\magnification=1200
\baselineskip=12pt
\nopagenumbers

\noindent
MATH 180\hfill FALL 1995

%\noindent
%10:00 SECTION\hfill FONG

\bigskip\noindent
\centerline{ASSIGNMENT \#29}

\bigskip\noindent
5.1, \#5.
$f(x)=2x^3+3x^2-36x+5$, $f'(x)=6x^2+6x-36=6(x+3)(x-2)$.
The critical points $-3$ and $x=2$ divide the domain
of $f$ into three intervals:
$(-\infty,-3)$, $(-3,2)$, $(2,\infty)$.
The sign of $f'$ on each interval can be determined by
evaluating $f'(x)=6(x+3)(x-2)$ at a suitable point,
say $-5$ in $(-\infty,-3)$, $0$ in $(-3,2)$, $3$ in
$(2,\infty)$.  We find
$f'(-5)=6(-2)(-7)>0$, $f'(0)=6(3)(-2)<0$, $f(3)=6(6)(1)>0$.
So $x=-3$ is a local maximum of $f$ and $x=2$ is a local minimum of $f$
by the first-derivative test.

\bigskip\noindent
5.1, \#9.
$\displaystyle f(x)={x\over x^2+1}$, $\displaystyle f'(x)=
{(x^2+1)\cdot1-x\cdot2x\over (x^2+1)^2}={1-x^2\over (x^2+1)^2}$.
$f'(x)$ is defined everywhere and is zero at $x=-1$ and 1.
The sign of
$f'(x)$ is the sign of the numerator $1-x^2$, since
the denominator $(x^2+1)^2$ is always positive.   $1-x^2$
is negative on $(-\infty,-1)$, positive on
$(-1,1)$, and negative on $(1,\infty)$.  Thus $x=-1$ is a local minimum
of $f$ and $x=1$ is a local maximum of $f$.

\bigskip\noindent
5.1, \#13.
$f(x)=x-\ln x$, $f'(x)=1-(1/x)=(x-1)/x$.  Thus $f$ has one critical
point $x=1$ on the given domain $[0.1,2]$.  The sign of $f'(x)$ is
negative on $[0.1,1)$, positive on $(1,2]$.
Thus $x=1$ is a local minimum.  For global extrema
we compute $f(0.1)=2.40$, $f(1)=1$, $f(2)=1.31$.
Thus $f$ has a global maximum
at $x=0.1$ and a global minimum at $x=1$.

\bigskip\noindent
5.2, \#1.
The inflection points of $f$ are the points where $f''$ changes
sign, or equivalently, where $f'$ has a local extremum.
So the tops and bottoms of the graph of $f'$ are inflection points of
$f$.

\bigskip\noindent
5.2, \#4.  $y=0.5xe^{-10x}$, $\displaystyle y'=({1\over2}-5x)e^{-10x}$,
$y''=(50x-10)e^{-10x}$.  The root $x=1/10$ of $y'$ is
the critical
point of $y$.  The sign of $y'$ is that of $(1/2)-5x$ since
$e^{-10x}$ is always positive.  Thus $x=1/10$ is a local maximum of
$y$ by the first-derivative test.  The root $x=1/5$ of $y''$
is an inflection point of $y$ since $y''$ changes sign at $x=1/5$.

\bigskip\noindent
5.2, \#7.
$f(x)=\sin x+\cos x$, $f'(x)=\cos x-\sin x$, $f''(x)=-\sin x-\cos x$.
The critical points of $f$ in the interval $[0,6\pi]$ are
$x=\pi/4$, $5\pi/4$, $9\pi/4$, $13\pi/4$, $17\pi/4$, $21\pi/4$.
The graphs of $\sin x$ and $\cos x$ show $f'(x)$
changes sign at these critical points.  The points alternate
between local maxima and local minima with $x=\pi/4$ being a local
maximum.  The zeros of $f''(x)$ in $[0,6\pi]$ are
$x=3\pi/4$, $7\pi/4$, $11\pi/4$, $15\pi/4$, $19\pi/4$, $23\pi/4$.
These are inflection points since $f''(x)$ changes sign at these points.

%\bigskip\noindent
%Comments:

\end
