2.4 Arithmetic Combinations of Functions,
Composition of Functions,
and their Domains
2.5 Inverse Functions
Tim Busken
Graduate Teacher
Department of Mathematics
San Diego State University
Dynamical Systems Program
September 30, 2011
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Combinations and Compositions of Functions
In this lecture we want to construct more complicated or uglier
looking functions by combining and composing familiar functions
together. Then we want to learn how to find the domains of these
new functions. For that we need to recall a definition that was
presented in 1.7, the lecture on inequalities.
Definition
The intersection of two sets A and B, written A ∩ B, is the set of
all elements (numbers) that are in both A and B. The ∩ symbol
means the word “and.”
Example: Suppose A = {1,2,3,4} and B = {2,4,20}.
Then A ∩ B = {2,4}
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Arithmetic Combinations of Functions
Definition (Sum, Difference, Product and Quotient Functions)
Let f and g be any (continuous) functions. We can make a new
function h(x) by combining the two functions (using the arithmetic
operators) in one of the following ways:
1
h(x) = f (x) + g (x)
notation
= (f + g ) (x)
2
h(x) = f (x) − g (x)
notation
= (f − g ) (x)
3
4
h(x) = f (x) · g (x)
h(x) =
f (x)
g (x)
notation
notation
=
f
g
= (f · g ) (x)
(x)
provided that g (x) 6= 0.
√
EXAMPLE Suppose f (x) = 3 x − 2 and g (x) = x 2 + 5. Use the
above defns to determine the following
1
(f + g ) (4)
2
(f − g ) (1)
3
(f · g ) (0)
f
4
g (9)
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Theorem (Intersection Rule for Function Combinations)
Suppose h(x) is an arithmetic combination of the functions f and
g (previous slide). The domain of h(x) is the intersection of the
domains of f (x) and g (x). In addition, we exclude from the
domain of f /g any value of x that makes g (x) = 0.
√
Example: Suppose f (x) = x, g (x) = x 2 and
√
h(x) = (f + g )(x) = x + x 2 . We know from previous discussion
that dom(f ) ≡ [0, ∞) and dom(g ) ≡ (−∞, ∞) The domain of h is
the set of x ∈ [0, ∞) ∩ (−∞, ∞) ≡ [0, ∞).
s
- >x ∈ [0, ∞)
xaxis
0
xaxis <
xaxis
b
s
0
Tim Busken
- > (−∞, ∞)
dom(h)
- > ≡ x ∈ [0, ∞)
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Arithmetic Compositions of Functions
Suppose f (x) = x 2 − 14x and g (x) = x + 14. Determine the
following four function combinations, and their domains.
1
(f + g ) (x)
2
(f − g ) (x)
3
(f · g ) (x)
4
Dom(f + g ), Dom(f − g ), Dom(f · g )
5
f
g
6
Dom(f /g )
(x)
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Arithmetic Combinations of Functions
Suppose f (x) =
following:
√
1
(f + g ) (x)
2
(f − g ) (x)
3
(f · g ) (x)
4
f
g
7 − x and g (x) =
√
7 + x. Determine the
(x)
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Classroom Examples:
√
√
Suppose f (x) = 7 − x and g (x) = 7 + x. Determine the
following:
1
Dom(f + g ), Dom(f − g ), Dom(f · g )
2
Dom(f /g )
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Classroom Examples:
Suppose f (x) =
7
x−14
1
(f + g ) (x)
2
(f − g ) (x)
3
(f · g ) (x)
4
f
g
and g (x) =
15
x+8 .
Determine the following:
(x)
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Classroom Examples:
7
15
Suppose f (x) = x−14
and g (x) = x+8
. Determine the following
four function combinations, and their domains.
1
Dom(f + g ), Dom(f − g ), Dom(f · g )
2
Dom(f /g )
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Compositions of Functions
Definition
If f and g are two functions, the composition of f and g , written
f ◦ g is defined by the equation
f ◦ g = f (g (x)),
provided that g (x) is in the domain of f . Armed with two
functions f and g , in order to make up this new function, call it
h(x) = f (g (x)) (or just h), we simply
replace the x in f(x) with the function g(x).
Classroom Example: Suppose f (x) =
Find the composite function f ( g (x) ).
√
x and g (x) = 2 x + 1.
Soln: replace the x in f(x) with the function g(x),
√
f (g (x)) = f (2 x + 1) = 2 x + 1
Question: What is ? What is ?
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
2.4 Compositions of Functions
Classroom Examples: Suppose
f (x) = x 2 + 7 and
g (x) = x − 5.
Determine the following:
1
f (g (3))
2
f (g (2))
3
f ◦g
4
the domain of f ◦ g
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Definition
The composition of f and g , written f ◦ g is defined by the
equation
f ◦ g = f (g (x)),
provided that g (x) is in the domain of f .
Example: Suppose g ≡ {(1, 2), (3, 4), (5, 6)} and
f ≡ {(2, 8), (4, 9), (1, 1)}. Find f ◦ g .
Since g (1) = 2 and f (2) = 8, then f (g (1)) = 8, and (1, 8) is an
ordered pair in f ◦ g . Also since g (3) = 4 and f (4) = 9, then
f (g (3)) = 9, and (3, 9) is an ordered pair in f ◦ g . Now g (5) = 6
but 6 is not in the domain of f . So there are only two ordered
pairs in f ◦ g , namely f ◦ g ≡ {(1, 8), (3, 9)}
Comment: the domain of g is {1, 3, 5} while the domain of f ◦ g is
{1, 3}. In order to find the domain of f ◦ g we remove from
the domain of g any number x such that g (x) is not in the
domain of f .
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
1
Classroom Examples: Suppose f (x) =
and g (x) = 4x − 9.
x
Determine the following composite functions, and their respective
domains.
1
h(x) = f ◦ g
2
h(x) = g ◦ f
3
h(x) = f ◦ f
4
h(x) = g ◦ g
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Classroom Examples: Suppose h(x) = f (g (x)) for the following
functions. Determine what functions f and g were used to makeup
these compositions. (there are more than one right answer.)
1
h(x) = (x 3 − 2 x + 1)6
2
h(x) =
1
(x 4 −2 x+1)3
3
h(x) =
p
(x − 2)5
4
h(x) = |x 2 − 2 x| + 1
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Theorem
Suppose n is any real number. The domain of h(x) = n ·
the solution set of x values of f (x) ≥ 0.
p
f (x) is
√
Example: Consider the function h(x) = 3 x − 1 (i.e. in this case
f (x) = x − 1). The domain of h is all x values that satisfy
f (x) ≥ 0, or equivalently x − 1 ≥ 0, which after adding one to
both sides renders x ≥ 1. Therefore the domain of h is all x values
that live on the interval [1, ∞)
Theorem
The domain of h(x) = p
n
f (x)
is the solution set of x values of
f (x) > 0.
3
(i.e. in this case
Example: Consider the function h(x) = √x−1
f (x) = x − 1). The domain of h is x ∈ (1, ∞).
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Definition
A function f with domain D and range R is a one to one function if
either of the following equivalent conditions is satisfied:
1
2
Whenever a 6= b in D, then f (a) 6= f (b) in R.
Whenever f (a) = f (b) in R, then a = b in D.
So in plain English, this means that a one to one function f (x) has
the following characteristic: for each functional value f (x) in the
range R there corresponds EXACTLY ONE element in the domain
D.
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Definition
A function f with domain D and range R is a one to one function if
either of the following equivalent conditions is satisfied:
1
2
Whenever a 6= b in D, then f (a) 6= f (b) in R.
Whenever f (a) = f (b) in R, then a = b in D.
Example 1:
Prove that f (x) = x 2 is NOT a one to one function.
y
(−2, 4)
(2, 4)
x
Soln 1: Counterexample: f (x) = x 2 is not one to one since for a
= -2 and b = 2, it is true that a 6= b and f (a) = f (b) = 4.
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Theorem (The Horizontal Line Test)
A function f is one to one if and only if every horizontal line
intersects the graph of f in at most one point.
This means that if we can draw a horizontal line on the graph of a
function, and the line intersects the graph at two or more points,
well then the function is not one to one. For example
y
y
(−2, 4)
(2, 4)
x
x
Y = X2
is not one to one
Tim Busken
but Y = X 3
is one to one.
2.4 Arithmetic Combinations of Functions,Composition of Functio
Definition (Inverse Function)
Suppose f is a one to one function, with domain D and range R.
The inverse function of f is the function denoted f −1 with domain
R and range D provided that
f −1 (f (x)) = x
Note:
one.
A function has an inverse (function) only when it is one to
Classroom Example Suppose that f (x) = 59 x + 32 has the
inverse function f −1 (x) = 95 (x − 32). Prove the following:
f −1 (f (x)) = x
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
CAUTION: f −1(x) 6= f (x)−1
f −1 (x) is notation for the function inverse of a one to one function f
f (x)−1 = (f (x))−1 =
1
is the multiplicative inverse of the
f (x)
number f (x).
Example: Suppose f is one-to-one and f (−9) = 15, then
f −1 (15) =
and (f (−9))−1 =
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Theorem (Properties of Inverse Functions)
Suppose that f is a one to one function with domain D and range
R. Then
The inverse function f −1 is unique.
The domain of f −1 is the range of f .
The range of f −1 is the domain of f .
The statement f (x) = y is equivalent to f −1 (y ) = x
Note: We customarily express the relationship for f −1 using the
variable x to represent the values in the domain of f −1 . In this
way we can give the graph of f −1 the same framework as the
graph of f , that is, with its domain along the x-axis.
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Note: As a result of this construction, and because of the
properties of inverse functions, it turns out that the graph of
y = f −1 (x) is the reflection of the graph of y = f (x) about the
line y = x. For every point (a, b) on the graph of f (x) there is a
corresponding point (b, a) on the graph of f −1 (x).
y
f (x)
(a, b)
f −1(x)
(b, a)
x
y=x
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
How to find the inverse of a one to one function:
1 Replace f (x) with y. Then interchange x and y.
2 Solve the resulting equation for y.
3 Replace y with f −1 (x).
Classroom Example: Find f −1 (x) for the following one to one
function:
1
f (x) = x 2 , with domain [0,∞)
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
How to find the inverse of a one to one function:
1 Replace f (x) with y. Then interchange x and y.
2 Solve the resulting equation for y.
3 Replace y with f −1 (x).
Classroom Example: Find f −1 (x) for the following one to one
function:
1
f (x) = x+3
2−x
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio
Classroom Example:
function:
1
f (x) = 2 − x 3
Find f −1 (x) for the following one to one
Tim Busken
2.4 Arithmetic Combinations of Functions,Composition of Functio