\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 4}
\centerline{November 3, 2000}
\bigskip
\noindent
Section 3.5/7.  The function $f(x) = {x^3\over x^2 - 1}$
has infinite discontinuities (vertical asymptotes) at $x = \pm 1$.
We have 
$$\eqalign{
\lim_{x\to -1^-} f(x) = -\infty \quad & \lim_{x\to -1^+} f(x) = +\infty\cr
\lim_{x\to 1^-} f(x) = -\infty \quad & \lim_{x\to 1^+} f(x) = +\infty\cr}$$
Furthermore $f(0) = 0$ and $f(2) = 8/3$.  

We get an $x_1 \in (-1,0)$ with $f(x_1) = 10$.  Reason:
Since $\lim_{x\to -1^+} f(x) = +\infty$, there exists 
$0> b > -1$ such that $f(b) = k > 10$.  Then the IVT applies to 
$f(x)$ on the interval $[b,0]$ since that interval does not 
contain any of the discontinuities of $f$.  Since $0 = f(0) < 10 < f(b) = k$,
there exists $x_1 \in [b,0]$ such that $f(x_1) = 10$.  

Similarly, since $\lim_{x\to 1^+} f(x) = +\infty$ and $f(2) = 8/3 < 10$, 
we get a second point $x_2 \in (1,2)$ such that $f(x_2) = 10$.

Finally, we also have $\lim_{x\to+\infty}f(x) = +\infty$.  So
there exists some $c > 2$ such that $f(c) > 10$.  The IVT also 
applies on the interval $[2,c]$ (because it misses
both discontinuities of $f$, and hence there exists
$x_3 > 2$ such that $f(x_3) = 10$.

\bigskip
\noindent
14b.  (This proof follows the same ideas as the proofs of 13a and 14a.)
We assume that $f$ is continuous on $[0,1]$, with $f(0) < 0$ and $f(1) > 1$.
Also assume that $g$ is continuous on $[0,1]$ with $g(0) \ge 0$ and 
$g(1) \le 1$.  Consider the function $h(x) = f(x) - g(x)$.  Since
$f,g$ are continuous, so is $h$.  We also have by hypothesis:
$$\eqalign{
h(0) = f(0) - g(0) < 0\cr
h(1) = f(1) - g(1) > 0\cr}$$
If $m$ is the minimum of $h$ on $[0,1]$ and $M$ is the maximum of 
$h$ on $[0,1]$, then $m < 0$ and $M > 0$.  
Hence by the IVT, there exists $c \in [0,1]$ such that $h(c) = f(c) - g(c) = 0$,
or $f(c) = g(c)$.

\bigskip
\noindent
17.

a) This statement is {\it true}.  Proof:  Aiming for a contradiction,
suppose that $f$ is continuous on $[a,b]$, that $f(x) \ne 0$ for all
$x \in [a,b]$, but $f(x)$ takes both positive and negative values
on the interval $[a,b]$.  Then the minimum $m$ of $f$ on the interval
$[a,b]$ must be negative
and the maximum $M$ must be positive, so we get a contradiction to the 
IVT:  $m < 0 < M$, but there is no $x \in [a,b]$ such that $f(x) = 0$.
\bigskip

b) This is also {\it true}.  Proof:  By the EVT, if $m$ is the 
minimum value of $f$ on $[a,b]$, then $m = f(c)$ for some $c \in [a,b]$.
But then $f(c) > 0$ and $f(x) \ge f(c) = m > 0$ for 
all $x \in [a,b]$.
\bigskip
\noindent
18.  

a) This statement is {\it true}.  By the IVT and the EVT, if 
$m$ is the minimum of $f$ on $[a,b]$ and $M$ is the maximum, then 
$f([a,b]) = [m,M]$ (every $y$ with $m \le y \le M$ is $f(x)$ for 
some $x \in [a,b]$).
\bigskip

b)  This statement is {\it false}.  Counterexample:  $f(x) = x^2$, 
$(a,b) = (-1,1)$.  The image $f((-1,1))$ equals $[0,1)$ which is 
not an open interval.  (The open interval $(-1,1)$ gets ``folded
over'' at its midpoint under the mapping $f$, so the image is 
not an open interval.)
\bigskip
\noindent
19.
This statement is {\it false}.  We will construct a 
function that is continuous on the half-open interval $(0,1]$, but
that has {\it neither a maximum nor a minimum on that interval}.  The
idea is to look for a function that has oscillations of greater
and greater amplitude as $x \to 0^+$, but with shorter and shorter
distance between each successive max and min as $x \to 0^+$, so there are
infinitely many local maximum and minimum points in all intervals
$(0,b]$.  An example of this type is given by the function
$$f(x) = {1\over x} \sin\left({1\over x}\right)$$  
(see graph on page 74, and 
multiply that by ${1\over x}$ so that the amplitudes of the oscillations
increase without bound as $x \to 0^+$.)
\bye


