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Next: 13.2 Optimization Problems. Up: 13. Applications Previous: 13. Applications   Index

13.1 Curve Sketching

13.1   Example. Let $\displaystyle {f(x)={{x^3}\over {1-x^2}}}$. Here $\mbox{{\rm dom}}(f)=\mbox{{\bf R}}\setminus \{\pm
1\}$ and $f$ is an odd function. We have

\begin{displaymath}f^\prime (x)={{(1-x^2)3x^2-x^3(-2x)}\over
{(1-x^2)^2}}={{3x^2-x^4}\over{(1-x^2)^2}}={{x^2(3-x^2)}\over {(1-x^2)^2}}.\end{displaymath}

From this we see that the critical set for $f$ is $\{0,\sqrt 3, -\sqrt 3\}$. We can determine the sign of $f^\prime (x)$ by looking at the signs of its factors: Since $f$ is odd, I will consider only points where $x>0$.

\begin{displaymath}\begin{array}{\vert c\vert c\vert c\vert c\vert}\hline
& 0<x<...
...ex] \hline
f^\prime (x) & + & + & -  [2ex] \hline
\end{array}\end{displaymath}

Thus $f$ is strictly increasing on $(0,1)$ and on $(1,\sqrt 3)$, and $f$ is strictly decreasing on $(\sqrt 3,\infty )$. Also

\begin{displaymath}f(\sqrt 3)={{3\sqrt 3}\over {1-3}}=-{3\over 2}\sqrt3 \mbox{ and } f(0)=0.\end{displaymath}

We see that $\vert f(x)\vert$ is unbounded on any interval $(1-\delta,1)$ or $(1,1+\delta)$, since the numerator of the fraction is near to $1$, and the denominator is near to $0$ on these intervals. Also

\begin{displaymath}f(x)={{x^3}\over {1-x^2}}=x\Big( {{x^2}\over {1-x^2}}\Big)=x\Big(
{1\over {-1+{1\over {x^2}} }}\Big),\end{displaymath}

so $\vert f(x)\vert$ is large when $x$ is large. ($f(x)$ is the product of $x$ and a number near to $-1$.) Using this information we can make a reasonable sketch of the graph of $f$.
\psfig{file=ch13a.eps,width=3.5in}

Here $f$ has a local maximum at $\sqrt 3$ and a local minimum at $-\sqrt 3$. It has no global extreme points.

13.2   Definition (Infinite limits.) Let $\{x_n\}$ be a real sequence. We say

\begin{displaymath}\lim\{x_n\}=+\infty \mbox{ or } \{x_n\}\to +\infty\end{displaymath}

if for every $B\in\mbox{{\bf R}}$ there is an $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that for all $n\in\mbox{{\bf Z}}_{\geq N}$ $\Big(
x_n>B\Big)$.

We say

\begin{displaymath}\lim\{x_n\}=-\infty \mbox{ or } \{x_n\}\to -\infty\end{displaymath}

if for every $B\in\mbox{{\bf R}}$ there is an $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that for all $n\in\mbox{{\bf Z}}_{\geq N}$ $\Big(
x_n<B\Big)$. Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$, and let $a\in\mbox{{\bf R}}$. We say

\begin{displaymath}\lim_{x\to a^+} f(x)=+\infty\end{displaymath}

if $\mbox{{\rm dom}}(f)$ contains an interval $(a,a+\epsilon)$ and for every sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)\cap (a,\infty)$

\begin{displaymath}(\{x_n\}\to a)\hspace{1em}\mbox{$\Longrightarrow$}\hspace{1em}\Big( \{ f(x_n)\}\to +\infty \Big).\end{displaymath}

We say

\begin{displaymath}\lim_{x\to a^-}f(x)=+\infty\end{displaymath}

if $\mbox{{\rm dom}}(f)$ contains an interval $(a-\epsilon ,a)$ and for every sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)\cap (-\infty ,a)$

\begin{displaymath}(\{x_n\}\to a)\hspace{1em}\mbox{$\Longrightarrow$}\hspace{1em}\Big( \{ f(x_n)\}\to +\infty \Big).\end{displaymath}

Similar definitions can be made for

\begin{displaymath}\lim_{x\to a^+}f(x)=-\infty,\; \lim_{x\to a^-} f(x)=-\infty.\end{displaymath}

We say $\displaystyle { \lim_{x\to +\infty}f(x)=+\infty}$ if $\mbox{{\rm dom}}(f)$ contains some interval $(a,\infty)$ and for every sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)$

\begin{displaymath}\{x_n\}\to +\infty \hspace{1em}\mbox{$\Longrightarrow$}\hspace{1em} \{f(x_n)\}\to +\infty.\end{displaymath}

Similarly if $c \in \mbox{{\bf R}}$ we can define

\begin{displaymath}\lim_{x\to +\infty}f(x)=-\infty,\; \lim_{x\to +\infty}f(x)=c,\;
\lim_{x\to-\infty}f(x)=+\infty,\; {\rm etc.}\end{displaymath}

13.3   Example. If $f$ is the function in the previous example (i.e. $\displaystyle {f(x) = {x^3 \over 1 - x^2}}$) then

\begin{eqnarray*}
{\displaystyle {\lim_{x\to 1^+}}} f(x) &= &-\infty,\\
{\dis...
...fty,\\
{\displaystyle {\lim_{x\to +\infty}}} f(x) &=& -\infty,
\end{eqnarray*}



and

\begin{eqnarray*}
{\displaystyle {\lim_{x\to -\infty}}} f(x) &=& +\infty.
\end{eqnarray*}



Also,

\begin{eqnarray*}
\lim_{x\to \infty} {1\over x} & =& 0,\\
\lim_{x\to 0^+} {x\ov...
...rt}} & =& 1,\\
\lim_{x\to 0^-} {x\over {\vert x\vert}} & =& -1,
\end{eqnarray*}



and

\begin{displaymath}
\lim_{x\to +\infty}{x^2 + 1 \over x^2 + 3x} = \lim_{x\to +\infty}
{{1+{1\over {x^2}}}\over {1+{3\over x}}}=1.
\end{displaymath}

The situation here is very similar to the situation in the case of ordinary limits, and we will proceed without writing out detailed justifications.

13.4   Exercise. Write out definitions for

\begin{displaymath}\Big( \lim_{x\to +\infty}f(x)=-\infty\Big) \mbox{ and for } \Big(\lim_{x\to
a^-}f(x)=-\infty \Big).\end{displaymath}

13.5   Exercise. Find one function $f$ satisfying all of the following conditions:

\begin{eqnarray*}
\displaystyle {\lim_{x\to + \infty}} f(x) & = & 3,\\
\display...
...\infty ,\\
\displaystyle {\lim_{x\to 3^-}} f(x) & = & +\infty .
\end{eqnarray*}



13.6   Example. Let $f(x)=\sin (2x) + 2\sin (x).$ Then $f(x+2\pi) = f(x)$ for all $x \in \mbox{{\bf R}}$, so I will restrict my attention to the interval $[-\pi ,\pi ]$. Also $f$ is an odd function, so I will further restrict my attention to the interval $[0,\pi]$. Now

\begin{eqnarray*}
f^\prime (x) &=& 2\cos 2x +2\cos x = 2(2\cos ^2x-1)+2\cos x ...
...x-1)(\cos x+1)\\
&=& 4\Big( \cos x -{1\over 2}\Big) (\cos x+1).
\end{eqnarray*}



Hence $x$ is a critical point for $f$ if and only if $\displaystyle { \cos x\in\Big\{
{1\over
2},-1\Big\}}$. The critical points of $f$ in $[0,\pi]$ are thus $\pi$ and $\displaystyle {{\pi\over 3}}$, and the critical points in $[-\pi ,\pi ]$ are $\displaystyle { \Big\{
-\pi,\pi,{\pi\over 3},-{\pi\over 3}\Big\} }$. Now $f(\pi)=f(0)=0$ and

\begin{displaymath}f\Big({\pi\over 3}\Big)=\sin\Big({{2\pi}\over 3}\Big) +2\sin ...
...\sqrt 3}\over 2}={{3\sqrt 3}\over 2}=2.6
(\rm {approximately}),\end{displaymath}

and $\displaystyle { f\Big(-{\pi\over 3}\Big)=-f\Big( {\pi\over 3}\Big) }$. Also note $f^\prime
(0)=4$. Since $f$ is continuous on $[-\pi ,\pi ]$, we know that $f$ has a maximum and a minimum on this interval, and since $f(x+2\pi) = f(x)$ for all $x \in \mbox{{\bf R}}$, the maximum (or minimum) of $f$ on $[-\pi ,\pi ]$ will be a global maximum (or minimum) for $f$. Since $f$ is differentiable everywhere, the extreme points are critical points and from our calculations $f$ has a maximum at $\displaystyle {{\pi\over 3}}$ and a minimum at $\displaystyle {
-{\pi\over 3}}$. I will now determine the sign of $f^\prime$ on $[0,\pi]$:

\begin{displaymath}\begin{array}{\vert l\vert l\vert l\vert}\hline
& 0<x<{\pi\ov...
... [2ex] \hline
f^\prime (x) & + & -  [2ex] \hline
\end{array}\end{displaymath}

Thus $f$ is strictly increasing on $\Big(\displaystyle {0,{\pi\over 3}}\Big)$ and $f$ is strictly decreasing on $\displaystyle {\Big( {\pi\over 3},\pi\Big)}$. We can now make a reasonable sketch for the graph of $f$.
\psfig{file=ch13b.eps,width=4in}

13.7   Exercise. Sketch and discuss the graphs of the following functions. Mention all critical points and determine whether each critical point is a local or global maximum or minimum.
a)
$f(x)=(1-x^2)^2$.
b)
$g(x)=\displaystyle { {x\over {1+x^2}}}$.
c)
$h(x)=x+\sin (x)$.
d)
$k(x)=x\ln (x)$.
(The following remark may be helpful for determining $\displaystyle {\lim_{x \to 0} k(x)}$. If $0 < t < 1$, then $\displaystyle {{1 \over t} < {1 \over t^{3\over 2}}}$. Hence if $0 < x < 1$, then

\begin{eqnarray*}
\vert \ln (x)\vert &=& \left\vert \int_1^x {1\over t} dt \righ...
... = 2\left( {1\over \sqrt{x}} - 1
\right) \leq {2\over \sqrt{x}}.
\end{eqnarray*}



Thus,

\begin{displaymath}\vert x \ln(x)\vert \leq 2 \sqrt{x} \mbox{ for }0 < x < 1).\end{displaymath}


next up previous index
Next: 13.2 Optimization Problems. Up: 13. Applications Previous: 13. Applications   Index
Ray Mayer 2007-09-07