13.1 Curve Sketching

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

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

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 $x>0$.

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

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

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

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

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

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 $f$.

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

$$\lim\{x_n\}=+\infty \mbox{ or } \{x_n\}\to +\infty$$

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

$$\lim\{x_n\}=-\infty \mbox{ or } \{x_n\}\to -\infty$$

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

$$\lim_{x\to a^+} f(x)=+\infty$$

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

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

We say

$$\lim_{x\to a^-}f(x)=+\infty$$

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

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

Similar definitions can be made for

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

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

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

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

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

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

$$\lim_{x\to +\infty}{x^2 + 1 \over x^2 + 3x} = \lim_{x\to +\infty} {{1+{1\over {x^2}}}\over {1+{3\over x}}}=1.$$

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

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

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

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

and $${ f\Big(-{\pi\over 3}\Big)=-f\Big( {\pi\over 3}\Big) }$$. 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 $${{\pi\over 3}}$$ and a minimum at $${ -{\pi\over 3}}$$. I will now determine the sign of $f^\prime$ on $[0,\pi]$:

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

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

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 $${\lim_{x \to 0} k(x)}$$. If $0 < t < 1$, then $${{1 \over t} < {1 \over t^{3\over 2}}}$$. 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,

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