§ 2.3

The Algebra of Functions – Finding the Domain

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.3

Domain of a Function 117

Finding a Function’s DomainIf a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f(x) is a real number.

Exclude from a function’s domain real numbers that cause division by zero.

Exclude from a function’s domain real numbers that result in a square root of a negative number.

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.3

Domain of a Function

• Consider the function

5

1)(

xxf

Because division by 0 is undefined (and not a real number), the denominator, x – 5, cannot be 0. Then x cannot be 5, and 5 is not in the domain of the function.

5 x andnumber real a is | ofDomain xxf

117

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.3

Domain of a Function

• Now consider the function:

7)( xxgThe equation tells us to take the square root of x – 7. Because only nonnegative numbers have square roots that are real numbers, the expression under the square root must be nonnegative. Then x must be greater than or equal to 7.

7 x andnumber real a is | ofDomain xxg

117

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.3

Domain of a Function

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the domain of the function: . 724 2 xxxf

number real a is | ofDomain xxf

Since the function f has no denominator or square root, there are no real numbers that when plugged into the function for x would cause the value of the function to yield something other than a real number. Therefore, the domain is:

This is set notation and it is read: “the set of all x such that x is a real number.”. Using this notation, the rule stating the conditions for x follows the vertical bar which just means “such that.”

118

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.3

Domain of a Function

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the domain of the function: .

The function has no square roots so we don’t have to worry about pursuing that avenue. However the function does have x in two different denominators. Therefore I do the following:

xx

xf

5

6

4

2

118

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.3

Domain of a Function

04 x

5 and 4 andnumber real a is | ofDomain xxxxf

4x

Therefore, a denominator of the function is equal to zero when x = 4 or x = -5. Then the domain is:

CONTINUECONTINUEDD

05 x5x

Set a denominator equal to zero

Solve

Set a denominator equal to zero

Solve

118

xx

xf

5

6

4

2

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.3

Domain of a Function

Check Point 1aCheck Point 1a

SOLUTIONSOLUTION

Find the domain of the function: . 32

1 xxf

Since the function f has no denominator or square root, there are no real numbers that when plugged into the function for x would cause the value of the function to yield something other than a real number. Therefore, the domain is:

number real a is | ofDomain xxf

118

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.3

Domain of a Function

Because division by 0 is undefined (and not a real number), the denominator, x + 5, cannot be 0. Then x cannot be -5, and -5 is not in the domain of the function.

5- x andnumber real a is | g ofDomain xx

Check Point 1bCheck Point 1b

Find the domain of the function: . 5

47

x

xxg

117-118

DONE