\magnification=\magstep1
\centerline{Mathematics 242--Principles of Analysis}
\centerline{Problem Set 5 -- Solutions}
\centerline{November 11, 2000}
\bigskip
\noindent
Section 4.1/
\bigskip
\noindent
1. 
\item{a)} For $f(x) = {1\over x}$, for all $c \ne 0$,
$$\eqalign{
f'(c) &= \lim_{x\to c} {{1\over x} - {1\over c}\over x-c}\cr
&= \lim_{x\to c}{c-x\over cx(x-c)}\cr
&= \lim_{x\to c}{-1\over cx}\cr
&= {-1\over c^2}\cr}$$
The derivative function is $f'(x) = {-1\over x^2}$.
\item{b)}  For $g(x) = {1\over x^2}$, for all $c \ne 0$,
$$\eqalign{
g'(c) &= \lim_{x\to c} {{1\over x^2} - {1\over c^2}\over x-c}\cr
&= \lim_{x\to c}{c^2-x^2\over c^2 x^2(x-c)}\cr
&= \lim_{x\to c}{-(c+x)\over c^2x^2}\cr
&= {-2c\over c^4}\cr
&={2\over c^3}\cr}$$
The derivative function is $g'(x) = {-2\over x^3}$.
\item{c)}  For $h(x) = ax^2 + bx + c$, 
$$\eqalign{
h'(d) &= \lim_{x\to d} {(ax^2+bx+c) - (ad^2 +bd+c) \over x-d }\cr
&= \lim_{x\to d}{a(x^2-d^2)+b(x-d)\over (x-d)}\cr
&= \lim_{x\to d}{a(x+d)+b}\cr
&= {2ad+b}\cr}$$
The derivative function is $h'(x) = 2ax + b$.
\item{d)}  For $i(x) = {x\over x+1}$,
$$\eqalign{
i'(c) &= \lim_{x\to c} {{x\over x+1} - {c\over c+1}\over x-c}\cr
&= \lim_{x\to c}{x-c\over (c+1)(x+1)(x-c)}\cr
&= \lim_{x\to c}{1\over (c+1)(x+1)}\cr
&= {1\over (c+1)^2}\cr}$$
The derivative function is $i'(x) = {1\over (x+1)^2}$.
\item{e)}  For $j(x) = \sqrt{x^2 + 1}$,
$$\eqalign{
j'(c) &= \lim_{x\to c} {\sqrt{x^2+1} - \sqrt{c^2+1}\over x-c}\cr
&= \lim_{x\to c}{(\sqrt{x^2+1} - \sqrt{c^2+1})(\sqrt{x^2+1} + \sqrt{c^2+1})\over (x-c)(\sqrt{x^2+1} + \sqrt{c^2+1})}\cr
&= \lim_{x\to c}{x^2-c^2\over(x-c)(\sqrt{x^2+1} + \sqrt{c^2+1})}\cr
&= \lim_{x\to c}{x+c\over (\sqrt{x^2+1} + \sqrt{c^2+1})}\cr
&={2c\over 2\sqrt{c^2+1}}\cr
&={c\over\sqrt{c^2+1}}\cr}$$
The derivative function is $j'(x) = {x\over \sqrt{x^2+1}}$.
\bigskip
\noindent
15.  For $g(x) = (f(x))^2$,
$$\eqalign{
g'(c) &= \lim_{x\to c} {(f(x))^2 - (f(c))^2\over x-c}\cr
&= \lim_{x\to c}{(f(x) - f(c))(f(x) + f(c))\over (x-c)}\cr
&= \lim_{x\to c}{f(x) - f(c)\over(x-c)}\cdot \lim_{x\to c} (f(x) + f(c))\cr
&=f'(c)\cdot 2f(c) \cr}$$
To get to the last line, we used the fact that $f$ differentiable at $c$
implies $f$ is continuous at $c$.
The derivative function is $g'(x) = 2f(x) f'(x)$.

\bigskip
\noindent
16.  For $g(x) = \sqrt{f(x)}$,
$$\eqalign{
g'(c) &= \lim_{x\to c} {\sqrt{f(x)} - \sqrt{f(c)}\over x-c}\cr
&= \lim_{x\to c}{(\sqrt{f(x)} - \sqrt{f(c)})(\sqrt{f(x)} + \sqrt{f(c)})\over (x-c)(\sqrt{f(x)} + \sqrt{f(c)})}\cr
&= \lim_{x\to c}{f(x) - f(c)\over(x-c)(\sqrt{f(x)} + \sqrt{f(c)})}\cr
&= \lim_{x\to c}{f(x) - f(c)\over(x-c)}\cdot \lim_{x\to c}{1\over (\sqrt{f(x)} + \sqrt{f(c)})}\cr
&=f'(c)\cdot {1\over 2\sqrt{f(c)}}\cr}$$
To get to the last line, we used the fact that $f$ differentiable at $c$
implies $f$ is continuous at $c$.
The derivative function is $g'(x) = {f'(x)\over 2 \sqrt{f(x)} }$.
\bigskip
\noindent
Section 4.3/
\bigskip
\noindent
1.  Our hypothesis is that
$f(x)$ has a local minimum at $c$ and $f'(c)$ exists.  So 
$f(x) \ge f(c)$ for all $x$ in some neighborhood of $c$.
Taking $x < c$, we have $f(x) - f(c) \ge 0$, so
${f(x) - f(c) \over x - c} \le 0$.  Therefore 
$$\lim_{x\to c^-} {f(x) - f(c) \over x - c} \le 0.$$ 
On the other hand, taking $x > c$, we have $f(x) - f(c) \ge 0$, so
${f(x) - f(c) \over x - c} \ge 0$.  Therefore 
$$\lim_{x\to c^+} {f(x) - f(c) \over x - c} \ge 0.$$ 
Since we assume $\lim_{x\to c} {f(x) - f(c) \over x - c}$ exists,
the two one-sided limits must exist and be equal.  Therefore
$\lim_{x\to c} {f(x) - f(c) \over x - c} = 0$.
\bigskip
\noindent
5.
\item{a)}  $f(x) = x^{33}-33x$.  We have $f'(x) = 33x^{32} - 33 = $
which is zero only when $x = \pm 1$.  $x = -1$ is a local maximum
and $x = 1$ is a local minimum since $f'(x) > 0$ for $x < -1$, $f'(x) < 0$
for $-1 < x < 1$ and $f'(x) > 0$ for $x > 1$.
\item{b)}  $f'(x) = 34(x^{33}-1)$ is zero only at $x = 1$, which is a local 
minimum.
\item{c)}  $f'(x) = 35x^{34} - 4480x^{27} = 35x^{27}(x^7 - 128)$
which is zero at $x = 0$ and $x = 2$.  $x = 0$ is a local maximum, 
$x = 2$ is a local minimum.
\item{d)}  $f'(x) = 36x^{35} - 576 x^{31} = 36x^{31}(x-2)(x+2)(x^2+4)$.
$x = -2$ is a local minimum, $x = 0$ is a local maximum, $x = 2$ is a 
local minimum.
\vfill\eject
\noindent
6.  Here the idea is to compute $f'(x)$, find critical points, 
then compare the values of $f$ at the critical points and the
endpoints of the interval.
\item{a)} $f(x) = x^4 - 4x^3 + 1$ on $[-1,4]$.  
$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$.  The derivative is 
zero at $x = 0$, $x = 3$.  $f(-1) = -2$, $f(0) = 1$, $f(3) = -26$, $f(4) = 1$.
So the maximum on the interval is $M = 1$, attained at $x = 0$ (local max),
and also at $x = 4$ (endpoint).  The minimum is $m = -26$, attained at
$x = 3$.
\bigskip
\noindent
12.
\item{a)} The MVT does apply since $f(x) = \sqrt{x}$ is continuous
on $[0,b]$ and differentiable on $(0,b)$  (not at $x = 0$, but that
is not necessary).  We have $f'(c) = {f(b) - f(0) \over b - 0} = 1$, 
when $c = {b\over 4}$.
\item{b)}  The MVT does not apply since $g(x) = [x]$ (greatest
integer function) is not continuous on $[1,2]$ (it has a jump at the
right endpoint).
\item{c)}  The MVT does not apply since $h(x) = 2x + |x-2|$ is 
not differentiable at $x = 2$.  This is true because
$\lim_{x\to 2^-}{h(x) - h(2)\over x-2} = 2 - 1 = 1$ but
$\lim_{x\to 2^+}{h(x) - h(2)\over x-2} = 2 + 1 = 3$.
\bigskip
\noindent
17.  If $f(x)$ has another zero $x \in [a,b]$, then apply Rolle's Theorem
to $f$ on the interval $[a,x]$.  We get that $f'(c) = 0$ for 
some $c$ between $a$ and $x$.  But this contradicts the hypothesis
that $f'(x) \ne 0$ for all $x \in (a,b)$.  Therefore $f$ can have only 
one zero.
\bigskip
\noindent
20.  
\item{a)}  Since $\cos(\pi/x)$ is undefined at $0$, we must use
the definition to compute $f'(0)$:
$$\eqalign{f'(0) &= \lim_{x\to 0} {x + x^2\cos(\pi/x)\over x}\cr
                 &= \lim_{x\to 0} {1 + x\cos(\pi/x)}\cr
                 &=1,\cr}$$
since $\cos(\pi/x)$ is bounded as $x\to 0$.
\item{b)}  
$$\eqalign{
f\left({1\over 2n}\right) &= {1\over 2n} + {1\over (2n)^2}\cos(2n\pi) = {1\over 2n} + {1\over (2n)^2}\cr
f\left({1\over 2n-1}\right) &= {1\over 2n-1} + {1\over (2n-1)^2}\cos((2n-1)\pi) = {1\over 2n-1} - {1\over (2n-1)^2}\cr
  f\left({1\over 2n}\right) -     f\left({1\over 2n-1}\right) &=
    {1\over 2n} + {1\over (2n)^2} - {1\over 2n-1} + {1\over (2n-1)^2}\cr
&= {2n(2n-1)^2 + (2n-1)^2 - (2n)^2 (2n-1) + (2n)^2\over (2n)^2 (2n-1)^2} \cr
&={8n^3 - 8n^2 + 2n + 4n^2 - 4n + 1 - 8n^3 + 4n^2 + 4n^2\over (2n)^2 (2n-1)^2}\cr
&={4n^2 - 2n + 1\over (2n)^2 (2n-1)^2}\cr
&={3n^2 + (n-1)^2 \over (2n)^2 (2n-1)^2}\cr
&>0\cr}$$
\noindent
This shows$ f\left({1\over 2n}\right)>  f\left({1\over 2n-1}\right)$.

This problem shows the surprising result that it is not sufficient to 
have $f'(c) > 0$ to conclude that $f$ is increasing on an interval 
containing $c$.  In this case, {\it there is no interval containing 0 
on which $f(x)$ is increasing, even though $f'(0) = 1$}.  (!)
\bigskip
\noindent
Section 5.1
\bigskip
\noindent 
9.
\item{a)}  Isolate the discontinuity by choosing a suitable partition.
For example, let $\delta < 1$, and let
$P$ be the partition $P = \{0,1-\delta,1+\delta,2\}$.  Then on 
the first subinterval we have minimum value $m_1 = 1$ and maximum 
value $M_1 = 1$.  The second interval has $m_2 = 1$ and $M_2 = 2 = f(1)$.
The third interval has $m_3 = 1 = M_3$.  Therefore the 
upper sum is 
$$U_P(f) = M_1(1-\delta) + M_2 (2\delta) + M_3(1-\delta) = 2 + 2\delta$$
and the lower sum is 
$$L_P(f) = m_1(1-\delta) + m_2 (2\delta) + m_3(1-\delta) = 2$$
The difference $U_P(f) - L_P(f) = 2\delta$ can be made less than 
$\epsilon$ by taking $\delta < \epsilon/2$.  By Theorem 5.1.5 in the 
text, $f$ is integrable.  The value of the integral is the 
greatest lower bound of the upper sums, which equals the least upper bound
of the lower sums, namely $\int_0^2 f(x)\ dx = 2$.
\item{b,c)}  The idea is similar -- take partitions isolating the 
discontinuities. For example, in c, 
$$[2x] = \cases{0 &if $0\le x < 1/2$\cr
               1 &if $1/2 \le x < 1$\cr
               2 &if $1 \le x < 3/2$\cr
               3 &if $3/2\le x < 2$\cr
               4 &if $x = 2$ \cr}$$
               Let $\delta < 1/4$ and
$$P = \{0,1/2-\delta,1/2+\delta,1-\delta,1+\delta,3/2-\delta,3/2+\delta,
2-\delta,2\}$$
Then
$$\eqalign{
U_P(f) &= 0\cdot(1/2-\delta) + 1\cdot(2\delta) + 1\cdot(1/2 - 2\delta) 
2\cdot(2\delta)\cr
       &\quad + 2\cdot(1/2 - 2\delta) + 3\cdot(2\delta) + 
                3\cdot(1/2 - 2\delta) +4\delta\cr
&=3 + 4\delta\cr}$$
The lower sum $L_P(f) = 3$.  Hence, letting $\delta \to 0$,
the value of the integral is 3.
\bigskip
\noindent
10.  Let $P_n$ be the regular partition of $[0,2]$ into $n$ equal
subintervals -- $x_i={2i\over n}$, and $\Delta x = {2\over n}$.  Then
since $f(x) = 2x + 1$ is increasing on the whole interval we have 
$$L_P(f) = \sum_{i=1}^n \left(2{2(i-1)\over n} + 1\right){2\over n}$$
and 
$$U_P(f) = \sum_{i=1}^n \left(2{2i\over n} + 1\right){2\over n}$$
When we subtract, all but the last term in $U_P(f)$
cancel with all but the first term in $L_P(f)$ and the difference is
$$U_P(f) - L_P(f) = {2\over n}(4 + 1 - 1) = {8\over n}.$$
This can be made $< \epsilon$ by chosing $n > {8\over\epsilon}$.
Hence Theorem 5.1.5 in the text shows that $f(x)$ is integrable
over $[0,2]$.  

To find the value of the integral, we can take
the limit of the upper sums as $n \to \infty$.  (Since the 
difference between the upper and lower sums for the regular
partitions goes to 0 an $n\to\infty$, this shows
that $\lim_{n\to\infty} U_{P_n}(f)$ equals the greatest
lower bound of the upper sums, which equals the integral.)
 The summation
formula $1 + 2 + \cdots + n = \sum_{i=1}^n i = {n(n+1)\over 2}$ will
be used at a key step:
We have 
$$\eqalign{U_{P_n}(f) 
&= \sum_{i=1}^n \left(2{2i\over n} + 1\right){2\over n}\cr
&= {8\over n^2} \sum_{i=1}^n i + {2\over n}\sum_{i=1}^n 1\cr
&= {8\over n^2}{n(n+1)\over 2} + 2\cr
&= 6 + {4\over n}\cr}$$
As $n\to \infty$, this goes to $\int_{0}^2 2x + 1\ dx = 6$.
\bigskip
\noindent
11.  Similar to 10, using the summation formula for
$\sum_{i=1}^n i^3$ given in the problem to compute the integral.
$$\eqalign{
\int_0^1 x^3\ dx &= 
\lim_{n\to\infty} \sum_{i=1}^n \left({i\over n}\right)^3 {1\over n}\cr
&= \lim_{n\to\infty} {1\over n^4}\sum_{i=1}^n i^3\cr
&= \lim_{n\to\infty} {1\over n^4} {n^2(n+1)^2\over 4}\cr
&= \lim_{n\to\infty} {1\over 4} + {1\over 2n} + {1\over 4n^2}\cr
&= {1\over 4}\cr}$$
\bigskip
\noindent
12.  The idea for all three parts
here is similar to what we did in problem 9.  For 
instance, in part a, if you let $\delta < 1/2$, and 
$$P = \{0,1-\delta,1+\delta, 2-\delta, 2+\delta, 3-\delta, 3+\delta,
4-\delta, 4+\delta, 5 - \delta\}.$$
Then $U_P(f) = 9\delta$, and $L_P(f) = 0$.  By picking $\delta < \epsilon/9$,
the difference can be made less than any $\epsilon > 0$.  Hence
$f$ is integrable by Theorem 5.1.5 (and the value of the integral is 0, since
all lower sums for this function equal 0).
\bigskip
\noindent
13.  Our hypothesis is that $\int_a^b f(x)\ dx$ exists and equals some 
number $I > 0$.  Since the integral is the least upper bound of the
lower sums for the function, for every $\epsilon > 0$, there exists
some partition $P$ such that $L_P(f) > I - \epsilon$.  If we take
$0< \epsilon < I$, this shows that $L_P(f) > 0$ for some $P$.  But then
some term $m_i \Delta x_i$
in the lower sum must be strictly positive.  As a result, $m_i > 0$
for at least one of subintervals from $P$.  Since
$m_i = {\rm glb}\{f(x) : x \in [x_{i-1},x_i]\}$, this shows that 
$f(x) > m_i > 0$ for all $x \in J = [x_{i-1},x_i]$, the interval
from the partition.

 


\bye