If |x-1| < 1 then Prove |x^2 -4x + 3| < 3.
The Attempt at a Solution
proof: Assume |x-1| < 1. Then X has to be between 0 and 2.Because X has to be between 0 and 2 then |x-3|<3,and |x-1||x-3|<3 by multiplication of inequalities.
|x-1||x-3|=|x^2 - 4x + 3| by distribution. Thus, |x^2 -4x + 3| < 3. (QED)
I was wondering if this is sufficient. I was a little unsure if what I did in the second sentence, and the start of the third was ok