Find an upper bound to the limit of a function

Click For Summary

Homework Help Overview

The problem involves finding an upper bound M for the function f(x) = |(x-2) / (x+(1/2))| under the condition that |x+1| < 1/4. Participants are exploring the implications of the given inequality and the behavior of the function within the specified range.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to understand the implications of the inequality |x+1| < 1/4 and its relation to the function's behavior. There are questions about whether this condition implies other inequalities, such as |x-1| < 1/4. Additionally, there is inquiry into the meaning of "upper bound M" and how it relates to the function's values.

Discussion Status

The discussion is active, with participants clarifying the conditions of the problem and questioning the definitions involved. Some have provided examples to illustrate the concept of upper bounds, while others are still seeking to understand the implications of the given inequalities.

Contextual Notes

Participants note that the condition |x+1| < 1/4 leads to specific constraints on the values of x, and there is a concern about the denominator approaching -1/2. The discussion reflects a need for clarity on the definitions and implications of the terms used in the problem.

zeion
Messages
455
Reaction score
1

Homework Statement



Find an upper bound M for f(x) = |x-2 / x+(1/2)| if |x+1| < 1/4


Homework Equations





The Attempt at a Solution



I'm confused about this |x+1| < 1/4. Does this mean that |x-1| < 1/4?

|x-2/x+(1/2)| = x-2/(2x+1)/2 = 2(x-2)/(2x+1) = 2x - 4/2x + 1 = x-2/x+(1/2) < M

Given -1/4 < |x+1| < 1/4
-5/4 < x < -3/4
-13/4 < x - 2 < -11/4

Also
-3/4 < x + (1/2) < -1/4
then
-4 < 1/ x + (1/2) < -4/3
3 < x - 2 / x + (1/2) < 1/3 ?
 
Physics news on Phys.org
"I'm confused about this |x+1| < 1/4. Does this mean that |x-1| < 1/4?"

No. Why would it mean that? You have an (x + 1/2) in the denominator. (At least I suppose you do; it should have parentheses around it.) So you don't want x to get too close to -1/2. Doesn't |x+1| < 1/4 help you with that?
 
If |x+1| < 1/4
-1/4 < x+1 < 1/4
-5/4 < x < -3/4
x cannot = -1/2

What does "upper bound M" mean?
 
zeion said:
If |x+1| < 1/4
What does "upper bound M" mean?
A number M such that f(x) <= M for all x in some set. For example, M = 1 is an upper bound for g(x) = sin(x), for all real numbers x. Also, 2 is an upper bound, as is 1.7. 1 is the least upper bound.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
4K
Replies
8
Views
5K
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
11
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K