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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Midterm Exam 3}
\centerline{December 2, 2005}
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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).
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I.  All parts of this question refer to the function
$f(x) = x^2 - 6x + 3.$
\item{A)}  (15)  Show using the $\varepsilon,\delta$
definition that 
$$\lim_{x\to 1} f(x) = -2.$$
\item{B)}  (15)  Show that $f(x)$ is integrable on 
$[3,4]$ using any applicable method.  (Don't work too hard!) 

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II.
\item{A)}  (10) State the Intermediate Value Theorem.  
\item{B)}  (10) Let $f(x) = a_1 + a_2\cos(x)$, where $a_1,a_2$ are constants
with $0 < a_1 < a_2$.  Show that there exists some $c \in [0,\pi]$
such that $f(c) = 0$.

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\noindent  
III.  (20)  Show that if $f$ is continuous on $[a,b]$
and $s = \sup f([a,b])$, then there exists $c \in [a,b]$ where
$f(c) = s$.    (Note: This is the portion of the Extreme
Value Theorem showing that $f$ attains a maximum.)

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\noindent
IV.  True-False.  Say whether each of the following statements is 
true or false, and give a detailed reason for your assertion. 
Do any three parts.  If you submit
solutions for all four, then I will consider the other one for
Extra Credit.  
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\noindent
All parts refer to the function $f$ defined by:
$$f(x) = \cases{ x^2 & if $x\in {\bf Q}$\cr
                  0 & if $x\notin {\bf Q}$\cr}$$

\item{A)} (10) $\lim_{x\to 3} f(x)$ exists.
\item{B)} (10) $f$ is {\it differentiable} at $x = 0$.
\item{C)} (10) $f$ {\it integrable} on $[1,2]$.
\item{D)} (10) The {\it conclusion} of the Extreme Value 
Theorem is true for $f$ on the interval $[-\sqrt{2},\sqrt{2}]$.
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(Note:  The {\it conclusion} of 
theorem phrased as ``if $p$, then $q$'' is the statement $q$; the 
{\it hypotheses} are the statement(s) $p$.  The conclusion of
a theorem can be true in a particular example even if the 
hypotheses are not satisfied.)
\bye

