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\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
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\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large Optimization 4}
\rhead{\small page   \ \thepage}
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\lfoot{\footnotesize   \copyright $\;$  Hidegkuti,  Powell,  2011}
\rfoot{\footnotesize  Last revised: March 28, 2019}
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\begin{document}


\begin{center}
{\Large Sample Problems}
\end{center}

\begin{enumerate}
\item A company determines that if $n$ is the number of items\vspace{0.03in}
produced, they can all be sold at a price of \newline
$p\left( n\right) =\sqrt{1200-0.2n}$ \ What is the greatest revenue possible?

\item A company has $\$120\,000$ to spend on the development and promotion
of a new product. \ \ The company estimates that if $x$ is spent on the
development and $y$ is spent on promotion, then approximately $\dfrac{%
x^{1/2}y^{3/2}}{400\,000}$ \ items of new product will be sold. Based on
this estimate, what is the maximum number of products that the company can
sell?

\item An underground telephone cable is to be laid between two boat docks on
opposite banks of a straight river. \ \ One boathouse is $600$ meters
downstream from the other. \ The river is $200$ meters wide. \ If the cost
of laying the cable is $\$50$ per meter under water and $\$30$ per meter on
land, how should the cable to be laid to minimize cost? \ 

\begin{enumerate}
\item[a)] Find the lowest cost possible.

\item[b)] Use the second derivative test to prove that we found a relative
minimum in part a).

\item[c)] Prove that the relative minimum we found in part a) \ is also an
absolue minimum on the domain $\left[ 0,600\right] .$

\item[d)] Prove that if we define $C$ on the set of all real numbers, then
the minimum we found in part a) \ is still an absolute minimum.
\end{enumerate}

\item The location function of an object is $s\left( t\right) =\dfrac{120}{%
1+2e^{-t}}$ where $t$ is time, measured in seconds. \ Where is the object
when it is moving with the greatest speed? \ When is that greatest speed
achieved?

\item Consider the function $g\left( x\right) =\dfrac{-2x}{\left(
x^{2}+1\right) ^{2}}$.

\begin{enumerate}
\item[a)] Find all relative extrema of $g$.

\item[b)] Find all absolute extrema of $g$.

\item[c)] Find all values of $c$ for which the function $f\left( x\right) =%
\dfrac{1}{x^{2}+1}+cx$ \ is increasing on its entire domain.\pagebreak
\end{enumerate}
\end{enumerate}

\begin{center}
{\Large Sample Problems - Answers\bigskip }
\end{center}

\begin{enumerate}
\item $\$80\,000$

\item \ $11\,691$

\item a) \ $\,\$26\,000$ when the cable is laid $450$ meters on the ground \
\ \ \ b) \ see solutions \ \ c) \ see solutions \ d) \ see solutions

\item $s=60$ when $t=\ln 2$

\item a) $\ \left( -\dfrac{1}{\sqrt{3}},\dfrac{3\sqrt{3}}{8}\right) \ $%
relative maximum and a relative minimum $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3%
\sqrt{3}}{8}\right) $ \ \ \ \ b) \ see solutions \ \ c) \ $c\geq \dfrac{3%
\sqrt{3}}{8}${\Large \pagebreak }
\end{enumerate}

\begin{center}
{\Large Sample Problems - Solutions}
\end{center}

\begin{enumerate}
\item A company determines that if $n$ is the number of items produced,%
\vspace{0.03in} they can all be sold at a price of \newline
$p\left( n\right) =\sqrt{1200-0.2n}$ \ What is the greatest revenue possible?

Solution: \ If we sell all $n$ products, the revenue is $R\left( n\right) =n%
\sqrt{1200-0.2n}$ Since the price should be positive, the domain of this
function is determined by $n>0$ and $1200-0.2n>0.$ \ We solve these
inequalities and obtain the domain. $\left( 0,6000\right) .$ \ \ We will
find the maximum of this function by finding the zeroes of the derivative.%
\begin{eqnarray*}
R^{\prime }\left( n\right) &=&\sqrt{1200-0.2n}+n\dfrac{1}{2\sqrt{1200-0.2n}}%
\left( -0.2\right) =\sqrt{1200-0.2n}-\dfrac{0.1n}{\sqrt{1200-0.2n}} \\
&=&\dfrac{1200-0.2n}{\sqrt{1200-0.2n}}-\dfrac{0.1n}{\sqrt{1200-0.2n}}=\dfrac{%
1200-0.2n-0.1n}{\sqrt{1200-0.2n}}=\dfrac{-0.3n+1200}{\sqrt{1200-0.2n}} \\
&=&\dfrac{-0.3\left( n-4000\right) }{\sqrt{1200-0.2n}}
\end{eqnarray*}%
$\allowbreak $The derivative has only one zero, at $n=4000$. \ The
denominator is positive for all $n$ with $0<n<6000$ and the numerator is
positive before $4000$ and negative after $4000.$ \ Consequently, $R\left(
n\right) $ \ has a relative and absolute maximum at $n=4000$. \ The maximal
possible revenue is then $R\left( 4000\right) $.%
\begin{equation*}
R\left( 4000\right) =4000\sqrt{1200-0.2\left( 4000\right) }=4000\sqrt{%
1200-800}=4000\sqrt{400}=4000\left( 20\right) =80\,000
\end{equation*}

\item A company has $\$120\,000$ to spend on the development and promotion
of a new product. \ \ The company estimates that if $x$ is spent on the
development and $y$ is spent on promotion, then approximately $\dfrac{%
x^{1/2}y^{3/2}}{400\,000}$ \ items of new product will be sold. Based on
this estimate, what is the maximum number of products that the company can
sell?

Solution: \ Since $y=120\,000-x$, we can write the number sold as a function
of $x$.%
\begin{equation*}
N\left( x\right) =\dfrac{1}{400\,000}x^{1/2}\left( 120\,000-x\right) ^{3/2}
\end{equation*}%
The domain is clearly $\left[ 0,120\,000\right] $. \ We differentiate $N$.%
\begin{eqnarray*}
N^{\prime }\left( x\right) &=&\dfrac{1}{400\,000}\left[ \dfrac{1}{2}%
x^{-1/2}\left( 120\,000-x\right) ^{3/2}+x^{1/2}\left( \dfrac{3}{2}\right)
\left( 120\,000-x\right) ^{1/2}\left( -1\right) \right] \\
&=&\dfrac{1}{400\,000}\left( \dfrac{1}{2}\right) \left[ \dfrac{\left(
120\,000-x\right) ^{3/2}}{\sqrt{x}}-3\sqrt{x}\left( 120\,000-x\right) ^{1/2}%
\right]
\end{eqnarray*}%
\begin{eqnarray*}
N^{\prime }\left( x\right) &=&\dfrac{1}{800\,000}\left( \dfrac{\sqrt{%
120\,000-x}\left( 120\,000-x\right) }{\sqrt{x}}-3\sqrt{x}\sqrt{120\,000-x}%
\right) \text{ \ \ \ \ \ \ \ factor out }\sqrt{120\,000-x} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{\left( 120\,000-x\right) 
}{\sqrt{x}}-3\sqrt{x}\right) \text{ \ \ \ \ \ \ \ \ bring difference to
common denominator} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{\left( 120\,000-x\right) 
}{\sqrt{x}}-3\dfrac{x}{\sqrt{x}}\right) =\dfrac{\sqrt{120\,000-x}}{800\,000}%
\left( \dfrac{120\,000-x-3x}{\sqrt{x}}\right) \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{120\,000-4x}{\sqrt{x}}%
\right) =\dfrac{-4\sqrt{120\,000-x}\left( x-30\,000\right) }{800\,000\sqrt{x}%
} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}\left[ -4\left(
x-30\,000\right) \right]
\end{eqnarray*}%
\ \ \ \ 
\begin{equation*}
N^{\prime }\left( x\right) =\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}%
~~\left( -4\right) \left( x-30\,000\right)
\end{equation*}%
The first factor, $\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}$ is positive
for all $x$ in the domain, except for the endpoints. \ The second factor, \ $%
-4\left( x-30\,000\right) $ \ is positive before $30\,000$, zero at $%
x=30\,000,$ and negative after $30\,000$, indicating an absolute maximum at $%
x=30\,000$. \ Then $y=120\,000-30\,000=90\,000$ and the maximal number sold
is then 
\begin{equation*}
N\left( 30\,000\right) =\dfrac{x^{1/2}y^{3/2}}{400\,000}=\dfrac{%
30\,000^{1/2}90\,000^{3/2}}{400\,000}\approx 11691.\,\allowbreak 342\,95
\end{equation*}%
The integer nearest to this number is $11\,691$.

\item An underground telephone cable is to be laid between two boat docks on
opposite banks of a straight river. \ \ One boathouse is $600$ meters
downstream from the other. \ The river is $200$ meters wide. \ If the cost
of laying the cable is $\$50$ per meter under water and $\$30$ per meter on
land, how should the cable to be laid to minimize cost? \ 

a) \ Find the lowest cost possible.\newline
Solution: \ Let us denote by $x$ - as shown on the picture below - the
distance alongside the river of the part of the cable to be laid under the
water.\FRAME{dtbpF}{3.8778in}{1.6397in}{0pt}{}{}{image1.bmp}{\special%
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Then the length of the cable laid under water is $\sqrt{200^{2}+x^{2}}$ \
and $600-x$ on the ground. \ We can now express the cost as a function of $x$%
.%
\begin{equation*}
C\left( x\right) =50\sqrt{200^{2}+x^{2}}+30\left( 600-x\right) \text{ \ \ \
on domain }\left[ 0,600\right]
\end{equation*}%
We find the minimum of $C$ by differentiating $C$.%
\begin{equation*}
C^{\prime }\left( x\right) =50\dfrac{1}{2\sqrt{200^{2}+x^{2}}}\left(
2x\right) +30\left( -1\right) =\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30
\end{equation*}%
To find the extrema, we solve for the zeroes of the derivative.%
\begin{eqnarray*}
C^{\prime }\left( x\right) &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }25x^{2}=9\left( x^{2}+40\,000\right) \\
\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30 &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }25x^{2}=9x^{2}+360\,000 \\
\dfrac{50x}{\sqrt{200^{2}+x^{2}}} &=&30\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }16x^{2}=360\,000 \\
50x &=&30\sqrt{200^{2}+x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }%
x^{2}=22\,500 \\
5x &=&3\sqrt{200^{2}+x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }x=\pm 150 \\
25x^{2} &=&9\left( 200^{2}+x^{2}\right)
\end{eqnarray*}%
Since the domain is $\left[ 0,600\right] ,$ the only possibility is $x=150.$
\ Thus the minimum cost is $C\left( 150\right) $.\FRAME{dtbpF}{3.397in}{%
1.529in}{0pt}{}{}{image2.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
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0.0614in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'image2.bmp';file-properties "XNPEU";}}%
\begin{equation*}
C\left( 150\right) =\$50\sqrt{200^{2}+150^{2}}+\$30\left( 600-150\right)
=\$50\cdot 250+\$30\cdot 450=\$26\,000
\end{equation*}

b) \ Use the second derivative test to prove that we found a relative
minimum in part a).\newline
Solution: \ We already know that $C^{\prime }\left( 150\right) =0$. \ If $%
C^{\prime \prime }\left( 150\right) $ is positive, then $C$ has a relative
minimum at $x=150$. \ If $C^{\prime \prime }\left( 150\right) $ is negative,
then $C$ has a relative maximum at $x=150$. \ If $C^{\prime \prime }=0,$ the
second derivative test is inconclusive.%
\begin{eqnarray*}
C^{\prime }\left( x\right) &=&\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30 \\
C^{\prime \prime }\left( x\right) &=&\dfrac{50\sqrt{200^{2}+x^{2}}-50x\dfrac{%
1}{2\sqrt{200^{2}+x^{2}}}\left( 2x\right) }{200^{2}+x^{2}}=\dfrac{50\sqrt{%
200^{2}+x^{2}}-\dfrac{50x^{2}}{\sqrt{200^{2}+x^{2}}}}{200^{2}+x^{2}} \\
C^{\prime \prime }\left( x\right) &=&\dfrac{50}{200^{2}+x^{2}}\left( \sqrt{%
200^{2}+x^{2}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) \\
C^{\prime \prime }\left( 150\right) &=&\dfrac{50}{200^{2}+150^{2}}\left( 
\sqrt{200^{2}+150^{2}}-\dfrac{150^{2}}{\sqrt{200^{2}+150^{2}}}\right) =%
\dfrac{16}{125}>0
\end{eqnarray*}%
Since its second derivative is positive at $x=150$, the function $C$ has a
relative minimum at $x=150$.

c) \ Prove that the relative minimum we found in part a) \ is also an
absolue minimum on the domain $\left[ 0,600\right] .$\newline
Solution: \ Since the function $C\left( x\right) $ is continuous on the
closed interval $\left[ 0,600\right] ,$ it achieves the ansolute minimum and
absolute maximum. \ To find these, we only need to evaluate the function at
the relative extrema and the endpoints of the interval. \ We need to compute 
$C\left( 0\right) $, $C\left( 150\right) ,$ and $C\left( 600\right) $. \
These values are $C\left( 0\right) =\$28\,000$, \ $C\left( 150\right)
=\$26\,000$, and $C\left( 600\right) =\$10\,000\sqrt{10}\approx
\$31622.\,\allowbreak 776\,602$. \ Thus $C$ has an absolute minimum at $%
x=150 $.

d) \ Prove that if we define $C$ on the set of all real numbers, then the
minimum we found in part a) \ is still an absolute minimum.\newline
Solution: \ Finding absolute extrema on domains others than a closed
interval is often tricky. \ In this case, we can prove the statement by
looking at $C^{\prime \prime }$ more carefully.%
\begin{eqnarray*}
C^{\prime \prime }\left( x\right) &=&\dfrac{50}{200^{2}+x^{2}}\left( \sqrt{%
200^{2}+x^{2}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) =\dfrac{50}{%
200^{2}+x^{2}}\left( \dfrac{\left( \sqrt{200^{2}+x^{2}}\right) ^{2}}{\sqrt{%
200^{2}+x^{2}}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) \\
&=&\dfrac{50}{200^{2}+x^{2}}\left( \dfrac{200^{2}+x^{2}-x^{2}}{\sqrt{%
200^{2}+x^{2}}}\right) =\dfrac{50}{200^{2}+x^{2}}\left( \dfrac{200^{2}}{%
\sqrt{200^{2}+x^{2}}}\right)
\end{eqnarray*}%
Since $C^{\prime \prime }$ is positive for all $x$, $C^{\prime }$ is
increasing for all $x$. \ So the zero at $x=150$ is the only zero of $%
C^{\prime }$; \ $C^{\prime }$ is negative on $\left( -\infty ,150\right) $
and positive on $\left( 150,\infty \right) $. \ Consequently, $C$ is
decreasing on on $\left( -\infty ,150\right) $ and increasing on $\left(
150,\infty \right) $, and so $C$ has an absolute minimum at $x=150$.

\pagebreak

\item The location function of an object is $s\left( t\right) =\dfrac{120}{%
1+2e^{-t}}$ where $t$ is time, measured in seconds. \ Where is the object
when it is moving with the greatest speed? \ When is that greatest speed
achieved?\newline
Solution: 
\begin{eqnarray*}
s\left( t\right) &=&\dfrac{120}{1+2e^{-t}}=120\left( 1+2e^{-t}\right) ^{-1}
\\
v\left( t\right) &=&s^{\prime }\left( t\right) =120\left( -1\right) \left(
1+2e^{-t}\right) ^{-2}\left( 2e^{-t}\right) \left( -1\right) =\dfrac{%
240e^{-t}}{\left( 1+2e^{-t}\right) ^{2}}=240\dfrac{e^{-t}}{\left(
1+2e^{-t}\right) ^{2}}
\end{eqnarray*}%
We need to find the maximum of $v$ which means we need to differentiate
again. \ This time we will need to use the quotient rule.%
\begin{eqnarray*}
a\left( t\right) &=&v^{\prime }\left( t\right) =240\dfrac{e^{-t}\left(
-1\right) \left( 1+2e^{-t}\right) ^{2}-e^{-t}\left( 2\right) \left(
1+2e^{-t}\right) \left( 2e^{-t}\right) \left( -1\right) }{\left(
1+2e^{-t}\right) ^{4}}=\text{ \ \ \ simplify by }\left( 1+2e^{-t}\right) \\
&=&240\dfrac{e^{-t}\left( -1\right) \left( 1+2e^{-t}\right) -e^{-t}\left(
2\right) \left( 2e^{-t}\right) \left( -1\right) }{\left( 1+2e^{-t}\right)
^{3}}\text{ \ \ \ \ \ \ factor out }e^{-t} \\
&=&240e^{-t}\dfrac{-\left( 1+2e^{-t}\right) +4e^{-t}}{\left(
1+2e^{-t}\right) ^{3}}=240e^{-t}\dfrac{-1-2e^{-t}+4e^{-t}}{\left(
1+2e^{-t}\right) ^{3}}=240e^{-t}\dfrac{2e^{-t}-1}{\left( 1+2e^{-t}\right)
^{3}}
\end{eqnarray*}%
$240e^{-t}$ and the denominator are always positive. \ The only way $%
v^{\prime }\left( t\right) =0$ is when $2e^{-t}-1=0$. \ We solve this
equation for $t$.%
\begin{eqnarray*}
2e^{-t}-1 &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }-t=\ln \left( 
\dfrac{1}{2}\right) \\
2e^{-t} &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }t=-\ln \left( 
\dfrac{1}{2}\right) =\ln 2\text{\ } \\
e^{-t} &=&\dfrac{1}{2}
\end{eqnarray*}%
Thus $t=\ln 2$ is when the object is fastest. \ The location of the object
is then 
\begin{equation*}
s\left( \ln 2\right) =\dfrac{120}{1+2e^{-\ln 2}}=\dfrac{120}{1+2\left( 
\dfrac{1}{2}\right) }=\dfrac{120}{2}=60
\end{equation*}

\item Consider the function $g\left( x\right) =\dfrac{-2x}{\left(
x^{2}+1\right) ^{2}}$.

a) \ Find all relative extrema of $g$.\newline
Solution: \ We differentiate $g$ using the quotient rule.%
\begin{eqnarray*}
g^{\prime }\left( x\right) &=&\dfrac{-2\left( x^{2}+1\right) ^{2}-\left(
-2x\right) 2\left( x^{2}+1\right) \left( 2x\right) }{\left( x^{2}+1\right)
^{4}}\text{ \ \ \ \ \ \ simplify by }x^{2}+1 \\
&=&\dfrac{-2\left( x^{2}+1\right) -\left( -2x\right) 2\left( 2x\right) }{%
\left( x^{2}+1\right) ^{3}}=\dfrac{-2x^{2}-2+8x^{2}}{\left( x^{2}+1\right)
^{3}}=\dfrac{6x^{2}-2}{\left( x^{2}+1\right) ^{3}}
\end{eqnarray*}%
We now factor the numerator to see when $g^{\prime }$ is positive and
negative.%
\begin{equation*}
g^{\prime }\left( x\right) =\dfrac{6x^{2}-2}{\left( x^{2}+1\right) ^{3}}=%
\dfrac{6\left( x^{2}-\dfrac{1}{3}\right) }{\left( x^{2}+1\right) ^{3}}=%
\dfrac{6\left( x+\dfrac{1}{\sqrt{3}}\right) \left( x-\dfrac{1}{\sqrt{3}}%
\right) }{\left( x^{2}+1\right) ^{3}}
\end{equation*}%
The denominator is always positive, the numerator is a quadratic expression
with a positive leading coefficient, and so $g^{\prime }$ is positive on $%
\left( -\infty ,-\dfrac{1}{\sqrt{3}}\right) \cup \left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $ \ \ and negative on $\left( -\dfrac{1}{\sqrt{3}},\dfrac{1}{%
\sqrt{3}}\right) .$ \ Consequently, $g$ is increasing on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $, decreasing on $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $. \ That means that $g$ has a relative maximum at $x=-%
\dfrac{1}{\sqrt{3}}$ and a relative minimum at $x=\dfrac{1}{\sqrt{3}}$. \ We
compute the function values at $-\dfrac{1}{\sqrt{3}}$ and $\dfrac{1}{\sqrt{3}%
}.$ \ 
\begin{equation*}
g\left( -\dfrac{1}{\sqrt{3}}\right) =\dfrac{-2\left( -\dfrac{1}{\sqrt{3}}%
\right) }{\left( \left( -\dfrac{1}{\sqrt{3}}\right) ^{2}+1\right) ^{2}}=%
\dfrac{\dfrac{2}{\sqrt{3}}}{\left( \dfrac{1}{3}+1\right) ^{2}}=\dfrac{\dfrac{%
2}{\sqrt{3}}}{\left( \dfrac{4}{3}\right) ^{2}}=\dfrac{2}{\sqrt{3}}\cdot 
\dfrac{9}{16}=\dfrac{3\sqrt{3}}{8}
\end{equation*}%
This is a good time to notice that $g$ is an odd function%
\begin{equation*}
g\left( -x\right) =\dfrac{-2\left( -x\right) }{\left( \left( -x\right)
^{2}+1\right) ^{2}}=\dfrac{2x}{\left( x^{2}+1\right) ^{2}}=-g\left( x\right)
\end{equation*}%
and so $g$ has a relative maximum: $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3%
\sqrt{3}}{8}\right) $ and a relative minimum $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{3\sqrt{3}}{8}\right) $.

b) \ Find all absolute extrema of $g$.\newline
Solution: \ We will show that the relateive extrema we found in part a) are
in fact absolute extrema. \ This is not a fact that simply follows from the
signs of the derivative. \ For all we know, a decreasing, then increasing,
then decreasing function may look like an upside down cubic function that
has neither absolute minumum nor absolute maximum.\newline
Claim: \ $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3\sqrt{3}}{8}\right) $ is an
absolute maximum, i.e. for all real numbers $x$, $g\left( x\right) \leq 
\dfrac{3\sqrt{3}}{8}$.\newline
In part a) \ we determined that $g$ is increasing on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $, decreasing on $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $.\newline
Because $g$ is increasing on $\left( -\infty ,-\dfrac{1}{\sqrt{3}}\right) $, 
\begin{equation*}
\text{for all }x\leq -\dfrac{1}{\sqrt{3}}\text{, \ \ \ \ \ }g\left( x\right)
\leq g\left( -\dfrac{1}{\sqrt{3}}\right)
\end{equation*}%
Because $g$ is decreasing on $\left( -\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}%
\right) $,%
\begin{equation*}
\text{for all }-\dfrac{1}{\sqrt{3}}\leq x\leq \dfrac{1}{\sqrt{3}}\text{, \ \
\ \ \ }g\left( x\right) \leq g\left( -\dfrac{1}{\sqrt{3}}\right)
\end{equation*}%
We only have to prove that for all $x\geq \dfrac{1}{\sqrt{3}}$, \ \ \ \ \ $%
g\left( x\right) \leq g\left( -\dfrac{1}{\sqrt{3}}\right) $ is also true,
but the increasing/decreasing behavior does not help here, since $g$ is
increasing on $\left( \dfrac{1}{\sqrt{3}},\infty \right) $. \ Luckily, there
is a simple, elementary way to finish the proof. \ Recall that $g\left(
x\right) =\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}$. \ It is easy to see that
if $x$ is positive, then $g\left( x\right) $ is negative. \ Thus $g$ is
negative on $\left( \dfrac{1}{\sqrt{3}},\infty \right) $ and so 
\begin{equation*}
\text{for all }x\geq \dfrac{1}{\sqrt{3}}\text{, \ \ \ \ \ }g\left( x\right)
\leq g\left( -\dfrac{1}{\sqrt{3}}\right) =\dfrac{3\sqrt{3}}{8}
\end{equation*}%
Thus $g$ has an absolute maximum at $x=-\dfrac{1}{\sqrt{3}}$.\newline
Claim: \ $\left( \dfrac{1}{\sqrt{3}},-\dfrac{3\sqrt{3}}{8}\right) $ is an
absolute minimum, i.e. for all real numbers $x$, $g\left( x\right) \leq -%
\dfrac{3\sqrt{3}}{8}$.\newline
proof: \ a very similar argument could work. $\ g$ is decreasing on $\left( -%
\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( 
\dfrac{1}{\sqrt{3}},\infty \right) $ and positive on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $ \ Another way to prove this is to use the fact
that $g$ is an odd function.%
\begin{eqnarray*}
\text{for all }x\text{, }g\left( x\right) &\leq &\dfrac{3\sqrt{3}}{8}\text{
\ \ \ \ \ \ multiply by }-1 \\
\text{for all }x\text{, }-g\left( x\right) &\geq &-\dfrac{3\sqrt{3}}{8}\text{
\ \ \ \ since }g\text{ is odd, }-g\left( x\right) =g\left( -x\right) \\
\text{for all }x\text{, }g\left( -x\right) &\geq &-\dfrac{3\sqrt{3}}{8}
\end{eqnarray*}%
and so $g$ has an absolute minimum at $x=\dfrac{1}{\sqrt{3}}$.\FRAME{dtbpFX}{%
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c) \ Find all values of $c$ for which the function $f\left( x\right) =\dfrac{%
1}{x^{2}+1}+cx$ \ is increasing on its entire domain.\newline
Solution: \ $f\left( x\right) $ is always increasing if $f^{\prime }\left(
x\right) $ \ is always positive. \ $f^{\prime }\left( x\right) =\dfrac{-2x}{%
\left( x^{2}+1\right) ^{2}}+c$. \ We proved that the absolute minimum of $%
\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}$ is $-\dfrac{3\sqrt{3}}{8}$ and so
if we set $c\geq \dfrac{3\sqrt{3}}{8},$ then \ $f^{\prime }\left( x\right) =%
\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}+c$ will be non-negative for all $x$.
\end{enumerate}

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