Curve Sketching

Existence of symmetry

Domain of a function

Intersection of a curve with -axis

Asymptotic line

Incereasing/Decreasin, concavity, Critical point, inflection point

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. $\ \blacksquare$

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.

By quotient rule for differentiation,

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

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

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 $\ \blacksquare$

Next: Polar Coordinates Up: Extreme Value of Functions Previous: Extreme Value of Functions Contents Index

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
	$\begin{displaymath}\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...\rm IP} & {\rm up} & {\rm up} & {\rm up} \\ \hline \end{array} \end{displaymath}$

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.

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