# Analysis questions involving inequalities

## Homework Statement

Prove that -1< x < 0 implies |x^2 - 2x +1| < 1.25|x-1|

## The Attempt at a Solution

Attempt at 1st question:

|(x-1)(x^2 + x -1)| < 1.25|x-1|
|(x^2 + x -1)| < 1.25
-1.25 < (x^2 + x -1) < 1.25
-0.25 < x^2 + x < 2.25
-0.5 < (x + 0.5)^2 < 2.25 **
0 < (x + 0.5)^2 < 2.25
0 < x + 0.5 < 1.5 **
0 < x < 1
0 > x > -1

I don't have the answers to this in my book, but does this rough work for the proof look ok?
*edit: I just realized that this is completely wrong, that I didn't add correctly on the steps marked by **

## Homework Statement

Prove that -3 <= x <= 2 implies |x^2 + x - 6| <= 6|x-2|

## The Attempt at a Solution

Attempt at 2nd question:
|(x+3)(x-2)| <= 6|x-2|
|x+3| <= 6
-6 <= x + 3 <= 6
-9 <= x <= 3

how does this imply -3 <= x <= 2?? Not sure how to do this one.

Last edited:

Your first proof contains several algebraic mistakes, including in the first step. (x-1)(x^2 + x -1) is not the correct factorization of x^2-2x+1. The correct factorization is (x-1)^2.

Next, your have attempted to show that |x^2 - 2x +1| < 1.25|x-1| implies -1< x < 0. The problem you stated is to show the converse, that -1< x < 0 implies |x^2 - 2x +1| < 1.25|x-1|. In other words, you should try to run all your steps backwards.

This is also an issue for your second problem.

For example, a satisfactory proof for you second problem is as follows:

-3 <= x <= 2
-9 <= x <= 3
-6 <= x + 3 <= 6
|x+3| <= 6
|x-2||x+3| <= 6 |x-2|
|(x-2)(x+3)| <= 6 |x-2|