%fa07 problem set 1


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Now that we are usiing the chain rule, it became more complicated to sort
out how the derivative changes sign around its zero. \ \ For such cases, we
use the second derivative test.\vspace{0.1in}

Consider a function $f$ that is twice differentiable on an open interval
containing $c$. \ Suppose further that $f^{\prime }\left( c\right) =0$. \
Let us look at the function $f^{\prime }$ first. \ Since $f$ is twice
differentiable, $f^{\prime }$ is differentiable. \ Differentiable functions
have just a few ways in which they take a zero value.\bigskip

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Case 1. \ Suppose that $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) $ is negative.

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That indicates that $f^{\prime }$ is strictly decreasing on an interval
containing $c$. Taking a zero value while decreasing means that $f^{\prime } 
$ changes sign from positive to negative. \ That indicates that $f$ has a
relative maximum at $c.$ \vspace{0.1in}

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\textbf{Theorem:} \ If $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) <0$, then $f$ has a relative maximum at $c$.\vspace{0.06in}%
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Case 2. \ Suppose that $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) $ is positive. \ 

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That indicates that $f^{\prime }$ is strictly increasing on an interval
containing $c$. Taking a zero value while increasing means that $f^{\prime }$
changes sign from negative to positive. \ That indicates that $f$ has a
relative minimum at $c.$ \vspace{0.1in}

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\textbf{Theorem:} \ If $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) >0$, then $f$ has a relative minimum at $c$.\vspace{0.06in}%
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Case 3. \ Suppose that $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) =0$. \ Consider $f\left( x\right) =x^{8}$ and $g\left(
x\right) =x^{9}$ near $x=0$.

$f$ has a relative minimum at $x=0$ and $g$ has neither a maximum nor a
minimum at $x=0$, yet \vspace{0.07in}

\ $f^{\prime }\left( 0\right) =0$, $f^{\prime \prime }\left( 0\right) =0$
and $g^{\prime }\left( 0\right) =0$, $g^{\prime \prime }\left( 0\right) =0$%
\vspace{0.07in}

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Furthermore, the higher order derivatives of $f$ and $g$ will be zero for
quite a while. \ Therefore, the second derivative in this case did not
distinguish between maximums and minimums. \vspace{0.1in}

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\textbf{Theorem:} \ If $f^{\prime }\left( c\right) =0$ and $f^{\prime \prime
}\left( c\right) =0$, then the second derivative test did not yield for any
useful result. \ \vspace{0.06in}%
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In such cases, we need to apply other methods to tell maximums, minimums,
(or neither) apart.\vspace{0.15in}

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\textbf{Example 1.} Suppose that $f\left( x\right) =\sin \left( \dfrac{1}{x}%
\right) $. \ Prove that $f$ has a relative maximum at $x=\dfrac{2}{\pi }$.

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\textbf{Solution:} \ We differentiate $f$ and then $f^{\prime }$: \ \ \ \ \
\ \ $f^{\prime }\left( x\right) =-\dfrac{1}{x^{2}}\cos \dfrac{1}{x}$ \ \ \
and $f^{\prime \prime }\left( x\right) =\dfrac{2}{x^{3}}\cos \dfrac{1}{x}-%
\dfrac{1}{x^{4}}\sin \dfrac{1}{x}$\vspace{0.08in}

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Now we compute $f^{\prime }\left( \dfrac{2}{\pi }\right) $ and \ $f^{\prime
}\left( \dfrac{2}{\pi }\right) $\vspace{0.08in}

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$f^{\prime }\left( \dfrac{2}{\pi }\right) =-\dfrac{1}{\left( \dfrac{2}{\pi }%
\right) ^{2}}\cos \left( \dfrac{1}{\left( \dfrac{2}{\pi }\right) }\right) =-%
\dfrac{\pi ^{2}}{4}\cos \left( \dfrac{\pi }{2}\right) =0$\vspace{0.08in}

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and $f^{\prime \prime }\left( \dfrac{2}{\pi }\right) =2\left( \dfrac{\pi }{2}%
\right) ^{3}\cos \left( \dfrac{\pi }{2}\right) -\left( \dfrac{\pi }{2}%
\right) ^{4}\sin \left( \dfrac{\pi }{2}\right) =-\dfrac{1}{16}\pi ^{4}$%
\vspace{0.08in}

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So we have that $f^{\prime }\left( \dfrac{2}{\pi }\right) =0$ and $f^{\prime
\prime }\left( \dfrac{2}{\pi }\right) $\vspace{0.08in} is negative. \
Therefore, by the second derivative test, $f$ has a relative maximum at $%
\dfrac{2}{\pi }$.\vspace{0.08in}

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The second derivative test worked in this case.\vspace{0.08in} \
Furthermore, we needed it, because it is difficult to sort out how \newline
$f^{\prime }\left( x\right) =\allowbreak -\dfrac{1}{x^{2}}\cos \dfrac{1}{x}$
changes sign at $x=\dfrac{2}{\pi }$.

\textbf{Example 2. }\ \ A company wants to manufacture cylindrical aluminum
cans with a volume of $1000$ cubic centimeters (one liter). \ What
dimensions would guarantee the minimal amount of aluminum needed to produce
a can? \ 

\textbf{Solution:} \ \textbf{\vspace{1in}}

{\Large 
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}

\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.}

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