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How do you find the domain and range of #f(x) = sqrt(x-9)#?

Answer 1

Julia Adams

Domain #{x:x>=9}#
Range #{y:y>=0}#

The value of #x# cannot be less than 9, otherwise, the function would be complex ( not a real function).

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Answer 2

Jace Anderson

To find the domain of ( f(x) = \sqrt{x - 9} ), set the expression inside the square root greater than or equal to zero and solve for ( x ):

[ x - 9 \geq 0 ]

[ x \geq 9 ]

So, the domain of the function is all real numbers greater than or equal to 9, expressed as ( {x \in \mathbb{R} : x \geq 9} ).

To find the range, note that the square root function returns non-negative values. Since the expression inside the square root must be non-negative, the minimum value of ( f(x) ) occurs when ( x - 9 = 0 ), which gives ( f(9) = 0 ). Therefore, the range of the function is all real numbers greater than or equal to 0, expressed as ( {f(x) \in \mathbb{R} : f(x) \geq 0} ).

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Answer from HIX Tutor

When evaluating a one-sided 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.