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\centerline{MATH 135 -- Calculus 1}
\centerline{Answers for Sample Questions for Exam 1}
\centerline{September 19, 2019}
\vskip 10pt
\noindent
I.  Express the set of $x$ satisfying $|2x - 5| > 1$ as an interval or union of intervals.  
{\it Answer:} $|2x - 5| > 1$ is equivalent to $2x - 5 > 1$ or $2x - 5 < -1$.
The first says $2x > 6$, so $x > 3$.  The second says $2x < 4$, so $x < 2$.
Another way to write this is as the union of the two intervals:
$(-\infty,2) \cup (3,\infty)$.
\vskip 10pt
\noindent
II.  The following table contains values for
three different functions: $f(x),g(x),h(x)$.  
$$\begin{array}{ccccccc}
x  &\vline&0   &0.1  &0.2  &0.3   &0.4\\ \hline
       f(x)  &\vline&-4.2&-5.9 &-7.6 &-9.3  &-11.0\\
       g(x)  &\vline&10   &20    &40    &80 &160\\
       h(x)  &\vline&4   &2.3  &1.5  &2.1   &6.1
\end{array}$$
\begin{enumerate}
\item[A)]  One of these is a linear function. 
Explain how you can tell which one it is, and give a formula for it.
\item[] {\it Answer:}  $f(x)$ is the linear one, since each change
of $.1$ in $x$ changes $f(x)$ by $-1.7$.  The formula is $f(x) = -17x - 4.2$ 
\item[B)]  One of these functions is {\it neither linear nor exponential}.  
Explain which one that is and why.
\item[] {\it Answer:}  Exponential and linear functions are either increasing
for all $x$ or decreasing for all $x$.  That is not true for $h(x)$. 
\item[C)]  Give a possible formula for $g(x)$.  (Hint:  the values are doubling every
time $x$ increases by $.1$.)
\item[] {\it Answer:}  $g(x) = 10 2^{t/.1} = 10 (2^{10})^t = 10 (1024)^t$
\end{enumerate}
\vskip 10pt
\noindent
III. All parts refer to $f(x) = -3 x^2 + 12 x + 21$.
\begin{enumerate}
\item[A)] Where does the graph $y = f(x)$ cross the $x$-axis?
\item[] {\it Answer:} By the quadratic formula, when $-3x^2 + 12 x + 21 = 0$, 
$x = \frac{-12 \pm \sqrt{144 + 252}}{-6} = 2 \pm \sqrt{11} \doteq -1.317, 5.317$.   
\item[B)] Where does the graph $y = f(x)$ cross the $y$-axis.  
\item[] {\it Answer:}  This happens when $x = 0$, so $y = 21$.  
\item[D)] Sketch the graph $y = -3 x^2 + 12 x + 21$ for $x$ in
$[-4,4]$ and showing correct scales on both the $x$- and $y$-axes.
\item[]  {\it Answer:} The graph is a parabola opening down from 
the vertex $(2,33)$ like this:
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{para.png}
\end{center}
\caption{Figure for Question III, part D}
\end{figure}
\end{enumerate}
\vskip 10pt
\noindent
IV.  You start at $x = 0$ at time $t = 0$ (hours) and drive along the $x$-axis
($x$ values in miles) at 40 miles an hour for 2 hours.  At $t = 2$ you stop for one hour.  Then starting at $t = 3$, you retrace your earlier path 
and return to your starting position at 80 miles per hour. 
\begin{enumerate} 
\item[A)] Sketch the graph of your position as a function of time.
\item[] {\it Answer}:  See graph on next page.
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{trip.png}
\end{center}
\caption{Figure for Question IV,A}
\end{figure}
\item[B)] Give (piecewise) formulas for your function
on the appropriate $t$-intervals. 
\item[] {\it Answer:}
$$x(t) = \begin{cases} 40 t & {\rm if}\ 0 \le t \le 2\\
                       80 & {\rm if}\ 2 < t \le 3\\
                       80 - 80(t - 3) & {\rm if} 3 t \le 4.
         \end{cases}
$$
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{RatFun.png}
\end{center}
\caption{Figure for Question V}
\end{figure}

\noindent
V.  
\begin{enumerate}
\item[A)] Express the domain of the function $f(x) = \frac{x}{x^2 - 1}$ as a union of intervals.
\item[] {\it Answer:}  It is all $x \ne -1,1$, so 
$(-\infty,-1) \cup (-1,1) \cup (1,\infty)$
\item[B)] The figure for this problem shows the graph $y = \frac{x}{x^2 - 1}$.  Based on this, what can you say about the range of $f(x)$?
\item[] {\it Answer:}  Seems to be all real numbers: $\mathbb{R}$, or 
$(-\infty,\infty)$
\item[C)] Explain why $f(x)$ (on its default domain) \emph{fails} to have an inverse function.
\item[] {\it Answer:} The graph does not pass the horizontal line test, so 
$f(x)$ is not one-to-one.  
\item[D)] Give a restricted domain on which $f(x)$ \emph{does} have an inverse function, and sketch the graph of the inverse.  
\item[] {\it Answer:}  The interval of $x$-values $(-1,1)$ is one such.  (The 
intervals $(1,\infty)$ and $(-\infty,-1)$ would be others.)
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{inversefunction.png}
\end{center}
\caption{Figure for Question V,D}
\end{figure}
\end{enumerate}
\vskip 10pt
\noindent
VI.  
\begin{enumerate}
\item[A)] What are the \emph{amplitude} and \emph{period} of the sinusoidal function
$y = 3 \sin\left(\frac{x}{2}\right)$?
\item[] {\it Answer:}  Amplitude $= 3$, period $= 4\pi$.
\item[B)] 
What would change in your answer to B) if the formula was $y = \frac{1}{3} \sin(2x) + 2$?
\item[] {\it Answer:}  The amplitude would change to $\frac{1}{3}$ and the period
would change to $\pi$.
\end{enumerate}
\vskip 10pt
\noindent
VII.  
\begin{enumerate}
\item[A)]  Simplify:  $\log_3(27) + \ln(e^{-3})$.  
\item[] {\it Answer:}  $0$
\item[B)]  Solve for $x$:  $2^{x + 3} = 3^{x/2}$.  
\item[] {\it Answer:}  $x = \frac{6\ln(2)}{\ln(3) - 2 \ln(2)}$.
\item[C)]  The population of a city (in millions) at time $t$ (years) is $P(t) = 2.4 e^{0.06 t}$.  What is the population at $t = 0$?  When will the population reach $4$ million?  
\item[] {\it Answer:}  Population at time $t = 0$ is $P(0) = 2.4$ million.
The population reaches $4$ million at $t = \frac{\ln(4/2.4)}{.06} \doteq 8.5$ years.
\item[D)]  (Continuation of C)  How long will it take for the population to reach double the number at $t = 0$? 
\item[] {\it Answer:}  $t \doteq 11.6$ years.  
\end{enumerate}
VIII. Let $f(x)$ be the function tabulated below. 
$$\begin{array}{ccccccc}
x  &\vline&0   &0.1  &0.2  &0.3   &0.4\\ \hline
       f(x)  &\vline&-3&-5.9 &-7.6 &-8  &-12.0
\end{array}$$
\begin{enumerate}
\item[A)]  What is the \emph{average rate of change} of $f(x)$ over the interval $[0.1,0.2]$? 
\item[] {\it Answer:}  The average rate of change is 
$$\frac{f(.2) - f(.1)}{.2 - .1} = \frac{-7.6 - (-5.9)}{.1} = -17$$ 
\item[B)]  Same question for the interval $[0.2,0.3]$.
\item[] {\it Answer:} Compute:
$$\frac{f(.3) - f(.2)}{.3 - .2} = \frac{-8 - (-7.6)}{.1} = -4$$ 
\item[C)]  Given the information you have, what is your best estimate for
the \emph{instantaneous rate of change} at $t = 0.2$?
\item[] {\it Answer:} The best estimate would come by averaging the answers
from parts A and B, so $\frac{-17 + (-4)}{2} = -10.5$.  
\end{enumerate}
\vskip 10pt
\noindent
IX.  Investigate $$\lim_{x \to 0} \frac{2^x - 1}{3^x - 1}$$
numerically by computing the 
values of $f(x) = \frac{2^x - 1}{3^x - 1}$ 
at $x = -.1, -.01, -.001, .001, .01, .1$.  What's your estimate of the value of
this limit?
\vskip 10pt
\noindent
{\it Answer:} We compute the following values (with a calculator, rounding to 4 decimal places):
$$\begin{array}{cccccccc}
x  &\vline&-.1   &-.01  &-.001  &.001   &.01 & .1\\ \hline
       f(x)  &\vline&.6437 & .6322 &.6311 & .6308  & .6297 & .6181 
\end{array}$$ 
From this evidence it seems the limit should be between $.6311$ and $.6308$ something like $.6310$
or $.6309$.  {\it Note:}  With calculus, we will be able to show later that the exact value
is $\frac{\ln(2)}{\ln(3)} \doteq .6309$.  
\vskip 10pt
\noindent
X. Consider the function graphed in Figure 5.
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{M2I.png}
\end{center}
\caption{Figure for Question X}
\end{figure}

\begin{enumerate}
\item[A)]  What is $\lim_{x \to 0} f(x)$?
\item[] {\it Answer:}  $\lim_{x \to 0} f(x) = 3$
\item[B)]  What are $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$?
\item[] {\it Answer:}  $\lim_{x \to 2^-} f(x) = 0$ and $\lim_{x \to 2^+} f(x) = -1$
\item[C)]  What does your answer to part B say about $\lim_{x \to 2} f(x)$?
\item[] {\it Answer:}  It says that $\lim_{x\to 2} f(x)$ does not exist.  (The two one sided limits must exist and {\it be the same} for $\lim_{x\to 2} f(x)$ to exist.)
\end{enumerate}
\end{document}

