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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#36}

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5.7, \#1.  $f(x)=x^3+3x^2+3x-6$, $f'(x)=3x^2+6x+3=3(x+1)^2$.
In particular, $f'(x)$ is never negative so
that $f(x)$ is increasing.  Thus $f$ has only one root $r$.
Since $f(0)=-6$, $f(1)=1$, it follows that $r$ is in the interval
$(0,1)$.  We start the iterative
process $x_{n+1}=x_n-(f(x_n)/f'(x_n))$
with $x_0=1$.  The truncated values for the next three $x_i$ are
$$\vbox{
\halign{\hfil\quad# \quad\cr
  \noalign{\vskip5pt}
  0.9166667 \cr
  0.9129384 \cr
  0.9129311 \cr}}$$
A calculation shows that $f(0.9129311)>0$ and $f(0.9129300)<0$.
The sign change shows that
$0.9129300<r< 0.9129311$.  Thus
0.913 approximates $r$ to 2 decimal places since
$|0.913-r|<0.913-0.9129300<1/200$.


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5.6, \#2.  $\root 3\of {50}$ is a root of $f(x)=x^3-50$.  Since
$f'(x)=3x^2$, $f(x)$ is increasing and $f$ has only one root $r$.
Now $f(4)=14>0$, $f(3.5)=-7.125<0$, so
$r$ is in $(3.5,4)$.  We start the iterative process
with $x_0=3.5$.
The truncated values for the next three $x_i$ are
$$\vbox{
\halign{\hfil\quad# \quad\cr
  \noalign{\vskip5pt}
  3.693877 \cr
  3.684057 \cr
  3.684031 \cr}}$$
A calculation shows that $f(3.68406)>0$ and $f(3.68403)<0$.
The sign change shows that
$3.68403<r< 3.68406$.  Thus 3.684
approximates $r$ to 2 decimal places since
$|3.684-r|<3.68406-3.684 <1/200$.

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5.6, \#6.
$f(x)=\cos x-x$, $f'(x)=-\sin x-1$.  In particular, $f'(x)$ is
never positive.  So
$f(x)$ is decreasing and $f$ has only one root $r$.
Now $f(0)=1>0$, $f(1)=-0.45969\cdots<0$, so
$r$ is in $(0,1)$.  We start the iterative process with $x_0=1$.
The truncated values for the next three $x_i$ are
$$\vbox{
\halign{\hfil\quad# \quad\cr
  \noalign{\vskip5pt}
  0.750363\cr
  0.739112\cr
  0.739085\cr}}$$
A calculation shows that $f(0.73908)>0$ and $f(0.73911)<0$.
The sign change shows that
$0.73908<r< 0.73911$.  Thus 0.739
approximates $r$ to 2 decimal places since
$|0.739-r|<0.73911-0.739 <1/200$.


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