SOLUTION: Please find the domain of H(x)=√(25-x₂) Thank You

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Question 146271:
Please find the domain of H(x)=√(25-x₂)
Thank You

Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
I'm assuming you mean,

The domain is every x where H(x) makes sense or is defined.
The square root only makes sense or is defined when the argument,
that's the thing under the square root sign, is greater than or equal to zero.


Taking the square root of both sides, we get two answers.
1.
2.
From 2,

So the domain of H(x) is

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Please find the domain of
Thank You

Note that  means the same as ,
with the index  understood.

If a radical has an even index, that is:

, such as , , , etc.

then what is under the radical must not be negative.

Non-negative means "positive or 0", and "positive" means "greater that 0",
therefore what is under the radical must be greater than or equal to 0.

So we make an inequality that says  is greater than or equal to 0.

 

We find all critical values by setting the left side = 0:

                              
                               
                              ,

So we draw a number line and mark those numbers

<----------o-----------------------------o--------->
 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

Next we choose a number left of -5, say -6.
Substitute it in the original inequality:






This is false so we do not shade to the left of -5.
We still have this number line. 

<----------o-----------------------------o--------->
 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

Next we choose a number right of -5 but left of 5, say 0.
Substitute it in the original inequality:






This is true so we do shade to the line between -5 and +5,
like this:

<----------o=============================o--------->
 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

Next we choose a number right of 5, say 6.
Substitute it in the original inequality:






This is false so we do not shade to the right of 5.  

Next we must test the critical values in the original inequality

Testing 






This is true so we darken the circle at -5.

<----------@=============================o--------->
 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

Testing 






This is also true so we also darken the circle at 5.

<----------@=============================@--------->
 -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

The interval notation for that is the abbreviation 
for the above number line graph:  

               [-5,-5]

Edwin

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