1. The Algebra of Functions
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2. Using basic algebraic functions, what limitations
are there when working with real numbers?
A) You CANNOT divide by zero.
 Any values that would result in a zero denominator are NOT
allowed, therefore the domain of the function (possible x values)
would be limited.
B) You CANNOT take the square root (or any even root) of
a negative number.
 Any values that would result in negatives under an even radical
(such as square roots) result in a domain restriction.

3. Example
 Find the domain:
 There is an x under an even radical AND x terms
in the denominator, so we must consider both of
these as possible limitations to our domain.
65
2
2
xx
x
}3,2:{:
3,2,0)2)(3(
065
2,02
2
xxxDomain
xxx
xx
xx

4. Find the indicated function values and determine whether the given values
are in the domain of the function.
f(1) and f(5), for
f(1) =
Since f(1) is defined, 1 is in the domain of f.
f(5) =
Since division by 0 is not defined, the number 5 is not in the domain
of f.
1 1 1
1 5 4 4
1 1
5 5 0
1
( )
5
f x
x

5. Find the domain of the function
Solution:
We can substitute any real number in the numerator, but we
must avoid inputs that make the denominator 0.
Solve, x2 3x 28 = 0.
(x 7)(x + 4) = 0
x 7 = 0 or x + 4 = 0
x = 7 or x = 4
2
2
3 10 8
( )
3 28
x x
g x
x x
The domain consists of the set of all real numbers except
x= 4 and x= 7 or {x | x 4 and x 7}.
, 4 ( 4,7) (7, )

6. Rational Functions
 To find the domain of a function that has a variable in
the denominator, set the denominator equal to zero and
solve the equation. All solutions to that equation are
then removed from consideration for the domain.
Find the domain:
 Since the radical is defined only for values that are greater
than or equal to zero, solve the inequality
( ) 5f x x
5 0x
5x
5x
( ,5]

7. Visualizing Domain and Range
Keep the following in mind regarding the graph of a
function:
 Domain = the set of a function’s inputs; found on the x-axis
(horizontal).
 The domain of a function is the set of all “first coordinates”
of the ordered pairs of a relation
 Range = the set of a function’s outputs; found on the y-axis
(vertical).
 The range of a function is the set of all “second coordinates”
of the ordered pairs of a relation.

8. Example
Graph the function. Then
estimate the domain and range.
(Note: Square root
function moved one unit right)
Domain = [1, )
Range = [0, )
( ) 1f x x ( ) 1f x x

9. Algebra of functions
 (f + g)(x) = f(x) + g(x)
 (f - g)(x) = f(x) – g(x)
 (fg)(x) = f(x)g(x)
0)(,
)(
)(
)( xg
xg
xf
x
g
f

10. Example
Find each function and state its domain:
 f + g
 f – g
 f ·g
 f /g
;1 1g xf x x x
;1 : 11x Domainf xx xg x
;1 : 11x Domainf xx xg x
2
1; :1 1 1x x Domaing xx xf x
1
; : 1
1
x D
x
f omain x xg
x

11. BA
Their sum f + g is the function given by
(f + g)(x) = f(x) + g(x)
The domain of f + g consists of the numbers x that are in the
domain of f and in the domain of g.
Their difference f - g is the function given by
(f – g ) (x) = f(x) - g(x)
The domain of f – g consists of the numbers x that are in the
domain of f and in the domain of g.
BA
If f and g are functions with domains A and B:

12. Their product f g is the function given by
BA
The domain of f g consists of the numbers x that are in the
domain of f and in the domain of g.
Their quotient f /g is the function given by
(f / g ) (x) = f(x) / g(x) where g(x) ≠ 0;
(f g)(x) = f(x) g(x)
If f and g are functions with domains A and B:
The domain of f / g consists of the numbers x for which g(x) 0
that are in the domain of f and in the domain of g.
0)(xgBA

13. Example
Given that f(x) = x + 2 and g(x) = 2x + 5, find
each of the following.
a) (f + g)(x) b) (f + g)(5)
a) ( )( ) ( ) ( )
2 2 5
3 7
f g x f x g x
x x
x
b) We can find (f + g)(5) provided 5 is in the domain of each
function.
This is true.
f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15
(f + g)(5) = f(5) + g(5) = 7 + 15 = 22
(f + g)(5) = 3(5) + 7 = 22or

14. Example
Given that f(x) = x + 2 and g(x) = 2x + 5, find
each of the following
a) (f - g)(x) b) (f - g)(5)
a) ( )( ) ( ) ( )
2 (2 5)
2 2 5
3
f g x f x g x
x x
x x
x
b) We can find (f - g)(5) provided 5 is in the domain of each function.
This is true.
f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15
(f - g)(5) = f(5) - g(5) = 7 - 15 = -8
(f - g)(5) = -(5) - 3 = -8or

15. Example
Given that f(x) = x + 2 and g(x) = 2x + 5, find
each of the following.
a) (f g)(x) b) (f g)(5)
a)
2
( )( ) ( ) ( )
( 2)(2 5)
2 9 10
fg x f x g x
x x
x x
b) We can find (f g)(5) provided 5 is in the domain of each
function. This is true.
f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15
(f g)(5) = f(5)g(5) = 7 (15) = 105
or (f g)(5) = 2(25) + 9(5) + 10 = 105

16. ( )
( )
f x
g x 2
3
16
x
x
The domain of f / g is {x | x > 3, x 4}.
( ) 3f x x
2
( ) 16g x x
Given the functions below, find (f/g)(x) and give the
domain.
( / )( )f g x
The radicand x – 3 cannot be negative.
Solving x – 3 ≥ 0 gives x ≥ 3
We must exclude x = -4 and x = 4 from the domain
since g(x) = 0 when x = 4.

17. Composition of functions
 Composition of functions is the successive application of
the functions in a specific order.
 Given two functions f and g, the composite function
is defined by and is read
“f of g of x.”
 The domain of is the set of elements x in the
domain of g such that g(x) is in the domain of f.
 Another way to say that is to say that “the range of
function g must be in the domain of function f.”
 Composition of functions means the output from the inner
function becomes the input of the outer function.
f g f g x f g x
f g

18.  Composition of functions means the output from
the inner function becomes the input of the outer
function.
 f(g(3)) means you evaluate function g at x=3, then
plug that value into function f in place of the x.
 Notation for composition: ))(())(( xgfxgf 
f g
x
g(x)
f(g(x))
domain of g
range of
f
range of g
domain of f
g
f

19. f g x f g x 1
2
f
x
1
2x
1
2x
.
gf
x
xgxxf
Find
2
1
)(and)(Suppose
Suppose f x x( ) and g x
x
( )
1
2
. Find
the domain of f g .
The domain of f g consists of those x in the domain of g,
thus x = -2 is not in the domain of f g .
In addition, g(x) > 0, so
1
0
2x
2x
The domain of f g is {x | x > -2}.

20. 2
2
2 1 3
2 4x
xf g x
2
2
2
2 1
2 6 9 1
2 12 18 1
3g
x x
x
f x
x
x
Example
 Evaluate and :


f g x g f x
3f x x
2
2 1g x x
 2
2 4
(you check)
f g x x
2
2 12 17g f x x x
You can see that function composition is not commutative.
NOTE: This is not a formal proof of the statement.

21. (Since a radicand can’t be negative in the set of real numbers, x must be
greater than or equal to zero.)
Example
Find the domain of and :f g x g f x
1f x x
g x x
 1 : 0f g x x Domain x x
 1 : 1g f x x Domain x x
(Since a radicand can’t be negative in the set of real numbers, x – 1 must
be greater than or equal to zero.)

22. Solution to Previous Example :
 Determine a function that gives the cost of producing
the helmets in terms of the number of hours the
assembly line is functioning on a given day.
Cost C n C P t
2
75 2C t t
2
14 525 100 2 40 $5 8C t t
2
75 2n P t t t 7 1000C n n

23. 1. Suppose that and2
( ) 1f x x ( ) 3g x x
( ) ?g f x
( ) ( ( ))g f x g f x
2
( 1)g x 2
3( 1)x 2
3 3x
2. Suppose that and2
( ) 1f x x ( ) 3g x x
(2) ?g f
(2) 2g f g f
2
(2 1)g
(3)g
(3)(3) 9

24. The End
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