\magnification=\magstep1
\nopagenumbers
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Midterm Exam 3}
\centerline{May 3, 2013}
\bigskip
\noindent
{\it Directions}
\bigskip
Do all work in the blue exam booklet.  There are 100 possible 
regular points and 10 possible Extra Credit points.
Possibly useful information:
$$\sum_{i=1}^n i = {n(n+1)\over 2}, \qquad \sum_{i=1}^n i^2 = {n(n+1)(2n+1)\over 6}.$$
\bigskip
\noindent
I.  Both parts of this question refer to the function $f : {\bf R} \to {\bf R}$
defined by $f(x) = x^2$.
\medskip
\item{A)}  (25)  Show directly, using upper and lower sums,
that $f$ is integrable on $[0,1]$ and determine the value $\int_0^1 x^2\ dx$.
\bigskip
\item{} {\it Solution:}  Consider a regular partition ${\cal P}_n$ of $[0,1]$ with $n$ smaller intervals.  Then 
$\Delta x = {1\over n}$ and the endpoints
are $x_i =  i/n$ for $i = 0, \ldots, n$.  Since $x^2$ is increasing 
on the interval $[0,1]$ we have $m_i = ((i-1)/n)^2$ and $M_i = (i/n)^2$, so 
$$\eqalign{
L(f,{\cal P}_n) &= \sum_{i=1}^n ((i-1)/n)^2 (1/n)\cr
&= {(n-1)n(2n-1)\over 6n^3},\ {\rm and}\cr
U(f,{\cal P}_n) &= \sum_{i=1}^n (i/n)^2 (1/n)\cr
&= {n(n+1)(2n+1)\over 6n^3}.\cr}$$
Hence
$$U(f,{\cal P}_n) - L(f,{\cal P}_n) = {1\over n}$$
Given $\varepsilon > 0$, the difference will be $< \varepsilon$ whenever
$n > {1\over \varepsilon}$.  So the function is integrable.  
\medskip
\item{} {\it Note:}  The 
same result can be derived using the observation we made in the proof
of the general result that monotone functions are integrable.  Then 
$U(f,{\cal P}_n) - L(f,{\cal P}_n) = {(f(b) - f(a))(b - a)\over n} = {1\over n}$.
This is certainly ``cleaner!'')  
\medskip
\item{} The value of the integral is 
$$\lim_{n\to\infty} U(f,{\cal P}_n) = \lim_{n\to\infty} {n(n+1)(2n+1)\over 6n^3}
= \lim_{n\to\infty} {1\over 3} + {1\over 2n} + {1\over 6n^2} = {1\over 3}.$$
\bigskip
\item{B)}  (5) Compute $\int_0^1 x^2\ dx$ using the Fundamental Theorem
of Calculus to check your work.
\bigskip
\item{} {\it Solution:}  An antiderivative of $f(x) = x^2$ is $F(x) = {x^3\over 3}$,  so by part (2) of the FTC, the value is 
$$\int_0^1 x^2\ dx = \left. {x^3\over 3} \right|_0^1 = {1\over 3} - {0\over 3} = {1\over 3}.$$
\bigskip
\noindent
II.   Let $f(x) = {\sin(x)\over x}$ for $x \ne 0$.
\medskip
\item{A)}  (5) How should a value $f(0)$ be defined to make $f$ continuous
at $x = 0$?  (Note:  If you do not how to decide, you may 
``buy'' the answer for 5 points so that you can do the next part.)
\bigskip
\item{} {\it Solution:}  By results from Problem Set 7 and 8, we know 
$$\lim_{x \to 0} {\sin(x)\over x} = 1$$
so we should set $f(0) = 1$ to get continuity.  (Note that ${\sin(x) \over x}$
is differentiable, hence continuous at all $x \ne 0$.)
\bigskip
\item{B)}  (15) Using your value from part A for $f(0)$, use the limit
definition to show that the resulting 
function is differentiable and find the value $f'(0)$.  Say what method
you are using to compute this limit.
\bigskip
\item{} {\it Solution:}  We need to compute 
$$\lim_{x \to 0} {f(x) - f(0)\over x - 0} 
= \lim_{x\to 0} {{\sin(x)\over x} - 1\over x - 0} = \lim_{x \to 0} {\sin(x) - x\over x^2}.$$
This is a $0/0$ indeterminate form, so we can apply L'Hopital's Rule (twice) 
as follows:
$$\eqalign{
\lim_{x \to 0} {\sin(x) - x\over x^2} &= \lim_{x \to 0} {\cos(x) - 1\over 2x}\ ({\rm still\ } 0/0)\cr
&= \lim_{x \to 0} {-\sin(x)\over 2}\cr
&= 0\cr}$$
So $f'(0)$ exists and equals $0$.  (The graph $y = f(x)$ has a local maximum
at $x = 0$.)

\bigskip
\noindent  
III.  
\item{A)}  (10)  State the Mean Value Theorem.  
\bigskip
\item{} {\it Solution:}  Let $f$ be continuous on $[a,b]$ and 
differentiable on $(a,b)$.  Then there exists a $c \in (a,b)$ 
such that $f(b) - f(a) = f'(c)(b - a)$.  
\bigskip
\item{B)}  Determine
$${\rm lub} \left\{ {\ln(x)\over x} \mid x \in (e,+\infty)\right\}$$  
and justify your answer completely using facts we have developed in
this course.  
\bigskip
\item{} {\it Solution:}  Let $f(x) = {\ln(x)\over x}$.  By the 
quotient rule, we have 
$f'(x) = {1 - \ln(x)\over x^2}$.  
Therefore, $f'(x) > 0$ if $x < e$ and $f'(x) < 0$ for $x > e$.  By 
one of our corollaries of the MVT, 
$f$ is strictly decreasing on the given interval.  That means
that the value $f(e) = {1\over e}$ at the left endpoint
is the least upper bound of the set of values.  
\bigskip
\item{} {\it Note:}  It is not enough just to notice that 
$\lim_{x \to\infty} {\ln(x) \over x} = 0$.  To say the lub of the 
set of values equals $f(e)$, you must also show that $f$ does not
get larger anywhere between $e$ and $\infty$.  Showing $f$
is strictly decreasing implies that.  
%(Include the special
%case known as Rolle's Theorem and deduce the general case from that.)  
%\bigskip
%\noindent
%{\it Solution:}  Let $f$ be continuous on $[a,b]$ and differentiable on 
%$(a,b)$.  Then there exists a $c \in (a,b)$ such that $f'(c) = {f(b) - f(a)\over b %- a}$.
%\bigskip
%\noindent
%Proof:  First we assume $f(b) = f(a)$.  If $f$ is constant on $[a,b]$, then 
%any $c \in (a,b)$ will give $f'(c) = 0$.  Otherwise, then by the EVT, $f$
%must attain either a maximum value $M = f(c_1) > f(a) = f(b)$, or a 
%minimum value $m = f(c_2) < f(a) = f(b)$.  We claim that at any such 
%$c = c_1$ or $c_2$, we must have $f'(c) = 0$.  To see this, suppose 
%that $f(c)$ is a local maximum (the case of a minimum will be similar
%but with some signs reversed).  Then if $x < c$ we have
%$${f(x) - f(c) \over x - c} < 0 \Rightarrow 
%\lim_{x\to c^-} {f(x) - f(c) \over x - c} \le 0.$$
%But similarly,  if $x > c$ we have
%$${f(x) - f(c) \over x - c} > 0 \Rightarrow 
%\lim_{x\to c^+} {f(x) - f(c) \over x - c} \ge 0.$$
%Since we assume that $f'(c)$ exists, the two one-sided limits must be equal 
%and then $f'(c) = 0$.  

%Now we deduce the general statement from the special case above.  Let
%$g(x) = f(x) - {f(b) - f(a) \over b - a}(x - a)$.  Then $g(x)$ is a
%linear combination of continuous functions on $[a,b]$, therefore is 
%continuous on $[a,b]$.  It is also differentiable on $(a,b)$ since 
%$f(x)$ is and so is the linear function $ {f(b) - f(a) \over b - a}(x - a)$.
%Now $g(a) = f(a) - 0 = f(a)$ while 
%$$g(b) = f(b) - {f(b) - f(a) \over b - a} (b - a) = f(a).$$
%Therefore the special case applies and we get that there is some $c \in (a,b)$
%where $g'(c) = 0$.  But $g'(x) = f'(x) - {f(b) - f(a)\over b - a}$ so 
%$g'(c) = 0$ implies $f'(c) = {f(b) - f(a)\over b - a}$, and this is 
%what we wanted to prove.  
\bigskip
\noindent
IV.  True-False.  Say whether each of the following statements is 
true or false.  For true statements, give short proofs or reasons; for false ones
give reasons or counterexamples.  
\bigskip 
%\noindent
%\item{A)} (10)  Let $f(x) = {\ln(x) \over \ln(x) + 1}$.  
%There exists some $c \in (1,\infty)$ such that $f(c) = {997\over 1000}$.
%(Note: It is not necessary to find such a $c$ explicitly, but if you can, then
%that is one way to answer this!)
%\bigskip
%\item{} {\it Solution:}  TRUE.  Note that $f$ is differentiable on 
%$(1,\infty)$ because $\ln(x)$ is differentiable.  Hence $f$ is 
%also continuous.  Moreover $f(1) = 0$, while 
%$$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} {\ln(x) \over \ln(x) + 1} = 1.$$
%By results we proved earlier this semester, this implies that 
%$$ {\ln(b) \over \ln(b) + 1} > {997\over 1000} $$
%for some $b$ sufficiently large.  Applying the IVT on the interval
%$[1,b]$ we get that there is some $c$ with $f(c) = {997\over 1000}$.  
%(In this case, it is also possible to solve for $c$ explicitly:
%$${\ln(x) \over \ln(x) + 1} = {997\over 1000}$$
%is equivalent to $1000\ln(x) = 997\ln(x) + 997$, or 
%$3 \ln(x) = 997$ so $x = e^{997\over 3}$ (a very large number!)
%\medskip
\item{A)} (10)  $f(x) = \ln(x)$ is uniformly continuous on $(1,+\infty)$.  
\bigskip
\item{} {\it Solution:}  TRUE.  $f'(x) = {1\over x}$ satisfies $|f'(c)| < 1$
for all $c \in (1,+\infty)$.  Therefore, as we saw on PS 8, $f$ must
be uniformly continuous on the interval $(1,+\infty)$.  Here's the proof again: The MVT
implies that if $1 < x_1 < x_2$, then there is some $c \in (x_1,x_2)$ where 
$$|f(x_2) - f(x_1)| = |f'(c)||x_2 - x_1|.$$
But $|f'(c)| < 1$, so this shows
$$|f(x_2) - f(x_1)| < |x_2 - x_1|.$$ 
Hence $|f(x_2) - f(x_1)| < \varepsilon$ whenever $|x_2 - x_1| < \delta = \varepsilon$.  (Or alternatively, the bound on $|f'(x)|$ shows that $f$ 
is Lipschitz continuous on that interval, with Lipschitz constant
$k = 1$.)   
\medskip
\item{B)} (10) If $A \subset {\bf R}$ contains
no nonempty interval $(a,b)$, then $A$ is finite or countably infinite.  
\bigskip
\item{} {\it Solution:}  FALSE.  The Cantor set is a counterexample. It
contains no intervals since it is a subset of $[0,1]$, but the intervals
removed in its construction have total length $1$.  On the other hand, 
we saw in class that the Cantor set is uncountably infinite.  
\medskip
\item{C)} (10) Let $f$ be continuous on $[a,b]$ and assume $f(x) > 1$ for
all $x \in [a,b]$.  It is possible for ${\rm glb}\{f(x)\mid x \in [a,b]\}$ 
to equal $1$. 
\bigskip
\item{} {\it Solution:}  FALSE.  By the EVT, we have that $f$ attains
a minimum $m = f(c),$ for some $c\in [a,b]$.  This $m$ must equal the greatest lower bound of the set of values of $f$.   By the given information 
$f(c) > 1$.  Therefore the glb must also be strictly larger than $1$.  
\bigskip
\noindent
{\it Extra Credit} (10)  Show that if $f$ is differentiable on an open 
interval $I$ and $[a,b] \subset I$ where $f'(a) < 0$ and $f'(b) > 0$,
then there must be some $c \in (a,b)$ where $f'(c) = 0$.  
\bigskip
\noindent
{\it Solution:}  Since differentiability of $f$ implies continuity of $f$, 
the EVT applies to $f$ on the closed interval $[a,b]$ and $f$ attains a 
minimum value at some $c \in [a,b]$.  We claim first that $c \ne a,b$.  
Arguing by contradiction, suppose $f(a) \le f(x)$ for all $x \in [a,b]$.
Then we have ${f(x) - f(a)\over x - a} \ge 0$ for all $x \in [a,b]$, and
this implies
$$f'(a) = \lim_{x\to a^+} {f(x) - f(a) \over x - a} \ge 0.$$
But this contradicts the given information that $f'(a) < 0$.  Hence
the minimum is not attained at $a$.  A similar argument shows the
minimum is not attained at $b$ either.  Hence it must be attained
at some $c \in (a,b)$.  But now by the usual argument, we have 
$$f'(c) = \lim_{x \to c^-} {f(x) - f(c)\over x - c} \le 0,$$
while 
$$f'(c) = \lim_{x \to c^+} {f(x) - f(c)\over x - c} \ge 0.$$
So $f'(c) = 0$.  
\bigskip
\noindent
{\it Notes:} 
\item{1)}  You cannot apply Rolle's theorem directly here because
we are not given that $f(a) = f(b)$.  
\item{2)}  In addition, it is not correct to assume (even without saying so!)
that $f'$ is continuous, since that is not given either.  This means,
for instance, that the IVT does not necessarily apply to $f'$ here. 
\item{3)}  This is the main step in a proof of a result called 
{\it Darboux's Theorem}, which says essentially
that derivatives attain intermediate values, just as continuous functions do by 
the IVT.  The general statement says for instance that if $f'(a) < k < f'(b)$
then there is some $c \in (a,b)$ with $f'(c) = k$.  We can see that 
by applying the result here to the function $g(x) = f(x) - kx$.  
It is not necessarily true that the derivative function
$f'(x)$ is continuous, though, so this fact is {\it not a consequence of the IVT}.  
\bye

