Hence and
is convex on
and on
, while and
is concave on .
In summary, the graph of is roughly as follows:
The generation of this document was assisted by
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Dror Bar-Natan
2004-12-01