
\input epsf.def

\gdef\Localbox#1{\epsfscale=2400\vtop to3.25in{\null\vskip
-1in\epsfbox{:graphs:#1.eps}\vss}}

\magnification1200
\def\newp#1.{\par\noindent{\bf #1.}}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\medskip}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\ans{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}

\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\def\Lim{\displaystyle\lim}
\def\Int{\displaystyle\int}
\def\arcsec{\rm arcsec}

\def\frac#1#2{{\displaystyle{{\textstyle#1}\over{\textstyle#2}}}}

\newp 11.
You may compute the derivative from the definition either of two ways.

\noindent{\bf 1.}
$f^\prime(x)=\Lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=
\Lim_{h\to 0}\frac{\smfrac{1}{x+h+1}-\smfrac{1}{x+1}}{h}=\penalty-10000
\Lim_{h\to 0}\frac{\smfrac{(x+1)-(x+h+1)}{(x+1)(x+h+1)}}{h}=
\Lim_{h\to 0}\frac{\smfrac{x+1-x-h-1}{(x+1)(x+h+1)}}{h}=
\Lim_{h\to 0}\frac{\smfrac{-h}{(x+1)(x+h+1)}}{h}=
\Lim_{h\to 0}\frac{-1}{(x+1)(x+h+1)}=\frac{-1}{(x+1)(x+1)}=
\frac{-1}{(x+1)^2}$ .
\medskip
\noindent OR\par
\noindent{\bf 2.}
$f^\prime(x)=\Lim_{y\to x}\frac{f(y)-f(x)}{y-x}=
\Lim_{y\to x}\frac{\smfrac{1}{y+1}-\smfrac{1}{x+1}}{y-x}=
\Lim_{y\to x}\frac{\smfrac{(x+1)-(y+1)}{(x+1)(y+1)}}{y-x}=\penalty-10000
\Lim_{y\to x}\frac{\smfrac{(x-y)}{(x+1)(y+1)}}{y-x}=
\Lim_{y\to x}\frac{x-y}{(y-x)(x+1)(y+1)}=
\Lim_{y\to x}\frac{-1}{(x+1)(y+1)}=\frac{-1}{(x+1)^2}
$ .
\medskip
\noindent Also acceptable was

\noindent
$f^\prime(a)=\Lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=
\Lim_{h\to 0}\frac{\smfrac{1}{a+h+1}-\smfrac{1}{a+1}}{h}=\penalty-10000
\Lim_{h\to 0}\frac{\smfrac{(a+1)-(a+h+1)}{(a+1)(a+h+1)}}{h}=
\Lim_{h\to 0}\frac{\smfrac{a+1-a-h-1}{(a+1)(a+h+1)}}{h}=
\Lim_{h\to 0}\frac{\smfrac{-h}{(a+1)(a+h+1)}}{h}=
\Lim_{h\to 0}\frac{-1}{(a+1)(a+h+1)}=\frac{-1}{(a+1)(a+1)}=
\frac{-1}{(a+1)^2}$ .

\bigskip
For part b), you need to say where $f^\prime(x)$ does not exist.
The formula gives the derivative everywhere both the formula 
and the function are defined. 
Since there are no roots being taken, the only issue is when are we
dividing by $0$.
The function $f$ itself is undefined when $x+1=0$, or $x=-1$.
The derivative is therefore not defined at $x=-1$.
The formula for the derivative makes sense everywhere except where
$(x+1)^2=0$, or again $x=-1$.
Hence the domain of $f^\prime$ is alll real numbers except $-1$.
In interval notation $(-\infty,-1)\cup (-1,\infty)$.
\endp

\newp 12.
\item{a)} $f^\prime(x)=\frac{x^\prime(1+x^2) - x(1+x^2)^\prime}
{(1+x^2)^2}=\frac{1\cdot (1+x^2)
- x\cdot 2x}{(1+x^2)^2}=\frac{(1-x^2)}{(1+x^2)^2}$ .

\item{b)} $f^\prime(x)=0$,$x=$ ?

\item{}$f^\prime(x)=\frac{(1-x^2)}{(1+x^2)^2}=0$
holds whenever $1-x^2=0$.
This last equation has two solutions $x_1=1$ and $x_2=-1$.

\endp


\newp 13.
$f(x)$ is continuous on $[1,2]$ since it is a rational
function and its denominator is not zero on $[1,2]$. By plugging in,
$f(1)=-1$ and $f(2)=\frac{4}{3}.$ Since $f(x)$ is continuous on $[1,2]$
and $f(1)$ is negative and $f(2)$ is positive, it follows from the
Intermediate Value Theorem that there must be some $c$ in $(1,2)$ such
that $f(c)=0.$
\endp
\newp 14.
\Localbox{BlankGraphSol}

Your graph needs to look something like the graph above. 
It must go through $(0,3)$ and have slope $-1$ there, so the
green line must be a tangent line to your graph.
The tangent lines to your graph at $x=\pm2$ must be parallel
to the red lines, but your graph may pass through different points,
either up or down.
Finally, your graph must extend to the edges of the graph in the
direction of the $x${--}axis.
\endp
\newp 15.
\item{(a)} Velocity is the instantaneous change in position with respect
to time, i.e., the derivative of $s(t)$.  Thus $v(t) = s'(t) =
\frac{d}{dt}(4t - 3t^2 - t^3) = 4 - 3\cdot 2 t^{2-1} - 3t^{3-1} = 4 - 6t
- 3t^2$.

\item{(b)} When the particle hits the ground, it will be zero meters above the
ground, so set $s(t) = 0 = 4t - 3t^2 - t^3 = t(4-3t-t^2) =
t(4+t)(1-t)$.   Thus $s(t) = 0$ when $t = -4, 0$ or $1$.  The root $t =
-4$ is not physically valid since time is moving forward; the root $t =
0$ corresponds to the initial time when the ball is tossed into the air;
and thus the remaining root $t = 1$ must be when the ball hits the ground
after being thrown.

\item{}Therefore the velocity of the ball when it hits the ground is $v(1) = 4 -
6(1) - 3(1)^2 = -5$ m/s.  (Note that the velocity is negative indicating
that the ball is traveling towards the surface of planet X when it hits
the ground.)

\endp


\end