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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 7 -- {\bf due:} Friday, April 4}
\bigskip
\noindent
{\it `A' Section}
\bigskip
\noindent
1. Let $f(x) = {15 x \over x^4 + 3x^2 + 1}$.  Use the Intermediate
Value and/or Extreme Value Theorems to show the following:
\item{A)}  For all $k \in [-3,3]$, there exist
$c \in [-1,1]$ such that $f(c) = k$.  
\medskip
\item{} {\it Solution:}  First, $f(x)$ is a rational function and $x^4 + 3x^2 + 1 \ge 1$
for all $x \in {\bf R}$, so it follows that $f(x)$ is continuous at all $c \in {\bf R}$.  
We have $f(-1) = -3$ and $f(1) = 3$.  Therefore, by the IVT, for all $k \in [-3,3]$, there
exist $c \in [-1,1]$ such that $f(c) = k$.
\medskip  
\item{B)}  For all $k$ with $0 < k < 3$, there exist some $c \in (1,\infty)$
such that $f(c) = k$.
\medskip  
\item{} {\it Solution:}  We see by the ``Big Theorem'' on function limits that
$\lim_{x\to \infty} f(x) = 0$.  Hence if we have any $k$ with $0 < k < 3$, then
there exists a $B > 1$ such that $f(B) < k < 3 = f(1)$.  Applying the IVT on the interval
$[1,B]$, we see there is a $c \in (1,B) \subset (1, \infty)$ such that $f(c) = k$.  
\medskip
\item{C)}  There is a $c \in (0,1)$ where $f(c) = 3$.  
\medskip
\item{} {\it Solution:}  Consider the function 
$$g(x) = f(x) - 3 = {15x - 3x^4 - 9x^2 - 3 \over x^4 + 3x^2 + 1}$$
We see that $g(0) = -3 < 0$ and $g(1/2) = {33\over 29} > 0$.  $g(x)$ is also continuous
on $[0,1/2]$.  Therefore, by the IVT, there exists $c \in (0,1/2) \subset (0,1)$
where $g(x) = 0$.
\medskip  
\item{D)}  There is a $d \in (0,1)$ where $f'(d) = 0$. 
\medskip
\item{} {\it Solution:}  Consider the number $c$ from part c.  We have $f(c) = f(1) = 3$, 
but $f$ is certainly not constant between $c$ and $1$.  
Therefore by the Extreme Value Theorem, there must exist a maximum value $f(d) > 3$ or 
a minimum value $f(d) < 3$ for $f$ on the interval $[c,1]$ attained at some $d \in (c,1)$.  
By plotting the function $f$, we can see that in fact the first statement is the one 
that is true -- there is a maximum.  We claim that $f'(d) = 0$.  First note that 
$$\lim_{x \to d^-} {f(x) - f(d)\over x - d} \ge 0$$ 
since $f$ has a maximum at $d$.  Similarly, 
$$\lim_{x \to d^+} {f(x) - f(d)\over x - d} \le 0$$ 
Since $f$ is a rational function whose denominator never equals zero, $f$
is differentiable at $d$, so these one-sided limits must both exist and be
equal.  Therefore $f'(d) = 0$.  
\bigskip
\noindent
2. Show that there is a solution of the equation $\tan(x) = x$ in the interval
$\left({(2k - 1)\pi\over 2}, {(2k + 1)\pi\over 2}\right)$ for every $k \in {\bf Z}$.  
\bigskip
\noindent
{\it Solution:}  We know that $\tan(x)$ has infinite discontinuities at
${(2k - 1)\pi\over 2}$ and ${(2k + 1)\pi\over 2}$ and hence 
$$\lim_{x\to {(2k - 1)\pi\over 2}^+} \tan(x) = -\infty \quad 
\lim_{x\to {(2k + 1)\pi\over 2}^-} \tan(x) = +\infty$$
It follows that we also have
$$\lim_{x\to {(2k - 1)\pi\over 2}^+} \tan(x) - x = -\infty \quad 
\lim_{x\to {(2k + 1)\pi\over 2}^-} \tan(x) - x = +\infty$$
Therefore, $\tan(x) - x$ must change sign on some interval 
$[a,b] \subset \left({(2k - 1)\pi\over 2}, {(2k + 1)\pi\over 2}\right)$ and 
the Intermediate Value Theorem implies $\tan(x) - x = 0$ somewhere in
that interval.  
\bigskip 
\noindent
3.  Using the definition of the derivative, find the value of $f'(c)$, or 
say why $f$ is not differentiable at $x = c$:
\bigskip
\item{A)}  $f(x) = x^3 + 2x - 4$ at $c = 2$.
\medskip
\item{} {\it Solution:}  We have 
$$\eqalign{f'(2) &= \lim_{x\to 2} {x^3 + 2x - 12 \over x - 2}\cr
&=\lim_{x\to 2} {(x - 2) (x^2 + 2x + 6) \over x - 2}\cr
&=\lim_{x\to 2} {x^2 + 2x + 6}\cr
&= 14.}$$  
\item{B)}  $f(x) = \sin(|x|)$ at $c = 0$.  Hint: Look back at Problem Set 6, B 2.   
\medskip
\item{} {\it Solution:} $f'(0)$ does not exist for this function because
$$\lim_{x\to 0^+} {\sin(|x|) \over x} = +1$$
by the indicated problem on Problem Set 6, while
$$\lim_{x\to 0^-} {\sin(|x|) \over x} = \lim_{x\to 0^-} {-\sin(x) \over x} = -1.$$
Since the one-sided limits are not equal, the derivative at $0$ does not exist.
\medskip
\item{C)}  The function defined by
 $$f(x) = \cases{x^2 & if $x > 1$\cr
                          2x - 1 & if $x \le 1$\cr}$$
at $c = 1$.  
\medskip
\item{} {\it Solution:}  We have 
$$\lim_{x \to 1^+} {f(x) - f(1)\over x - 1} = \lim_{x\to 1^+} {x^2 - 1\over x - 1} = 
\lim_{x \to 1^+} x + 1 = 2.$$
On the other hand, 
$$\lim_{x \to 1^-} {f(x) - f(1)\over x - 1} = \lim_{x\to 1^-} {2x - 2\over x - 1} = 
\lim_{x \to 1^-} 2 = 2.$$
Since the one-sided limits exist and are equal, $f'(1)$ exists and equals $2$.  
\medskip
\item{D)}  The function defined by
$$f(x) = \cases{x^2 & if $x \in {\bf Q}$\cr
                  0 & if $x \in {\bf Q}^c$\cr}$$
at $c = 0$. 
\medskip
\item{} {\it Solution:}  We have for $x \ne 0$, 
$${f(x) - f(0) \over x - 0} = \cases{x & if $x \in {\bf Q}$\cr 
                                     0 & if $x \in {\bf Q}^c$\cr}$$
Given any $\varepsilon > 0$, if we take $\delta = \varepsilon$, then 
for all $x$ in the deleted neighborhood defined by $0 < |x| < \varepsilon$, 
$$\left| {f(x) - f(0) \over x - 0} - 0 \right| < \varepsilon.$$
It follows that
$$\lim_{x \to 0} {f(x) - f(0) \over x - 0} = 0 = f'(0).$$
(It is not too hard to show that $f'(c)$ exists only for this one $c = 0$.
This function is not differentiable anywhere else.)

\bigskip
\noindent
{\it `B' Section}
\bigskip
\noindent
1.  Let $f$ be continuous on $[0,1]$ with $f(0) < 0$ and $f(1) > 1$.  
Suppose that $g$ is another continuous function on $[0,1]$ such that
$g(0) \ge 0$ and $g(1) \le 1$.  Show that there exists some $c \in (0,1)$
such that $f(x) = g(x)$.  
\bigskip
\noindent
{\it Solution:}  Let $h(x) = f(x) - g(x)$.  Since $f,g$ are continuous on 
$[0,1]$, the same is true for $h$.  By the given information, 
$h(0) = f(0) - g(0) < 0$ and $h(1) = f(1) - g(1) > 0$.  Therefore, the IVT
implies that $h(c) = 0$ for some $c \in (0,1)$.  But then $0 = h(c) = f(c) - g(c)$, 
so $f(c) = g(c)$.   
\bigskip 
\noindent
2. Let $f$ be continuous on $[a,b]$ with $f(a) < k < f(b)$.  Here
is a variation on our proof of the Intermediate Value Theorem.  
\item{A)}  Let 
$$T = \{x \in [a,b] \mid f(x) > k\}.$$
Show that $T$ has a greatest lower bound and that $f({\rm glb}(T)) = k$.
\bigskip 
\item{} {\it Solution:}  $T$ is contained in the interval $[a,b]$, so 
it is a bounded subset of ${\bf R}$.  Then $c = {\rm glb}(T)$ exists 
by the LUB Axiom.  Note that $a < c$ since $f(a) < k$.  Hence the interval
$[a,c)$ is contained in the complement of $T$.  If we let 
$\{x_n\}$ be any sequence contained in $[a,c)$ converging to $c$, then 
since $f$ is continuous, $\lim_{n\to\infty} f(x_n) = f(c)$.  But $f(x_n) \le k$ 
for all $n$, so 
$$f(c) = \lim_{n\to\infty} f(x_n) \le k\leqno(1)$$
 also (by Corollary 2.2.8 in
the text).  On the other hand, given any $\varepsilon > 0$, $c + \varepsilon$
is not a lower bound for $T$, so there exists some $x\in T$ such that
$c \le x < c + \varepsilon$.  Apply this for $\varepsilon = {1\over n}$
for each natural number.  Then we get a sequence $x_n'$ such that
$x_n' \in T$ for all $n$ and $c \le x_n' < c + {1\over n}$.  It follows
easily that $x_n' \to c$ as $n \to \infty$.  Therefore since $f$ is 
continuous at $c$, $\lim_{n\to\infty} f(x_n') = f(c)$.  But $x_n' \in T$
for all $n$, so $f(x_n') > k$.  Hence 
$$f(c) = \lim_{n\to\infty} f(x_n') \ge k.\leqno(2)$$  
The two inequalities (1) and (2) show that $f(c) = k$.  
\bigskip
\item{B)}  Will this ${\rm glb}(T)$ always be the same as the $c$ we found
in our proof of the IVT with $f(c) = k$?  If so, prove they are the same;
if not, give a counterexample.  
\bigskip
\item{} {\it Solution:}  In the proof we did in class we considered
$$S = \{x \in [a,b]\mid f(x) \le k\}$$
and we showed that if $c' = {\rm lub}(S)$, then $f(c') = k$.  The $c$ found
in part A and the $c'$ here do not have to be the same.  For instance, let
$f(x) = x^3 - 2x + 1$ on $[-2,2]$. We have $f(-2) = -3$ and $f(2) = 5$.  So
the IVT will apply for any $k$ with $-3 < k < 5$.  Consider $k = 0$.  The
equation $x^3 - 2x + 1 = 0$ actually has three different roots in the interval
$[-2,2]$:  One between $-2$ and $-1$ (call this one $\alpha$), a second between $1/2$ and $1$
(call this one $\beta$), and a third 
at $x = 1$.  The set $T$ as in part A is the union $T = (\alpha,\beta) \cup (1,2)$, 
so $c = {\rm glb}(T) = \alpha$.  On the other hand, the set $S$ as in the 
proof we did in class is $S = [-2,\alpha] \cup [\beta,1]$, so $c' = {\rm lub}(S) = 1$.  
\bigskip
\noindent
3.  This property deals with another property of real-valued functions of a real variable
sometimes called {\it Lipschitz continuity}.  
\bigskip
\item{A)}  Let $f$ be a function on an interval $I$ with the property that there exists
a strictly 
positive constant $k$ such that $|f(x) - f(x')| \le k |x - x'|$ for all $x,x' \in I$ (this is 
the definition of Lipschitz continuity).  Show 
that $f$ is uniformly continuous on $I$.
\bigskip  
\item{} {\it Solution:} Given $\varepsilon > 0$, let $\delta = \varepsilon/k$.
Then for any $x,x'\in I$ such that $|x - x'| < \delta = \varepsilon/k$, it follows
that 
$$|f(x) - f(x')| \le k |x - x'| < k \cdot \varepsilon/k =\varepsilon.$$
This shows that the definition of uniform continuity is satisfied for $f$ on $I$. 
\bigskip
\item{B)}  The converse of the statement in part A is not true:  
Show that $f(x) = x^{1/3}$ is uniformly continuous on $[-1,1]$, but there is
no constant $k$ such that $|f(x) - f(x')| \le k |x - x'|$ for all $x,x' \in [-1,1]$.
Hint:  Think slopes of secant lines to the graph $y = x^{1/3}$. 
\bigskip
\item{} {\it Solution:}  First, $f(x)$ is continuous on $[-1,1]$, hence it is 
uniformly continuous by the result of Theorem 3.6.8 (proved in class before 
Easter break).  
Let $x' = 0$ and take arbitrary $x > 0$ we have
$${f(x) - f(0) \over x - 0} = {x^{1/3} \over x} = {1\over x^{2\over 3}}.$$
But $\lim_{x \to 0^+} {1\over x^{2\over 3}} = +\infty$.  
In other words, the value of the difference quotient 
will get unboundedly large as $x \to 0^+$.  Hence there is no 
single $k$ such that 
$$\left|{f(x) - f(0) \over x - 0}\right| \le k$$
for all $x$ in $[-1,1]$.  But that shows that there is no 
$k$ such that $|f(x) - f(0)| \le k |x - 0|$ for all $x$ in $[-1,1]$.  

\bye
