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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 7 -- {\bf due:} Friday, April 4}
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{\it `A' Section}
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\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 $f(c) = k$.  
\item{B)}  For all $k$ with $0 < k < 3$, there exist some $c \in (1,\infty)$
such that $f(c) = k$.  
\item{C)}  There is a $c \in (0,1)$ where $f(c) = 3$
\item{D)}  There is a $d \in (0,1)$ where $f'(d) = 0$.  
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\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 $k \in {\bf Z}$.  
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\noindent
3.  Using the definition of the derivative, find the value of $f'(c)$, or 
say why $f$ is not differentiable at $x = c$:
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\item{A)}  $f(x) = x^3 + 2x - 4$ at $c = 1$.
\item{B)}  $f(x) = \sin(|x|)$ at $c = 0$.  Hint: Look back at Problem Set 6, B 2.   
\item{C)}  The function defined by
 $$f(x) = \cases{x^2 & if $x > 1$\cr
                2x - 1 & if $x \le 1$\cr}$$
at $c = 1$.  
\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 $c = 0$. 
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\noindent
{\it `B' Section}
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\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 $f(x) = g(x)$.  
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\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 $f({\rm glb}(T)) = k$.
\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.  
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\noindent
3.  This property deals with another property of real-valued functions of a real variable
sometimes called {\it Lipschitz continuity}.  
\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 $I$.  
\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 $y = x^{1/3}$.  
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