How do you find the domain of #f @ g# given #f(x)= sqrt(x-1)# and #g(x)= 1/(x-7)#?

Answer 1

#x in (7, 8]#

#f(1/(x-7)) = sqrt{1/(x-7) - 1}#
#x ne 7 and 1/(x-7) - 1 >= 0#
#frac{1 - x + 7}{x - 7} >= 0#
#(x - 7)(x - 8) <= 0#
#7 <= x <= 8#
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Answer 2

To find the domain of ( f @ g ) where ( f(x) = \sqrt{x - 1} ) and ( g(x) = \frac{1}{x - 7} ), we need to consider the domain restrictions of both functions. The domain of ( f @ g ) is the set of all ( x ) values for which both ( f(x) ) and ( g(x) ) are defined. Since ( f(x) ) involves taking the square root of ( x - 1 ), the expression under the square root must be non-negative, i.e., ( x - 1 \geq 0 ). This implies ( x \geq 1 ). Additionally, since ( g(x) ) involves division by ( x - 7 ), the denominator cannot be zero, so ( x - 7 \neq 0 ), or ( x \neq 7 ). Combining these conditions, the domain of ( f @ g ) is ( x \geq 1 ) and ( x \neq 7 ).

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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.

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.

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.

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.

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