Curve Sketching

Existence of symmetry
Domain of a function
Intersection of a curve with -axis
Asymptotic line
Incereasing/Decreasin, concavity, Critical point, inflection point

Asymptote Asymptote is a line for which a graph of function is getting close. Thus we have two cases. One is a line for which the denominator of a function is 0. The other one is that $\vert x\vert$ approaches infinity.

Example 2..19 Sketch the graph of function $\displaystyle{f(x) = x^5 - 5x^4 + 1}$ .

SOLUTION

1.	implies no symmetry.
2.	${\rm Dom}(f) = (-\infty,\infty)$
3.	implies this function cross -axis at least once
4.	$\lim_{x \to \infty}f(x) = \infty$ , $\lim_{x \to -\infty}f(x) = -\infty$
5.	By example2.22
	$\begin{displaymath}\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...\rm IP} & {\rm up} & {\rm up} & {\rm up} \ \hline \end{array} \end{displaymath}$

We now sketch the graph of function. $\blacksquare$

**Figure 2.11:** Example2-19
$\includegraphics[width=4.0cm]{CALCFIG/Fig2-5-2.eps}$

Exercise 2..19 Sketch the graph of a function $\displaystyle{y = \frac{x^3}{x^2 - 2}}$ .

SOLUTION

1.	$f(-x) = \frac{(-x)^3}{(-x)^2 - 2} = -f(x)$ implies that is symmetric
	with respect to the origin.
2.	${\rm Dom}(f) = \{x:x \neq \sqrt{2}\}$ .
3.	To find an intersection with -axis, set .
	Then we have .
4.	When the denominator is 0, we have $x = \pm \sqrt{2}$
	as asymptotes. Next we can express the function as
	$\displaystyle{y = \frac{x^3}{x^2 - 2} = \frac{x(x^2 -2) + 2x}{x^2-2} = x + \frac{2x}{x^2 - 2} }$
	Thus is asymptote.

**Figure 2.12:** Division
$\includegraphics[width=3cm]{SOFTFIG-2/longdivision.eps}$

By quotient rule for differentiation,

**Figure 2.13:** Exercise2-19
$\includegraphics[width=4.0cm]{CALCFIG/Fig2-6-1.eps}$

$\displaystyle y^{\prime}$	$\displaystyle =$	$\displaystyle \frac{(x^3)'(x^2 -2) - x^3(x^2 - 2)'}{(x^2 - 2)^2} = \frac{x^2 (x^2 - 6)}{(x^2 - 2)^2},$
$\displaystyle y''$	$\displaystyle =$	$\displaystyle \frac{(x^2)'(x^2 -6) - x^2(x^2 - 6)'}{(x^2 -2)^4} = \frac{4x(x^2 + 6)}{(x^2 - 2)^3}$

implies $x = 0, \pm \sqrt{6}$ are the candidates for a critical point. Now write a concavity table, we have

$\begin{displaymath}\begin{array}{c\vert ccccccccccccc} x & & -\sqrt{6} & & - \sq... ...rrow & & \searrow & \frac{3\sqrt{6}}{2} & \nearrow \end{array} \end{displaymath}$

Thus, the sketch of graph is as follows $\blacksquare$