CRASH COURSE IN PRECALCULUS
Shiah-Sen Wang
The graphs are prepared by Chien-Lun Lai
Based on : Precalculus: Mathematics for Calculus
by J. Stuwart, L. Redin & S. Watson,
6th edition, 2012, Brooks/Cole
Chapter 2. Section 2.1, 2.2, 2.6 and 2.7.
LECTURE 4. FUNCTIONS
The purpose of this lecture is to introduce the concept of
functions.
What is a Function?
Definition
A function f is a rule that assigns each element x in a set A
exactly one element, called f (x) in a set B.
f (x) is read: “f of x”, “the value of f at x” or “the image of x
under f ”.
The set A is called the domain of the function f .
The range of f is the set of all possible value of f (x) as x varies
throughout the domain, that is
range of f = {f (x) ∣ x ∈ A} = f (A) ⊂ B
Some notations and terminologies for a function:
f ∶ A Ð→ B or simply y = f (x) for x ∈ A, y ∈ B .
We call x a independent variable of f and y a dependent
variable of f when y ∈ f (A).
How to Represent a Function
● By explicit formula. e.g., f (x) = x 2 .
● In words. e.g., The area of any disc in the Cartesian plane
with fixed center at the origin is π times the square of its
radius. So the domain A of the function f is (0, ∞)
because the radius of a disc must be positive. The
√ function
f in√explicit formula is f (x) = πx 2 for x ∈ A. (⇒ f ( 2) =
π( 2)2 = 2π.) We can choose B = R, noticing that when
y < 0 it can not be a dependent variable for this choice.
But if we choose B = f (A) = (0, ∞)=the range of f , then
any y ∈ B is a dependent variable in this case.
x f (x)
● By a table. e.g., 1
3
5
8
How to Represent a Function
● Visually (by a diagram or a graph). e.g.,
The Domain of a Function
Examples
1. (Explicitly Given) f (x) = x 2
2. (Implicitly Given)
0 ≤ x ≤ 5.
√
(i) f (x) = x.
The maximal possible domain of f = [0, ∞)
1
(ii) f (x) = √
2
x −x
First
√ of all, because the denominator of the function is
x 2 − x > 0. Secondly, to take square root of x 2 − x, x 2 − x
must be nonnegative. Together, when 0 < x 2 − x, the
function f is well-defined. Solve 0 < x 2 − x
= x(x − 1) ⇒ x < 0 or x > 1. Hence (−∞, 0) ∪ (1, ∞) is the
maximal
√ possible domain of f .
2x − 1
(iii) f (x) = 2
x −1
The Domain of a Function
Examples
√
⎧
⎪ 2x − 1 ≥ 0
2x − 1
⎪
, we must have ⎨ 2
To make sense of 2
⎪
x −1
x −1≠0
⎪
⎩
⎧
⎧
1
⎪
⎪
1
⎪2x − 1 ≥ 0
⎪x ≥ 2
⇔ [ , 1) ∪ (1, ∞) is
⇔⎨
⇔⎨
⎪
⎪
2
x ≠ ±1
(x − 1)(x + 1) ≠ 0
⎪
⎪
⎩
⎩
the maximal possible domain of f .
(iv) Is there any difference between the functions f (x) = 1 and
x −1
g(x) =
?
x −1
First we notice that f is defined on the whole real line R, but
the maximal possible domain of g = (−∞, 1) ∪ (1, ∞) = A
= R ∖ {1}.
For x ∈ A, g(x) = 1 = f (x) and they are identical. However, f
is defined when x = 1, but g is not defined at 1. So they are
different functions in this sense.
√
Graphs of Functions
Definition
If f is a real-valued function with domain A ⊂ R, then the graph
of f is the set of ordered pairs
{(x, f (x)) ∣ x ∈ A}
plotted in a Cartesian plane. In other words, the graph of f is
the set of all points (x, y) such that y = f (x); that is the graph of
f is the graph of equation y = f (x).
Examples
⎧
⎪
⎪x
1. (The absolute Value Function) ∣x∣ = ⎨
⎪
⎪
⎩−x
if x ≥ 0,
if x < 0.
Graphs of Functions
Examples
The graph of y = ∣x∣ is
⎧
⎪
⎪x 2 , if x < 1;
2. (Piecewise Defined Function) f (x) = ⎨
⎪
⎪
⎩4, if x ≥ 1.
Graphs of Functions
Examples
⎧
⎪
⎪x 2 ,
The graph of f (x) = ⎨
⎪
⎪
⎩4,
if x < 1;
is
if x ≥ 1.
Graphs of Functions
Question: Which curves in the xy-plane are graphs of
functions?
Vertical Line Test: A curve in the xy -plane is the graph of a
function if and only no vertical line intersects the curve more
than once.
Example
Combining Functions
● Algebra of Functions: Let f and g be functions with
domains A and B. Then f + g, f − g, fg and f /g are defined
as follows:
(i) (f + g) (x) = f (x) + g(x)
for x ∈ A ∩ B.
(ii) (f − g) (x) = f (x) − g(x)
for x ∈ A ∩ B.
(iii) (fg) (x) = f (x)g(x)
for x ∈ A ∩ B.
f (x)
f
for x ∈ {x ∈ A ∩ B ∣ g(x) ≠ 0}.
(iv) ( ) (x) =
g
g(x)
Examples
1
, (f ± g)(x) =?
x
√
The domain of f (x) = x is A = [0, ∞) and the domain of
1
g(x) = is B = (−∞, 0) ∪ (0, ∞). When x ∈ A ∩ B = (0, ∞),
x
√
3
√
1
x2 ±1 x x ±1
(f ± g)(x) = f (x) ± g(x) = x ±
=
=
.
x
x√
√x
√
1
2 2 1
2 2±1
(f ± g)(2) = f (2) ± g(2) = 2 ±
=
±
=
.
2
2
2
2
(i & ii) Let f (x) =
√
x, g(x) =
Combining Functions
Examples
√
(ii) Let f (x) = g(x) = x for x ≥ 0. (f − g)(x) =?
Then (f − g)(x) = f (x) − g(x) = 0 only when x ≥ 0 , even
though the√function h(x) =√
0 is defined for x ∈ R.
2
(iii & iv) Let f (x) = x√ − x, g(x) = x 2 , (fg)(x) =?, and (f /g)(x) =?
2 = ∣x∣ is defined for all x ∈ R = B and
First, g(x)
√ = x √
f (x) = x 2 − x = x(x − 1) is defined if and only if when
x(x − 1) ≥ 0 ⇔ x ∈ (−∞, 0] ∪ [1, ∞) √
= A. Hence,
for
√
2 − x x2 =
x
∈
A
∩
B
=
A,
(fg)(x)
=
f
(x)g(x)
=
x
√
√
√
√
(x 2 − x)x 2 = x 4 − x 3 = x 3 (x − 1) = ∣x∣ x 2 − x. Before
computing (f /g)(x), we must eliminate the possibility that
g(x) = 0 for x ∈ B that is when x = 0. Thus,
√ for
f
f (x)
x2 − x
x ∈ (−∞, 0) ∪ [1, ∞), ( ) (x) =
= √
=
2
g
g(x)
x
√
√
√
x2 − x
x −1
x2 − x
=
=
. Exercise: Rationalize the
x2
x
∣x∣
f
numerator of ( ) (x).
g
Combining Functions
● Composition of Functions: Given two functions f and g,
the composite function f ○ g (also called the
composition of f and g) is defined by
(f ○ g)(x) = f (g(x))
See the diagram
and noticing that j is not in the range of g even though it is
in the domain of f .
Combining Functions
Examples
√
√
(i) f (x) = x, g(x) = 2 − x, find the following functions f ○ g,
g ○ f , f ○ f and g ○ g and their corresponding domains.
First we note that the domain of f is [0, ∞) and the domain
of g is (−∞, 2]. Now be compute
√√
√
2−x
(f ○ g)(x) = f (g(x)) = f ( 2 − x) (for x ≤ 2) =
(for x ≤ 2) = (2 − x)1/4 (for x ≤ 2)
⇒ domain of f ○ g = (−∞, 2].
√
√
√
x)
(for
x
≥
0)
=
2 − x for
(g ○ f ) = g(f (x))
=
g(
√
√
x ≥ 0 and 2 − x ≥ 0 ⇔ 2 ≥ x ≥ 0 ⇔ 4 ≥ x ≥ 0. Hence
the domain of g ○ f is [0, 4].
(f ○ f )(x) = x 1/4 and its domain is [0, ∞).( The details are
left as an exercise.)
√
√
√
(g ○√g)(x) = g(g(x))
√ = g( 2 − x) = 2 − 2 − x ⇒ 0 ≤
2 − 2 − x ⇔ 0 ≤ 2 − x ≤ 2 ⇔ 0 ≤ 2 − x ≤ 4 ⇔ −2 ≤ x ≤ 2.
Hence the domain of g ○ g is [−2, 2].
Ô⇒ f ○ g ≠ g ○ f in general.
Combining Functions
Before we give the next example, we will try to compose
three different functions f , g and h (by assuming no
ambiguities in the domains) for f ○ (g ○ h) and (f ○ g) ○ h:
[f ○ (g ○ h)] = f ((g ○ h) (x)) = f (g (h(x))).
On the other hand,
((f ○ g) ○ h) (x) = (f ○ g)(h(x)) = f (g (h(x))) ⇒
f ○ (g ○ h) = (f ○ g) ○ h
i.e., the associative law hold true for composition of
functions.
Combining Functions
Examples
√
x
and h(x) = 3 x. Find f ○ g ○ h
x −1
and its domain.
¿ √
√
3
Á 3x
√
x
3
À√
f (g (h(x))) = f (g ( x)) = f ( √
) =Á
.
3
3
1
2
x −1 3
x −1
We shall find its domain: First of all, = holds true ⇔ x ∈ R.
1√
√
Then, = holds true ⇔ 3 x − 1 ≠ 0 ⇔ 3 x ≠ 1 ⇔ x ≠ 1, =
(ii) Let f (x) =
√
x, g(x) =
2
1
and = hold true ⇔ x ≠ 1. Finally, = holds true
2
3
√
3
x
x 1/3
⇔0≤ √
=
3
x − 1 x 1/3 − 1
x 1/3
(x 1/3 )2 + x 1/3 + 1
⇔ 0 ≤ 1/3
⋅ 1/3 2
x − 1 (x ) + x 1/3 + 1
⇔0≤
x 1/3 [(x 1/3 )2 + x 1/3 + 1]
2
x 1/3 [(x 1/3 + 1/2) + 3/4]
=
x −1
x −1
⇔ 0 ≤ x 1/3 (x − 1) and x ≠ 1 ⇔ 0 ≤ x(x − 1) and x ≠ 1
⇔ x ∈ (−∞, 0] ∪ (1, ∞), which is the domain of f ○ g ○ h.
Combining Functions
Examples
(iii) (Recognizing a Composition of Functions) Given
2
√
F (x) =
, find f , g and h so that F = f ○ g ○ h.
(3 + x)2
√
2
Choose f (x) = , g(x) = x 2 and h(x) = 3 + x. Compute
x
√
√ 2
(f ○ g ○ h)(x) = f (g (h(x))) = f (g (3 + x)) = f ((3 + x) )
2
√
=
= F (x), as required.
(3 + x)2
One-to-One Functions and Their Inverses
Definition
A function with domain A is called a one-to-one function(or
1 − 1 function) if no two elements of A have the same images,
that is
f (x1 ) ≠ f (x2 )
whenever x1 ≠ x2 .
Equivalently,
If f (x1 ) = f (x2 ), then x1 = x2 .
For functions represented by diagrams:
The first two are 1 − 1 and the third is not.
One-to-One Functions and Their Inverses
For functions represented by graphs, we have
Horizontal Line Test: A function is 1 − 1 if and only if no
horizontal line intersects the graph more than once.
The first one is 1 − 1 and the second is not.
Deciding Whether Functions are One-to-One
Examples
1. f (x) = x 3
For any x1 , x2 ∈ R, we compute f (x1 ) − f (x2 ) = x13 − x23 =
x 2 3
(x1 − x2 )(x12 + x1 x2 + x22 ) = (x1 − x2 ) [(x1 − 2 ) + x22 ].
2
4
x2 2 3 2
0 = f (x1 ) − f (x2 ) ⇔ x1 − x2 = 0 or (x1 − ) + x2 = 0.
2
4
x1 − x2 = 0 ⇔ x1 = x2 .
x 2 3
x 2
3
(x1 − 2 ) + x22 = 0 ⇔ (x1 − 2 ) = 0 and x22 = 0 ⇔
2
4
2
4
x2
x1 =
and x2 = 0 ⇔ x1 = x2 = 0.
2
Hence f is one-to-one.
2. f (x) = x 2 .
It is not one-to-one, since f (1) = 12 = 1 = (−1)2 = f (−1).
The Inverse of a Function
Definition
Let f be a one-to-one function with domain A and range B.
Then the inverse function f −1 has domain B and range A and
is define by
f −1 (y) = x
⇐⇒ f (x) = y
for any y ∈ B.
Remarks
1. When we write f −1 (y ) = x, y is the independent variable
and x is the dependent variable for the inverse function.
2. f −1 (x) means the value of f −1 at x, but f (x)−1 means the
1
when f (x) ≠ 0.
reciprocal number of f (x); i.e.,
f (x)
Inverse Function Properties
Let f be a one-to-one function with domain A and range B. The
inverse function f −1 satisfying the following properties:
1. f −1 (f (x)) = x = (f −1 ○ f )(x) for all x ∈ A
⇔ f −1 ○ f = 1A , the identity function of A, that is 1A (x) = x
for any x ∈ A.
2. f (f −1 (x)) = x = (f ○ f −1 )(x) for all x ∈ B
⇔ f ○ f −1 = 1B , the identity function of B, that is 1B (x) = x
for any x ∈ B.
3. The graphs of y = f (x) and y = f −1 (x) are symmetric with
respect to the = y line in the plane. See
Inverse Function Properties
How to Find the Inverse of a 1 − 1 Function
Step (a) Write y = f (x).
Step (b) Solve the equation for x in terms of y (if possible).
Step (c) Interchange x and y. The resulting equation is y = f −1 (x).
Examples
1. f (x) = 2x + 5.
(a) y = 2x + 5. We note that its a linear equation and its
graph is a line in the coordinate plane with slope 2, so
x −5
x −5
y −5
. (c) y =
. Hence f −1 (x) =
.
1 − 1. (b) x =
2
2
2
2. f (x) = x 3 .
(a) y = x 3 . By the horizontal
line test, it is 1 − 1. (b)
√
√
y 1/3 = 3 y = x. (c) y = 3 x = x 1/3 ⇒ f −1 (x) = x 1/3 .
√
3. f (x) = x 2 . f −1 (x) = x, under the assumption that both
the domains and ranges are chosen to be [0, ∞).
How to Find the Inverse of a 1 − 1 Function
Examples
√
4. f (x) = 1 + 1 + x.
We first note that the domain of f = [−1, ∞) = A and the
range B = [1, ∞). Secondly, we note that f is 1 − 1. Indeed,
for x1 , x2 ∈ A, we compute
√
√
√
) = 1 + x1 −
f (x1 ) − f (x2 ) = (1 + 1 + x1 ) − (1√+ 1 + x1√
√
√
√
1 + x1 + 1 + x2
√
=
1 + x2 = [ 1 + x1 − 1 + x2 ] ⋅ √
1 + x1 + 1 + x2
(1 + x1 ) − (1 + x2 )
x1 − x2
√
√
√
=√
. Hence
1 + x1 + 1 + x2
1 + x1 + 1 + x2
f (x1 ) = √
f (x2 ) ⇔ x1 = x2 and
√ so f is 1 − 1. 2
y = 1 + 1 + x ⇔ y − 1 = 1 + x ⇔ (y − 1) = 1 + x ⇔ x =
(y − 1)2 − 1 = y 2 − 2y . Hence f −1 (x) = x 2 − x.