%
\def\whichexam{Exam I}
\def\examdate{September 25, 2001}

\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle
\font\sm=cmr7
\magnification1200
\centerline{\whichexam}
\centerline{\examdate}
\def\Int{\displaystyle\int}
\def\Lim{\displaystyle\lim}
\def\frac#1#2{{{#1}\over{#2}}}
\bigskip
\newp 11.
The formula from Newton's method is $x_1=x_0-\frac{f(x_0)}{f^\prime(x_0)}$ which in our case
amounts to
$$x_1=x_0-\frac{x_0^3+x^2_0+1}{3x_0^2+2x_0}\ .$$
Compute $f(-2)=(-2)^3+(-2)^2+1=-8+4+1=-3<0$ and
$f(-1)=(-1)^3+)-1)^2+1=-1+1+1=1>0$ so by the Intermediate Value Theorem, there is at least
one zero in this interval.
({\bf Parenthetical Remark}: It is not hard to check that $3x^2+2x>0$ on the interval $[-2,-1]$
so $f$ is increasing on this interval and so there is precisely one zero there.)
\endp
\newp 12.
If the two sides of the rectangle are $x$ and $y$, then the area is $A=xy$ and the cost of fence
is $C=10x+20(x+2y)=30x+40y$ if we let $x$ denote the length of the side with the fence.
Since $A=80$, $y=\frac{80}{x}$ and
$C=30x+40\cdot\frac{80}{x}$.
We need to minimize $C$ so compute $\frac{dC}{dx}=30-40\cdot\frac{80}{x^2}$.
The critical points are $x=0$ and $30-40\cdot\frac{80}{x^2}=0$, or
$40\cdot\frac{80}{x^2}=30$, or $x^2=\frac{40\cdot 80}{30}=\frac{320}{3}$.
Hence $x=\pm\sqrt{\frac{320}{3}}$.
In our problem, $x>0$ so $x=\sqrt{\frac{320}{3}}$ is a critical point.
By the first derivative test, $x=\sqrt{\frac{320}{3}}$ is a local maximum and is the only critical
point on the interval $(0,\infty)$ and so there it is a global minimum.

Hence you should make the side with the fence $\sqrt{\frac{320}{3}}$ feet long and the other
side of the rectangle should be $\frac{80}{\sqrt{\frac{320}{3}}}$ feet long.
The total cost in dollars is
$30\cdot\sqrt{\frac{320}{3}}+40\cdot\frac{80}{\frac{80}{\sqrt{\frac{320}{3}}}}=
30\cdot\sqrt{\frac{320}{3}}+40\cdot \sqrt{\frac{320}{3}}=70\cdot \sqrt{\frac{320}{3}}$.
\endp
\newp 13.
You are being asked to minimize the distance from the point $(x,x^2)$ on the parabola to the
point$(0,1)$.
A formula for this distance is $d=\sqrt{(x-0)^2+(x^2-1)^2}$. 
You can simplify your work by minimizing $d^2$ but we will stick with $d$.
Well, $d=\sqrt{x^2+x^4-2x^2+1}=\sqrt{x^4-x^2+1}$.
The only critical points in this case occur where the derivative,
$\frac{4x^3-2x}{2\sqrt{x^2+x^4-2x^2+1}}$,
vanishes.
But this happens if and only if $4x^3-2x=0$ or when $x=0$ and $x=\pm\sqrt\frac12$.
The sign of the derivative alternates as we pass through each critical point, so $\pm\sqrt\frac12$
are local minima and $0$ is a local maximum.
Since $\Lim_{x\to\pm\infty} d=\infty$, it follows that $x=\pm\sqrt\frac12$ are global minima.
The corresponding points on the parabola are $\Bigl(\pm\sqrt\frac12, \frac12\Bigr)$.
\endp
\newp 14.
Since $\Int_a^b f(x)\>dx=\Lim_{n\to\infty} \sum_{i=1}^nf(x_i^\ast)\>\Delta x$ 
in general we proceed as follows.
In our case, $a=0$, $b=1$ and $f(x)=3x^2+2$.
Moreover, $x_i^\ast$ is to be the right{--}hand end point of the $i$th interval.
Then 
$\Delta x=\frac{b-a}{n}=\frac{1}{n}$ and the right{--}hand end point of the $i$th interval is
given by $x_i^\ast=a+i\Delta x=\frac{i}{n}$.

Hence we get 
 $\Int_0^1 (3x^2+2)\>dx=
\Lim_{n\to\infty} \sum_{i=1}^n\biggl(3\Bigl(\frac in\Bigr)^2+2\biggr)\frac{1}{n}=
\Lim_{n\to\infty} \sum_{i=1}^n\biggl(\frac {3i^2}{n^3}+\frac{2}{n}\biggr)=
\Lim_{n\to\infty} \sum_{i=1}^n\biggl(\frac {3i^2+2n^2}{n^3}\biggr)=
\Lim_{n\to\infty} \frac {3\sum_{i=1}^ni^2+2n^2\sum_{i=1}^n1}{n^3}=
\Lim_{n\to\infty} \frac {3\frac{n(n+1)(2n+1)}{6}+2n^3}{n^3}=$\par\noindent$
\Lim_{n\to\infty} \frac{n(n+1)(2n+1)}{2n^3} +\Lim_{n\to\infty} 2=
1+2=3$.
\endp
\end