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2.4 Arithmetic Combinations of Functions,Composition of Functions,

and their Domains2.5 Inverse Functions

Tim Busken

Graduate TeacherDepartment of MathematicsSan Diego State UniversityDynamical Systems Program

September 30, 2011

Tim Busken 2.4 Arithmetic Combinations of Functions,Composition of Functions,and

2.4 Combinations and Compositions of Functions

In this lecture we want to construct more complicated or uglierlooking functions by combining and composing familiar functionstogether. Then we want to learn how to find the domains of thesenew functions. For that we need to recall a definition that waspresented in 1.7, the lecture on inequalities.

Definition

The intersection of two sets A and B, written A ∩ B, is the set ofall elements (numbers) that are in both A and B. The ∩ symbolmeans the word “and.”

Example: Suppose A = {1,2,3,4} and B = {2,4,20}.Then A ∩ B = {2,4}


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 newfunction h(x) by combining the two functions (using the arithmeticoperators) 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 h(x) = f (x) · g(x) notation = (f · g) (x)4 h(x) = f (x)

g(x) notation =fg(x) provided that g(x) 6= 0.

EXAMPLE Suppose f (x) = 3√x − 2 and g(x) = x2 + 5. Use the

above defns to determine the following1 (f + g) (4)2 (f − g) (1)3 (f · g) (0)4 f

g(9)


Theorem (Intersection Rule for Function Combinations)

Suppose h(x) is an arithmetic combination of the functions f andg (previous slide). The domain of h(x) is the intersection of thedomains of f (x) and g(x). In addition, we exclude from thedomain of f /g any value of x that makes g(x) = 0.

Example: Suppose f (x) =√x , g(x) = x2 and

h(x) = (f + g)(x) =√x + x2. We know from previous discussion

that dom(f ) ≡ [0,∞) and dom(g) ≡ (−∞,∞) The domain of h isthe set of x ∈ [0,∞) ∩ (−∞,∞) ≡ [0,∞).

-

0

x ∈ [0,∞)xaxis s >

-

b(−∞,∞)xaxis < >

-

0

dom(h) ≡ x ∈ [0,∞)xaxis s >


2.4 Arithmetic Compositions of Functions

Suppose f (x) = x2 − 14x and g(x) = x + 14. Determine thefollowing 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 fg(x)

6 Dom(f /g)Tim Busken 2.4 Arithmetic Combinations of Functions,Composition of Functions,and

2.4 Arithmetic Combinations of Functions

Suppose f (x) =√7− x and g(x) =

√7 + x . Determine the

following:

1 (f + g) (x)

2 (f − g) (x)

3 (f · g) (x)

4 fg(x)


Classroom Examples:

Suppose f (x) =√7− x and g(x) =

√7 + x . Determine the

following:


2 Dom(f /g)


Classroom Examples:

Suppose f (x) = 7x−14 and g(x) = 15

x+8 . Determine the following:

1 (f + g) (x)

2 (f − g) (x)

3 (f · g) (x)

4 fg(x)


Classroom Examples:

Suppose f (x) = 7x−14 and g(x) = 15

x+8 . Determine the followingfour function combinations, and their domains.


2 Dom(f /g)


2.4 Compositions of Functions

Definition

If f and g are two functions, the composition of f and g , writtenf ◦ g is defined by the equation

f ◦ g = f (g(x)),

provided that g(x) is in the domain of f . Armed with twofunctions f and g , in order to make up this new function, call ith(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) =√x and g(x) = 2 x + 1.

Find the composite function f ( g(x) ).

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 Functions,and

2.4 Compositions of Functions

Classroom Examples: Suppose

f (x) = x2 + 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


Definition

The composition of f and g , written f ◦ g is defined by theequation

f ◦ g = f (g(x)),

provided that g(x) is in the domain of f .

Example: Suppose g ≡ {(1, 2), (3, 4), (5, 6)} andf ≡ {(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 anordered pair in f ◦ g . Also since g(3) = 4 and f (4) = 9, thenf (g(3)) = 9, and (3, 9) is an ordered pair in f ◦ g . Now g(5) = 6but 6 is not in the domain of f . So there are only two orderedpairs 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 fromthe domain of g any number x such that g(x) is not in thedomain of f .


Classroom Examples: Suppose f (x) =1

xand g(x) = 4x − 9.

Determine the following composite functions, and their respectivedomains.

1 h(x) = f ◦ g

2 h(x) = g ◦ f

3 h(x) = f ◦ f

4 h(x) = g ◦ g


Classroom Examples: Suppose h(x) = f (g(x)) for the followingfunctions. Determine what functions f and g were used to makeupthese compositions. (there are more than one right answer.)

1 h(x) = (x3 − 2 x + 1)6

2 h(x) = 1(x4−2 x+1)3

3 h(x) =√

(x − 2)5

4 h(x) = |x2 − 2 x |+ 1


Theorem

Suppose n is any real number. The domain of h(x) = n ·√

f (x) isthe solution set of x values of f (x) ≥ 0.

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 satisfyf (x) ≥ 0, or equivalently x − 1 ≥ 0, which after adding one toboth sides renders x ≥ 1. Therefore the domain of h is all x valuesthat live on the interval [1,∞)

Theorem

The domain of h(x) =n

√

f (x)is the solution set of x values of

f (x) > 0.

Example: Consider the function h(x) = 3√

x−1(i.e. in this case

f (x) = x − 1). The domain of h is x ∈ (1,∞).Tim Busken 2.4 Arithmetic Combinations of Functions,Composition of Functions,and

Definition

A function f with domain D and range R is a one to one function ifeither of the following equivalent conditions is satisfied:

1 Whenever a 6= b in D, then f (a) 6= f (b) in R.

2 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) hasthe following characteristic: for each functional value f (x) in therange R there corresponds EXACTLY ONE element in the domainD.


Definition

A function f with domain D and range R is a one to one function ifeither of the following equivalent conditions is satisfied:

1 Whenever a 6= b in D, then f (a) 6= f (b) in R.

2 Whenever f (a) = f (b) in R, then a = b in D.

Example 1: Prove that f (x) = x2 is NOT a one to one function.

y

x

(−2, 4) (2, 4)

Soln 1: Counterexample: f (x) = x2 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.


Theorem (The Horizontal Line Test)

A function f is one to one if and only if every horizontal lineintersects the graph of f in at most one point.

This means that if we can draw a horizontal line on the graph of afunction, and the line intersects the graph at two or more points,well then the function is not one to one. For example

y

x(−2, 4) (2, 4)

y

x

Y = X 2 but Y = X 3

is not one to one is one to one.


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 domainR and range D provided that

f −1(f (x)) = x

Note: A function has an inverse (function) only when it is one toone.

Classroom Example Suppose that f (x) = 95 x + 32 has the

inverse function f −1(x) = 59 (x − 32). Prove the following:

f −1(f (x)) = x


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

f (x)is the multiplicative inverse of the

number f (x).

Example: Suppose f is one-to-one and f (−9) = 15, then

f −1(15) = and (f (−9))−1 =


Theorem (Properties of Inverse Functions)

Suppose that f is a one to one function with domain D and rangeR. 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 thevariable x to represent the values in the domain of f −1. In thisway we can give the graph of f −1 the same framework as thegraph of f , that is, with its domain along the x-axis.


Note: As a result of this construction, and because of theproperties of inverse functions, it turns out that the graph ofy = f −1(x) is the reflection of the graph of y = f (x) about theline y = x . For every point (a, b) on the graph of f (x) there is acorresponding point (b, a) on the graph of f −1(x).

y

x(b, a)

(a, b)

f (x)

f−1(x)

y = x


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 onefunction:

1 f (x) = x2, with domain [0,∞)


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).


1 f (x) = x+32−x



1 f (x) = 2− x3


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