trouble spots for the domain may occur where the denominator is ? or where the expression under a square root symbol is negative

Question

The denominator of f(x) = sq root x-5 / x-19

the denominator is 0 when x=

Answer 1

the value of "x" that makes the denominator equal to 0 is a trouble spot because it's not possible to divide by 0 so it makes the function undefined.

in this case you can find that by setting the denominator equal to 0

x - 19 = 0

x = 19

meaning that x = 19 is a trouble spot for this function

Answer 2

f(x) = sqrx -5/x-19

may mean

(sqr(x)-5)/(x-19) if so x can't =19 as it makes the denominator zero

and x can't be <0 as it makes the square root imaginary

domain is the interval [0,19)U(19, infinity)

but it may also mean

f(x) = sqrx -(5/x) - 19

then x can't =0 as it make 5/x's denominator = 0

so domain becomes x>0 or the interval (0, infinity)

or it may also mean

f(x) = sqr(x-5)/(x-19)

then x can't be <5 or 19

domain is then (5,19)U(19, infinity)

or it may mean

[sqr(x-5)/x] -19 then x can't be less than 5

domain then is (5, infinity) or x>5

odds are you probably want the 1st solution above

Answer 3

Case: 1) f(x) =sqrt(x-5)/(x-19)  (If square root is only for numerator)

ANSWER: The function F tells us to take the square root of x-5 and divide this result by x-19.

This requires that for the numerator => x-5≥0,so x≥5 , and also that for the denominator x-19≠0, so x≠19.

Combining these two restrictions, the domain of F is [5,19)U(19,infinity)

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Case: 2) f(x)=sqrt(x-5/x-19) (If the square root is for both numerator and denominator)

ANSWER:

The function F tells us to take the square root of x-5 and divide this result by square root of x-19.

This requires that for the numerator => x-5≥0,so x≥5 , and also that  for the denominator x-19>0, so x>19.

Combining these two restrictions, the domain of F is (19,infinity)

Answer 4

You're on the right track. Trouble spots for the domain occur in both places. Because nothing can be divided by 0, the first trouble spot you want to look for is when the denominator is equal to 0 (as you suggested). Therefore, one trouble spot occurs when x - 19 = 0. Simplifying this by isolating the 'x' will give you one value that 'x' cannot equal. 

The second troublespot occurs for values of 'x' that makes the following expression an imaginary number sqrt(x-5). If the number under the sqrt is negative, the number is imaginary and is no longer in the domain of the functino. Therefore, determining the critical point at which this becomes negative is crucial. By setting the expression x - 5 =0, you can determine the point at which it will become negative. 

Note: it will be helpful to use interval notation when writing the domain.

Answer: (5,19)U(19,infinity)