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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Extreme Values}
\centerline{November 20, 2015}
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\noindent
{\it Background}
\vskip 10pt
Recall from today's video and yesterday's class:
\begin{itemize}
\item A continuous function on a closed interval attains a 
\emph{maximum value} and a \emph{minimum value} on that interval. 
\item Those maximum and minimum values are attained either at endpoints 
of the interval or at interior points -- nothing deep going on there, just
logic(!)  
\item If the maximum or minimum value is attained at a point $x = c$
\emph{other than an endpoint}, then $c$ must be a \emph{critical point}
of $f$ -- either a solution of $f'(c) = 0$, or else a place where
$f'(c)$ does not exist.  
\item So to find the maximum and minimum value of a continuous function
of a continuous $f(x)$ on a closed interval $[a,b]$, we can:
\begin{enumerate}
\item[(i)] Compute $f'(x)$ and find all critical points $c$ in $[a,b]$.
\item[(ii)] Compute $f(a),f(b)$, and $f(c)$ for all critical points found
in the first step.  
\item[(iii)] Then the maximum value will be the largest of the numbers
found in the previous step and the minimum value will be the smallest
of those numbers.   
\end{enumerate}
\end{itemize}
 
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\noindent
{\it Questions}
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\noindent
For each of the following functions, 
\begin{enumerate}
\item[(i)] determine all critical points in the given interval, 
\item[(ii)] compute the values of $f$ at the critical points in the interval,
and compute the values at the endpoints of the interval, 
\item[(iii)] determine the maximum and minimum values of $f$ on the interval.
\end{enumerate}
This is the process sketched above(!)
\begin{enumerate}
\item[1.]  $f(x) = x^3 - 12x^2 + 21x$ on $[0,11]$.
\item[2.]  $f(x) = (x^2 + 2x)e^{-x}$ on $[1,5]$
\item[3.]  $f(x) = 5\tan^{-1}(x) - x$ on $[-5,5]$.
\item[4.]  $f(x) = (t - t^2)^{2/3}$ on $[0,2]$.  (Note:  Careful on this one -- you
should find critical points where $f'(x)$ does not exist.)
\end{enumerate}

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