# What reasoning error am I making here?

1. Jun 7, 2010

### blunkblot

I apologize for a misclick double post:

Mentor action: double posted thread merged. No harm, no foul.

Last edited by a moderator: Jun 7, 2010
2. Jun 7, 2010

### blunkblot

Apparently I cannot read. I figure this is the wrong thread. Please move or kill as needed :(

You need to find M where

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

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: Jun 7, 2010
3. Jun 7, 2010

### Staff: Mentor

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.

4. Jun 7, 2010

### blunkblot

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

5. Jun 7, 2010

### Tedjn

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.

6. Jun 7, 2010

### blunkblot

The book got it by merging the -5 into the fraction, giving

$$\left| \frac{-4x + 2}{x} \right|$$

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.

7. Jun 7, 2010

### Tedjn

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.

8. Jun 10, 2010

### blunkblot

Makes sense, thank you.