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222.tex

@@ -0,0 +1,23 @@
+\parindent=0pt
+\nopagenumbers
+
+18.6. Prove that $x=\cos x$ for some $x$ in $(0,\pi/2)$.
+
+\bigskip
+We have $\cos(0)=1$ and $\cos(\pi/2)=0$.
+The problem is that the interval is open, not closed as required by IVT.
+However, since $\cos(0)\ne0$ and $\cos(\pi/2)\ne\pi/2$ we know that
+$\cos x=x$ does not occur at the endpoints.
+Therefore it is safe to use the closed interval $[0,\pi/2]$.
+%(Is this o.k. or b.s.?)
+Use the trick from Example 1: $f(x)=x-\cos x$.
+We have $f(0)=0-1=-1$ and $f(\pi/2)=\pi/2-0=\pi/2$ and
+therefore $f(0)<0<f(\pi/2)$.
+By IVT there exits an $x_0$ such that $f(x_0)=0$.
+Because $f(x_0)=x_0-\cos x_0=0$, we have a point $x_0$
+at which $x=\cos x$.
+We have already shown that $x\ne\cos x$ at either $0$ or $\pi/2$ so
+$x=\cos x$ must occur on the interval $(0,\pi/2)$.
+
+\vfill
+\end

223.tex

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+\parindent=0pt
+\nopagenumbers
+
+18.7. Prove that $x2^x=1$ for some $x$ in $(0,1)$.
+
+\bigskip
+
+We define the function $f(x)=x2^x$.
+For this function we have $f(0)=0$ and $f(1)=2$.
+Since $f(0)<1<f(1)$, by IVT $f(x)=1$ must occur somewhere in the
+closed interval $[0,1]$.
+We have $f(0)\ne1$ and $f(1)\ne1$ so $x2^x=1$ must occur somewhere in the
+open interval $(0,1)$.
+
+\vfill
+\end

224.tex

@@ -0,0 +1,17 @@
+\parindent=0pt
+\nopagenumbers
+
+18.8. Suppose that $f$ is a real-valued continuous function in $R$ and that
+$f(a)f(b)<0$ for some $a,b\in R$.
+Prove that there exists $x$ between $a$ and $b$ such that $f(x)=0$.
+
+\bigskip
+
+From $f(a)f(b)<0$ we conclude that
+$f(a)$ and $f(b)$ have opposite signs and neither is zero.
+Therefore either $f(a)<0<f(b)$ or $f(b)<0<f(a)$.
+In both cases we have the existence of $f(x)=0$ and $x\ne a,b$ by
+the intermediate value theorem.
+
+\vfill
+\end

ross.html

@@ -89,6 +89,26 @@ for at least one <i>x</i><sub>0</sub> in [<i>a</i>, <i>b</i>].
 18.5. (b) Show that Example 1 can be viewed as a special case of part (a).
 </a>
 
+<p>
+<a href="222.pdf">
+18.6. Prove that <i>x</i> = cos <i>x</i> for some <i>x</i> in
+(0, <i>&pi;</i> / 2).
+</a>
+
+<p>
+<a href="223.pdf">
+18.7. Prove that <i>x</i>2<sup><i>x</i></sup> = 1 for some <i>x</i>
+in (0, 1).
+</a>
+
+<p>
+<a href="224.pdf">
+18.8. Suppose that <i>f</i> is a real-valued continuous function on R
+and that <i>f</i>(<i>a</i>)<i>f</i>(<i>b</i>) &lt; 0 for some
+<i>a</i>, <i>b</i> &isin; R. Prove that there exists <i>x</i> between
+<i>a</i> and <i>b</i> such that <i>f</i>(<i>x</i>) = 0.
+</a>
+
 </body>
 </html>