\magnification=1200
\baselineskip=12pt
\nopagenumbers

\noindent
MATH 180\hfill FALL 1995

\noindent
10:00 SECTION\hfill FONG

\bigskip\noindent
\centerline{ASSIGNMENT \#13}

\bigskip\noindent
2.6, \#4.  The tangent line to $y=f(x)$ at $x=0.6$ is
the line through $(0.6,3.9)$ with slope $f'(0.6)$.
An approximate value for $f'(0.6)$ is given by the average of the
average rates of change on the intervals $[0.4,0.6]$ and
$[0.6,0.8]$, that is, the average rate of change
on the larger interval $[0.4,0.8]$:
$$f'(0.6)\approx{f(0.8)-f(0.4)\over 0.8-0.4}=1.25$$
The equation of the tangent line is then $y-3.9=1.25(x-0.6)$
or $y=1.25x+3.15$.  Using the tangent line to estimate
$f(x)$ near $0.6$, we find that
$$f(0.7)\approx 4.025,\quad f(1.2)\approx 4.65,\quad
   f(1.4)\approx 4.9.$$
$0.7$ is much closer to $0.6$ than
1.2 or 1.4 in the scale of the problem,
so one might ``feel most confident about'' the estimate for
$f(0.7)$.


\bigskip\noindent
2.6, \#5.  $x=7.0$ and $x=8.5$ are inputs
in the table, so we approximate $f'(7.0)$ and $f(8.5)$
by the average of the average rates of changes on two
adjacent intervals as in \#4:
$$\eqalign{
  f'(7.0)&\approx {f(7.5)-f(6.5)\over 7.5-6.5}=6.5-10.3=-3.8,\cr\quad
  f'(8.5)&\approx {f(9.0)-f(8.0)\over 9.0-8.0}=3.2-5.2=-2.\cr}$$
$x=6.75$ is between the inputs 6.5 and 7.0, so we
approximate $f'(6.75)$ by the average rate of change on $[6.5,7.0]$:
$$f'(6.75)\approx {f(7.0)-f(6.5)\over 7.0-6.5}=
   {8.2-10.3\over 0.5}=-4.2.$$
Since $f'(7)=-3.8$,
the tangent line at $x=7$ is approximately $y-8.2=(-3.8)(x-7.0)$
or $y=-3.8x+34.8$. In particular, the tangent line gives
$f(6.8)\approx (-3.8)(6.8)+34.8=8.96$.

%Comments:

\end