Download Chapter Two 2.3

§ 2.3
The Algebra of Functions – Finding the
Domain
Domain of a Function
117
Finding a Function’s Domain
If 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 #2 Section 2.3
Domain of a Function
117
• Consider the function
1
f ( x) 
x 5
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.
Domain of f  x | x is a real number and x  5
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.3
Domain of a Function
117
• Now consider the function:
g ( x) 
x7
The 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.
Domain of g  x | x is a real number and x  7
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.3
Domain of a Function
118
EXAMPLE
2
Find the domain of the function: f x   4 x  2 x  7 .
SOLUTION
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:
Domain of f  x | x is a real number 
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.”
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.3
Domain of a Function
118
EXAMPLE
Find the domain of the function: f x  
2
6

x4 5 x
.
SOLUTION
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:
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.3
Domain of a Function
118
f x  
CONTINUED
2
6

x4 5 x
x4  0
x4
Set a denominator equal to zero
Solve
5 x  0
x  5
Set a denominator equal to zero
Solve
Therefore, a denominator of the function is equal to zero
when x = 4 or x = -5. Then the domain is:
Domain of f  x | x is a real number and x  4 and x  5
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.3
Domain of a Function
118
Check Point 1a
Find the domain of the function:
1
f x  
x  3.
2
SOLUTION
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:
Domain of f  x | x is a real number 
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.3
Domain of a Function
117-118
Check Point 1b
Find the domain of the function:
7x  4
.
g x  
x5
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.
Domain of g  x | x is a real number and x  - 5
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.3
DONE