How do you find the domain and range of #sqrt(x^2  9)#?
Domain =
Range =
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To find the domain and range of ( \sqrt{x^2  9} ):

Domain: The expression under the square root, (x^2  9), must be nonnegative to ensure that the square root is defined for real numbers. Thus, we solve the inequality (x^2  9 \geq 0).
[x^2  9 \geq 0]
[(x  3)(x + 3) \geq 0]
The critical points are (x = 3) and (x = 3). We test the intervals created by these points:
 For (x < 3), both factors are negative, so the inequality is positive.
 For (3 < x < 3), the factor (x  3) is negative, while (x + 3) is positive, so the inequality is negative.
 For (x > 3), both factors are positive, so the inequality is positive.
Therefore, the domain is (x \leq 3) or (x \geq 3).

Range: Since the square root of a number is always nonnegative, the range of ( \sqrt{x^2  9} ) is all nonnegative real numbers.
Thus, the domain is (x \leq 3) or (x \geq 3), and the range is all nonnegative real numbers.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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