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Question 50932: what is the domain of y=sq.root of (9-x^2) i htink the answer may be x is less than or equal to 3 that is what i got when i worked it out
Found 2 solutions by stanbon, AnlytcPhil:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! what is the domain of y=sq.root of (9-x^2)
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9-x^2 must be greater than or equal to zero.
9-x^2>=0
(3-x)(3+x)>=0
Put 3 and -3 on a number line and figure out where
the rest of the solution set lies.
There are three intervals on the line. Check each interval, as follows:
Interval I is (-infinity,-3), so check with x=-10.
You get (3-(-10))(3+(-10)) which is <0 so no solutions there.
Interval II is (-3,3), so check x=0.
(3-0)(3+0) is >0 so this interval is part of the solution set.
Interval III is (3,+infinity), so check with x=10.
You get (3-10)(3+10) which is <0 so no solutions there.
SOLUTION:
All x values between -3 and +3 including 3 and -3: or -3<=x<=0
Cheers,
Stan H.
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
What is the domain of y=sq.root of (9-x^2)
I think the answer may be x is less than
or equal to 3 that is what I got when I
worked it out
No that is incorrect.
For
____
y = √9-x²
we require that what's under the radical
not be negative, that is, greater than
or equal to 0.
9 - x² > 0
Factoring,
(3 - x)(3 + x) > 0
We find the critical numbers by setting
each factor = 0
3 - x = 0 gives critical number 3
3 + x = 0 gives critical number -3
Therefore we mark these critical values
on a number line
-----------o-----------------o----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Then we test any value of x less than -3,
to see whether it satisfies the inequality
(3 - x)(3 + x) > 0
or not. Let's pick -4 to test
( 3 - (-4) )( 3 + (-4) ) < 0
(3 + 4)(3 - 4) < 0
(7)(-1) < 0
-7 < 0
This is false. So we do not shade the
number line to the left of -3.
-----------o-----------------o----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Next we test any value of x between -3 and 3,
to see whether it satisfies the inequality
(3 - x)(3 + x) > 0
or not. Let's pick 0 to test
(3 - 0)(3 + 0) > 0
(3)(3) > 0
9 > 0
This is true. So we DO shade the
number line between -3 and 3.
-----------o=================o----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Now we test any value of x greater than 3,
to see whether it satisfies the inequality
(3 - x)(3 + x) > 0
or not. Let's pick 4 to test
(3 - 4)(3 + 4) > 0
(-1)(7) > 0
-7 > 0
This is false. So we do not shade the
number line to the right of 3.
-----------o=================o----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Now we test the critical numbers themselves
to see whether they satisfy the inequality
(3 - x)(3 + x) > 0
or not.
Testing x = -3, we get 0 > 0 which is true,
so we shade -3
-----------l=================o----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Testing x = 3, we get 0 > 0 which is true,
so we shade 3
-----------l=================l----------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
So the answer is -3 < x < 3
Which in set-builder notation is {x|-3 < x < 3}
or in interval notation [-3, -3]
Edwin
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