# Uniform Continuity

• Mar 1st 2010, 03:43 PM
summerset353
Uniform Continuity
prove g(x)=x^2 + 2x - 5 is uniformly continuous on [0,3].

proof: Let e>0 and choose delta=e/3. If |y-x|< delta and x,y e[0,3] then |g(y)-g(x)| = |(y^2 +2y -5)-(x^2 +2x -5)|=|y^2 -x^2 +2y-2x|=
|y(y+2) - x(x+2)| < 3|y-x| < (3)delta = e.

Did i do the proof right? I get confused when an interval is given such as [0,3]. What is the difference if the interval was [0,3)? How does that effect the proof?
• Mar 1st 2010, 04:45 PM
Black
I don't think that inequality is going to work (take x=2, y=3).

Ok, so we have

$|y^2+2y-5-x^2-2x+5|=|y^2-x^2+2y-2x|=|(y+x+2)(y-x)|.$

Since 8 is the maximum number you can get out of y+x+2 on the interval [0,3], let $\delta=\varepsilon/8$.
• Mar 1st 2010, 06:03 PM
Krizalid
Quote:

Originally Posted by summerset353
Did i do the proof right? I get confused when an interval is given such as [0,3]. What is the difference if the interval was [0,3)? How does that effect the proof?

being continuous on a compact set, then it's uniformly continuous, and obviously uniformly continuous on $[0,3)$ since this set is a subset of $[0,3].$