\magnification=\magstep1
\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Final Examination -- Solutions}
\centerline{December 13, 2005}
\bigskip
\noindent
I.  $f(x)$ is approximately exponential.  With $f(x) = ca^x$, we have
$$\eqalign{c a^{.3} &= 4.12\cr
           c a^{.6} &= 3.74\cr}$$
So dividing, $a^{.3} \doteq .908$ and $a \doteq (.908)^{10/3} = .724$.
Then $c \doteq 4.12/(.724)^{.3} = 4.54$:
$$f(x) \doteq 4.54 (.724)^x.$$

$g(x)$ is linear since the slope is constant:  ${g(.6) - g(.3)\over .6 - .3} = -.8$
so $g(x) = (-.8)(x - .3) + 4.31 = -.8x + 4.55$:
$$g(x) = -.8x + 4.55.$$
\bigskip
\noindent
II. A) The graph is sinusoidal with period $4$, amplitude $5$ and vertical
shift $+3$:
$$f(x) = 5\sin\left({2\pi x\over 4}\right) + 3 = 5\sin\left({\pi x\over 2}\right) + 3$$
\item{B)}  The function has an inverse function if we restrict to $-1 \le x \le 1$
because that portion of the graph passes the horizontal line test:  horizontal
lines $y = k$ for $-2 \le k \le 8$ intersect that section of the graph exactly
once.  The domain of the inverse function is $-2 \le x \le 8$.
\bigskip
\noindent
III.  
\item{A)}
$$\eqalign{f'(x) &= \lim_{h\to 0} {5(x+h)^2 - (x+h) + 3 - 5x^2 + x - 3\over h}\cr
 &= \lim_{h\to 0} {5x^2 + 10 xh + 5h^2 - x - h + 3 - 5x^2 + x - 3\over h}\cr
 &= \lim_{h\to 0} 10 x - 1 + 5h\cr
 &= 10 x - 1.\cr}$$
\item{B)}  By the chain rule and the rule for exponential functions:
$$g'(x) = 3^x \ln(3) - {1\over x^3 + x}\cdot (3x^2 + 1) = 3^x\ln(3) - {3x^2 +1 \over x^3+x}$$
\item{C)}  Rewrite $h(x)$ using powers: $h(x) = 3 x^{1/5} + 7 x^{-1/3} + 3x^8$.
Then 
$$h'(x) = {3\over 5} x^{-4/5} - {7\over 3} x^{-4/3} + 24 x^7$$
\item{D)}  By the quotient rule:
$$i'(x) = {(\sin^2(x) - 1)(-\sin(x)) - \cos(x)\cdot 2\sin(x)\cos(x)\over (\sin^2(x) - 1)^2}$$
\item{E)} 
$$j'(x) = {1\over 1 + (3x + 2)^2}\cdot 3$$
\item{F)}  Using implicit differentiation:
$$2y{dy\over dx}\sqrt{x} + y^2 {1\over 2\sqrt{x}} - e^y {dy\over dx} = 0$$
Solving for ${dy\over dx}$:
$${dy\over dx} = {-y^2{1\over 2\sqrt{x}}\over 2y\sqrt{x} - e^y} = {-y^2\over 4xy - 2\sqrt{x}e^y}$$
\bigskip
\noindent
IV. 
\item{A)}  $f$ is increasing on intervals where $f'(x) \ge 0$, so here for all $x > -.5$.
$f$ is decreasing where $f'(x) < 0$, so $x < -.5$.
\item{B)}  $y = f(x)$ is concave up on intervals where $f''(x) > 0$, 
or equivalently, where $f'(x)$ is increasing:  $x < 0$ and $x > 1$.  $y = f(x)$
is concave down on intervals where $f'(x)$ is decreasing: $0 < x < 1$.
\item{C)}  The graph should be parabolic in shape with $x$-axis intercepts
at $x = 0, x = 1$, opening up.  The vertex of the parabola should be
at $x = .5$.
\bigskip
\noindent
V.  A)  The temperature at time $t = 0$ is $H(0) = 70 - 50 = 20$ degrees.
\item{B)}  The rate of change of the temperature at time $t = 10$ is
$H'(10) = 5e^{-10/10} \doteq 1.84$ degrees F per minute.
\item{C)}  The temperature reaches 60 degrees when
$$60 = 70 - 50 e^{-t/10} \Rightarrow {-10\over -50} = e^{-t\over 10} \Rightarrow
t = -10\ln(1/5) \doteq 16.1$$ minutes.
\bigskip
\noindent
VI.  By the product and chain rules, $f'(x) = -e^{-x} \sin(x) + e^{-x}\cos(x)
= e^{-x}(\cos(x) - \sin(x))$.  For $0 \le x \le \pi$, the only solution of
$f'(x) = 0$ is $x = \pi/4$.  Then 
$$f(0) = 0 \qquad f(\pi/4) = e^{-\pi/4} \sin(\pi/4) \doteq .322 \qquad f(\pi) = 0$$
so the minimum value is $0$ and the maximum value is $e^{-\pi/4}\sin(\pi/4)$.
\bigskip
\noindent
VII. A)  The cost to build the pipeline is 3 times the distance under water,
plus 2 times the distance along the shore:
$$C(x) = 3\sqrt{4 + x^2} + 2(6 - x)$$
\item{B)}  We find the critical points of $C$:
$$C'(x) = {3\over 2}(4 + x^2)^{-1/2} (2x) - 2 = {3x\over \sqrt{x^2 + 4}} - 2$$
This is zero when 
$$3x = 2\sqrt{x^2 + 4} \Rightarrow 9x^2 = 4x^2 + 16 \Rightarrow 5x^2 = 16$$
So $x =  {4\over \sqrt{5}}\doteq 1.79$ (ignore the negative root since
it clearly cannot give a minimium of the cost!).  This is a minimum
by the 1st derivative test:
$C'(1) = {3\over \sqrt{5}} - 2 \doteq -.66 < 0$
while 
$C'(2) = {6\over \sqrt{8}} - 2 \doteq .12 > 0$.
The pipeline of minimum cost goes from the island to the point 
with $x = 1.79$ miles along the shore.
\bigskip
\noindent
VIII.  The volume of the cube is $V = x^3$ where $x$ is the length
of the side.  So 
$${dV\over dt} = 3x^2 {dx \over dt}$$
when $V = 216$, $x = {\root 3 \of {216}} = 6$, and ${dV\over dt} = -10$,
so 
$$-10 = 3(6)^2 {dx \over dt}\Rightarrow {dx\over dt} = {-10\over 108} = {-5\over 54}$$
The side of the cube is decreasing at $\doteq .093$ cm/min.
  

\bye