Finding Domain & Range of $$y=\sqrt{x^2+y^2}$$

[Maged Saeed]

I was thinking about a function whose domain and range are absolutely one point , then I'm stuck in finding the domain and range of the following function .. could anyone help

$$y=\sqrt{x^2+y^2}$$

 

[MarkFL]

Well, first we see that:

$$0\le y$$

Square both sides...what do you find?

 

[Maged Saeed]

Yeah I got it now ..

You mean that y is greater than zero so the range is from zero to infinity

x is all real number !

right ?

 

[MarkFL]

Yeah I got it now ..

You mean that y is greater than zero so the range is from zero to infinity

x is all real number !

right ?

$y$ and be any non-negative real number, including zero. Squaring, we obtain:

$$y^2=x^2+y^2$$

What does this tell you about $x$?

 

[Maged Saeed]

$y$ and be any non-negative real number, including zero. Squaring, we obtain:

$$y^2=x^2+y^2$$

What does this tell you about $x$?

This tell me that x can be any real number and must be zero in order to solve the equation.Thanks Mr MarkFL

 

[MarkFL]

Well, it does tell you $x=0$...and so the original equation becomes:

$$y=\sqrt{y^2}=|y|$$

So, what is the locus of points satisfying the equation?

 

[Maged Saeed]

It will be the equation of the y-axis since x is always zero
the points are (0,y)

I found that it is an alternative way to express the equation of the axis.

Similarly the equation of the x-axis in this way is

$$x=\sqrt{x^2+y^2}$$

 

[MarkFL]

Well, it's only the non-negative $y$-axis, isn't it?

 

[Maged Saeed]

Yes, It is