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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 5B Solutions, October 27, 2017}
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\noindent
Section 3.1
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\noindent
51.  The derivative is positive on the intervals $1.0 < x < 2.5$ and again on $x > 3.5$.
These are the points on the graph where the slope of the tangent line is \emph{positive}. 
The slopes of the tangent lines are negative or zero at all other points.  
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\noindent
52.  The graph is in Figure 1 at the top of the next page (for typographical reasons).  
$x = \pi/2$ is at the midpoint of the interval, where the tangent line
to $y = f(x) = \sin(x)$ appears to be a \emph{horizontal} line.  Hence we 
expect $f'(\pi/2) = 0$.  We check this by taking the difference quotient
$$\frac{f(\pi/2 + h) - f(\pi/2)}{h}$$
for $h = 0.001$ and $h = -0.001$:
$$\frac{\sin(\pi/2 + 0.001) - \sin(\pi/2)}{.001} \doteq -0.0005$$
and 
$$\frac{\sin(\pi/2 - 0.001) - \sin(\pi/2)}{-0.001} \doteq 0.0005$$
Note that both are close to zero.  Moreover, the difference quotient
for $h = +0.001$ gives a negative value, while $h = -0.001$ gives
a positive value.  This is expected from the graph, since the values
above are slopes of secant lines.  The one with $h > 0$ is sloping 
slightly downward because $f(x)$ is starting to decrease from the 
maximum at $x = \pi/2$, while the one with $h < 0$ is positive, because
that secant line is sloping upward.  
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\noindent
Section 3.2 
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\noindent
44.  Looking at Figure 14 in the text, we see $y = g(x) < 0$ when $y = f(x)$ is sloping down, 
while $y = g(x) > 0$ when $y = f(x)$ is sloping up.  Moreover $g(x) = 0$
at the same $x = 0$ where $f(x)$ has a minimum.  Therefore $g(x) = f'(x)$.
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46.   
\begin{enumerate}
\item[(a)]  See plot in Figure 2 which gives the rough shape of $Q'(t)$.  
$Q'(t)$ represents the rate of change of the cumulative total amount of crude oil 
as a function of time.  
\item[(b)]  The maximum derivative occurs roughly at $t = 2005$ (about 1/5
of the way between $t = 2000$ and $t = 2025$).  The question did not ask for this, but
I estimated the  
value $Q'(2005)$ to generate the plot in Figure 2 like this:
Draw in a line tangent to the graph at $t = 2005$ by hand and see where
that line crosses the lines $Q = 0.5$ and $Q = 1.5$.  That happens
roughly at $t = 1985$ and $t = 2020$, so the slope of the tangent is 
approximately 
$$Q'(2005) \doteq \frac{1.5 - 0.5}{2020 - 1985} \doteq .029$$
(units are trillions of barrels per year).  
\item[(c)]  $\lim_{t \to \infty} Q(t) = 2.3$ trillion barrels.
That is the total cumulative amount of crude oil produced 
over all time.  
\item[(d)]  $\lim_{t \to \infty} Q'(t) = 0$, because the 
slope of the graph $Q(t)$ goes to zero as $t$ increases without
bound.  The 
interpretation is that eventually, the cumulative amount
of crude oil produced will not be changing any more.  \emph{We
will have found and used up all of the easily accessible crude
oil, so no more will be produced.}
\end{enumerate} 
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\noindent
66.  
\begin{enumerate}
\item[] (A) has derivative (III) (Note:  (A) has three points where 
the derivative is zero; (III) is the only graph that crosses the 
$x$-axis three times) 
\item[] (B) has derivative (I)
\item[] (C) has derivative (II)
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=3.0in]{3point1fifty-two.png} 
\caption{$y = \sin(x)$ on $[0,\pi]$.}
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 \end{figure}

\begin{figure}
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\includegraphics[width=3.0in]{PeakOil.png} 
\caption{$Q'(t)$ for Problem 46.}
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 \end{figure}

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1^+} f(x) = 1$.
This implies $\lim_{x\to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
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