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\noindent
{\bf Math 105} \hfil {\bf Answers to test 3}

\medskip
\noindent {\bf Part I.  Multiple Choice}

\mm
{\bf 1.} $\ds{\frac{d}{dx}(x^{3}e^{4x})} = 3x^{2}e^{4x} + 4x^{3}e^{4x}$ by the product rule.

\mm
{\bf 2.} Suppose $f(x)$ is a function which satisfies $f'(3) = 0$, $f'(5) = 0$,
$f''(3) = -4$, and $f''(5) = 5$.  Which of the following statements is true?
\ {\bf Ans.} $f(x)$ has a relative maximum at $x = 3$ and a relative minimum at $x =
5$.  (The second derivative test)

\mm
{\bf 3.} Let $y = \sqrt{\ln x}$.  Then $\ds y' = {\frac{1}{2x \sqrt{\ln x}}}$ by the chain rule.

\mm
{\bf 4.} Let $h(x) = \ds{\frac{f(x)}{g(x)}}$.  Then $\ds h'(0) = \frac{g(0)f'(0) - f(0)g'(0)}{[g(0)]^2} = -\frac{5}{8}$. (Use the given
table.)

\mm
{\bf 5.} Let $k(x) = f(g(x))$. Then $k'(0) = f'\left(g\left(0\right)\right)\cdot g'(0) = f'(2)\cdot g'(0) = 1.$ (Use the given table.)

\mm
{\bf 6.} Let $f(x)$ be a function with $f'(2) = 0$.  Which of the following
{\bf must} be true about $f$? \ {\bf Ans.} The tangent line to the graph of $f$ at $(2, f(2))$ is horizontal.

\mm
{\bf 7.} Suppose that $f(x)$ is increasing and concave up everywhere. Further,
$f(2) = 2$ and the tangent line at the point $(2,2)$ is the line $y = x$. Which
of the following can you conclude?  \ {\bf Ans.} $f(5) > 5$.  If the graph is concave up its slope is increasing, so it
is above the tangent line $y=x$ for all $x > 2$.

\mm
{\bf 8.} The graph of a function $f(x)$ is sketched below.  Which of the
following
statements is true? \ {\bf Ans.} $f(x)$ is increasing and $f'(x)$ is decreasing.

\vskip1.5in \noindent
\begin{minipage}[t]{3.5in}
{\bf 9.} Suppose that $f(x)$ is a function with $f''(x) > 0$ on the interval
$(2,\infty)$ and $f''(x) < 0$ on the interval $(-\infty,2)$.  Which of the
following could be the graph of $f(x)$? (Use the letter to the upper left
of each graph.) Note that the answer is the only graph that is concave up for $x > 2$ and down for $x < 2$.
\end{minipage} \  \begin{minipage}[t]{2.5in} \ \quad{\bf Ans.} \end{minipage}

\page \noindent
{\bf 10.}  Suppose that $f(x)$ is a function whose {\bf derivative} is $f'(x) =
x\ln x$.  Identify the $x$-value(s) of all inflection point(s) of $f(x)$. \ {\bf Ans.} $1/e$, because $f''(x) = \ln x + 1$. The only
solution to $f''(x) = 0$ is $x = 1/e$, and one checks that $f''(x)$ changes sign there.

\mm
{\bf Part I.  Partial Credit} 

\smallskip \noindent
{\bf 11.} A pot of boiling water ($100^\circ$ Celsius) is
left to
cool in a room whose temperature is $20^\circ$ Celsius so that after $t$
minutes, the temperature of the water is $H = 20 + Ae^{kt}$ for some constants
$A$ and $k$.  After 30 minutes, the water is $60^\circ$.  At what {\bf
rate} (with
respect to time) is the temperature of the water changing after 90 minutes?


\noindent
 {\bf Ans.} $-(\ln 2)/3 \approx - 0.231.$  First, substitute $t=0$, and $H = 100$ to get $A = 80.$ Then, substitute $t=30$ and
$H = 60$ and solve for $k$ to get $k = -(\ln 2)/30$.  The rate at which the temperature changes is
$H'(t) = kAe^{kt}$.  Using the values of $A$ and $k$ just determined and setting $t=90$ gives the answer.

\mm
{\bf 12.}  The half-life of carbon-14 is 5730 years, so carbon
decays according to the equation $y = Ae^{-kt}$, where $k = .000121$.  An
ancient
document written on papyrus is found to have 37.5 $\%$ of the carbon-14 that a
freshly cut sample has.  How old is the document?

\noindent
{\bf Ans.} $-(\ln(0.375))/0.000121 \approx 8106$ years.  Because 
$\ds \ 0.375 = \frac{y(t)}{y(0)} = \frac{Ae^{-kt}}{A} = e^{-0.000121t}.$
Solving for $t$ gives the answer.

\mm
{\bf 13.} Find the the equation of the line tangent to $y =
\ds\frac{e^{x^{2} + 2x}}{x^{3} + 1}$ at $(0, 1)$.
\ 
{\bf Ans.} $y = 2x + 1$.  First find $y'$ by using the quotient and chain rules.  Plugging in $x=0$ gives $y'(0) = 2$, which is the
slope of the tangent line.

\mm
{\bf 14.} The period $P$ in seconds of a simple pendulum is
completely determined by its length $L$ according to the formula
$\ds P = \frac{2\pi}{\sqrt{10}}\sqrt{L}$, where $\pi \approx 3.14$.
Therefore, if
the length of a pendulum increases, its period will increase.  (That is why
clocks
with metal pendulums tend to run slow in the summertime.)  Find the rate (with
respect to time) at which a simple pendulum's period is increasing when the
pendulum is 4m long and its length is increasing at a rate of .001m/day.
\ {\bf Ans.} $\ds \frac{0.001\pi}{2\sqrt{10}} \approx 0.0004967.$  Use the chain rule to get
$$\frac{dP}{dt}  = \frac{2\pi}{\sqrt{10}}\cdot\frac{1}{2\sqrt{L}}\cdot\frac{dL}{dt} =
\frac{\pi}{\sqrt{10}}\cdot\frac{1}{\sqrt{L}}\cdot\frac{dL}{dt}.$$
Then plug in $L=4$ and $dL/dt = 0.001$ to get the answer.

\page \noindent
{\bf 15.} The demand $q$ for Widgets is a function
of the price $p$, i.e.\ $q = f(p)$. When the price $p = \$10$, we have $q =
f(10)
= 100$ and $f'(10) = -4$.

\mm
 (a) The revenue $R$ is determined by $R = pq$.  Find
$dR/dp$ when
$p = 10$. \ {\bf Ans.} 60.  Write $R = pf(p)$, so by the product rule $R'(p) = pf'(p) + f(p)$.  Then plug in $p=10$.


\mm
 (b) The cost of producing $q$ Widgets is $C(q) = 100q + 5000$. Find
$dC/dp$ when $p = 10$. 

\noindent
 {\bf Ans.} $-400$. By the chain rule, $dC/dp = (dC/dq)\cdot(dq/dp) = 100f'(p)$.
Then plug in $p=10$.

\mm
(c) If $P$ represents profit, find $dP/dp$ when $p = 10$. \ {\bf Ans.} $460$.  Use $P = R - C$, so that
$dP/dp = dR/dp - dC/dp$.

\next
{\bf 16.}  Consider the function $f(x) = x - 3x^{\frac23}$.\

\mm
(a) Identify the intervals on which $f(x)$ is increasing and those on which
$f(x)$ is decreasing.

\mm
 (b) Identify the $x$-value(s) of all relative extreme point(s) of
$f(x)$ and
specify whether each is a maximum or a minimum.

\smallskip \noindent 
{\bf Ans.} (a) $f(x)$ is increasing on $(-\infty, 0)$ and on $(8, \infty)$, decreasing on $(0,8)$. \ (b) There is a relative max at
$x=0$ and a relative min at $x=8$.  First compute the derivative, $f'(x) = 1 - 2x^{-{1\over3}}$. There are two critical points:
at $x=0$, where $f'(x)$ does not exist, and at $x=8$, where $f'(x) = 0$.  Checking the sign of $f'(x)$ on the intervals
$(-\infty, 0)$,  $(0,8)$, and $(8,\infty)$ gives the answer.

\next
{\bf 17.}  Consider the function $f(x) = x^4 - 54x^2 + 27x
- 17$.

\smallskip \noindent 
 (a) Identify the intervals on which $f(x)$ is concave up and those on which
$f(x)$ is concave down.

\smallskip \noindent 
(b) Identify the $x$-value(s) of all inflection point(s) of $f(x)$.

\smallskip \noindent 
{\bf Ans.} (a) Concave up on $(-\infty, - 3)$ and on
$(3,\infty)$, concave down on $(-3,3)$. (b) Inflection points at $x=3$ and $x=-3$.  Compute the second derivative,
$f''(x) = 12x^2 - 108$, and solve $f''(x) = 0$ to get $x=\pm 3$.  Checking the sign of $f''(x)$ on the intervals $(-\infty, -3)$,
$(-3,3)$, and $(3,\infty)$ gives the answer.

\page \noindent
{\bf 18.} Suppose that $f(x)$ is a function which
satisfies the
following list of properties:
\begin{tabbing}
\hs \=$\bullet$\ $f(-3)$ \= $= -1$, $f(0) = -3$, $f(2) = -5$, and $f(4) = 0$.\\
\>$\bullet$\ The only solutions to $f'(x) = 0$ are $x = -3$ and $x = 2$.\\
\>\>$\bullet$\ $f'(x) > 0$ on the intervals $(-\infty,-3)$, $(2,\infty)$.\\
\>\>$\bullet$\ $f'(x) < 0$ on the interval $(-3,2)$.\\
\>$\bullet$\ The only solution to $f''(x) = 0$ is $x = 0$.\\
\>\>$\bullet$\ $f''(x) > 0$ on $(0,\infty)$.\\
\>\>$\bullet$\ $f''(x) < 0$ on $(-\infty,0)$.
\end{tabbing}

\mm
(a) List all the relative extreme points and inflection points of $f(x)$.

\mm
(b) Sketch a possible graph of $f(x)$, labeling the points you listed in
part (a).

\mm {\bf Ans.} Relative maximum at $x=-3$, relative minimum at $x=2$, inflection point at $x=0$.  The graph is drawn below.





 
 





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