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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Midterm Exam 3}
\centerline{May 2, 2014}
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%{\it Directions}
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%Do all work in the blue exam booklet.  There are 100 possible 
%regular points and 10 possible Extra Credit points (see question IV below).
%Possibly useful information:
%$$\sum_{i=1}^n = {n(n+1)\over 2}, \qquad \sum_{i=1}^n i^2 = {n(n+1)(2n+1)\over 6}.$$
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%\noindent
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\noindent
I. (15)  Let
$$f(x) = \cases{ x^{4/5} \sin(1/x) & if $x \ne 0$\cr
                       0 & if $x = 0$.\cr}$$
Is $f$ continuous at $x = 0$?  Is $f$ differentiable at $x = 0$?  Give complete
reasons for your assertions.
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\noindent
{\it Solution:}  Since $|\sin(1/x)| \le 1$ for all $x$, we have 
$$-x^{4/5} \le f(x) \le x^{4/5}$$
for all $x \ne 0$.  Also $\lim_{x\to 0} \pm x^{4/5} = 0$.  Therefore, by the 
limit squeeze theorem, $\lim_{x\to 0} f(x) = 0 = f(0)$.  Therefore, $f$ is 
continuous at $x = 0$.  However, $f$ is not differentiable at $x = 0$ because
$$\lim_{x \to 0} {f(x) - f(0)\over x - 0} = \lim_{x\to 0} {\sin(1/x)\over x^{1/5}}$$
does not exist.  
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\noindent
II.  Both parts of this question refer to the function $f : {\bf R} \to {\bf R}$
defined by $f(x) = 1 - x^2$.
\medskip
\item{A)}  (20)  Consider the regular partitions ${\cal P}_n$ of the
interval $[1,3]$ and show directly, using the upper and lower sums,
that $f$ is integrable on $[1,3]$.  
\medskip
\item{} {\it Solution:}  Note that $f$ is {\it decreasing} on $[1,3]$
since $f'(x) = -2x < 0$ for all $x$ with $1\le x \le 3$. 
The partition is 
$${\cal P}_n = \{1,1 + 2/n, 1 + 4/n,\ldots,3\},$$
with $x_i = 1 + 2i/n$ for $i = 0,1,\ldots, n$.   
Hence, since $f$ is smallest at the right endpoint in each subinterval,
$$\eqalign{
L_{{\cal P}_n}(f) &= \sum_{i=1}^n \left(1 - (1 + 2i/n)^2 \right){2\over n}\cr
&= -{8\over n^3} \sum_{i=1}^n i^2 - {8\over n^2}\sum_{i=1}^n i \cr
&= -{8\over n^3}\cdot {n(n+1)(2n+1)\over 6} - {8\over n^2}\cdot {n(n+1)\over 2} \cr
&= - {20\over 3} - {8\over n} - {4\over 3 n^2}.\cr}$$
Similarly, $f$ is largest at the left endpoint in each subinterval, so
$$\eqalign{
U_{{\cal P}_n}(f) &= \sum_{i=1}^n \left(1 - (1 + 2(i-1)/n)^2\right){2\over n}\cr
&= - {20\over 3} + {8\over n} - {4\over 3 n^2}.\cr}$$
Therefore, for any given $\varepsilon > 0$, if $n > 16/\varepsilon$,
$$U_{{\cal P}_n}(f) - L_{{\cal P}_n}(f) = {16\over n} < \varepsilon.$$
This shows that $f$ is integrable.  
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\item{B)}  (15)  Explain why the hypothesis of the Mean
Value Theorem is satisfied for $f$ on the interval $[1,3]$
and find the number $c$ mentioned in the conclusion.
\item{} {\it Solution:}  $f$ is a polynomial function, so it is differentiable,
hence continuous everywhere.  On the interval $[1,3]$, $f(3) - f(1) = -8 - 0 = -8$.
The MVT says that there is some $c \in (1,3)$ where $-8 = f'(c) \cdot (3 - 1)$.
Since $-8 = 2f'(c) = 2 - 4c$, this is true for $c = 5/2$.
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\noindent  
III.  (20)  Show that if $f$ is continuous on $[a,b]$,
then $f$ is integrable on $[a,b]$.  
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\noindent
{\it Solution:}  By a previous result we know that $f(x)$ continuous 
on $[a,b]$ implies that $f(x)$ is uniformly continuous on $[a,b]$.  
Therefore, given $\varepsilon > 0$, there exists a $\delta > 0$ such 
that $|f(x) - f(x')| < \varepsilon/(b - a)$ whenever $|x - x'| < \delta$
(with $x,x' \in [a,b]$, of course).  Now, let ${\cal P}$ be any partition 
of $[a,b]$ with $\Delta x_i <$ this $\delta$ for all $i$.  By the EVT, on 
the interval $[x_{i-1},x_i]$ from the partition ${\cal P}$, $f$ attains
a maximum $M_i = f(c_i)$ and a minimum $m_i = f(d_i)$ at some $c_i, d_i \in [x_{i-1},x_i]$.
But then $M_i - m_i < \varepsilon/(b - a)$ since $|c_i - d_i| < \delta$.  
Hence 
$$U(f,{\cal P}) - L(f,{\cal P)) < \sum_{i=1}^n \varepsilon/(b - a) \Delta x_i = \varepsilon.$$
(This follows since $\sum_i \Delta x_i = b - a$.)  Therefore $f$ is integrable on 
$[a,b]$.  
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\noindent
IV.  True-False.  Say whether each of the following statements is 
true or false.  For true statements, give short proofs; for false ones
give reasons or counterexamples.  Do any {\it three} parts.  If you submit
solutions for all four, then I will consider the other one for
Extra Credit.  
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\noindent
\item{A)} (10)  Let $f(x) = e^{2x} - e^{x}$.  There exists some $c \in (0,\ln(2))$
such that $f(c) = 1$.
\medskip
\item{} {\it Solution:} The statement is TRUE.  We apply the IVT. First, 
$f$ is continuous everywhere since the exponentials
$e^x$ and $e^{2x}$ are differentiable everywhere.  On the interval $[0,\ln(2)]$, we
have $f(0) = 1 - 1 = 0$ and $f(\ln(2)) = 4 - 2 = 2$.  Since $1$ is
in the range between the endpoint values, the (``weak form'' of) the IVT
implies that there exists $c \in (0,\ln(2))$ such that $f(c) = 1$. 
\medskip
\item{B)} (10)  The function $f(x) = \arctan(x)$ is uniformly continuous
on the interval $(-1,1)$.  
\medskip
\item{} {\it Solution:}  This is TRUE.  We apply the MVT to $f$ on the 
interval $[x,x']$ where $-1 < x < x' < 1$ are arbitary.  Then there
exists a $c \in (x,x')$ such that $f(x) - f(x') = f'(c)(x - x')$.
But $f'(x) = {1\over 1 + x^2}$ is bounded above by $1$ on this interval.
Therefore $|f(x) - f(x')| \le |x - x'|$, so $f$ is Lipschitz continuous, hence
uniformly continuous.
\medskip
\item{C)} (10) There are continuous functions $f(x)$ on $[a,b]$ for which there exist
no differentiable function $F(x)$ on $[a,b]$ with $F'(x) = f(x)$. 
\medskip
\item{} {\it Solution:}  This is FALSE.  The Fundamental Theorem of Calculus
(part 1) implies that $F(x) = \int_a^x f(t)\ dt$ is always an antiderivative
of $f(x)$ on $[a,b]$.  
\medskip
\item{D)} (10) Let $f$ be differentiable on an open interval $I$ with $[a,b] \subset I$.
If $f'(a) > 0$ and $f'(b) < 0$, then there must exist some $c \in (a,b)$ where $f'(c) = 0$.  
\medskip
\item{} {\it Solution:}  This is TRUE. Since $f$ is differentiable 
everywhere on $[a,b]$, it is also continuous on that interval.
By the Extreme Value Theorem, $f$ reaches a maximum value $M = f(c)$ for $c$ somewhere in 
that interval.  The inequalities $f'(a) > 0$ and $f'(b) > 0$ imply that 
$f$ must take values larger than both $f(a)$ and $f(b)$ in the interval.
(To get 
$$f'(a) = \lim_{x \to a} {f(x) - f(a) \over x - a} > 0,$$
it must be true that $f(x) > f(a)$ on some interval $[a,a + \delta)$  for $\delta > 0$.)  
Hence the location where the maximum is attained is $c \in (a,b)$.  
It follows that $f(c)$ must be a local maximum (not an endpoint maximum) and hence $f'(c) = 0$.  

\bye

