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\centerline{MATH 135 -- Calculus 1}
\centerline{Extreme Values}
\centerline{November 11, 2019}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Recall from today's video:
\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}
\vskip 10pt
\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[] {\it Solution:}  $f'(x) = 3x^2 - 24x + 21 = 3(x - 7)(x - 1)$, so 
$f$ has two critical points $x = 1, 7$; both are in the interval $[0,11]$.  (Note:  $f$ and $f'$ are
polynomial functions, so they are defined for all real $x$.)  Comparing the values 
\begin{eqnarray*}
f(0) &=& 0\\
f(11) &=& 110\ \leftarrow \text{ maximum}\\
f(1) &=& 10\\
f(7) &=& -98\ \leftarrow \text{ minimum}\\  
\end{eqnarray*}
\item[2.]  $f(x) = (x^2 + 2x)e^{-x}$ on $[1,5]$
\item[] {\it Solution:} $f'(x) = (2 - x^2)e^{-x}$.  Thus $f'(x) = $ at $x = \pm \sqrt{2}$,
but $x = \sqrt{2}$ is the only critical point in the interval $[1,5]$.  As in (1), both $f$ and 
$f'$ are defined for all real $x$.  Comparing the values 
\begin{eqnarray*}
f(1) &\doteq& 1.104\\
f(\sqrt{2}) &\doteq& 1.174\ \leftarrow \text{ maximum}\\
f(5) &\doteq& .0236\ \leftarrow \text{ minimum}
\end{eqnarray*}
\item[3.]  $f(x) = 5\tan^{-1}(x) - x$ on $[-5,5]$.
\item[] {\it Solution:} We have
$$f'(x) = \frac{5}{1 + x^2} - 1 = \frac{4 - x^2}{1 + x^2}$$
This (and $f(x)$) is defined for all real $x$.  The critical points are $x = \pm 2$. Comparing the values 
\begin{eqnarray*}
f(-5) &\doteq& -1.867\\
f(-2) &\doteq& -5.536\ \leftarrow \text{ minimum}\\
f(2) &\doteq& 3.536\ \leftarrow \text{ maximum}\\
f(5) &\doteq& 1.867  
\end{eqnarray*}
\item[4.]  $f(x) = (x - x^2)^{2/3}$ on $[0,2]$.  (Note:  Careful on this one -- you
should find critical points where $f'(x)$ does not exist.)
\item[] {\it Solution:}  By the Chain Rule have $f'(x) = \frac{2}{3}(x - x^2)^{-1/3}(1 - 2x)$.
This is undefined at $x = 0,1$ and it is equal to zero at $x = 1/2$.  These are all critical points.
In addition, $x = 0$ is one of the endpoints of the interval.  
Comparing the values: 
\begin{eqnarray*}
f(0) &=& 0\ \leftarrow \text{ minimum}\\
f(1/2) &\doteq& .397\\
f(1) &=& 0\ \leftarrow \text{ minimum}\\
f(2) &\doteq& 1.587\ \leftarrow \text{ maximum}
\end{eqnarray*}

\end{enumerate}

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