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Question 84464: I do not fully understand what domain is and I am having trouble with the following problem:
y= square root of(9-x^2) + square root of(1-x^2)
The answer on my key is domain=[1,3] but I do not understand why. Can you please explain?
Found 3 solutions by venugopalramana, ikleyn, MathTherapy:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I do not fully understand what domain is and I am having trouble with the following problem:
y= square root of(9-x^2) + square root of(1-x^2)
The answer on my key is domain=[1,3] but I do not understand why. Can you please explain?
HERE X IS CALLED INDEPENDENT VARIABLE AND Y DEPENDENT VARIABLE.
OUR UNIVERSE IS REAL NUMBERS FOR THIS PURPOSE.
NOW , THIS DOES NOT MEAN THAT X CAN TAKE ANY VALUE IN REAL NUMBERS.
IT CAN TAKE ONLY SUCH VALUES WHICH MAKE THE GIVEN FUNCTION F(X) OR Y IN SHORT MEANINGFUL.THAT IS A MATHEMATICALLY ALLOWED OPERATION.
CERTAIN OPERATIONS LIKE DIVISION BY ZERO , SQUARE ROOT OF A NEGATIVE NUMBER,LOG OF ZERO OR NEGATIVE NUMBER ARE NOT DEFINED.EITHER THEY ARE CALLED INFINITY(PLUS OR MINUS)OR IMAGINARY.
SO WE SAY X CAN NOT TAKE SUCH VALUES WHICH REQUIRES US TO PERFORM SUCH OPERATIONS.
X CAN TAKE VALUES OF ALL OTHER REAL NUMBERS.THE SET OF THESE REAL NUMBERS IS CALLED DOMAIN OF THE FUNCTION.
IF YOU HAVE
Y=1/(X-3)...DOMAIN IS ALL REAL VALUES EXCEPT 3 AS IT WILL LEAD TO DIVISION BY ZERO .
IN YOUR EXAMPLE THE FIRST TERM BECOMES MEANING LESS IF
9-X^2<0.HENCE
9-X^2>=0
9>=X^2
X^2<=9
|X|<=3......THAT IS X SHOULD BE BETWEEN -3 AND +3
WE SHOW IT AS X = [-3,3]......[..]..BRACKETS INDICATE X CAN EQUAL THE BORDER VALUES
THAT IS X=3 AND X=-3 ARE ALLOWED.
NOW THERE IS A SECOND TERM.SQRT.(1-X^2)
SIMILAR ARGUMENT LEADS US TO CONCLUDE X = [-1,1]
THE COMBINATION OF THE I AND II CONDITIONS MAKE US TO CONCLUDE THAT X = [-1,1]
YOUR ANSWER IS [1,3]WHICH IS NOT CORRECT.
ON THE OTHER HAND,IF SECOND TERM IS SQ.RT(X^2-1)....THEN |X| >=|
X<=-1......OR............X>=1...COMBINING THIS WITH I CONDITION , WE GET
X=[1,3] AND [-3,-1]
Answer by ikleyn(54034)  (Show Source):
You can put this solution on YOUR website! .
I do not fully understand what domain is and I am having trouble with the following problem:
y= square root of(9-x^2) + square root of(1-x^2)
The answer on my key is domain=[1,3] but I do not understand why. Can you please explain?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The answer in your answer key is incorrect.
The solution in the post by @venugopalramana is incorrect, too.
I came to bring a correct solution.
In order for the given function 'y' to be defined, the expressions under the square roots both must be non-negative.
So, both inequalities 9-x^2 >= 0 and 1-x^2 >= 0 must hold.
Equivalently, both inequalities x^2 <= 9 and x^2 <= 1 must be satisfied at the same time.
It is possible if and only if x^2 <= 1, or -1 <= x <= 1.
This is the ANSWER to this problem: the domain of the given function is -1 <= x <= 1, or the interval [-1,1].
Solved correctly.
Answer by MathTherapy(10865)  (Show Source):
You can put this solution on YOUR website!
I do not fully understand what domain is and I am having trouble with the following problem:
y= square root of(9-x^2) + square root of(1-x^2)
The answer on my key is domain=[1,3] but I do not understand why. Can you please explain?
*****************************
Domain indicates the value(s) of x.
The smaller of the 2 SQUARE ROOTS is  .
As you may know, a RADICAND CANNOT be negative (< 0). So, we SET the 
This gives us: 
 ---- Subtracting 1 for each side
 -- Dividing each sides by - 1
 ------ Notice that the INEQUALITY SIGN changed, since the INEQUALTY was divided by a negative expression
- 1 and 1 are considered the CRITICAL POINTS for x, the domain. And, since we have an equals sign
attached to the INEQUALITY, the domain, or x, will contain the values between - 1 and 1, INCLUSIVE.
So, in interval notation, the domain, or x, is [- 1, 1]
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