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\noindent {\bf Question}

\noindent For each of the following functions, the same as in
Exercise 8.11, use Rolle's theorem or the mean value theorem to
determine whether the solutions described in Exercise 8.11 are the
only ones.
\begin{enumerate}
\item $f(x) = 0$ on the interval $[-a,a]$, where $a$ is an arbitrary
positive real number and $f(x) = x^{1995} + 7654 x^{123} + x$;
\item $\tan(x)=e^{-x}$ for $x$ in $[-1,1]$;
\item $3\sin^2(x)=2\cos^3(x)$ for $x>0$;
\item $3+x^5-1001x^2=0$ for $x>0$;
\end{enumerate}


\medskip

\noindent {\bf Answer}

\noindent \begin{enumerate}
\item we know that there is one solution to $f(x) =0$ in $[-a,a]$,
namely $x =0$ (which can be found with using the intermediate
value theorem or by inspection).  To see that there are no others,
we again use Rolle's theorem: if there were $b$ in $[-a,a]$, $b\ne
0$, with $f(b) =0$, then there would exist some point $c$ between
$b$ and $0$ with $f'(c) =0$.  However, $f'(x) = 1995 x^{1994} +
941442x^{122} + 1$ and so $f'(c) \ge 1 > 0$ for all $c \in {\bf
R}$.  Hence, by Rolle's theorem, there is no second solution to
$f(x) =0$.
\item again working with $g(x) = \tan(x) -e^{-x}$, we saw earlier that
there is a solution to $g(x) =0$ in the interval $[-1,1]$.
However, since $g'(x) = \sec^2(x) +e^{-x} >0$ for all $x\in
(-1,1)$, Rolle's theorem implies that there can be no second
solution to $g(x) =0$ in the interval $[-1,1]$.  (It is the same
reasoning as before: if there were two solutions to $g(x) =0$,
then there would exist a point $c$ between them at which $g'(c)
=0$; however, the calculation above shows that $g'(c)\ne 0$ for
all $c$ in $(-1,1)$)
\item we don't have enough information to decide whether we've found
all the solutions to $f(x) =0$.  With $f(x) = 3\sin^2(x)
-2\cos^3(x)$, we have that $f'(x) = 6\sin(x)\cos(x) +
6\cos^2(x)\sin(x) = 6\sin(x)\cos(x) (1+\cos(x)) = 0$ when $x =
k\pi$ for $k\in {\bf N}$ (since $\sin(k\pi) =0$) and when $x = (k
+\frac{1}{2})\pi$ (since $\cos((k +\frac{1}{2})\pi) =0$ for $k\in
{\bf N}$).  Note that $f(k\pi) = -2\cos^3(k\pi) = (-1)^{k+1}2\ne
0$ and that $f((k+\frac{1}{2})\pi) = 3\sin^2((k+\frac{1}{2})\pi) =
3\ne 0$.  So, for any $m\in {\bf N}$, consider the interval
$(m\pi, (m+2)\pi)$.

\medskip
\noindent So, there exist three points in this interval at which
$f'(x) =0$, namely at $(m+\frac{1}{2})\pi$, $(m+1)\pi$, and
$(m+\frac{3}{2})\pi$, and our earlier analysis using the
intermediate value theorem found only two points in this interval
at which $f(x) =0$.  However, while Rolle's theorem yields that
two points at which $f(x) =0$ yields one point at which $f'(x)
=0$, we are unable to argue the other way: there may be many
points at which $f'(x) =0$ and still no points at which $f(x) =0$.
This example shows the limitations of this sort of analysis.
\item for $f(x) =3+x^5-1001x^2$ on $x>0$, again differentiate: $f'(x)
= 5x^4 -2002x = x(5x^3 - 2002)$, and so there is only one point in
$(0,\infty)$ at which $f'(x) =0$, namely the solution $c$ of $5c^3
-2002 =0$.  By calculation, we have that $c = 7.3705 ...$, and so
if there is a second solution to $f(x) =0$ in $(0,\infty)$, it
must lie in the interval $(0,c)$ (since by Rolle's theorem, if
there are two solutions to $f(x) =0$, then there exists at least
one solution to $f'(x) =0$ between them).

\medskip
\noindent Since $f(0) =3$ and since $f(c) = -32624.3179...$, the
intermediate value property implies that that there is a solution
to $f(x) =0$ in the interval $(0,c)$.  Since the only solution to
$f'(x) =0$ on $(0,\infty)$ occurs at $c$, Rolle's theorem implies
that there can be at most two solutions to $f(x) =0$ in
$(0,\infty)$, and we have found them both.
\end{enumerate}


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