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

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Discussion Overview

The discussion revolves around finding the domain and range of the function $$y=\sqrt{x^2+y^2}$$. Participants explore the implications of the equation and its geometric interpretation.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant notes that since $$y$$ must be non-negative, the range could be from zero to infinity.
  • Another participant confirms that $$x$$ can be any real number, suggesting that it must be zero to satisfy the equation.
  • A later reply indicates that the locus of points satisfying the equation corresponds to the y-axis, specifically the non-negative y-axis.
  • There is a mention of an alternative expression for the x-axis, $$x=\sqrt{x^2+y^2}$$, but its implications are not fully explored.

Areas of Agreement / Disagreement

Participants generally agree that $$y$$ is non-negative and that $$x$$ must be zero, leading to the conclusion that the points lie on the non-negative y-axis. However, the implications of the alternative expression for the x-axis remain less clear.

Contextual Notes

The discussion does not resolve the implications of the alternative expression for the x-axis, and there may be additional assumptions regarding the interpretation of the function that are not fully articulated.

Maged Saeed
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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

$$y=\sqrt{x^2+y^2}$$
 
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Well, first we see that:

$$0\le y$$

Square both sides...what do you find?
 
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 ?
 
Maged Saeed said:
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 $x$?
 
MarkFL said:
$y$ 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
 
Last edited:
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?
 
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

$$x=\sqrt{x^2+y^2}$$
 
Last edited:
Well, it's only the non-negative $y$-axis, isn't it?
 
Yes, It is
 

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