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\markright{Chapter 3: Answers 1}
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\Bf{Chapter 3: Answers 1 \hfill Jack K. Cohen \hfill Colorado School of Mines}

\begin{enumerate}
\item The slope of $f(x)$ at $x = a$ is the  \Em{derivative} of the function at $x =a$.   If $f(x)$ is increasing on the interval $[x_1, x_2]$ then the derivative is  \Em{positive}  on this interval.   If the slope is negative on an interval, then the function is  \Em{decreasing} there.  If the slope is negative on $[a, c)$ and positive on $(c, b]$ and the function is continuous at $c$, then the function has a  \Em{minimum} at $x = c$.    If the derivative is positive on $[a, c)$ and negative on $(c, b]$ and the function is continuous at $c$, then the function has a  \Em{maximum} at $x = c$.

\item  The derivative of $f(x) = x^3 - 2x + 1$ is $f'(x) = 3x^2 - 2 = 3(x^2 - 2/3) = 2 (x - \sqrt{ 2/3})(x + \sqrt{2/3})$.   Thus on the closed interval $[-1, 1]$, we have:

\begin{tabular}{l|c|c|c}
$x$ & $ [-1, -\sqrt{2/3})$ & $(-\sqrt{2/3}, \sqrt{2/3})$ & $(\sqrt{2/3}, 1]$ \\  \hline
$f'$ & $-$ & $+$ & $-$ \\  \hline
$f$ & $\searrow$ & $\nearrow$ & $\searrow$
\end{tabular}

The arrows in the table indicate whether $f$ is increasing/decreasing (we get the third row from the second of course).  We see that the candidates for maximum are $x = -1$ and $x = \sqrt{2/3}$, evaluating $f$ at these points shows that the maximum is at $x = \sqrt{2/3}$.  Similarly $x = -\sqrt{2/3}$ gives the minimum.  See Figure 3.

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans1p3.eps}}
\caption{Plot of $x^3 - 2x + 1$ on $[-1, 1]$.} 
\end{figure}


\item The table and candidates are the same as in the last problem, except replace -1 by -2 and 1 by 2.  This time the endpoints win out, see Figure 2.

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans1p4.eps}}
\caption{Plot of $x^3 - 2x + 1$ on $[-2, 2]$.} 
\end{figure}


\item  See Figure 3 (not a hand sketch admittedly).  A nice way to display the increasing/decreasing intervals is to make a table is like that shown in problem 3. 

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans1p5.eps}}
\caption{Derivative of  $x^3 - 2x + 1$.} 
\end{figure}

\item  See Figure 4.

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans1p6.eps}}
\caption{Result of the \Tt{Plot} in Problem 5.} 
\end{figure}

Since we know that the derivative is $2x$, and the graph is apparently a straight line through the points (0, 0) and (1, 2), the approximate slope function seems to be a fine approximation to the derivative---at least on the crude graphics level.

\item  See Figure 5.  The approximate derivative function (the lighter graph) is a shifted version of the $\sin$ function.  Since the shifted function peaks at $x = 0$, it would seem that the derivative of $\sin x$ is $\cos x$.  You could also measure the shift as about 1.5 from the graph as compared to the correct value $\pi/2 \approx 1.57$ to further verify this guess.  Later this semester, we will prove that  this result is correct. 

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans1p7.eps}}
\caption{Graphs for Problem 6.} 
\end{figure}

\item Re which is the function and which is the derivative.
	\begin{enumerate}
	\item  See Figure 6.   The zero at $b$ is a dead give-away.  If the light curve were the derivative, then the function curve would have a horizontal tangent at $b$, but it does not.  Thus, the light curve must be the function and the dark one is the derivative.  Let's see that everything is consistent with this.  The dark curve has zeroes at $b$ and $d$, sure enough, the light curve has horizontal tangents at these points as a ``good'' function should.  On $[a, 0]$ the dark curve is positive and here, the light curve is increasing as a ``good'' function should.  On $[0, d]$, the dark curve is negative and, sure enough, the light one is decreasing.  Finally, on $[d, f]$, the dark curve is positive and the light function curve increases again.   Open and shut case: the dark curve is the derivative of the light curve.
	
	\begin{figure}[htb]
	\epsfysize 90pt
	\centerline{\epsffile{ans1p8a.eps}}
	\caption{Which is the derivative?} 
	\end{figure}
	
	
	\item See Figure 7.   The zeroes of the dark curve at $b, 0$ correspond to horizontal tangents of the dark curve, so it likely is the derivative curve.  Let's make sure that the other evidence is consistent with the light curve being the function and the dark one being the derivative.  The dark curve is positive in the intervals $[a, b], [0, e]$ and, indeed, the light curve is increasing in these intervals.  Likewise,  the dark curve is negative in $[b, 0]$ and, indeed, the light curve is decreasing in this interval.  As evidence that the light curve can not be the derivative function, we can note that the light curve is always positive, so its function curve would have to always increase---however, the dark curve is initially decreasing---case closed: the dark curve is the derivative of the light curve.
	
	\begin{figure}[htb]
	\epsfysize 90pt
	\centerline{\epsffile{ans1p8b.eps}}
	\caption{Which is the derivative?} 
	\end{figure}
	
	\item See Figure 8.  This one is a stumper.  Assume that the dark curve is the derivative: it is negative on $[a, 0]$ and the light curve decreases on this interval; it is positive on $[0, b]$ and the light curve increases here.  Seems OK.   Assume that the light curve is the derivative: it is always positive and, yes, the dark curve is always increasing. This seems OK, too!  When we learn how to interpret the slope of the derivative (this function is called the second derivative), we will be able to tell that actually the light curve is the function and the dark one its derivative. 
	
	\begin{figure}[htb]
	\epsfysize 90pt
	\centerline{\epsffile{ans1p8c.eps}}
	\caption{Which is the derivative?} 
	\end{figure}
	
	\end{enumerate}
	
\end{enumerate}
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