2x2 Use the guidelines to sketch the curve y = x2 - 1 A. The domain is {x | x2 - 1 + 0} {x | x + +1} (-0, – 1) U (-1,1)U(1,00) (using interval notation). В. The x- and y-intercepts are both 0. C. Since f(-x) = f(x), the function f is even 8. The curve is symmetric about the y-axis. 2x2 D. lim lim 1 X- to 1 - x2 X - t00 x2 - 1 Therefore, the line y =| 2 is a horizontal asymptote. Since the denominator is 0 when x = ±1, we compute the following limits. 2x2 lim = 00 X→1+ x2 2x2 - 1 lim х> 1" х* —1 2x2 lim x → -1+ x² - 1 2x2 lim X → -1- x2 - 1 Therefore, the lines x = 1 and x = -1 are vertical asymptotes. This information about limits and fall.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question
100%

Pls do parts E,F,G,H only.

 

2x2
Use the guidelines to sketch the curve y =
x2
- 1
A.
The domain is
{x | x2 - 1 + 0} = {x | x + ±1}
|(-0, – 1) U (-1,1) U (1,00) (using interval notation).
%D
В.
The x- and y-intercepts are both 0.
C.
Since f(-x) = f(x), the function f is
The curve is symmetric about the y-axis.
even
2x2
2
D.
lim
X → +0 x2
lim
1
1
x2
Therefore, the line y
2
is a horizontal asymptote.
Since the denominator is 0 when x =
±1, we compute the following limits.
2x2
lim
X → 1+ x2
2x2
1
lim
X → 1
x2
1
2x2
lim
%D
x →-1+ x2
2x2
1
lim
X → -1- X'
1
Therefore, the lines x = 1 and x = -1
are vertical asymptotes. This information about limits and
asymptotes enables us to draw the asymptotes in the following figure.
y=2
x=-1j
I = x||
8.
Transcribed Image Text:2x2 Use the guidelines to sketch the curve y = x2 - 1 A. The domain is {x | x2 - 1 + 0} = {x | x + ±1} |(-0, – 1) U (-1,1) U (1,00) (using interval notation). %D В. The x- and y-intercepts are both 0. C. Since f(-x) = f(x), the function f is The curve is symmetric about the y-axis. even 2x2 2 D. lim X → +0 x2 lim 1 1 x2 Therefore, the line y 2 is a horizontal asymptote. Since the denominator is 0 when x = ±1, we compute the following limits. 2x2 lim X → 1+ x2 2x2 1 lim X → 1 x2 1 2x2 lim %D x →-1+ x2 2x2 1 lim X → -1- X' 1 Therefore, the lines x = 1 and x = -1 are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure. y=2 x=-1j I = x|| 8.
4x(x² – 1) – 2x² - 2x
(x² – 1)2
E. f'(x) =
%3D
%3D
Since f'(x) > 0 when x < 0 (x * -1) and f'(x) < 0 when x > 0 (x + 1), f is increasing on the intervals (-o, -1) and
and decreasing on the intervals (0, 1) and
F. The only critical number is x = 0. Since f' changes from positive to negative at
is a local maximum by the First Derivative Test.
G. f"(x) = -4(x² – 1)² + 4x · 2(x² - 1)2x
(x² – 1)ª
Since the simplified numerator,
> O for all x, we have the following.
f"(x) > 0 + x2 - 1 > 0 +
|x| >
and f"(x) < 0 +
|x| <
Thus. the curve is concave upward on the intervals (-0, -1) and
. It has no point of inflection since 1 and
and concave downward on
are not in the domain of f.
H. Using the information in parts (E)-(G), we finish the sketch in the following figure.
y= 2
x=-1|
Transcribed Image Text:4x(x² – 1) – 2x² - 2x (x² – 1)2 E. f'(x) = %3D %3D Since f'(x) > 0 when x < 0 (x * -1) and f'(x) < 0 when x > 0 (x + 1), f is increasing on the intervals (-o, -1) and and decreasing on the intervals (0, 1) and F. The only critical number is x = 0. Since f' changes from positive to negative at is a local maximum by the First Derivative Test. G. f"(x) = -4(x² – 1)² + 4x · 2(x² - 1)2x (x² – 1)ª Since the simplified numerator, > O for all x, we have the following. f"(x) > 0 + x2 - 1 > 0 + |x| > and f"(x) < 0 + |x| < Thus. the curve is concave upward on the intervals (-0, -1) and . It has no point of inflection since 1 and and concave downward on are not in the domain of f. H. Using the information in parts (E)-(G), we finish the sketch in the following figure. y= 2 x=-1|
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,