Curve Sketching

1. Existence of symmetry
2. Domain of a function
3. Intersection of a curve with $x$-axis
4. Asymptotic line
5. 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 $${f(x) = x^5 - 5x^4 + 1}$$.

SOLUTION

1.	$f(-x) = - x^5 - 5x^4+ 1$ implies no symmetry.
2.	${\rm Dom}(f) = (-\infty,\infty)$
3.	$f(-1) = -5, f5. = 1$ implies this function cross $x$-axis at least once
4.	$\lim_{x \to \infty}f(x) = \infty$, $\lim_{x \to -\infty}f(x) = -\infty$
5.	By example2.22
	$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...\rm IP} & {\rm up} & {\rm up} & {\rm up} \ \hline \end{array}$$

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

Figure 2.11: Example2-19

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

SOLUTION

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

Figure 2.12: Division

By quotient rule for differentiation,

Figure 2.13: Exercise2-19

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

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

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

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

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