vergent, then so is Ja g(x) dx. (2) If fa g(x) dx is divergent, then so is f f (x) dx. To work with this theorem, we need to practice working with inequalities. In particular, we would like to develop a technique for constructing and justifying useful inequalities. Exercise 1. Let's review some very VERY basic inequality manipulation. (1) Recall that 2 < 3. So which is larger, 1/2 or 1/3? 글>ㅎ (2) If 0 < x < y, which is larger, 1/x or 1/y? How do you justify this? > because 2 <3 (a) If 0 < x > 1/3 y is bigger than x so both is posi (b) Divide both sides of x
vergent, then so is Ja g(x) dx. (2) If fa g(x) dx is divergent, then so is f f (x) dx. To work with this theorem, we need to practice working with inequalities. In particular, we would like to develop a technique for constructing and justifying useful inequalities. Exercise 1. Let's review some very VERY basic inequality manipulation. (1) Recall that 2 < 3. So which is larger, 1/2 or 1/3? 글>ㅎ (2) If 0 < x < y, which is larger, 1/x or 1/y? How do you justify this? > because 2 <3 (a) If 0 < x > 1/3 y is bigger than x so both is posi (b) Divide both sides of x
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Excerice 2 how do the inequalities in part 1
![vergent, then so is Ja g(x) dx.
(2) If fa g(x) dx is divergent, then so is f f (x) dx.
To work with this theorem, we need to practice working with inequalities. In particular, we
would like to develop a technique for constructing and justifying useful inequalities.
Exercise 1. Let's review some very VERY basic inequality manipulation.
(1) Recall that 2 < 3. So which is larger, 1/2 or 1/3?
글>ㅎ
(2) If 0 < x < y, which is larger, 1/x or 1/y? How do you justify this? >
because
2 <3
(a) If 0 < x <y, then ry is positive. Why?ble x is bigger than 0
but 1½ 2 >> 1/3
y is bigger than x so both is posi
(b) Divide both sides of x <y by xy, and simplify. This does not change the direction of
the inequality, because xy is positive.
Exercise 2. For 2< x, it is true that
*
x y<x
(1.1)
1
1+
•
X
X
x4
Let's work through the proof to see why (1.1) is true. We will start with the assumption that
2≤ x. We then use a more simple, true inequality, along with a step-by-step process, to build up
to (1.1).
(1) To start, we know that when 2x we must have 0 < (Why? Convince yourself!)
(2) adding 1 to both sides of 0 < we obtain
L
(3) after taking the square root of both sides of your answer in (2.) (since they're positive
(why?)), we have 1 < √itu
(4) Now multiply both sides of your answer in (3.) by 1, which is positive (why?) to obtain
the desired result
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd244322d-d59e-4d22-bc6d-7acbb17461df%2Fd88c9e92-dc86-4184-affd-591080dbd0f5%2Fquftye_processed.jpeg&w=3840&q=75)
Transcribed Image Text:vergent, then so is Ja g(x) dx.
(2) If fa g(x) dx is divergent, then so is f f (x) dx.
To work with this theorem, we need to practice working with inequalities. In particular, we
would like to develop a technique for constructing and justifying useful inequalities.
Exercise 1. Let's review some very VERY basic inequality manipulation.
(1) Recall that 2 < 3. So which is larger, 1/2 or 1/3?
글>ㅎ
(2) If 0 < x < y, which is larger, 1/x or 1/y? How do you justify this? >
because
2 <3
(a) If 0 < x <y, then ry is positive. Why?ble x is bigger than 0
but 1½ 2 >> 1/3
y is bigger than x so both is posi
(b) Divide both sides of x <y by xy, and simplify. This does not change the direction of
the inequality, because xy is positive.
Exercise 2. For 2< x, it is true that
*
x y<x
(1.1)
1
1+
•
X
X
x4
Let's work through the proof to see why (1.1) is true. We will start with the assumption that
2≤ x. We then use a more simple, true inequality, along with a step-by-step process, to build up
to (1.1).
(1) To start, we know that when 2x we must have 0 < (Why? Convince yourself!)
(2) adding 1 to both sides of 0 < we obtain
L
(3) after taking the square root of both sides of your answer in (2.) (since they're positive
(why?)), we have 1 < √itu
(4) Now multiply both sides of your answer in (3.) by 1, which is positive (why?) to obtain
the desired result
1
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