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\noindent {\bf Question}

\noindent For each of the following functions described below, use
the Intermediate value property for continuous functions to
determine whether there is a solution to the given equation in the
specified set.
\begin{enumerate}
\item $f(x) = x$, where $f(x)$ is continuous on the closed interval
$[a,b]$ and satisfies $f(a)<a<b<f(b)$ for all $x\in[a,b]$;
\item $g(x) = 0$, where $g(x) = x^2 - \cos(x)$;
\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 $x^3+2x^5+(1+x^2)^{-2}=0$ 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 as before, consider the continuous function $g(x) = f(x) -x$.
Since $f(a) <a$, we have that $g(a) = f(a) -a < 0$.  Since $f(b)
>b$, we have that $g(b) =f(b) -b >0$.  Hence, the intermediate
value property applied to $g$ yields that there exists $c$ in
$(a,b)$ with $g(c) =0$.  That is, $f(c) -c =0$, and so $f(c) =c$.
Hence, the equation $f(x) =x$ has a solution in $[a,b]$.
\item first of all, note that $g(x) =x^2 -\cos(x)$ is continuous on
all of ${\bf R}$, and so is continuous on every closed interval
$[a,b]$ in ${\bf R}$.  In order to apply the intermediate value
property to find a point $c$ at which $g(c) =0$, we need to find
$a$ and $b$ so that $g(a) >0$ and $g(b) <0$ (or vice versa), and
the intermediate value property then implies the existence of such
a number $c$ between $a$ and $b$.

\medskip
\noindent So, let's start plugging numbers into $g$:  $g(0) =
-\cos(0) =- 1 <0$ and $g(2) = (2)^2  -\cos(2) =  4.6536 ... > 0$,
and so there exists a number $c_1$ between $0$ and $2$ with
$g(c_1) = 0$.  (Note that since $(2)^2 = (-2)^2$ and $\cos(2) =
\cos(-2)$, we also have that there exists $c_2$ between $-2$ and
$0$ with $g(c_2) =0$.)
\item for $f(x) = x^{1995} + 7654 x^{123} + x$ on the closed interval
$[-a,a]$, start by verifying continuity; actually, $f$ is
continuous on all of ${\bf R}$ being a polynomial, and hence is
continuous on $[-a,a]$.  Now, check the sign of $f$ on the
endpoints of the given interval: $f(a) = a^{1995} + 7654 a^{123} +
a > 0$ (since $a >0$) and $f(-a) = (-a)^{1995} + 7654 (-a)^{123} +
(-a) = -f(a) < 0$, and so the intermediate value property implies
that there exists some $c$ in $(-a,a)$ with $f(c) =0$.  (And
actually, casual inspection reveals that $f(0) =0$.)
\item for $\tan(x)=e^{-x}$ for $x$ in $[-1,1]$, start by defining
$g(x) = \tan(x) -e^{-x}$, so that $\tan(c) =e^{-c}$ if and only if
$g(c) =0$, as was done above.  Note that $g$ is continuous on
$[-1,1]$, since $e^{-x}$ is continuous on all of ${\bf R}$ and
$\tan(x)$ is continuous as long as its denominator $\cos(x)$ is
non-zero, which holds true on $[-1,1]$.  Since we are working on
the closed interval $[-1,1]$, check the values of $g$ on the
endpoints: $g(1) = \tan(1) -e^{-1} = 1.1895 ... >0$ and $g(-1) =
-4.2757 ... <0$, and so there exists some $c$ in $(-1,1)$ with
$g(c) =0$, and hence with $\tan(c) = e^{-c}$.
\item as above, $f(x) = x^3+2x^5+(1+x^2)^{-2}$ is continuous on
$[-1,1]$, as it is the sum of a polynomial and a rational function
whose denominator is non-zero on $[-1,1]$.  As always, check the
endpoints of the interval first: $f(1) = \frac{13}{4}$ and $f(-1)
= -\frac{11}{4}$, and so by the intermediate value property, there
is some $c$ in $(-1,1)$ at which $f(c) = 0$.
\item consider $f(x) = 3\sin^2(x) - 2\cos^3(x)$.  Since both $\sin(x)$
and $\cos(x)$ are continuous on all of ${\bf R}$, we have that $f$
is continuous on all of ${\bf R}$.  Since no specific closed
interval is given, we need to find an appropriate interval on
which to apply the intermediate value property for $f$, if in fact
such an interval exists.  Fortunately, we remember that
$\sin(k\pi) =0$ for all integers $k$, and so we may consider the
interval $[k\pi, (k+1)\pi]$ for any integer $k\ge 1$, so that the
interval lies in $(0,\infty)$. At the endpoints of this interval,
$f(k\pi) = -2\cos^3(k\pi)$ and $f((k+1)\pi) = -2\cos^3((k+1)\pi)$.
Since $\cos(k\pi)$ and $\cos((k+1)\pi)$ are equal to $\pm 1$ and
have opposite signs, $f(k\pi)$ and $f((k+1)\pi)$ are both non-zero
and have opposite signs, and so by the intermediate value
property, there is a point $c_k$ in $(k\pi, (k+1)\pi)$ at which
$f(c_k) =0$, that is, at which $3\sin^2(c_k) = 2\cos^3(c_k)$, as
desired.
\item first, note that $f(x)= 3+x^5-1001x^2$ is a polynomial and so is
continuous on all of ${\bf R}$, and in particular is continuous
for $x>0$.  As above, we need to choose a closed interval on which
to apply the intermediate value property.  Let's start by
evaluating $f$ at some of the natural numbers: $f(1) = -997$;
$f(2) = -3969$; $f(10) = -90097$; $f(11) = 880$.  Hence, the
intermediate value property implies that there is a number $c$ in
the open interval $(10,11)$ at which $f(c) =0$.
\end{enumerate}


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