finding delta

[architectofthefuture]

I have a problem where i have to find delta. here are some steps to solving it:

sqrt(3.5) < |x| < sqrt(4.5) --> -sqrt(4.5) < x < -sqrt(3.5)

what was done here that that values were switched on either side, and that the signs changed to negative?

 

[pka]

I have a problem where i have to find delta. here are some steps to solving it:
sqrt(3.5) < |x| < sqrt(4.5) --> -sqrt(4.5) < x < -sqrt(3.5)
what was done here that that values were switched on either side, and that the signs changed to negative?

Here is an easy example.
Say we knows that \(\displaystyle 4<x^2<9\) then isn't easy to see that \(\displaystyle -3<x<-2\text{ or }2<x<3~?\)
You see \(\displaystyle 4<x^2<9\) is the same as \(\displaystyle 2<|x|<3\), just take the square root.

 

[architectofthefuture]

Here is an easy example.
Say we knows that \(\displaystyle 4<x^2<9\) then isn't easy to see that \(\displaystyle -3<x<-2\text{ or }2<x<3~?\)
You see \(\displaystyle 4<x^2<9\) is the same as \(\displaystyle 2<|x|<3\), just take the square root.

i understand how to take the square root, but why would you want to switch sides with the numbers, and make them negative? why not just leave it as \(\displaystyle 2<|x|<3\)?

i just left mine as sqrt(3.5) < x < sqrt(4.5), and got the wrong answer...

 

[pka]

i understand how to take the square root, but why would you want to switch sides with the numbers, and make them negative? why not just leave it as \(\displaystyle 2<|x|<3\)?
i just left mine as sqrt(3.5) < x < sqrt(4.5), and got the wrong answer...

You really have us at a disadvantage. We do not really the problem.
If it were to solve \(\displaystyle 4<x^2<9\) or even \(\displaystyle 2<|x|<3\) and you gave \(\displaystyle 2<x<3\) as the answer it is not complete, thus wrong.

Why not post the entire problem in exact wording? Then maybe we can help.

 

[architectofthefuture]

You really have us at a disadvantage. We do not really the problem.
If it were to solve \(\displaystyle 4<x^2<9\) or even \(\displaystyle 2<|x|<3\) and you gave \(\displaystyle 2<x<3\) as the answer it is not complete, thus wrong.

Why not post the entire problem in exact wording? Then maybe we can help.

sure!

they want you to find epsilon and delta. f(x) = x^2, L = 4, x(subscript 0) = -2, and e = 0.5

 

[pka]

find epsilon and delta. f(x) = x^2, L = 4, x(subscript 0) = -2, and e = 0.5

Find \(\displaystyle \delta>0\) such that \(\displaystyle |x+2|<\delta\implies|x^2-4|<0.5\).
If \(\displaystyle |x+2|<1\) then \(\displaystyle -3<x<-1\) so \(\displaystyle -5<x-2<-3\) or \(\displaystyle |x-2|<5\).

Take \(\displaystyle 0<\delta<0.1\) then \(\displaystyle |x^2-4|=|x+2||x-2|<(0.1)(5)=0.5\).