\documentclass[a4paper,12pt]{article} \begin{document} \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 $f(a)0$; \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 $f(a) b$, 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} \end{document}