This document discusses algebraic functions and their properties. Some key points:
- When working with real numbers, you cannot divide by zero or take the square root or even root of a negative number. These restrictions limit the domain of a function.
- To find the domain of a rational function, set the denominator equal to zero and exclude those values.
- The domain of a sum, difference, product, or quotient of functions f and g consists of values that are in the domains of both f and g, except for quotients where the denominator cannot be zero.
- Composition of functions means applying one function to the output of another. The domain of f∘g consists of values where g(x) is
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
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The Algebric Functions
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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