# Finding delta Algebraically

1. Sep 14, 2009

Given f(x) = x2, L = 4, xo = -2, e = 0.5 find delta.

-0.5 < x2 - 4 < 0.5

3.5 < x2 < 4.5

(3.5)1/2 - (-2) < x - (-2) < (4.5)1/2 - (-2)

=>|x - xo| < (3.5)1/2 - (-2) ~ 3.87

My text says the answer is 0.12 ?

I was convinced that I have been doing these right. Am I?

2. Sep 14, 2009

### lanedance

you have a skipped an important step here, when you take the square root, do you take the positive or negative squareroot.... thoughts?

ie. assuming
a2 < x2 < b2
you need to be careful about whether you treat a & b as positive or negative numbers

As a test try using you delta & see what e it gives you?

As an aside I would tray & start a little more explicit:
so there exists d>0 such that for all |x-x0|< d, |f(x)-f(x0)|<e

Now
|f(x)-f(x0)|= |x^2-x0^2|= |(x+x0)(x-x0)| < e

Any ideas how to re-write this in terms of delta?

Last edited: Sep 15, 2009
3. Sep 15, 2009

### njama

You did a mistake.

It should be:

√3.5 < |x| < √4.5

4. Sep 15, 2009

Okay then!

I am not really sure what you are doing here? Could you please explain?

Thanks :)

5. Sep 15, 2009

### Staff: Mentor

Starting with your inequality 3.5 < x2 < 4.5, it must be either that
1) $\sqrt{3.5}$ < x < $\sqrt{4.5}$, OR
2) $-\sqrt{4.5}$ < x < $-\sqrt{3.5}$

Since x0 in your problem is -2, which of the inequalities above is the one that should be used for this problem? For your limit, x should be reasonably close to x0.