5.6 Sketching Graphs Of Functions
| Calculus Of One Real Variable – By Pheng Kim Ving |
| 5.6 |
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Horizontal Asymptotes
| The horizontal line y = L is called a horizontal asymptote of the graph of y = f(x) if: |
Vertical Asymptotes
| The vertical line x = a is called a vertical asymptote of the graph of y = f(x) if one of the following is satisfied: |
Oblique Asymptotes
The graph of f(x) = (x2 + 5x – 4)/(2x – 2) is sketched in Fig. 1.4. Long division of the numerator x2 + 5x – 4 by the
denominator 2x – 2 gives: quotient = (1/2)x + 3 and remainder = 2. So:
Remark 1.1
The graph of a function can intersect a horizontal or oblique asymptote, but can never intersect a vertical asymptote (why?
hint: definition of a function).
One- And Two-Sided Asymptotes
iii. The graph of a function f may have two one-sided horizontal asymptotes. For example, see Fig. 1.5.
iv. The notions of one- and two-sided asymptotes also apply to vertical and oblique asymptotes.
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| 2. Asymptotes Of Rational Functions |
Let f(x) = P(x)/Q(x) be a rational function, ie, a ratio or fraction of two polynomials P(x) and Q(x). Let’s denote the
degree of a polynomial p by deg( p). Then:
| The graph of f has a vertical asymptote at every point x where Q(x) = 0. |
| If deg(P) < deg(Q), then the line y= 0 (the x-axis) is a horizontal asymptote of the graph of f. |
iii. Suppose deg(P) = deg(Q). Let a and b be the coefficients of the dominating terms of P and Q respectively; see
| If deg(P) = deg(Q), then the line y = a/b, where a and b are the coefficients of the dominating terms of P and |
iv. Suppose deg(P) = deg(Q) + 1. For an example, see the discussion of the function f(x) = (x2 + 5x – 4)/(2x – 2)
above and its graph sketched in Fig. 1.4. Divide P by Q using long division to get:
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| 3. Sketching Graphs Of Functions |
When sketching the graph of a function y = f(x), we have three sources of useful information: f itself, f ‘, and f ”. We’ll
follow these suggested steps:
i. Information From f: Determine each of the following if any:
a. The domain of f.
b. Intercepts: x-intercepts and y-intercept.
c. Symmetry of the graph (is f even, odd, or neither?).
d. Limits at points of discontinuity and at positive and negative infinity.
e. Asymptotes: determine vertical and horizontal ones from the limits above, and oblique ones by long division if
appropriate.
ii. Information From f ‘:
a. Find points x where f ‘(x) = 0 or f ‘(x) doesn’t exist (critical points). Calculate the value of f at each of them if
defined.
b. Determine the signs of f ‘; if it’s not obvious to do so by simply examining the expression of f ‘ , draw a chart for
f ‘; see Section 5.3 Part 3 and Part 5.
iii. Information From f ”:
a. Find points x where f ”(x) = 0 or f ”(x) doesn’t exist. Calculate the value of f at each of them if defined.
b. Determine the signs of f ”; if it’s not obvious to do so by simply examining the expression of f ” , draw a chart for
f ”; see Section 5.4 Part 4.
iv. Draw the chart for f, where the intervals of increase, decrease, and concavity are determined; see Section 5.4 Part 4. Indicate local maxima, local minima, and inflection points if any.
v. Determine additional points of the graph if helpful.
vi. Sketch the graph using all the information obtained above.
Example 3.1
Sketch the graph of the function:
Solution
Asymptotes:
long division yields:
vertical: lines x = –1 and x = 1, horizontal: none,
oblique: line y = x.
First Derivative:
The chart for y” is shown in Fig. 3.1. The chart for y is shown in Fig. 3.2. The graph of y is shown in Fig. 3.3.
| Fig. 3.1 Chart For y”. |
| Fig. 3.3 Graph of:
|
EOS
Remark 3.1
In the charts, the double vertical line below a point x and on a row means that the function on that row isn’t defined or
doesn’t exist at that point.
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1. Sketch the graph of the function y = x2(x2 – 1), making use of any suitable information you can obtain from the
function and its first and second derivatives.
Solution
Domain: R.
y = x2(x2 – 1) = x2(x + 1)(x – 1);
y = x2(x2 – 1) = x4 – x2.
Intercepts:
x-intercepts: x = 0, –1, and 1,
y-intercept: y = 0.
Symmetry:
y(–x) = (–x)4 – (–x)2 = x4 – x2 = y(x);
function is even; graph is symmetric about y-axis.
Limits:
Second Derivative:
from y‘ = 4x3 – 2x we get:
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2. Sketch the graph of the function:
using information from the function and its first and second derivatives.
Solution
Domain: R – {–1, 1}.
Intercepts:
x-Intercepts: x = 0,
y-Intercept: y = 0.
Symmetry:
Second Derivative:
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3. Sketch the graph of the function:
using information from the function and its first and second derivatives.
Solution
Domain: R – {1}.
Intercepts:
x-intercepts: x = – 2,
y-intercept: y = – 4.
Symmetry:
y‘ is never 0;
y‘ doesn’t exist at x = 1 where y doesn’t exist either;
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4. Sketch the graph of the function:
utilizing any useful information you can get from f , f ‘, and f ”.
Solution
Domain: R – {–1}.
Asymptotes:
long division yields:
vertical: line x = –1,
horizontal: none,
oblique: line y = x – 1.
First Derivative:
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5. Sketch the graph of the function:
employing any useful information you can get from y , y ‘, and y ”.
Solution
Domain: R – {1}.
Intercepts:
x-Intercepts: x = 0,
y-Intercept: y = 0.
Symmetry:
Asymptotes:
vertical: line x = 1,
horizontal: none,
oblique: none.
First Derivative:
y” = 0 at x = 0,
y” doesn’t exist at x = 1 where y doesn’t exist either;
if x = 0 then y = 0, point (0, 0);
sign of y” is same as that of x/(x – 1)3;
if x < 0 then x < 1 and so (x – 1)3 < 0, thus y” > 0,
if 0 < x < 1 then y” < 0,
if 1 < x then y” > 0.
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