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Question Video: Finding the Domain of a Rational Function with an Absolute Value under a Square Root in the Denominator Mathematics

Find the domain of the function 𝑓(π‘₯) = 9/√(9 βˆ’ |π‘₯ + 5|) .

04:16

Video Transcript

Find the domain of the function 𝑓 of π‘₯ equals nine over the square root of nine minus the absolute value of π‘₯ plus five.

We recall that the domain of any function 𝑓 of π‘₯ is the set of all possible values of π‘₯ such that 𝑓 of π‘₯ is defined. In particular, we are looking for any restriction imposed by part of the function. The function we have been given is not a simple one. It is a composition of an absolute value function inside a square root function in the denominator of a rational function. Let’s consider what we know about the domain of each type of function.

First, we recall the domain of the simple rational function 𝑦 equals one over π‘₯ is the set of all real numbers minus the set containing zero. For a rational function containing another function in the denominator, we need to use a more general definition for the domain of one over 𝑔 of π‘₯. The domain of the composite function 𝑦 equals one over 𝑔 of π‘₯, or in this case nine over 𝑔 of π‘₯, can be identified by finding the values of π‘₯ such that 𝑔 of π‘₯ equals zero and subtracting those values from the set of real numbers.

The domain of the square root function is the set of real numbers greater than or equal to zero, in other words, all real numbers in the left-closed, right-open interval from zero to ∞. For a square root containing a function, we will need to use the more general definition for the domain of the square root of β„Ž of π‘₯. The domain of the composite function square root of β„Ž of π‘₯ can be identified by finding the values of π‘₯ satisfying β„Ž of π‘₯ greater than or equal to zero. The domain of an absolute value function is the set of all real numbers. So the absolute value alone imposes no restrictions on the domain.

Having reviewed the domain of each type of function, we see that the denominator of the rational function and the square root will both place restrictions on the domain of 𝑓 of π‘₯. We begin by identifying the denominator of our function as 𝑔 of π‘₯. Whatever values of π‘₯ that cause 𝑔 of π‘₯ to equal zero will be removed from our domain. Therefore, we must ensure that the square root of nine minus the absolute value of π‘₯ plus five does not equal zero. The expression under the square root is our β„Ž of π‘₯ function. And the restriction placed on β„Ž of π‘₯ is that it must be greater than or equal to zero.

We will now go back to consider the implications of the first restriction. We recall that if a square root cannot equal zero, then the expression under the square root cannot equal zero either. Therefore, nine minus the absolute value of π‘₯ plus five cannot equal zero. When we combine these two restrictions, we find that nine minus the absolute value of π‘₯ plus five must be strictly greater than zero. We determine that the solutions to this inequality compose the set of values for which 𝑓 of π‘₯ is defined.

To solve this inequality, we need to first get the absolute value isolated. We can do this by adding the absolute value of π‘₯ plus five to both sides of the inequality. Then, we have nine is greater than the absolute value of π‘₯ plus five. We can rewrite this as the absolute value of π‘₯ plus five is less than nine. This absolute value statement means that all values of π‘₯ plus five must be less than nine but also greater than negative nine.

Finally, we solve for π‘₯ by subtracting five from all parts of the compound inequality. We find that π‘₯ is greater than negative 14 and less than four. We may also write our answer in interval notation, specifically the open interval from negative 14 to positive four.

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