\magnification=\magstep1
\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions -- Final Exam}
\bigskip
\noindent
I.  The function $g(x)$ is approximately linear since all differences
between successive entries are $-.71$, so
all the slopes are equal to ${-.71\over .2} = -3.55$.  Using the first point,
the equation is $g(x) - 2.30 = (-3.55)(x - .2)$, so 
$$g(x) = -3.55 x + 3.01$$
The function $f(x)$ is therefore approximately exponential.  (This
can also be seen since the ratios between successive entries are
approximately the same.)  To find an approximate equation 
$f(x) = ba^x$, we can use the first two table entries, which say
$$\eqalign{2.30 &= ba^{.2}\cr
           2.12 &= ba^{.4}\cr}$$
Hence dividing the second equation by the first, we get
$$.92 = {2.30\over 2.12} = {ba^{.4}\over ba^{.2}} = a^{.2} = a^{1\over 5}.$$
Taking fifth powers to solve for $a$:  $a \doteq (.92)^5 \doteq .66$.
Then from $2.3 = b (.66)^{.2}$, we get $b \doteq {2.3\over (.66)^{.2}}
\doteq 2.5$.  Therefore $f(x) \doteq (2.5)(.66)^x$.
\bigskip
\noindent 
II. 
\item{A)}  The graph looks like a sinusoid with 
average value $C = -2$, amplitude $A = 2$, and period 
$4 = {2\pi\over B}$, so $B = {\pi\over 2}$.  The equation is
$$y = A \sin(Bx) + C = 2\sin\left({\pi x\over 2}\right) - 2$$
\item{B)}  The part of the graph for $x$ with $-1 \le x\le 1$ 
meets horizontal lines $y = c$ with $-4 \le c \le 0$ exactly
once.  Therefore the restricted function has an inverse function.
The domain of $f^{-1}$ is the range of $f$ -- the interval
$[-4,0]$.
\bigskip
\noindent
III.  
\item{A)}  The graph $y = f'(x)$ looks approximately like:
\vglue 2.0truein
(Note: three zeroes at $x = 1,2,3$ where $y = f(x)$ has 
horizontal tangents, zero slope as $x \to \pm\infty$ since
$y = f(x)$ is approaching a horizontal asymptote, and $f'$
is positive where $f$ is increasing, negative where $f$ is
decreasing.)
\item{B)} The graph $y = f''(x)$ looks approximately like:
\vglue 2.0truein
\bigskip
\noindent
IV.  
\item{A)}
$$\eqalign{
f'(x) &= \lim_{h\to 0} {f(x+h) - f(x)\over h}\cr
      &= \lim_{h\to 0} {(x+h)^2 + 2(x+h) + 3 - x^2 - 2x - 3\over h}\cr
      &= \lim_{h\to 0} {2xh + h^2 + 2h\over h}\cr
      &= \lim_{h\to 0} 2x + h + 2\cr
      &= 2x + 2\cr}$$
\item{B)}  By the power rule,
$$g'(x) = 14x^6 + {3\over 2} x^{-1\over 2} + {2\over 3}x^{-{4\over 3}}$$
\item{C)}  By the exponential and logarithm rules, plus the chain rule,
$$h'(x) = \ln(5)5^x -{4x^3\over x^4 + 3}$$
\item{D)}  By the quotient rule,
$$i'(x) = {(\sin(x) - 1)(-\sin(x))-(\cos(x))(\cos(x))\over (\sin(x)-1)^2} 
= {1\over \sin(x) - 1}$$
\item{E)}  By the rule for arctangent and the chain rule, 
$$j'(x) = {1\over 1 + (3x+2)^2}\cdot 3 = {3\over 9x^2 + 12x + 5}$$
\vfill\eject
\noindent
V.  
\item{A)}  The temperature at time $t = 0$ is 
$$F(0) = 70-50e^{-0/10} = 70 - 50 = 20$$
degrees,
\item{B)}  The rate at which the temperature is changing at
$t = 10$ is the derivative $F'(10)$.  By the chain rule, 
$F'(t) = -50 e^{-t/10}\cdot{-1\over 10} = 5e^{-t/10}$.
So $F'(10) = 5e^{-1} \doteq 1.8$ degrees per minute.  (This
is positive since the frozen dinner is warming up as it sits
in the shopping cart.)
\item{C)}  Solve $60 = 70 - 50e^{-t/10}$ for $t$:
$$\eqalign{-10 &= -50e^{-t/10}\cr
            .2 &= e^{-t/10}\cr
           \ln(.2) &= -t/10\cr
          -10\ln(.2) &= t\cr}$$
Thus $t = -10\ln(.2)\doteq 16$ minutes.
\bigskip
\noindent
VI.
\item{A)}  $f$ is increasing on intervals where $f'$ is positive
and decreasing on intervals where $f'$ is negative.  
Increasing:  all $x$ with $a < x < c$;  decreasing:  $x < a$ and $x > c$.
\item{B)}  The critical points of $f$ are the points where 
$f'(x) = 0$ --  $x = a,c$.
\item{C)}  $y = f(x)$ is concave up when $f'(x)$ is increasing
and concave down when $f'(x)$ is decreasing.  Concave
up:  all $x < b$ and $x > d$;  concave down for all $x$ with $b < x < d$.
\bigskip
\noindent
VII.  Suppose the pipeline is laid from the island to a point
$x$ miles along the shore toward the town, then along the 
shore to the town.  The length under water is found by the
Pythagorean theorem:  $\sqrt{x^2 + 4}$.  The length along
the shore is then $6 - x$.  So the cost is
$$C(x) = 3\sqrt{x^2 + 4} + 2(6 - x).$$
To minimize this, we take the derivative (using the 
chain rule), set equal to zero, and
solve for $x$:
$$\eqalign{
0 &= C'(x)\cr
  &= {3\over 2}(x^2 + 4)^{-1\over 2} (2x) - 2\cr
  &= {3x\over \sqrt{x^2+4}} - 2\cr}$$.
Taking the 2 to the other side of the equation and clearing 
denominators, this gives:
$$3x = 2\sqrt{x^2 + 4}$$
Square both sides:  $9x^2 = 4x^2 + 16$, so $5x^2 = 16$,
and $x = \sqrt{16/5} \doteq 1.7$.
The minimum cost pipeline joins the island to a point
approximately 1.7 miles along the shoreline toward the town, then 
proceeds along the shoreline to the town.
\bigskip
\noindent
VIII.
\item{A)}  The left and right-hand sums are
$$\eqalign{
LHS &= f(0)\cdot{1\over 2}+f({1\over 2})\cdot{1\over 2}+f(1)\cdot{1\over 2}+f({3\over 2})\cdot{1\over 2}
+f(2)\cdot{1\over 2}+f({5\over 2})\cdot{1\over 2}+f(3)\cdot{1\over 2}\cr
&\qquad +f({7\over 2})\cdot{1\over 2}\cr
&= 0 + {1\over 2}\cdot{1\over 2} + 1\cdot {1\over 2} + {3\over 2}\cdot {1\over 2} + 2\cdot {1\over 2}
+0\cdot{1\over 2} + (-2)\cdot{1\over 2} + (-1)\cdot{1\over 2}\cr
&= {1\over 4} + {1\over 2} + {3\over 4} + 1 - 1 - {1\over 2}\cr
&=1\cr
RHS &= f({1\over 2})\cdot{1\over 2}+f(1)\cdot{1\over 2}+f({3\over 2})\cdot{1\over 2}
f(2)\cdot{1\over 2}+f({5\over 2})\cdot{1\over 2}+f(3)\cdot{1\over 2}+f({7\over 2})\cdot{1\over 2}\cr
&\qquad+f(4)\cdot{1\over 2}\cr
&={1\over 2}\cdot{1\over 2} + 1\cdot {1\over 2} + {3\over 2}\cdot {1\over 2} + 2\cdot {1\over 2}
+ 0\cdot{1\over 2} + (-2)\cdot{1\over 2} + (-1)\cdot{1\over 2} +0\cdot {1\over 2}\cr
&= 1\cr}$$
\item{B)}  The integral is the area of the triangle formed by the graph
over the interval $[0,5/2]$, minus the area of the triangle
formed by the graph over the interval $[5/2,4]$ (since that lies
below the $x$-axis).  We have
$$\int_0^4 f(x)\ dx = {1\over 2}\cdot{5\over 2}\cdot 2 - {1\over 2}\cdot {3\over 2}\cdot 2 = 1.$$
\bye

