Tom K. answered • 10/09/20
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For the square root to exist, 9 - x^2 >= 0, or x^2 <= 9, or - 3 <= x <= 3
Then, sqrt(9 - x^2) , as 0 <= 9 - x^2 <= 9, may range from 0 to 3.
Then, sqrt(9 - x^2) + 3 ranges from 3 to 6.
Then, 3/(sqrt(9-x^2) + 3) ranges from 3/3 to 3/6, or from 1/2 to 1
The domain is [-3, 3]
The range is [1/2, 1]
Tom K.
The square root cannot be both positive and negative, it can only be positive. If it can be both positive and negative, then we are giving y two values for the same x.
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10/09/20
Bobosharif S.
I think x cannot be 0. if x=0, then Sqrt( 9 - x^2 ) cab be -3 ( or +3). In this case denominator equal to zero10/09/20