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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Exam 1 Solutions}
\centerline{October 3, 2005}
\bigskip
\noindent
I.  First note that $f$ cannot be either linear or exponential
because it is increasing for some $x$ and decreasing for 
other $x$.
\item{A)}  $g(x)$ is the only linear function, because it
is the only one with constant slope.  Using the first two
points for instance
$m = {28 - 23\over 0 - (-2)} = {5\over 2}$.  The 
$y$-intercept is $28$ from the table so $g(x) = {5\over 2}x + 28$.
\item{B)}  $h(x)$ is exponential.  This can be seen because the
ratios of successive values is constant:  
$${18\over 2} = {2\over 2/9} = {2/9\over 2/81} = {2/81\over 2/729}=9.$$
To find the formula, we try to determine $a,b$ to match the table
with $h(x) = ba^x$.  From $h(0) = 2$, we get $2 = ba^0 = b$.  Then
from $h(-2) = 18$ we get $18 = 2a^{-2}$ so $9 = a^{-2}$, so $a^2 = 1/9$,
and $a = 1/3$.  The formula is $h(x) = 2 \left({1\over 3}\right)^x$.
\bigskip
\noindent
II.  Because the ``Explosium'' decays exponentially, we know
the amount is given by a formula $Q(t) = Q_0 e^{kt}$ or $Q(t) = Q_0 a^t$.
Either form can be used, and we will show how to do the problem
using both of them.
\bigskip
\noindent
{\it Method 1}.  Using $Q(t) = Q_0 e^{kt}$, the initial amount is $Q_0 = 10$, and 
the half-life is 12 years so 
$$\eqalign{
5 &= 10 e^{k\cdot 12}\cr
1/2 &= e^{12k}\cr
\ln(1/2) &= 12k\cr
{\ln(1/2)\over 12} &= k}$$
Then we want to determine the time $t$ when $Q(t) = 1$:
$$\eqalign{1 &= 10 e^{{\ln(1/2)\over 12}t}\cr
1/10 &= e^{{\ln(1/2)\over 12}t}\cr
\ln(1/10) &=  {\ln(1/2)\over 12}t\cr
{12\ln(1/10)\over \ln(1/2)} &= t\cr}$$

\bigskip
\noindent
{\it Method 2}.  Using $Q(t) = Q_0 a^t$, we have $Q = 10 a^t$ and substituting
$t = 12$ (the half-life) we get $5 = 10 a^{12}$, so $a = (1/2)^{1\over 12}$
and $Q = 10 \left(({1\over 2})^{1\over 12}\right)^t = 10 ({1\over 2})^{t/12}$.  Then 
as before we want to solve for $t$ in the equation $1 = 10 \left({1\over 2}\right)^{t/12}$
We get:
$$\eqalign{1/10 &= \left({1\over 2}\right)^{t/12}\cr
         \ln(1/10) &= (t/12)\ln(1/2)\cr
{12\ln(1/10)\over \ln(1/2)} &= t\cr}$$
(which agrees with the other method, of course!)

\bigskip
\noindent
III.  A)  Because this plot is sinusoidal, with a maximum at $x = 0$, it is
simplest to use the form $y = A \cos(Bx) + C$.  The amplitude
$A = (5 - (-3))/2 =4$.  The period is 1, so $2\pi/B = 1$ and $B = 2\pi$.
The graph is also shifted vertically by $C = -1$:  
$$y = 4\cos(2\pi x) - 1.$$
\bigskip
\noindent
B)  This graph looks like a polynomial of degree 3 since there 
are 2 turning points, a double root at $x = 1$, and a root at
$x = 2$.  This gives $y = k(x-1)^2(x-2)$ for some constant
$k$.  Since the graph ``starts high and finishes low'' we know
$k < 0$.  The exact value can be determined because
the graph shows the $y$-intercept at $y = 6$:
$$6 = k(0-1)^2(0-2) = -2k$$
So $k = -3$ and the formula is
$$y = -3(x-1)^2(x-2).$$  
\bigskip
\noindent
IV.  The graph $y = 10^{x+1}$ is an exponential growth 
curve (shifted left by 1).  So it does pass the 
horizontal line test (horizontal lines $y = c > 0$
intersect the graph exactly once).  This says there
is an inverse function.  To find the formula, interchange
$x$ and $y$ then solve for $y$ taking logarithms:
$$\eqalign{x &= 10^{y+1}\cr
         \ln(x) &= (y+1)\ln(10)\cr      
         {\ln(x)\over \ln(10)} &= y+1\cr
         {\ln(x)\over \ln(10)} - 1 = y\cr}$$
The inverse function is $f^{-1}(x) = {\ln(x)\over \ln(10)} - 1$.
\bigskip
\noindent
V. A)  $r(x)$ has vertical asymptotes where the {\it bottom}
is zero:  $16 - x^2 = 0$, so $x = \pm 4$.
\item{B)}  $r(x)$ has a horizontal asymptote at $y = -2$ 
because as $x\to \pm \infty$, 
$${2x^2 - x\over 16- x^2} = {2 - {1\over x}\over {16\over x^2} - 1} \to -2$$
\item{C)}  The graph $y = r(x)$ crosses the $x$-axis where 
the {\it top} is zero:  $0 = 2x^2 - x = x(2x - 1)$, so $x = 0$ 
or $x = 1/2$.  
\bye

