What reasoning error am I making here?

  • Thread starter Thread starter blunkblot
  • Start date Start date
  • Tags Tags
    Error
Click For Summary

Discussion Overview

The discussion revolves around finding the upper bound M for the expression \(\left|\frac{x+2}{x}-5\right|\) over the interval \(x \in (1, 4)\). Participants explore different approaches to determine M, including evaluating the function at specific points and manipulating the expression algebraically.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially proposes to find M by evaluating the expression at the endpoints of the interval, suggesting that the maximum value is 11.
  • Another participant points out that the function \(\frac{x+2}{x}\) is decreasing on the interval and that the maximum occurs at \(x = 1\), leading to an upper bound of 8.
  • A different participant mentions that the book provides an upper bound of 18 by merging -5 into the fraction and maximizing it, questioning the validity of this method compared to their own approach.
  • Some participants express confusion about how the book arrived at 18, with one stating that while it is technically correct, it seems to be an unexpected result.
  • Another participant defends the method of maximizing the modified fraction, suggesting that finding any upper bound is valid, while acknowledging that the original method is also acceptable.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to find the upper bound M, with multiple competing views on the validity of different approaches and the correctness of the book's answer.

Contextual Notes

The discussion highlights the complexity of determining upper bounds and the potential for different interpretations of the problem, as well as the impact of algebraic manipulation on the results.

blunkblot
Messages
7
Reaction score
0
I apologize for a misclick double post:

Mentor action: double posted thread merged. No harm, no foul.[/color]
 
Last edited by a moderator:
Mathematics news on Phys.org
Apparently I cannot read. I figure this is the wrong thread. Please move or kill as needed :(

You need to find M where

[tex]\left|\frac{x+2}{x}-5\right| \le M, x \in (1, 4)[/tex]

To do this, I put 4 for the top x, 1 for the bottom x, to give the greatest quotient possible, plus |-5|, which finally gives 11. The correct answer is to combine 5 into the fraction with x's, and that gives a different answer.

What did I overlook by NOT combining them?

You can view the entry on google books preview here:

http://books.google.com/books?id=10... by first writing the given function"&f=false

Gratitude for enlightenment.
 
Last edited:
Look at the range of values for (x + 2)/x for x in [1, 4]. This function is defined at all points in this interval, and is decreasing on this interval. The largest value of (x + 2)/x on this interval comes when x = 1.
 
Mark44 said:
Look at the range of values for (x + 2)/x for x in [1, 4]. This function is defined at all points in this interval, and is decreasing on this interval. The largest value of (x + 2)/x on this interval comes when x = 1.

This I understand. If you substitute x with 1, you can get | 3 | + | 5 | = 8 as an upper bound.

The solution as linked actually has M = 18. What makes 18 a better upper bound than 8? Or, is the book wrong?

Thank you
 
It shouldn't be a better upper bound. I don't know how the book got 18, but it is a M that works, so it's technically correct as well.
 
The book got it by merging the -5 into the fraction, giving

[tex]\left| \frac{-4x + 2}{x} \right|[/tex]

Then it maximizes this fraction within the allowed values for x by using 4 for the top x, 1 for the bottom x, and applying absolute value to each term since |-4x + 2| <= |4x| + |2|

This gives you 18. What makes this a good method?

It's really confusing, because by turning 5 into 5x/x, and plugging DIFFERENT values into the top x and bottom x, it greatly magnifies the result, apparently without good reason.
 
Magnifying the result is a good enough reason, since you're looking for any upper bound. But your way is another perfectly fine way to find an upper bound.
 
Makes sense, thank you.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
6K
Replies
37
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
17
Views
2K
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K