How do you find the domain and range of #sqrt(1sinx) #?
With a radical expression we have to make sure the argument is never negative.
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To find the domain and range of ( \sqrt{1  \sin(x)} ):

Domain: Since the square root function is defined for nonnegative real numbers, we need to find the values of ( x ) for which ( 1  \sin(x) \geq 0 ). This is because the expression inside the square root must be nonnegative. Solve the inequality ( 1  \sin(x) \geq 0 ) to find the domain.

Range: The range of ( \sqrt{1  \sin(x)} ) depends on the values of ( x ) that make ( 1  \sin(x) ) take on different values. Since ( \sin(x) ) varies between 1 and 1, the range of ( 1  \sin(x) ) is from 0 to 2. Therefore, the range of ( \sqrt{1  \sin(x)} ) is from ( \sqrt{0} = 0 ) to ( \sqrt{2} ).
In summary:
 Domain: Find the values of ( x ) such that ( 1  \sin(x) \geq 0 ).
 Range: The range is from 0 to ( \sqrt{2} ).
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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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