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

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

the denominator is 0 when x=

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2 Answers

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)


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


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)

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)



I may have interpreted this question wrong as it seems that the denominator is the whole expression sqrt(x-5/x-19). In that case, the domain is determined by what wil make the expression 0 on the denominator, as well as negative. First, again, we set x - 19 = 0. However, it gets a little more tricky when we try to see what makes this function negative. 

For example, when you plug in 6, for 'x', the expression is negative as you have 1/(-13). But when you plug in a number like 4, 4 - 5 produces -1 while 4-19 produces -15 where as -1/-15 is going to give you a positive number. Therefore, by setting x -5 = 0, we find another critical point. Therefore our domain changes a little. Numbers smaller than 5 (even negative numbers) will ultimately give us a positive number under the square root while numbers in between 5 and 19 give us  a negative number under the square root. Therefore your new domain is expressed in interval notation below. 



Case 2:

Whenever you have an irrational function, its domain is determined by the inequality obtained by placing the radicand greater than or equal to zero (=0). Therefore, the domain of this function is given by the solutions of the following inequality: [(x-5)/(x-19)]=0. Solving this algebraic inequality means to find all the x-values that makes the statement positive or equal to zero. Keep in mind that the procedure for solving this algebraic inequality excludes that the denominator is equal to zero. So, the domain will be

]-8,5] U ]19,+8[

where the value 19 doesn't belong to the domain because it makes zero the denominator.