For finding the roots,

like

root(x^2),

What's the value?

Is that if

x > 1, ans=x

x<1, ans=-x

x=0, ans=0?

But when we do quandratic equations,

(x^2)=4

x =2 or x=-2?

Should we rearrange the steps as

x^2=4

x=2 or -x=2?

It doesn't really matter, since [math]x=-2[/math] and [math]-x=2[/math] are the same equation. To show that this is true, just multiply both sides of the first equation by [math]-1[/math].

However, there is the surds are defined as positive.

Is that only for calculating roots?

Reading your post I would assume that you mean the following

 

First Part

 

[math]f(x) = \sqrt {x^2}[/math]

 

Hence [math]f(x) = x}[/math]

 

Therefore x has the domain of [math](-\infty , \infty )[/math]

 

Second Part

 

The problem is to find the real zeros of a linear function f(x)=0

 

[math]f(x) = x^2 - 4[/math]

 

[math](x -2)(x + 2) = 0[/math]

 

[math]x = 2, x = -2[/math]

if f has the domain you claim, Gauss, then it isn't the identity function, since roots are always taken to be positive.

[math]f(x) = \sqrt {x^2}[/math]

 

Hence [math]f(x) = x}[/math]

 

Could be wrong here' date=' but I think that [math']f(x) = |x|[/math].

Could be wrong here, but I think that [math]f(x) = |x|[/math'].

If I remember, there were two sides to this. I was doing a problem and trying to be a little creative. It asked you to write [math]|x|+|x+3|[/math] but without the absolute values. I knew they wanted me to restrict the domains and such to get three different equations that would all produce the desired result. But aplying the example problems right from the chapter to the homework questions is so boring. So I decided to not read the chapter and try and derive some of the solutions on my own. For this one, I rewrote the equation as [math]\sqrt{x^2}+\sqrt{x^2+3}[/math]. When graphed, made into a table or whatever you want, they always yeild the same solution. So I asked the teacher why that was not an equaly valid answer. She said that I found the technical definition of absolute values, but that writing the equation that way causes problems because people would assume that you should reduce the power and the root to leave you with just [math]x+x+3[/math](or not, but you get the point) which doesn't work. I still don't know exactly what's going on with this stuff. Does anyone know whether mine should be equaly valid or should be discarded?

The function f(x) = x is an identity function.

 

When expressions have more than one operation, we follow rules for the order of operations. PEMDAS

 

1. Parentheses

2. Exponents

3. Multiplication and Division

4. Addition and Subtraction

 

[math]f(x) = \sqrt{x^2}[/math]

[math]f(x) = x^{\frac {2}{2}[/math]

[math]f(x) = x[/math]

 

Case when x = -3

[math]f(x) = \sqrt{-3^2}[/math]

[math]f(x) = -3^{\frac {2}{2}[/math]

[math]f(x) = -3[/math]

 

Case when x = 3

[math]f(x) = \sqrt{3^2}[/math]

[math]f(x) = 3^{\frac {2}{2}[/math]

[math]f(x) = 3[/math]

 

Another way

 

Case when x = -3

 

[math]f(x) = \sqrt{-3^2}[/math]

[math]f(x) = \sqrt{9}[/math]

[math]f(x) = 3[/math]

 

Case when x = 3

 

[math]f(x) = \sqrt{3^2}[/math]

[math]f(x) = \sqrt{9}[/math]

[math]f(x) = 3[/math]

Case when x = -3

 

[math]f(x) = \sqrt{-3^2}[/math]

[math]f(x) = \sqrt{9}[/math]

[math]f(x) = 3[/math]

That's not the same as saying that [math]\sqrt{x^2}[/math] is equal to [math]x[/math]

So, Gauss, f(x)=x and then you show by example that f(-3) is not equal to -3.