\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.3in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \Large The Mean Value Theorem} \rhead{\small page \ \thepage} \cfoot{} \lfoot{\footnotesize \copyright $\;$ Hidegkuti, 2014} \rfoot{\footnotesize Last revised: March 13, 2019} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{27pt} \setlength{\parindent}{0in} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} } \setstretch{1.04} \begin{document} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{% %\TeXButton{\defcont}{\defcont %\setlength{\hangindent}{85pt} %\setstretch{1.2}}}% %BeginExpansion \defcont \setlength{\hangindent}{85pt} \setstretch{1.2}% %EndExpansion \textbf{Rolle's Theorem:} Suppose that $f$ is continuous on $\left[ a,b% \right] $ and differentiable on $\left( a,b\right) $. \ If $f\left( a\right) =f\left( b\right) ,$ then there exists $c$ in $\left( a,b\right) $ with $% f^{\prime }\left( c\right) =0$. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.05in}\vspace{0.05in} %TCIMACRO{% %\TeXButton{\setstretch{1.15}}{\setlength{\hangindent}{32pt} %\setstretch{1.15}}}% %BeginExpansion \setlength{\hangindent}{32pt} \setstretch{1.15}% %EndExpansion Proof: \ Suppose that that $f$ is continuous on $\left[ a,b\right] $, differentiable on $\left( a,b\right) $, and $f\left( a\right) =f\left( b\right) $. \ By the extreme value theorem, $f$ has an absolute maximum \ and minimum on $\left[ a,b\right] $. \ If that maximum or minimum is at an interior point $c$, then $f^{\prime }\left( c\right) =0$. \ If both the maximum and minimum are at an endpoint, then $f\left( a\right) =f\left( b\right) $ means that the function is constant and so at all interior point $% c$ we have $f^{\prime }\left( c\right) =0$. $\blacksquare $\bigskip %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{% %\TeXButton{\defcont}{\defcont %\setlength{\hangindent}{85pt} %\setstretch{1.2}}}% %BeginExpansion \defcont \setlength{\hangindent}{85pt} \setstretch{1.2}% %EndExpansion \textbf{The Mean Value Theorem}: \ Suppose that $f$ is continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $.\vspace{0.05in} \ Then there exists $c$ in $\left( a,b\right) $ with $f^{\prime }\left( c\right) =\dfrac{f\left( b\right) -f\left( a\right) }{b-a}$. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.05in}\vspace{0.05in} %TCIMACRO{\TeXButton{\begin{minipage}{3.9in}}{\begin{minipage}{3.9in}}}% %BeginExpansion \begin{minipage}{3.9in}% %EndExpansion %TCIMACRO{% %\TeXButton{\setstretch{1.15}}{\setlength{\hangindent}{32pt} %\setstretch{1.15}}}% %BeginExpansion \setlength{\hangindent}{32pt} \setstretch{1.15}% %EndExpansion Proof: Suppose that $f$ is continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $. \ Define $g$ a linear function to be the line connecting $f\left( a\right) $ and $f\left( b\right) $. \ The equation of this line is\vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $y-f\left( a\right) =m\left( x-a\right) $\vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $y=m\left( x-a\right) +f\left( a\right) $\vspace{0.05in}\vspace{0.05in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{3.5in}}{\begin{minipage}{3.5in}} }% %BeginExpansion \begin{minipage}{3.5in} %EndExpansion \ \ \ \ \ \ \ \ \ \FRAME{itbpF}{2.8349in}{1.2401in}{0in}{}{}{mvtpic1.bmp}{% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.8349in;height 1.2401in;depth 0in;original-width 3.2197in;original-height 1.3932in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'MVTpic1.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \ \ \ \ \ \ \ \ \ \ \ where \ \ $m=\dfrac{f\left( b\right) -f\left( a\right) }{b-a}$,\ and so $g\left( x\right) =\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) +f\left( a\right) $.\vspace{0.05in} Clearly $g$ is differentiable on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $. \ Now define the difference function $h$ of $f$ and $g$.% \begin{eqnarray*} h\left( x\right) &=&f\left( x\right) -g\left( x\right) =f\left( x\right) - \left[ \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) +f\left( a\right) \right] \\ h\left( x\right) &=&f\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) -f\left( a\right) \end{eqnarray*}% We will be able to apply Rolle's Theorem to this function. \ Since both $f$ and $g$ are continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $, so is $h$. \ Furthermore, $h\left( a\right) =h\left( b\right) =0$.% \begin{eqnarray*} h\left( a\right) &=&f\left( a\right) -g\left( a\right) =f\left( a\right) -\left( \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( a-a\right) +f\left( a\right) \right) =f\left( a\right) -\left( \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\cdot 0+f\left( a\right) \right) \\ &=&f\left( a\right) -f\left( a\right) =0\text{ \ \ \ \ \ and} \end{eqnarray*}% \begin{eqnarray*} h\left( b\right) &=&f\left( b\right) -g\left( b\right) =f\left( b\right) -\left( \dfrac{f\left( b\right) -f\left( a\right) }{% %TCIMACRO{\TeXButton{\cancel{ b-a}}{\cancel{ b-a}}}% %BeginExpansion \cancel{ b-a}% %EndExpansion }% %TCIMACRO{% %\TeXButton{\cancel{\left( b-a\right)}}{\cancel{\left( b-a\right)}}}% %BeginExpansion \cancel{\left( b-a\right)}% %EndExpansion +f\left( a\right) \right) =f\left( b\right) -\left( f\left( b\right) -f\left( a\right) +f\left( a\right) \right) ~~~~~~~~~~ \\ &=&f\left( b\right) -f\left( b\right) =0 \end{eqnarray*}% so the conditions hold for Rolle's Theorem. \ By this theorem, there exists $% c$ in $\left( a,b\right) $ so that $h^{\prime }\left( c\right) =0$. \ We differentiate $h$:% \begin{eqnarray*} h\left( x\right) &=&f\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) -f\left( a\right) \\ && \\ h^{\prime }\left( x\right) &=&f^{\prime }\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\text{ and for some }c\text{ in }\left( a,b\right) \text{, \ \ }h^{\prime }\left( c\right) =0\text{ } \end{eqnarray*}% \begin{eqnarray*} 0 &=&h^{\prime }\left( c\right) =f^{\prime }\left( c\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a} \\ 0 &=&f^{\prime }\left( c\right) -\dfrac{f\left( b\right) -f\left( a\right) }{% b-a} \\ f^{\prime }\left( c\right) &=&\dfrac{f\left( b\right) -f\left( a\right) }{b-a% } \end{eqnarray*}% This completes our proof.$\blacksquare $\vspace{0.12in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{% %\TeXButton{\defcont}{\defcont %\setlength{\hangindent}{85pt} %\setstretch{1.2}}}% %BeginExpansion \defcont \setlength{\hangindent}{85pt} \setstretch{1.2}% %EndExpansion \textbf{Corollary 1.} \ If $f^{\prime }\left( x\right) =0$ at each point of an open interval $\left( a,b\right) $, then $f\left( x\right) =C$ for all $x$ in $\left( a,b\right) $ for some constant $C$. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.05in}\vspace{0.05in} %TCIMACRO{% %\TeXButton{\setstretch{1.15}}{\setlength{\hangindent}{32pt} %\setstretch{1.15}}}% %BeginExpansion \setlength{\hangindent}{32pt} \setstretch{1.15}% %EndExpansion Proof: \ Suppose that $f^{\prime }\left( x\right) =0$ at each point of an open interval $\left( a,b\right) $. \ \ Let $x_{1}$ and $x_{2}$ be any two different points in $\left( a,b\right) $. \ Since $f$ is differentiable, it is also continuous on $\left[ x_{1},x_{2}\right] $. \ We will apply the Mean Value Theorem. \ The slope $\dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}}$ \ is achieved as a derivative somewhere in the interval $\left[ x_{1},x_{2}\right] $. \ There exists $c$ between $x_{1}$ and $x_{2}$ so that \begin{equation*} f^{\prime }\left( c\right) =\dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}} \end{equation*}% By the hypotheses, $f^{\prime }\left( c\right) =0$. \ This gives us% \begin{eqnarray*} \dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}} &=&0 \\ f\left( x_{2}\right) -f\left( x_{1}\right) &=&0 \\ f\left( x_{2}\right) &=&f\left( x_{1}\right) \end{eqnarray*}% Since this is true for any pair $x_{1}$ and $x_{2}$, $f$ is constant on $% \left( a,b\right) $. $\blacksquare $\vspace{0.12in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{% %\TeXButton{\defcont}{\defcont %\setlength{\hangindent}{85pt} %\setstretch{1.2}}}% %BeginExpansion \defcont \setlength{\hangindent}{85pt} \setstretch{1.2}% %EndExpansion \textbf{Corollary 2.} \ If $f^{\prime }\left( x\right) =g^{\prime }\left( x\right) $ at each point $x$ in an open interval $\left( a,b\right) $, then there exists a constant $C$ such that $f\left( x\right) =g\left( x\right) +C$ for all $x$ in $\left( a,b\right) $. \ That is, $f-g$ is constant on $\left( a,b\right) $. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.05in}\vspace{0.05in} %TCIMACRO{% %\TeXButton{\setstretch{1.15}}{\setlength{\hangindent}{32pt} %\setstretch{1.15}}}% %BeginExpansion \setlength{\hangindent}{32pt} \setstretch{1.15}% %EndExpansion Proof: \ Suppose that $f^{\prime }\left( x\right) =g^{\prime }\left( x\right) $ on $\left( a,b\right) $. \ Define $h\left( x\right) =f\left( x\right) -g\left( x\right) $ on $\left( a,b\right) $. \ Then $h$ is differentiable and \begin{equation*} h^{\prime }\left( x\right) =\left( f\left( x\right) -g\left( x\right) \right) ^{\prime }=f^{\prime }\left( x\right) -g^{\prime }\left( x\right) =0 \end{equation*}% By the previous corollary, $h^{\prime }\left( x\right) =0$ on $\left( a,b\right) $ inplies that $h\left( x\right) =C$ on $\left( a,b\right) $ for some constant $C$. \ Thus \begin{eqnarray*} h\left( x\right) &=&C\text{ \ \ on }\left( a,b\right) \\ f\left( x\right) -g\left( x\right) &=&C \\ f\left( x\right) &=&g\left( x\right) +C\text{ \ on }\left( a,b\right) \end{eqnarray*}% This completes our proof.$\blacksquare $\pagebreak \begin{center} {\LARGE Practice Problems\bigskip }\bigskip \end{center} \begin{enumerate} \item Find all values of $c$ that satisfies the conclusion of the Mean Value Theorem.% %TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}% %BeginExpansion \begin{multicols}{2}% %EndExpansion a) \ $f\left( x\right) =5x^{2}-3x+8$ \ on $\left[ -1,4\right] $ b) $\ f\left( x\right) =x^{2}+2x-1$ \ on\ \ \ $\left[ 0,1\right] $ \ c) $\ f\left( x\right) =x^{2/3}$ \ on\ \ \ $\left[ 0,1\right] $ d) $\ f\left( x\right) =x+\dfrac{1}{x}$ \ \ on\ \ $\left[ \dfrac{1}{2},2% \right] $ \ \ e) \ $f\left( x\right) =\ln x$ \ \ on $\left[ 1,10\right] $ f) $\ f\left( x\right) =\sqrt{x-1}$ \ on\ \ \ $\left[ 1,3\right] $ g) \ $f\left( x\right) =2x^{3}-5x+7$\ \ on $\left[ -2,2\right] $ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Suppose that $3\leq f^{\prime }\left( x\right) \leq 5$ for all values of $x$. \ Show that $18\leq f\left( 8\right) -f\left( 2\right) \leq 30$. \bigskip \end{enumerate} \vspace{0.5in} \begin{center} {\LARGE Answers - Practice Problems\bigskip } \end{center} \begin{enumerate} \item a) \ $\dfrac{3}{2}\qquad $b) $\ \dfrac{1}{2}\qquad $c) \ $\dfrac{8}{27}% \qquad $d) $\ 1\qquad $e) \ $\dfrac{9}{\ln 10}\qquad $f) $\ \dfrac{3}{2}$ \ \ \ \ \ \ g) \ $\pm \dfrac{2\sqrt{3}}{3}$ \item see solutions\bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Solutions - Practice Problems\bigskip } \end{center} \begin{enumerate} \item Find all values of $c$ that satisfies the conclusion of the Mean Value Theorem. a) \ $f\left( x\right) =5x^{2}-3x+8$ \ on $\left[ -1,4\right] $ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( -1\right) =\allowbreak 16$ and $f\left( 4\right) =76$. \ The slope of secant line connecting the two points is $\dfrac{76-16}{4-\left( -1\right) }=12$. \ So we are looking for all values of $c$ for which $% f^{\prime }\left( c\right) =12$. \ $f^{\prime }\left( x\right) =10x-3$, so we solve $10x-3=12$ and obtain $x=\dfrac{3}{2}$ b) $\ f\left( x\right) =x^{2}+2x-1$ \ on\ \ \ $\left[ 0,1\right] $ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( 0\right) =-1$ and $f\left( 1\right) =2$. \ The slope of secant line connecting the two points is $\dfrac{2-\left( -1\right) }{1-0}=3$. \ So we are looking for all values of $c$ for which $f^{\prime }\left( c\right) =3 $. \ $f^{\prime }\left( x\right) =2x+2$, so we solve $2x+2=3$ and obtain $% x=\dfrac{1}{2}$ c) $\ f\left( x\right) =x^{2/3}$ \ on\ \ \ $\left[ 0,1\right] $ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( 0\right) =0$ and $f\left( 1\right) =1$. \ The slope of secant line connecting the two points is $\dfrac{1-0}{1-0}=1$. \ So we are looking for all values of $c$ for which $f^{\prime }\left( c\right) =1$. \ $% f^{\prime }\left( x\right) =\allowbreak \dfrac{2}{3\sqrt[3]{x}}$, so we solve $\dfrac{2}{3\sqrt[3]{x}}=1$ and obtain $x=\dfrac{8}{27}$.\pagebreak d) $\ f\left( x\right) =x+\dfrac{1}{x}$ \ \ on\ \ $\left[ \dfrac{1}{2},2% \right] $ \ \ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( \dfrac{1}{2}\right) =\allowbreak \dfrac{5}{2}$ and $f\left( 2\right) =\allowbreak \dfrac{5}{2}$. \ The slope of secant line connecting the two points is $0$. \ So we are looking for all values of $c$ for which $% f^{\prime }\left( c\right) =0$. \ $f^{\prime }\left( x\right) =1-\dfrac{1}{% x^{2}}$, so we solve $1-\dfrac{1}{x^{2}}=0$ and obtain $x=\pm 1$. \ Recall that we are looking for numbers tha fall in the interval $\left[ \dfrac{1}{2}% ,2\right] $. \ That rules out $-1$ and so the only solution is $x=1$. e) \ $f\left( x\right) =\ln x$ \ \ on $\left[ 1,10\right] $ \ \ \ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( 1\right) =0$ and $f\left( 10\right) =\ln 10$. \ The slope of secant line connecting the two points is $\dfrac{\ln 10-0}{10-1}=\dfrac{\ln 10}{9}$. \ So we are looking for all values of $c$ for which $f^{\prime }\left( c\right) =\dfrac{\ln 10}{9}$. \ $f^{\prime }\left( x\right) =\allowbreak \dfrac{1}{x}$, so we solve $\dfrac{1}{x}=\dfrac{\ln 10}{9}$ and obtain $x=\dfrac{9}{\ln 10}$. \ This number is approximately $% 3.\,\allowbreak 908\,65$ and so it is in the iterval $\left[ 1,10\right] $. f) $\ f\left( x\right) =\sqrt{x-1}$ \ on\ \ \ $\left[ 1,3\right] $ \ Solution: \ We first evaluate the function at the endpoints of the interval. \ $f\left( 1\right) =0$ and $f\left( 3\right) =\sqrt{2}$. \ The slope of secant line connecting the two points is $\dfrac{\sqrt{2}-0}{3-1}% =\allowbreak \dfrac{\sqrt{2}}{2}$. \ So we are looking for all values of $c$ for which $f^{\prime }\left( c\right) =\dfrac{\sqrt{2}}{2}$. \ $f^{\prime }\left( x\right) =\dfrac{1}{2\sqrt{x-1}}=\allowbreak $, so we solve $\dfrac{1% }{2\sqrt{x-1}}=\dfrac{\sqrt{2}}{2}$, and obtain $x=\dfrac{3}{2}$. $\left( x\right) =\dfrac{1}{2\sqrt{x-1}}=\dfrac{1}{\sqrt{2}}$ g) \ \ $f\left( x\right) =2x^{3}-5x+7$\ \ on $\left[ -2,2\right] $\vspace{% 0.08in} %TCIMACRO{% %\TeXButton{\begin{minipage}{3.9in}}{\begin{minipage}{3.9in} %\setlength{\rightskip}{15pt}}}% %BeginExpansion \begin{minipage}{3.9in} \setlength{\rightskip}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.15}}{\setstretch{1.15}}}% %BeginExpansion \setstretch{1.15}% %EndExpansion Solution: \ We first evaluate the function at the endpoints of the interval. \ \begin{equation*} f\left( -2\right) =\allowbreak 1\text{ and }f\left( 2\right) =\allowbreak 13 \end{equation*}% The slope\vspace{0.08in} of secant line connecting the two points is $\dfrac{% 13-1}{2-\left( -2\right) }=3$.\vspace{0.08in} \ So we are looking for all values of $c$ for which $f^{\prime }\left( c\right) =3$. \ $f^{\prime }\left( x\right) =\allowbreak 6x^{2}-5\allowbreak $, so we solve $6x^{2}-5=3$% , and obtain $x=\pm \dfrac{2\sqrt{3}}{3}$.\vspace{0.08in} \ %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{3in}}{\begin{minipage}{3in}}}% %BeginExpansion \begin{minipage}{3in}% %EndExpansion \ \FRAME{itbpFX}{2.469in}{1.6457in}{0in}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.469in;height 1.6457in;depth 0in;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-2";xmax "2";xviewmin "-2";xviewmax "2";yviewmin "-5";yviewmax "15";viewset"XY";rangeset"X";plottype 4;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$2x^{3}-5x+7$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "-2,2";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";function \TEXUX{$3\left( x-\dfrac{2\sqrt{3}}{3}\right) +7-\dfrac{14}{9}\sqrt{3}$};linecolor "black";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "0.3,2";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$3\left( x+\dfrac{2\sqrt{3}}{3}\right) +7+\dfrac{14}{9}\sqrt{3}$};linecolor "black";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "-2,-0.4";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$\left( \left( -2,1\right) ,\left( 2,13\right) \right) $};linecolor "gray";linestyle 2;pointstyle "point";linethickness 1;lineAttributes "Dash";curveColor "[flat::RGB:0x00c0c0c0]";curveStyle "Line";valid_file "T";tempfilename 'POBWBT01.wmf';tempfile-properties "XPR";}} \ %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion Both of these numbers satisfy the conclusion of the Mean Value Theorem.\ \ \ \ \item Suppose that $3\leq f^{\prime }\left( x\right) \leq 5$ for all values of $x$. \ Show that $18\leq f\left( 8\right) -f\left( 2\right) \leq 30$. Proof: \ By the mean value theorem, there exists $c$ in $\left[ 2,8\right] $ with $f^{\prime }\left( c\right) =\dfrac{f\left( 8\right) -f\left( 2\right) }{8-2}=\dfrac{f\left( 8\right) -f\left( 2\right) }{6}$. Since $3\leq f^{\prime }\left( x\right) \leq 5$ for all values of $x$, we have that \begin{eqnarray*} 3 &\leq &\dfrac{f\left( 8\right) -f\left( 2\right) }{6}\leq 5\text{ \ \ \ \ multiply by }6 \\ 18 &\leq &f\left( 8\right) -f\left( 2\right) \leq 30 \end{eqnarray*} \vspace{0.08in} {\Large %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \end{enumerate} \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}