Question

Question #ce044

Answer 1

It's all about that the square root can't have negative numbers in it.

Explanation:

For example let's consider the function `f(x)=sqrt(x)`

in this function you can't plug in for example `-7` because then you get

`f(-7)=sqrt(-7)` and the `sqrt(-7)` is not a real number it's an imaginary number (learned in complex analysis)

That's why the function `f(x)=sqrt(x)` has a domain `D=x>=0`

Whereas the function `f(x)=sqrt(-x)` has a domain `D=x<=0`

Because if you plug in a positive number you then get a negative square root which is not accepted.

But if you plug a negative number for example `-4` you get

`f(-4)=sqrt(-(-4))=sqrt(4)=2` which is accepted.

I hope I helped you.

Answer 2

As we are looking at a real based function that is:

`y=sqrt(x) \ \` where `x,y in RR`

Then the above observation is absolutely correct, that is because the square root of any positive number is itself positive, and the square root of a negative number does not exist (for `x,y in RR)`. As observed by the graph of the function:

graph{sqrt(x) [-10, 10, -5, 5]}

Note: The observation that;

"since `sqrt(x)` can be both `+x` and`-x` "

is incorrect and confused.

I assume that you actually meant to say that

"since `sqrt(x)` can be both positive and negative "

As explained above `x ge 0`, so we must exclude `x lt0` from the domain of the function, which then gives a range `y ge 0`

The confusion is easily understood if we consider some examples:

`sqrt(0) = 0, sqrt(1)=1, sqrt(4)=3, sqrt(-1) notin RR`

However, consider the equation:

`x^2 = 4`

From which we conclude that there are two solutions:

`x = +-sqrt(2)`

Here we introduce the `+-` because of the square, and not because of the square root! This is because:

`(-2)^2 = (-2)(-2) = 4`
`(2)^2 \ \ \ \ \ = (2)(2) \ \ \ \ \ \ \ \ \ = 4`

But note that in both cases the value of `sqrt(4)` is `2`, and not `+-2`

However, if we consider the equation:

`y^2=x`

This would lead to possible solutions `x=+- sqrt(x)`; as

`(sqrt(x))(sqrt(x)) = x` and `(-sqrt(x))(-sqrt(x))=x`

And if we look at the graph of `y^2=x` we see that there are two branches:

graph{y^2=x [-10, 10, -5, 5]}

but in both case `sqrt(x) ge 0`