What is the Algebraic Method for Finding Delta?

[Saladsamurai]

Given f(x) = x², L = 4, x_o = -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 - x_o| < (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?

 

[lanedance]

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

-0.5 < x² - 4 < 0.5

3.5 < x² < 4.5

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

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

=>|x - x_o| < (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?

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?

 

[njama]

You did a mistake.

It should be:

√3.5 < |x| < √4.5

 

[Saladsamurai]

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

Okay then!

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

Thanks :)

 

[Mark44]

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.