%
\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.
By definition 
$f^\prime(1)=\Lim_{h\to0}\frac{\sqrt{1+h}-\sqrt{1}}{h}=
\Lim_{h\to0}\frac{\sqrt{1+h}-\sqrt{1}}{h}\cdot\frac{\sqrt{1+h}+\sqrt{1}}{\sqrt{1+h}+\sqrt{1}}
$\par\noindent
$=\Lim_{h\to0}\frac{\sqrt{1+h}^2-\sqrt{1}^2}{h}\cdot\frac{1}{\sqrt{1+h}+\sqrt{1}}=
\Lim_{h\to0}\frac{1+h-1}{h}\cdot\frac{1}{\sqrt{1+h}+\sqrt{1}}
\Lim_{h\to0}\frac{h}{h}\cdot\frac{1}{\sqrt{1+h}+\sqrt{1}}$\par\noindent
$=\Lim_{h\to0}\frac{1}{\sqrt{1+h}+\sqrt{1}}=\frac{1}{\sqrt{1}+\sqrt{1}}=\frac12
$, where the next to the last equality follows since $\sqrt{x+1}$ is continuous at $0$.

OR

By definition 
$f^\prime(1)=\Lim_{x\to1}\frac{\sqrt{x}-\sqrt{1}}{x-1}=
\Lim_{x\to1}\frac{\sqrt{x}-\sqrt{1}}{x-1}\cdot\frac{\sqrt{x}+\sqrt{1}}{\sqrt{x}+\sqrt{1}}
$\par\noindent
$=\Lim_{x\to1}\frac{\sqrt{x}^2-\sqrt{1}^2}{x-1}\cdot\frac{1}{\sqrt{x}+\sqrt{1}}=
\Lim_{x\to1}\frac{x-1}{x-1}\cdot\frac{1}{\sqrt{x}+\sqrt{1}}
\Lim_{x\to1}\frac{x-1}{x-1}\cdot\frac{1}{\sqrt{x}+\sqrt{1}}$\par\noindent
$=\Lim_{x\to1}\frac{1}{\sqrt{x}+\sqrt{1}}=\frac{1}{\sqrt{1}+\sqrt{1}}=\frac12
$, where the next to the last equality follows since $\sqrt{x}$ is continuous at $0$.
\endp
\newp 12.
\item{a.} Since $s=90 t^{1/2}-25 t^{3/2}+3 t^{5/2}$,
$v=90\cdot\frac{1}{2}t^{-1/2}-25\cdot\frac32 t^{1/2}+3\cdot\frac52 t^{3/2}$
or\par\item{}
$v=\frac{90}{2}t^{-1/2}-\frac{75}2 t^{1/2}+\frac{15}2 t^{3/2}=
\frac{15}{2}t^{-1/2}\bigl(
6-5t+t^2\bigr)$.
Hence $v(1)=\frac{15}{2}\cdot(6-5+1)=15$.
\item{b.}
We must solve $v(t)=0$ or $0=6-5t+t^2=(t-3)(t-2)$ so $t=2$ and $t=3$.
\item{c.} Since there are no zeros of the velocity function between $0$ and $1$ the
distance travelled is equal to the displacement which is $s(1)-s(0)=
\bigl(90-25+3)-(0)=68$.
\endp
\newp 13.
We know $f(x)$ is continuous at any point $x\neq1$ since there the function is given by a 
polynomial and polynomials are continuous.
By the same reasoning, it is easy to calculate one sided limits at $x=1$:
$\Lim_{x\to 1^+} f(x)=\Lim_{x\to1^+}x^3+cx=1+c$ since $x^3+cx$ is continuous;
$\Lim_{x\to 1^-} f(x)=\Lim_{x\to1^-}x=1$ since $x$ is continuous.
For $f$ to be continuous at $x=1$ we must have $\Lim_{x\to1} f(x)=f(1)=1$ so
$1+c=1$ or $c=0$.
\endp
\newp 14.
First we  write down the equation for the tangent line at $a$:
$y-f(a)=f^\prime(a)(x-a)$ and, since $f(x)=x^2+1$, we have $f^\prime(x)=2x$,
and or line is $y-(a^2+1)=(2a)(x-a)$.
This line passes through $(0,-3)$ whenever
$(-3)-(a^2+1)=(2a)(0-a)$.
This yields $-a^2-4=-2a^2$ or $a^2=4$ and $a=\pm2$.

A second approach proceeds as follows.
At the point $(x,x^2+1)$ on the curve, the slope of the tangent line is $2x$ and
the slope of the line from this point to $(0,-3)$ is
$\frac{x^2+1-(-3)}{x-0}=\frac{x^2+4}{x}$.
Equating these two slopes yields
$2x=\frac{x^2+4}{x}$ which again yields $x=\pm2$.
\endp
\end