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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions -- Final Examination}
\centerline{\it December 11, 2001}
\bigskip
\noindent
I.  
\item{A)} The function plotted looks most like a sinusoidal 
oscillation with amplitude $A = 5$, vertical shift $C = 2$, 
and period $8$, so $B = 2\pi/8 = \pi/4$.  One possible formula
is 
$$y = 5\sin\left({\pi x\over 4}\right) + 2$$
\noindent
{\it Note:}  It could also be argued that this looks like a polynomial
function, but then there would have be at least four roots
and four turning points, so the degree would have to be 5 at least.
\bigskip
\item{B)}  There is no inverse function for $f$ on $0\le x\le 4$
since that part of the graph does not pass the horizontal line test.
\bigskip
\noindent
II.  There are $x$-axis intercepts at $x = -2$ and $x = 3$, and
vertical asymptotes at $x = -1,1$.  So the rational function 
looks like:
$$r(x) = {c(x+2)(x-3)\over (x-1)(x+1)} = {c(x^2 - x - 6)\over x^2-1}$$
for some constant $c$.  To determine the value of $c$, we can look
at the $y$-intercept.  When $x =0$, $y = -6$, so
$$-6 = {c(0^2-0-6)\over 0^2-1} = {-6c\over -1} = 6c.$$
So $c = -1$, and 
$$r(x) = {-(x+2)(x-3)\over (x-1)(x+1)} = {-x^2 + x + 6\over x^2-1}$$
\bigskip
\noindent
III.  The function $f$ is approximately linear since the given 
$x$ values are equally spaced, and all the slopes are $.21/.2 = 1.05$.
For instance, between the first two points 
$$m = {\Delta y\over \Delta x} = {2.51-2.30\over .4-.2} = {.21\over .2} = 1.05$$
The equation of the line is $y = 1.05x + b$ for some $b$.  Substituting
$x = .2$, $y = 2.30$, we have $2.30 = 1.05\cdot .2 + b$, so 
$b = 2.09$.  The equation is $y = 1.05 x + 2.09$.
\bigskip
\noindent
IV.  
\item{A)} The temperature at time $t = 0$ is $T(0) = 70+80e^{-0/3} = 70+80 = 150$
degrees F.
\item{B)} The rate of change of the temperature at $t = 10$ is $T'(10)$ (the 
derivative).  We have $T'(t) = 80e^{-t/3}(-1/3)$ (by the chain rule).
So $T'(10) = -80e^{-10/3}/3 \doteq -.951$ degrees F per minute
(must be negative since the chocolate is cooling off and tending 
to the room temperature over time).
\item{C)} We must solve the equation $100 = 70+80e^{-t/3}$ for $t$.
Subtract 70 from both sides to get $30 = 80e^{-t/3}$.  Then
divide by $80$ to get $3/8 = e^{-t/3}$.  Take natural logs of both 
sides:  $\ln(3/8) = -t/3$, so $t=-3\ln(3/8) \doteq 2.94$ minutes.
\bigskip
\noindent
V.
\item{A)}  If $f(x) = {1\over x+1}$,
$$\eqalign{
f'(x) &= \lim_{h\to 0} {f(x+h) - f(x)\over h}\cr
      &= \lim_{h\to 0} {{1\over x + h + 1} - {1\over x + 1}\over h}\cr
      &= \lim_{h\to 0} {(x+1) - (x + h + 1) \over h(x+h+1)(x+1)}\cr
      &= \lim_{h\to 0} {-h\over h(x+h+1)(x+1)}\cr
      &= \lim_{h\to 0} {-1\over (x+h+1)(x+1)}\cr
      &= {-1\over (x+1)^2}\cr}$$
\item{B)}  The function $g(x)$ is 
$g(x) = 3x^4 + 3x^{-1/2} + 2x^{1/3}$.  So
$$g'(x) = 12x^3 - {3\over 2}x^{-3/2} + {2\over 3}x^{-2/3}$$
\item{C)} Using the quotient and chain rules:
$$\eqalign{
{d\over dx}\left({\tan(x) + x \over \cos(2x)}\right)
&= {\cos(2x) {d\over dx}(\tan(x) + x) - (\tan(x) + x){d\over dx}(\cos(2x))
\over (\cos(2x))^2}\cr
&= {\cos(2x)(\sec^2(x) + 1) -
(\tan(x) + x)(-\sin(2x)\cdot 2)\over \cos^2(2x)}\cr
&={\cos(2x)(\sec^2(x) + 1) + 2(\tan(x) + x)\sin(2x)\over \cos^2(2x)}\cr}$$
\item{D)} By the chain rule:
$${d\over dx}(\ln(x^2 + 3^x)) = {1\over x^2 + 3^x}(2x + 3^x\ln(3))$$
\item{E)} By the chain rule again:
$${d\over dx}(\arcsin(12x+2)) = {1\over \sqrt{1-(12x+2)^2}}\cdot 12$$
\vfill
\eject
\noindent
VI.  Qualitatively,
\vglue 2.0truein
\noindent
VII.  The given graph shows $y = f'(x)$, not $y = f(x)$.
\bigskip
\item{A)} $f$ is increasing on intervals where $f'(x) > 0$ (graph
of $f'$ above $x$-axis):  approximately
$-1.3 < x < -.5$, $0 < x < .5$, and $1.3 < x < 1.5$.
\item{B)} The critical points are approximately
$x = -1.3, -.5, 0, .5$, and $1.3$.  By the first derivative test,
$-1.3$ is a local minimum, $-.5$ is a local maximum, $0$
is a local minimum, $.5$ is a local maximum, and $1.3$ is another
local minimum of $f$.
\item{C)}  The graph of $f$ is concave up on intervals where
$f'$ is {\it increasing}, so approximately $-1.5 < x < -1$,
$-.3 < x < .3$, and $1 < x < 1.5$.
\item{D)}  The graph of $f$ has inflection points where the
concavity changes, or where $f'$ changes from increasing 
to decreasing or vice versa.  These are at approximately
$x = -1, -.3, .3, 1$ (the critical points of $f'$).
\bigskip
\noindent
VIII.  Here is a picture of the window as described, with dimensions
labeled:
\vglue 2.0truein
\noindent
Note that the radius of the semicircle is $x/2$, one half of 
that side of the rectangle.
The perimeter is the sum of the lengths of three sides of the 
rectangle, plus the length of the semicircle:
$$30 = x + 2y + {1\over 2}\cdot 2\pi {x\over 2} = 
\left(1 + {\pi\over 2}\right)x + 2y$$
From this we can solve for $y$ in terms of $x$:
$$y = {30 - \left(1 + {\pi\over 2}\right)x\over 2} = 15 - 
\left({1\over 2} +{\pi\over 4}\right)x\leqno(1)$$
The area of the window is the sum of the areas of the 
rectangle and the semicircle:
$$A = xy + {1\over 2}\cdot \pi \left({x\over 2}\right)^2$$
Substituting for $y$ from (1) and simplifying:
$$A = x\left(15 - \left({1\over 2} +{\pi\over 4}\right)x\right) + {\pi x^2\over 8}
 = 15x - \left({1\over 2} +{\pi\over 8}\right)x^2$$
We are now ready to maximize the area.  First, we find critical points:
$$A'(x) = 15 - \left(1 + {\pi\over 4}\right)x = 0$$
has only one solution:  
$$x = {15\over 1 + {\pi\over 4}} \doteq 8.4$$
Then from (1)
$$y =    15 - \left({1\over 2} +{\pi\over 4}\right)\cdot 8.4 \doteq 4.2$$
This is a local maximum since by the second derivative test,
$$A''(x) = -\left(1 + {\pi\over 4}\right) < 0$$
\bigskip
\noindent
IX.  This is a {\it true} statement.  The vertical stretching by a 
factor of 2 gives $y = 2\cdot 2^{-x} = 2^{-x + 1} = 2^{-(x-1)},$
which is the horizontal shift of $y = 2^{-x}$ by one unit to the 
right.


                 
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