### Video Transcript

Find the domain of the function π
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 π of π₯ is the set of all possible values of π₯ such that π 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. Letβs consider what we know about
the domain of each type of function.

First, we recall the domain of the
simple rational function π¦ 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 π of π₯. The domain of the composite
function π¦ equals one over π of π₯, or in this case nine over π of π₯, can be
identified by finding the values of π₯ such that π 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 β. 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 of π₯. The domain of the composite
function square root of β of π₯ can be identified by finding the values of π₯
satisfying β 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 π of π₯. We begin by identifying the
denominator of our function as π of π₯. Whatever values of π₯ that cause π
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 of π₯ function. And the restriction placed on β 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 π 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.