TABLE OF CONTENTS
1 FUNCTIONS
1.1 Definition and Examples . . . . . . . .
1.2 Domain and Range of a Function . . .
1.3 Operations on Functions . . . . . . . .
1.4 Functions as Mathematical Models . .
1.5 Linear Functions . . . . . . . . . . . .
1.6 Quadratic Functions . . . . . . . . . .
1.7 Rational Functions . . . . . . p
. . . . .
1.8 Functions of the Form f (x) = pg(x) .
1.9 Functions of the Form f (x) = 3 g(x) .
1.10 Exponential and Logarithmic Functions
1.11 Trigonometric Functions . . . . . . . .
1.12 Inverse Trigonometric Functions . . . .
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1
1
4
7
11
16
17
20
23
27
31
37
42
CHAPTER 1
FUNCTIONS
Learning Outcomes of the Lesson
At the end of the lesson, the student must be able to:
1. define function and find the value of a function at a real number.
2. define the sum, difference, product, and quotient of two functions; and find
their respective domains.
3. define the composite of two functions.
4. solve problems involving a function as a mathematical model.
5. find the domain and range of elementary functions.
6. define properties of elementary functions.
1.1
Definition and Examples
In this section, we limit our discussion to Cartesian product of real
numbers R × R = {(x, y) : x, y ∈ R}. We will define function as ordered pairs
of real numbers.
Definition 1.1.1 A function is a set f of ordered pairs in R × R = {(x, y) :
x, y ∈ R} such that no two distinct ordered pairs have the same first elements.
Example 1.1.2 The following sets are examples of functions in R × R.
1. f = {(1, 0), (2, 0), (3, 0), (4, 0)}.
2. f = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}.
3. f = {(x, y) ∈ R × R : 2x − y = 4}.
4. f = {(x, y) ∈ R × R : x2 −√y = 1}.
5. f = {(x, y) ∈ R × R : y = x − 1}.
6. f = {(x, y) ∈ R × R : 2x −√xy + y = 0}.
7. f = {(x, y) ∈ R × R : y = 3 x − 1}.
8. f = {(x, y) ∈ R × R : y = x3 }.
In the above definition, a function f is defined as a set of ordered pairs
(x, y) of real numbers. The numbers x and y are called variables. Since the
value of y is dependent on the value of x, we call x the independent variable
and y the dependent variable.
2
If (x, y) is an element of f , it is customary to write y = f (x) instead
of (x, y) ∈ f . We often refer to y as the value of f at the real number x, or
the image of the real number x under f .
Example 1.1.3 The functions in Example 2.1.2 can be written in the notation
y = f (x).
1. y = f (x) = 0, x ∈ {1, 2, 3, 4}.
2. y = f (x) = x2 , x ∈ {1, 2, 3, 4, 5, 6}.
3. y = f (x) = 2x − 4.
4. y = f (x) = √
x2 + 1.
5. y = f (x) = x − 1}.
2x
.
6. y = f (x) = x−1
√
3
7. y = f (x) = x − 1}.
8. y = f (x) = x3 .
Example 1.1.4 Let f be a function in R × R defined by f (x) = x2 − 3x + 4.
Find: (a) f (−1); (b) f (0); (c) f (2); (d) f (3a); (e) f (2x − 1); (f) f (x + h)
Solution: (a) f (−1) = (−1)2 − 3(−1) + 4 = 1 + 3 + 4 = 8;
(b) f (0) = (0)2 − (0) + 4 = 4;
(c) f (2) = (2)2 − 3(2) + 4 = 4 − 6 + 4 = 2;
(d) f (3a) = (3a)2 − 3(3a) + 4 = 9a2 − 9a + 4;
(e) f (2x−1) = (2x−1)2 −3(2x−1)+4 = 4x2 −4x+1−6x+3+4 = 4x2 −10x+8;
(f) f (x + h) = (x + h)2 − 3(x + h) + 4 = x2 − 2xh + h2 − 3x − 3h + 4. √
Example 1.1.5 Let f be a function in R × R defined by f (x) = x − 1.
Find: (a) f (1); (b) f (5); (c) f (9); (d) f (3a + 5); (e) f (2x − 1); (f) f (x + h)
√
√
Solution: (a)
f
(1)
=
1
−
1
=
0 = 0;
√
√
(b) f (5) = √ 5 − 1 = √ 4 = p
2;
√ √
√
(c) f (9) = √
9 − 1 = 8 = √4(2) = 4 2 = 2 2;
(d) f (3a) = 3a√+ 5 − 1 = 3a√+ 4;
(e) f (2x − 1) =√ 2x − 1 − 1 = 2x − 2;
(f) f (x + h) = x + h − 1. (
x + 4, if x ≤ −4
Example 1.1.6 Let f be a function in R×R defined by f (x) =
;
4 − x, if −4 < x
find (a) f (−6); (b) f (−4); (c) f (0); (d) f (4).
3
Solution: (a) If x ≤ −4, then f (x) = x + 4. Thus, f (−6) = −6 + 4 = −2;
(b) If x ≤ −4, then f (x) = x + 4. Thus, f (−4) = −4 + 4 = 0;
(c) If −4 < x, then f (x) = 4 − x. Thus, f (0) = 4 − 0 = 4;
(d) If −4 < x, then f (x) = 4 − x. Thus, f (4) = 4 − 4 = 0.


x + 3, if x < 2
Example 1.1.7 Let f be a function in R×R defined by f (x) = 4,
if x = 2 ;


2x − 1, if 2 < x
find (a) f (0); (b) f (2); (c) f (3); (d) f (−2).
Solution: (a) If x < 2, then f (x) = x + 3. Thus, f (0) = 0 + 3 = 3;
(b) If x = 2, then f (x) = 4. Thus, f (2) = 4;
(c) If 2 < x, then f (x) = 2x − 1. Thus, f (3) = 2(3) − 1 = 5;
(d) If x < 2, then f (x) = x + 3. Thus, f (−2) = −2 + 3 = 1.
Sample Problems 1.1.8 1. Given f (x) = 3x − 4, find (a) f (1); (b) f (−5);
(c) f (9); (d) f (3a + 5); (e) f (2x − 1); (f) f (x + h).
2. Given f (x) = 3x2 +2x−4, find (a) f (1); (b) f (−1); (c) f ( 21 ); (d) f (3a+5);
(e) f (2x − 1); (f) f (x + h).
x
3. Given f (x) =
, find (a) f (−2); (b) f (−1); (c) f ( 21 ); (d) f (3a + 5);
x−1
(e) f (2x − 1); (f)
√f (x + h).
4. Given f (x) = 9 − x, find (a) f (−2); (b) f (9−x); (c) f ( 21 ); (d) f (3a+5);
(e) f (2x − 1); (f) f (x + h).
(
x + 2, if x ≤ 0
5. Let f be a function defined by f (x) =
,
3x − 2, if 0 < x
find (a) f (−2); (b) f (−1); (c) f (0); (d)
( f (2).
x2 + 1, if x < 1
6. Let f be a function defined by f (x) =
,
3x − 1, if 1 ≥ x
find (a) f (−2); (b) f (−1); (c) f (0); (d)
 f (2).
2

if x < −1
x ,
7. Let f be a function defined by f (x) = 0,
if x = −1 ,


2x − 1, if −1 < x
find (a) f (−2); (b) f (−1); (c) f (0); (d) f (2).
4
1.2
Domain and Range of a Function
Definition 1.2.1 Let f be a function in R × R. Then the domain of f ,
denoted by D(f ), is the set of all real numbers x that occurs as first member
of the elements of f and the range of f , denoted by R(f ), is the set of all real
numbers y that ocuurs as second member of the elements of f .
To find the domain of a function y = f (x), solve D(f ) = {x ∈ R : f (x) ∈ R}.
To find the range of a function y = f (x), solve R(f ) = {y ∈ R : f −1 (y) ∈ R},
where
y = f (x)
⇔ x = f −1 (y).
Example 1.2.2 Let f be a function defined by f (x) = x2 − 1. Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= x ∈ R : x2 − 1 ∈ R
= {x ∈ R : x ∈ R}
= R.
f (x) = x2 − 1 ⇒ y = x2 − 1
⇒ x2 = y + 1
p
⇒x=± y+1
p
⇒ f −1 (y) = ± y + 1.
R(f ) = y ∈ R : f −1 (y) ∈ R
n
o
p
= y ∈R:± y+1∈R
= {y ∈ R : y + 1 ≥ 0}
= {y ∈ R : y ≥ −1}
= [−1, +∞). Example 1.2.3 Let f be a function defined by f (x) = 4 − x2 . Find D(f )
and R(f ).
5
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= x ∈ R : 4 − x2 ∈ R
= {x ∈ R : x ∈ R}
= R.
f (x) = x2 + 4 ⇒ y = 4 − x2
⇒ x2 = 4 − y
p
⇒x=± 4−y
p
⇒ f −1 (y) = ± 4 − y.
R(f ) = y ∈ R : f −1 (y) ∈ R
o
n
p
= y ∈R:± 4−y ∈R
= {y ∈ R : 4 − y ≥ 0}
= {y ∈ R : 4 ≥ y}
= {y ∈ R : y ≤ 4}
= (−∞, 4]. Example 1.2.4 Let f be a function defined by f (x) =
R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
2x
= x∈R:
∈R
x−3
= {x ∈ R : x − 3 6= 0}
= {x ∈ R : x 6= 3}
= (−∞, 3) ∪ (3, +∞).
2x
. Find D(f ) and
x−3
6
f (x) =
2x
2x
⇒y=
x−3
x−3
⇒ xy − 3y = 2x
⇒ xy − 2x = 3y
⇒ x(y − 2) = 3y
3y
⇒x=
y−2
3y
.
⇒ f −1 (y) =
y−2
R(f ) = y ∈ R : f −1 (y) ∈ R
3y
= y∈R:
∈R
y−2
= {y ∈ R : y − 2 6= 0}
= {y ∈ R : y 6= 2}
= (−∞, 2) ∪ (2, +∞). Example 1.2.5 Let f be a function defined by f (x) =
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
√
= x∈R: x−4∈R
= {x ∈ R : x − 4 ≥ 0}
= {x ∈ R : x ≥ 4}
= [4, +∞).
f (x) =
√
√
x−4⇒y = x−4
⇒ y2 = x − 4
⇒ x = y2 + 4
⇒ f −1 (y) = y 2 + 4.
√
x − 4. Find D(f )
7
R(f ) = y ∈ R : f −1 (y) ∈ R
√
= y ∈ R : y 2 + 4 ∈ R and y ≥ 0 , y = x − 4 ⇒ y ≥ 0
= {y ∈ R : y ∈ R and y ≥ 0}
= {y ∈ R : y ≥ 0}
= [0, +∞). Sample Problems 1.2.6 Find the domain and range of the following functions.
3x
1. f (x) = x2 − 5
2. f (x) = 2 − x2
3. f (x) =
x−2
√
√
x+4
4. f (x) = x − 3
6. f (x) = 5 − x.
5. f (x) =
x−5
1.3
Operations on Functions
Definition 1.3.1 Let f and g be functions in R × R. Then
(i) their sum, denoted by f + g, is the function defined by
(f + g)(x) = f (x) + g(x);
(ii) their difference, denoted by f − g, is the function defined by
(f − g)(x) = f (x) − g(x);
(iii) their product, denoted by f · g, is the function defined by
(f · g)(x) = f (x) · g(x);
f
, is the function defined by
g
f (x)
f
(x) =
.
g
g(x)
(iv) their quotient, denoted by
The domain of the resulting function consists of all real numbers x common
to the domain of f and g, with the additional requirement in case (iv) where
g(x) 6= 0 for all x.
8
Example 1.3.2 Let f and g be functions defined by f (x) = x2 − 1 and
g(x) = x − 1. Define the following functions and find their respective domains:
f
(a) f + g; (b) f − g; (c) f · g; and (d) .
g
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= x ∈ R : x2 − 1 ∈ R
= {x ∈ R : x ∈ R}
= R,
D(g) = {x ∈ R : g(x) ∈ R}
= {x ∈ R : x − 1 ∈ R}
= {x ∈ R : x ∈ R}
= R.
Hence,
D(f ) ∩ D(g) = R ∩ R = R.
(a) (f + g)(x) = x2 − 1 + x − 1 = x2 + x − 2 and
D(f + g) = D(f ) ∩ D(g) = R.
(b) (f − g)(x) = x2 − 1 − (x − 1) = x2 − x and
D(f − g) = D(f ) ∩ D(g) = R.
(c) (f · g)(x) = (x2 − 1)(x − 1) = x3 − x2 − x + 1 and
D(f · g) = D(f ) ∩ D(g) = R.
f
x2 − 1
(x + 1)(x − 1)
(d)
(x) =
=
= x + 1 and
g
x−1
x−1
f
D
= D(f ) ∩ D(g) ∩ {x ∈ R : g(x) 6= 0}
g
= R ∩ {x ∈ R : x − 1 6= 0}
= R ∩ {x ∈ R : x 6= 1}
= {x ∈ R : x 6= 1}
= (−∞, 1) ∪ (1, +∞). 9
x+4
and g(x) =
Example 1.3.3 Let f and g be functions defined by f (x) =
x−3
√
x − 1. Define the following functions: (a) f + g; (b) f − g; (c) f · g; and (d)
f
.
g
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x+4
= x∈R:
∈R
x−3
= {x ∈ R : x − 3 6= 0}
= {x ∈ R : x 6= 3}
= (−∞, 3) ∪ (3, +∞),
D(g) = {x ∈ R : g(x) ∈ R}
√
= x∈R: x−1∈R
= {x ∈ R : x − 1 ≥ 0}
= {x ∈ R : x ≥ 1}
= [1, +∞),
and
D(f ) ∩ D(g) = [(−∞, 3) ∪ (3, +∞)] ∩ [1, +∞) = [1, 3) ∪ (3, +∞).
√
+ x − 1 and
(a) (f + g)(x) = x+4
x−3
D(f + g) = D(f ) ∩ D(g) = [1, 3) ∪ (3, +∞).
(b) (f − g)(x) =
x+4 √
− x − 1 and
x−3
D(f − g) = D(f ) ∩ D(g) = [1, 3) ∪ (3, +∞).
√
x + 4√
(x + 4) x − 1
(c) (f · g)(x) =
x−1=
and
x−3
x−3
D(f · g) = D(f ) ∩ D(g) = [1, 3) ∪ (3, +∞).
10
x+4
x+4
f
√
√
(x) = x−3 =
and
(d)
g
x−1
(x − 3) x − 1
f
D
= [D(f ) ∩ D(g)] ∩ {x ∈ R : g(x) 6= 0}
g
√
= {[1, 3) ∪ (3, +∞)} ∩ {x ∈ R : x − 1 6= 0}
= {[1, 3) ∪ (3, +∞)} ∩ {x ∈ R : x − 1 6= 0}
= {[1, 3) ∪ (3, +∞)} ∩ {x ∈ R : x 6= 1}
= (1, 3) ∪ (3, +∞). Definition 1.3.4 Let f and g be functions in R × R. Then the the composite
function, denoted by f ◦ g, is the function defined by
(f ◦ g)(x) = f ((g(x)).
The domain of f ◦ g is the set of all real numbers x in the domain of g such
that g(x) is in the domain of f .
Example 1.3.5 Let f and g be functions defined by f (x) = x + 4 and g(x) =
√
x − 1. Define the following functions: (a) f ◦ g and (b) g ◦ f .
√
√
Solution: (a) (f ◦ g)(x) = f (g(x)) = f (√ x − 1) = x√− 1 + 4;
(b) (g ◦ f )(x) = g(f (x)) = g(x + 4) = x + 4 − 1 = x + 3. 2
Example
√ 1.3.6 Let f and g be functions defined by f (x) = x + 4 and
g(x) = x − 2. Define the following functions: (a) f ◦ g; (b) g ◦ f ; (c) f ◦ f
and; (d) g ◦ g.
√
√
Solution: (a) (f ◦ g)(x) = f (g(x)) = f ( x − 2) = ( x − 2)2 + 4 = x − 2 + 4 =
x + 2;
√
√
(b) (g ◦ f )(x) = g(f (x)) = g(x2 + 4) = x2 + 4 − 2 = x2 + 2.
2
2
4
2
(c) (f ◦ f )(x) = f (f (x)) = f (x2 + 4) = (x
p√+ 4) + 4 = x + 8x + 20;
√
(b) (g ◦ g)(x) = g(g(x)) = g( x − 2) =
x − 2 − 2. Sample Problems 1.3.7 1. Let f and g be functions defined by f (x) =
x2 − 4 and g(x) = x + 2. Define the following functions: (a) f + g; (b) f − g;
f
(c) f · g; and (d) .
g
√
2. Let f and g be functions defined by f (x) = x2 + 1 and g(x) = 2x + 3.
f
Define the following functions: (a) f + g; (b) f − g; (c) f · g; and (d) .
g
11
√
and g(x) = 3 x. Define the
f
following functions: (a) f + g; (b) f − g; (c) f · g; and (d) .
g
√
2
4. Let f and g be functions defined by f (x) = x − 3 and g(x) = x + 2.
Define the following functions: (a) f ◦ g; (b) g ◦ f√; (c) f ◦ f and; (d) g ◦ g.
5. Let f and g be functions defined by f (x) = x2 + 1 and g(x) = 3x − 4.
Define the following functions: (a) f ◦ g; (b) g ◦ f ; (c) f ◦ f and; (d) g ◦ g.
3. Let f and g be functions defined by f (x) =
1.4
x+2
x−5
Functions as Mathematical Models
In this lesson ,we will express a real-world situations using functions,
which is called a mathematical model of the situation. This will give
practice in obtaining functions as mathematical models and as preparation to
some of the applications in calculus such as extremum and related problems.
Suggestions for Solving Problems Involving Function as a Mathematical
Model
1. Draw a fugure if possible.
2. Determine the known and unknown quantities. Assign a symbol, say x,
for the independent variable and a symbol, say f (x), for the function to be
obtained.
3. Write any numerical facts about the variable and the function value and
determine two algebaic expressions for the same number. From these two
expressions form an equation that defines a function, which is the mathematical
model of the problem.
Example 1.4.1 A rectangular garden is to be fenced off with 30 m of fencing
material.
(a) Find a mathematical model expressing the area of the garden as a
function of its length.
(b) Find the domain of the function in (a).
(c) What is the area of the garden if the length is 12 m?
Solution: (a) Let x be the length of the garden, y be the width of the garden,
A be the area of the garden, and P = 30 be the perimeter of the garden. Then
A = xy
and
P = 2x + 2y.
Since P = 30, we have 30 = 2x + 2y. Solving for y, we get
12
A = xy
and
1
y = (30 − 2x = 15 − x.
2
Thus,
A = x(15 − x) = 15x − x2 .
Therefore, A is a function of x:
A(x) = 15x − x2 .
(b)D(A) = x ∈ R : 15x − x2 ≥ 0
= {x ∈ R : x2 − 15x ≤ 0}
= {x ∈ R : x(x − 15) ≤ 0}
= {x ∈ R : x ≥ 0 and x − 15 ≤ 0} ∪ {x ∈ R : x ≤ 0 and x − 15 ≥ 0}
= {x ∈ R : x ≥ 0 and x ≤ 15} ∪ {x ∈ R : x ≤ 0 and x ≥ 15}
= {x ∈ R : 0 ≤ x ≤ 15} ∪ { }
= {x ∈ R : 0 ≤ x ≤ 15}
= [0, 15].
(c) If x = 12 then A(12) = 15(12) − (12)2 = 180 − 144 = 36. Therefore,
if the length is 12 m, the area of the garden is 36 m2 . Example 1.4.2 A closed tin can of volume 16π in3 is to be made in the form
of a right circular cylinder.
(a) Find a mathematical model expressing the total surface area of the
can as a function of the radius.
(b) Find the domain of the function in (a).
(c) What is the surface area of the can when the radius is 10 in?
Solution: (a) Let r be the radius of the can, h be the height of the can, A be
the surface area of the can, and V = 60 be the volume of the can. Then
A = 2πrh + πr2 h
and
V = πr2 .
Since V = 60, we have 60 = πr2 h. Solving for h, we get
A = 2πrh + πr2
Thus,
and
h=
60
.
πr2
13
A = 2πr
60
r2
+ πr2 =
120π
+ πr2 .
r
Therefore, A is a function of r:
A(r) =
120π
+ πr2 .
r
120π
2
+ πr ∈ R and r > 0
(b)D(A) = r ∈ R :
r
= {r ∈ R : r 6= 0 and r > 0}
= {r ∈ R : r > 0}
= (0, +∞).
120π
+ π(10)r2 = 112π. Therefore, when
10r
the radius is 10 in., the surface area of the can is 112π in2 . (c) If r = 10 then A(10) =
Example 1.4.3 A sheet of cardboard 4 ft by 6 ft is to made into an open box
by cutting equal squares from each corner and turning up the sides.
(a) Find a mathematical model expressing the volume of the box as a
function of the length of the side of the square cut out.
(b) Find the domain of the function in (a).
(c) What is the volume of the box if the length of the square cut out is
0.5 ft?
Solution: (a) Let x be the length of the square cut out (this becomes the height
of the box, y be the width of the box, z be the length of the box, and V be
the volume of the box. Then
V = xy,
and
y = 4 − 2x,
z = 6 − 2x.
Thus,
V = x(4 − 2x)(6 − 2x).
Therefore, V is a function of x:
V (x) = x(4 − 2x)(6 − 2x).
14
(b)D(A) = {x ∈ R : x(4 − 2x)(6 − 2x) ≥ 0}
= {x ∈ R : x ≥ 0 and 4 − 2x ≥ 0 and 6 − 2x ≥ 0}, since x ≥ 0, y ≥ 0, z ≥ 0
= {x ∈ R : x ≥ 0 and x ≤ 2 and x ≤ 3}
= {x ∈ R : 0 ≤ x ≤ 2}
= [0, 2].
(c) If x = 0.5 then V (0.5) = 0.5[4 − 2(0.5)][6 − 2(0.5)] = 0.5[3][5] = 7.5.
Therefore, if the length of the square cut out is 6 in, the volume of the box is
7.5 ft3 . Example 1.4.4 An online seller sells a certain product by the kg; If not more
than 20 kg are ordered, the online seller charges 4 pesos per kg and if more
than 20 kg are ordered, the online seller charges 3 pesos per kg.
(a) Find a mathematical model expressing the total cost of the order
as a function of the amount of the product ordered.
(b) Find the domain of the function in (a).
(c) What is the total cost of an order of 16 kg?
(d) What is the total cost of an order of 25 kg?
Solution: (a) Let x be the number of kg of an order and C be the total cost
of an order. Then
(
4x, if 0 ≤ x ≤ 20
C=
.
3x, if 20 < x
Therefore, C is a function of x given by
(
4x, if 0 ≤ x ≤ 20
C(x) =
.
3x, if 20 < x
(b)D(C) = {x ∈ R : 0 ≤ x ≤ 20} ∪ {x ∈ R : 20 < x}}
= [0, 20] ∪ (20, +∞)
= [0, +∞).
(c) If 0 ≤ x ≤ 20, then C(x) = 4x. Thus, if x = 16, then C(16) =
4(16) = 64. Therefore, the total cost of an oder of 16 kg is 64 pesos.
(d) If 20 < x, then C(x) = 3x. Thus, if x = 25, then C(25) = 3(25) =
75. Therefore, the total cost of an oder of 25 kg is 75 pesos. 15
Sample Problems 1.4.5 In each of the following problem, find a mathematical
model of a particular situation.
1. A rectangular field is to be enclosed with 200 m of fence.
(a) Find a mathematical model expressing the total area of the field as
a function of its length.
(b) Find the domain of the function in (a).
(c) What is the area of the field if its length is 60 m?
2. A rectangular garden is to be fenced off with 100 m of fencing material.
(a) Find a mathematical model expressing the area of the garden as a
function of its length.
(b) Find the domain of the function in (a).
(c) What is the area of the garden if the length is 40 m?
3. A piece of tin with dimensions 8 in by 15 in will be made into an open box
by cutting equal squares from four corners and turning up the sides.
(a) Find a mathematical model expressing the volume of the box as a
function of the length of the side of the square cut out.
(b) Find the domain of the function in (a).
(c) What is the volume of the box if the length of the square cut out is
2 in?
4. A piece of cardboard with dimensions 40 cm by 60 cm will be made into an
open box by cutting equal squares from four corners and turning up the sides.
(a) Find a mathematical model expressing the volume of the box as a
function of the length of the side of the square cut out.
(b) Find the domain of the function in (a).
(c) What is the volume of the box if the length of the square cut out is
5 cm?
5. A seller sells a product by the grams; If less than 10 g are ordered, the seller
charges 50 pesos per gram and if at least 10 g are ordered, the seller charges
45 pesos per gram.
(a) Find a mathematical model expressing the total cost of the order
as a function of the amount of the product ordered.
(b) Find the domain of the function in (a).
(c) What is the total cost of an order of 9 g?
(d) What is the total cost of an order of 12 g?
6. The regular adult admission to an evening performance at a downtown
16
theater is 120 pesos, while the price for persons under 18 years of age is 100
pesos.
(a) Find a mathematical model expressing the admission price as a
function of the person’s age.
(b) Find the domain of the function in (a).
(c) What is the admission price if the person is 12 years old?
(d) What is the admission price if the person is 25 years old?
1.5
Linear Functions
Definition 1.5.1 A function f in R × R defined by f (x) = ax + b, where
a, b ∈ R and a 6= 0, is called a linear function.
Theorem 1.5.2 Let f (x) = ax + b, where a, b ∈ R and a 6= 0. Then
(i) D(f ) = R and R(f ) = R;
(ii) the graph of f is a line which is increasing if a > 0 and is decreasing
if a < 0.
Example 1.5.3 Let f be a function defined by f (x) = 2x + 4. Find D(f )
and R(f ). Identify the graph of f and determine whether it is increasing or
decereasing. Sketch the graph of f .
Solution: D(f ) = R and R(f ) = R.
The graph of f is a line which is increasing since a = 2 > 0.
Example 1.5.4 Let f (x) = −2x+2. Find D(f ) and R(f ). Identify the graph
of f and determine whether it is increasing or decereasing. Sketch the graph
of f .
Solution: D(f ) = R and R(f ) = R.
The graph of f is a line which is decreasing since a = −2 < 0.
Sample Problems 1.5.5 Given the following linear functions, find D(f ) and
R(f ); and determine whether the graph of f is increasing or decereasing.
Sketch the graph of f .
1. f (x) = 3x + 6
2. f (x) = 3x − 6
3. f (x) = −2x + 10
4. f (x) = −4x + 6
5. f (x) = 2x − 5
6. f (x) = −5x − 10
17
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1.6
Quadratic Functions
Definition 1.6.1 A function f in R×R defined by f (x) = ax2 +bx+c, where
a, b, c ∈ R and a 6= 0 is called a quadratic function.
Theorem 1.6.2 Let f (x) =ax2 + bx + c, where
a, b, c ∈ R and a 6= 0. Then
b 4ac − b2
;
(i) the vertex is V = − ,
2a
4a
18
(ii) D(f ) = R;
4ac − b2
4ac − b2
(iii) R(f ) =
, +∞ , if a > 0 and R(f ) = −∞,
,
4a
4a
if a < 0.
(iv) the graph of f is a parabola opening upward if a > 0 and is a
parabola opening downward if a < 0.
Example 1.6.3 Let f (x) = x2 − 2x. Find the vertex, D(f ), and R(f ).
Determine whether the graph of f is opening upward or opening downward.
Skecth the graph of f .
Solution: Let a = 1, b = −2, and c = 0. Then
−
−2
4ac − b2
4(1)(0) − (−2)2
b
=−
= 1 and
=
= −1.
2a
2(1)
4a
4(1)
Hence,
V = (1, −1),D(f ) = R, andR(f ) = [−1, +∞).
Since a > 0, the graph is a parabola opening upward. We may construct a
table of values:
0
0
x
y = f (x)
1
-1
2
0
Using these three points, we sketch a graph of the parabola.
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•
•
•
V = (1, −1)
19
Example 1.6.4 Let f (x) = −x2 + 2x + 2. Find the vertex, D(f ), and R(f ).
Determine whether the graph of f is opening upward or opening downward.
Sketch the graph of f .
Solution: Let a = −1, b = 2, and c = 2. Then
−
2
4ac − b2
4(−1)(2)
b
=−
= 1 and
=
= 3.
2a
2(−1)
4a
4(−1)
Hence,
V = (1, 3), D(f ) = R, and R(f ) = (−∞, 3].
Since a < 0, the graph is a parabola opening upward. Construct a table of
values.
x
y
0
2
1
3
2
2
Using these three points, sketch a graph of the parabola.
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V• = (1, 3)
•
•
Sample Problems 1.6.5 . Let f be a quadratic function. Find the vertex,
D(f ), and R(f ). Determine whether the graph of f is opening upward or
opening downward. Sketch the graph of f .
1. f (x) = x2 −2x+3.
2. f (x) = x2 +4x+5.
3. f (x) = −x2 −2x−3.
4. f (x) = −2x2 +4x.
5. f (x) = x2 −4x+1.
6. f (x) = −x2 +3x−3.
20
1.7
Rational Functions
Definition 1.7.1 A function f in R × R defined by f (x) = p(x)
, where p(x)
q(x)
and q(x) are polynomial functions and q(x) 6= 0, is called a rational function.
, where p(x) and q(x) are polynomial functions
Theorem 1.7.2 Let f (x) = p(x)
q(x)
and q(x) 6= 0. Then
(i) D(f ) = {x ∈ R : f (x) ∈ R} = {x ∈ R : q(x) 6= 0}.
(ii) R(f ) = {y ∈ R : f −1 (y) ∈ R}.
Example 1.7.3 Let f be a function defined by f (x) =
1
. Find D(f ) and
x−2
R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
1
∈R
= x∈R:
x−2
= {x ∈ R : x − 2 6= 0}
= {x ∈ R : x 6= 2}
= (−∞, 2) ∪ (2, +∞).
f (x) =
1
1
⇒y=
x−2
x−2
⇒ xy − 2y = 1
⇒ xy = 2y + 1
2y + 1
⇒x=
y
2y + 1
⇒ f −1 (y) =
.
y
R(f ) = y ∈ R : f −1 (y) ∈ R
2y + 1
= y∈R:
∈R
y
= {y ∈ R : y 6= 0}
= (−∞, 0) ∪ (0, +∞). 21
Example 1.7.4 Let f be a function defined by f (x) =
x+1
. Find D(f ) and
x+3
R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x+1
= x∈R:
∈R
x+3
= {x ∈ R : x + 3 6= 0}
= {x ∈ R : x 6= −3}
= (−∞, −3) ∪ (−3, +∞).
f (x) =
x+1
x+1
⇒y=
x+3
x+3
⇒ xy + 3y = x + 1
⇒ xy − x = 1 − 3y
⇒ x(y − 1) = 1 − 3y
1 − 3y
⇒x=
y−1
1 − 3y
.
⇒ f −1 (y) =
y−1
R(f ) = y ∈ R : f −1 (y) ∈ R
1 − 3y
= y∈R:
∈R
y−1
= {y ∈ R : y − 1 6= 0}
= {y ∈ R : y 6= 1}
= (−∞, 1) ∪ (1, +∞). Example 1.7.5 Let f be a function defined by f (x) =
R(f ).
x
. Find D(f ) and
x−2
22
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x
∈R
= x∈R:
x−2
= {x ∈ R : x − 2 6= 0}
= {x ∈ R : x 6= 2}
= (−∞, 2) ∪ (2, +∞).
f (x) =
x
x
⇒y=
x−2
x−2
⇒ xy − 2y = x
⇒ xy − x = 2y
⇒ x(y − 1) = 2y
2y
⇒x=
y−1
2y
⇒ f −1 (y) =
.
y−1
R(f ) = y ∈ R : f −1 (y) ∈ R
2y
= y∈R:
∈R
y−1
= {y ∈ R : y − 1 6= 0}
= {y ∈ R : y 6= 1}
= (−∞, 1) ∪ (1, +∞). Example 1.7.6 Let f be a function defined by f (x) =
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
4 − 3x
= x∈R:
∈R
2x
= {x ∈ R : 2x 6= 0}
= {x ∈ R : x 6= 0}
= (−∞, 0) ∪ (0, +∞).
4 − 3x
. Find D(f )
2x
23
f (x) =
4−x
4−x
⇒y=
2x
2x
⇒ 2xy = 4 − x
⇒ 2xy + x = 4
⇒ x(2y + 1) = 4
4
⇒x=
2y + 1
4
⇒ f −1 (y) =
.
2y + 1
R(f ) = y ∈ R : f −1 (y) ∈ R
4
= y∈R:
∈R
2y + 1
= {y ∈ R : 2y + 1 6= 0}
1
= y ∈ R : y 6= −
2
1
1
∪ − , +∞ . = −∞, −
2
2
Sample Problems 1.7.7 . Given the following rational functions, find D(f )
and R(f ).
3x
2
x−2
1. f (x) =
2. f (x) =
3. f (x) =
x−1
x+5
x+2
1
x+2
2x − 5
4. f (x) =
5. f (x) =
6. f (x) =
4−x
x−2
4x + 3
1.8
Functions of the Form f (x) =
p
g(x)
Theorem 1.8.1 Let f be a function in R×R defined by f (x) =
g(x) ≥ 0. Then
(i) D(f ) = {x ∈ R : f (x) ∈ R} = {x ∈ R : g(x) ≥ 0}.
(ii) R(f ) = {y ∈ R : f −1 (y) ∈ R and y ≥ 0}.
p
g(x), where
24
Example 1.8.2 Let f be a function defined by f (x) =
and R(f ).
√
x − 1. Find D(f )
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
√
= x∈R: x−1∈R
= {x ∈ R : x − 1 ≥ 0}
= {x ∈ R : x ≥ 1}
= [1, +∞).
f (x) =
√
√
x−1⇒y = x−1
⇒ y2 = x − 1
⇒ x = y2 + 1
⇒ f −1 (y) = y 2 + 1.
R(f ) = y ∈ R : f −1 (y) ∈ R and y ≥ 0
= y ∈ R : y 2 + 1 ∈ R and y ≥ 0
= {y ∈ R : y ∈ R and y ≥ 0}
= {y ∈ R : y ≥ 0}
= [0, +∞). √
Example 1.8.3 Let f be a function defined by f (x) = x2 − 4. Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
n
o
√
2
= x∈R: x −4∈R
= x ∈ R : x2 − 4 ≥ 0
= {x ∈ R : (x + 2)(x − 2) ≥ 0}, r1 = −2 and r2 = 2
= (−∞, −2] ∪ [2, +∞).
f (x) =
√
√
x2 − 4 ⇒ y = x2 − 4
⇒ y 2 = x2 − 4
⇒ x2 = y 2 + 4
p
⇒ x = ± y2 + 4
p
⇒ f −1 (y) = ± y 2 + 4.
25
R(f ) = y ∈ R : f −1 (y) ∈ R and y ≥ 0
o
n
p
= y ∈ R : ± y 2 + 4 ∈ R and y ≥ 0
= y ∈ R : y 2 + 4 ≥ 0 and y ≥ 0
= {y ∈ R : y ∈ R and y ≥ 0}
= {y ∈ R : y ≥ 0}
= [0, +∞). √
Example 1.8.4 Let f be a function defined by f (x) = 1 − x2 . Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
n
o
√
= x ∈ R : 1 − x2 ∈ R
= {x ∈ R : 1 − x2 ≥ 0}
= {x ∈ R : x2 − 1 ≤ 0}
= {x ∈ R : (x + 1)(x − 1) ≤ 0}, r1 = −1 and r2 = 1
= [−1, 1].
f (x) =
√
√
1 − x2 ⇒ y = 1 − x2
⇒ y 2 = 1 − x2
⇒ x2 = 1 − y 2
p
⇒ x = ± 1 − y2
p
⇒ f −1 (y) = ± 1 − y 2 .
R(f ) = y ∈ R : f −1 (y) ∈ R and y ≥ 0
n
o
p
2
= y ∈ R : ± 1 − y ∈ R and y ≥ 0
= {y ∈ R : 1 − y 2 ≥ 0 and y ≥ 0}
= {y ∈ R : y 2 − 1 ≤ 0 and y ≥ 0}
= {y ∈ R : (y + 1)(y − 1) ≤ 0 and y ≥ 0}
= [−1, 1] ∩ [0, +∞)
= [0, 1]. 26
r
Example 1.8.5 Let f be a function defined by f (x) =
x+1
. Find D(f )
x−2
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
(
)
r
x+1
= x∈R:y=
∈R
x−2
x+1
= x∈R:
≥ 0 , r1 = −1 and r2 = 2
x−2
= (−∞, −1] ∪ (2, +∞).
r
f (x) =
x+1
⇒y=
x−2
r
x+1
x−2
x+1
⇒ y2 =
x−2
⇒ xy 2 − 2y 2 = x + 1
⇒ xy 2 − x = 2y 2 + 1
⇒ x(y 2 − 1) = 2y 2 + 1
2y 2 + 1
⇒x= 2
y −1
2y 2 + 1
−1
⇒ f (y) = 2
.
y −1
R(f ) = y ∈ R : f −1 (y) ∈ R and y ≥ 0
2y 2 + 1
= y∈R: 2
∈ R and y ≥ 0
y −1
= y ∈ R : y 2 − 1 6= 0 and y ≥ 0
= {y ∈ R : (y + 1)(y − 1) 6= 0 and y ≥ 0}
= {y ∈ R : y + 1 6= 0 and y − 1 6= 0 and y ≥ 0}
= {y ∈ R : y 6= −1 and y 6= 1 and y ≥ 0}
= {y ∈ R : y 6= 1 and y ≥ 0}
= [0, +∞)\{1}
= [0, 1) ∪ (1, +∞). 27
Sample Problems 1.8.6 Find the domain and range of the following functions.
1. f (x) =
4. f (x) =
7. f (x) =
1.9
√
√
√
3x + 5
5x − 6
x2 − 9
√
x − 16
r
x+5
5. f (x) =
x+3
r
1
8. f (x) =
4−x
2. f (x) =
Functions of the Form f (x) =
p
3
√
9−x
r
4x
6.f (x) =
x−6
√
9. f (x) = 25 − x2 .
3.f (x) =
g(x)
p
Theorem 1.9.1 Let f be a function in R×R defined by f (x) = 3 g(x), where
g(x) is a function in R × R. Then
(i) D(f ) = {x ∈ R : f (x) ∈ R} = {x ∈ R : g(x) ∈ R}.
(ii) R(f ) = {y ∈ R : f −1 (y) ∈ R}.
√
Example 1.9.2 Let f be a function defined by f (x) = 3 x − 1. Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
√
= x∈R: 3x−1∈R
= {x ∈ R : x − 1 ∈ R}
= {x ∈ R : x ∈ R}
= R.
f (x) =
√
√
3
x−1⇒y = 3x−1
⇒ y3 = x − 1
⇒ x = y3 + 1
⇒ f −1 (y) = y 3 + 1.
R(f ) = y ∈ R : f −1 (y) ∈ R
= y ∈ R : y3 + 1 ∈ R
= {y ∈ R : y ∈ R}
= R. 28
Example 1.9.3 Let f be a function defined by f (x) =
and R(f ).
√
3
x2 − 1. Find D(f )
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
n
o
√
3
= x ∈ R : x2 − 1 ∈ R
= x ∈ R : x2 − 1 ∈ R
= {x ∈ R : x ∈ R}
= R.
f (x) =
√
√
3
3
x2 − 1 ⇒ y = x2 − 1
⇒ y 3 = x2 − 1
⇒ x2 = y 3 + 1
p
⇒ x = ± y3 + 1
p
⇒ f −1 (y) = ± y 3 + 1.
R(f ) = y ∈ R : f −1 (y) ∈ R
n
o
p
= y ∈ R : ± y3 + 1 ∈ R
= y ∈ R : y3 + 1 ≥ 0
= y ∈ R : y 3 ≥ −1
√
= y ∈ R : y ≥ 3 −1
= {y ∈ R : y ≥ −1}
= [−1, +∞). Example 1.9.4 Let f be a function defined by f (x) =
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
n
o
√
3
= x ∈ R : 1 − x2 ∈ R
= {x ∈ R : 1 − x2 ∈ R}
= {x ∈ R : x ∈ R
= R.
√
3
1 − x2 . Find D(f )
29
f (x) =
√
3
√
3
1 − x2 ⇒ y = 1 − x2
⇒ y 3 = 1 − x2
⇒ x2 = 1 − y 3
p
⇒ x = ± 1 − y3
p
⇒ f −1 (y) = ± 1 − y 3 .
R(f ) = y ∈ R : f −1 (y) ∈ R
o
n
p
= y ∈ R : ± 1 − y3 ∈ R
= {y ∈ R : 1 − y 3 ≥ 0}
= {y ∈ R : y 3 − 1 ≤ 0}
= {y ∈ R : y 3 ≤ 1}
√
3
= {y ∈ R : y ≤ 1}
= {y ∈ R : y ≤ 1}
= (−∞, 1]. r
Example 1.9.5 Let f be a function defined by f (x) =
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
(
)
r
x
+
1
∈R
= x∈R:y= 3
x−2
x+1
= x∈R:
∈R
x−2
= {x ∈ R : x − 2 6= 0}
= {x ∈ R : x 6= 2}
= (−∞, 2) ∪ (2, +∞).
3
x+1
. Find D(f )
x−2
30
r
f (x) =
3
x+1
⇒y=
x−2
r
x+1
x−2
x
+1
⇒ y3 =
x−2
3
⇒ xy − 2y 3 = x + 1
⇒ xy 3 − x = 2y 3 + 1
⇒ x(y 3 − 1) = 2y 3 + 1
2y 3 + 1
⇒x= 3
y −1
2y 3 + 1
.
⇒ f −1 (y) = 3
y −1
3
R(f ) = y ∈ R : f −1 (y) ∈ R
2y 3 + 1
= y∈R: 3
∈R
y −1
= y ∈ R : y 3 − 1 6= 0
= y ∈ R : y 3 6= 1
= {y ∈ R : y 6= 1}
= (−∞, 1) ∪ (1, +∞). r
Example 1.9.6 Let f be a function defined by f (x) =
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
(
)
r
2x
= x∈R: 3
∈R
4−x
2x
= x∈R:
∈R
4−x
= {x ∈ R : 4 − x 6= 0}
= {x ∈ R : x 6= 4}
= (−∞, 4) ∪ (4, +∞).
3
2x
. Find D(f )
4−x
31
r
f (x) =
3
2x
⇒y=
4−x
r
2x
4−x
2x
⇒ y3 =
4−x
3
⇒ 4y − xy 3 = 2x
⇒ xy 3 + 2x = 4y 3
⇒ x(y 3 + 2) = 4y 3
4y 3
⇒x= 3
y +2
4y 3
⇒ f −1 (y) = 3
.
y +2
3
R(f ) = y ∈ R : f −1 (y) ∈ R
4y 3
= y∈R: 3
∈R
y +2
= y ∈ R : y 3 + 2 6= 0
= y ∈ R : y 3 6= −2
√
= y ∈ R : y 6= 3 −2
n
o
√
3
= y ∈ R : y 6= − 2
√
√
3
3
= (−∞, − 2) ∪ (− 2, +∞). Sample Problems 1.9.7 Find the domain and range of the following functions.
√
3
1. f (x) = r
3x + 5
x+5
4. f (x) = 3
x+3
1.10
√
2. f (x) =r3 8 − x2
4x
5.f (x) = 3
x−6
√
3. f (x) =r3 x2 − 27
2x + 5
6. f (x) = 3
4 − 3x
Exponential and Logarithmic Functions
Definition 1.10.1 A function f in R × R defined by f (x) = bx , where b > 0
and b 6= 1, is called an exponential function.
32
Example 1.10.2 The following are examples of exponential functions:
f (x) = ex , f (x) = 2x , f (x) = 10x , f (x) = ( 21 )x , f (x) = ( 23 )x .
Theorem 1.10.3 Let f be a function defined by f (x) = bu(x) , where b > 0,
b 6= 1, and u(x) is a function in R × R. Then
(i) D(f ) = {x ∈ R : f (x) = bu(x) ∈ R} = {x ∈ R : u(x) ∈ R}.
(ii) R(f ) = {y ∈ R : f − (y) ∈ R}.
Definition 1.10.4 A function f in R × R defined by f (x) = logb x, where
b > 0 and b 6= 1, is called a logarithmic function.
Example 1.10.5 The following are examples of logarithmic functions:
f (x) = loge x, f (x) = log2 x, f (x) = log10 x, f (x) = log 1 x, f (x) = log 2 x.
2
3
Remark 1.10.6 For x > 0 and b > 0, b 6= 1, y = logb x is equivalent to
x = by .
Remark 1.10.7 For x > 0, y = loge x is equivalent to y = ln x.
Theorem 1.10.8 Let f be a function defined by f (x) = logb u(x), where b >
0, b 6= 1, u(x) is a function with u(x) > 0 for all x. Then
(i) D(f ) = {x ∈ R : f (x) = logb u(x) ∈ R} = {x ∈ R : u(x) > 0}.
(ii) R(f ) = {y ∈ R : f −1 (y) ∈ R}.
Example 1.10.9 Let f be a function defined by f (x) = log2 x. Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= {x ∈ R : log2 x ∈ R}
= {x ∈ R : x > 0}
= (0, +∞).
f (x) = log2 x ⇒ y = log2 x
⇒ 2y = x
⇒ x = 2y
⇒ f −1 (y) = 2y .
33
R(f ) = {y ∈ R : f −1 (y) ∈ R}
= {y ∈ R : 2y ∈ R}
= {y ∈ R : y ∈ R}
= R.
Example 1.10.10 Let f be a function defined by f (x) = log3 (3x − 6). Find
D(f ) and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= {x ∈ R : log3 (3x − 6) ∈ R}
= {x ∈ R : 3x − 6 > 0}
= {x ∈ R : x > 2}
= (2, +∞).
f (x) = log3 (3x − 6) ⇒ y = log3 (3x − 6)
⇒ 3y = 3x − 6
3y + 6
⇒x=
3
⇒ x = 3y−1 + 2
⇒ f −1 (y) = 3y−1 + 2.
R(f ) = {y ∈ R : f −1 (y) ∈ R}
= {y ∈ R : 3y−1 + 2 ∈ R}
= {y ∈ R : y − 1 ∈ R}
= {y ∈ R : y ∈ R}
= R.
Example 1.10.11 Let f be a function defined by f (x) = ln
D(f ) and R(f ).
√
4 − x. Find
34
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
√
= {x ∈ R : ln 4 − x ∈ R}
√
= {x ∈ R : 4 − x > 0}
= {x ∈ R : 4 − x > 0}
= {x ∈ R : x < 4}
= (−∞, 4).
√
√
f (x) = ln 4 − x ⇒ y = ln 4 − x
√
⇒ ey = 4 − x
⇒ e2y = 4 − x
⇒ x = 4 − e2y
⇒ f −1 (y) = 4 − e2y .
R(f ) = {y ∈ R : f −1 (y) ∈ R}
= {y ∈ R : 4 − e2y ∈ R}
= {y ∈ R : 2y ∈ R}
= {y ∈ R : y ∈ R}
= R. Example 1.10.12 Let f be a function defined by f (x) = ln
x−1
. Find D(f )
x−4
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x−1
= x ∈ R : ln
∈R
x−4
x−1
= x∈R:
> 0 , r1 = 1 and r2 = 4
x−4
= (−∞, 1) ∪ (4, +∞).
35
x−1
x−1
⇒ y = ln
x−4
x−4
x
−1
⇒ ey =
x−4
y
⇒ xe − 4ey = x − 1
⇒ xey − x = 4ey − 1
⇒ x(ey − 1) = 4ey − 1
4ey − 1
⇒x= y
e −1
4ey − 1
⇒ f −1 (y) = y
.
e −1
R(f ) = y ∈ R : f −1 (y) ∈ R
4ey − 1
∈R
= y∈R: y
e −1
= {y ∈ R : ey − 1 6= 0}
= {y ∈ R : ey 6= 1}
= {y ∈ R : y 6= 0}
= (−∞, 0) ∪ (0, +∞). f (x) = ln
Example 1.10.13 Let f be function defined by f (x) = 2x . Find D(f ) and
R(f )..
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= {x ∈ R : 2x ∈ R}
= {x ∈ R : x ∈ R}
= R.
f (x) = 2x ⇒ y = 2x
⇒ x = log2 y
⇒ f −1 (y) = log2 y.
R(f ) = y ∈ R : f −1 (y) ∈ R
= {y ∈ R : log2 y ∈ R}
= {y ∈ R : y > 0}
= (0, +∞).
36
Example 1.10.14 Let f be a function defined by f (x) = 3x+1 . Find D(f )
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= x ∈ R : 3x+1 ∈ R
= {x ∈ R : x + 1 ∈ R}
= {x ∈ R : x ∈ R}
= R.
f (x) = 3x+1 ⇒ y = 3x+1
⇒ x + 1 = log3 y
⇒ x = log3 y − 1
⇒ f −1 (y) = log3 y − 1.
R(f ) = y ∈ R : f −1 (y) ∈ R
= {y ∈ R : log3 y − 1 ∈ R}
= {y ∈ R : y > 0}
= (0, +∞).
Example 1.10.15 Let f be a function defined by f (x) = ex
and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
n
o
x2 +1
= x∈R:e
∈R
= x ∈ R : x2 + 1 ∈ R
= {x ∈ R : x ∈ R}
= R.
f (x) = ex
2 +1
2
⇒ y = ex +1
⇒ x2 + 1 = ln y
⇒ x2 = ln y − 1
p
⇒ x = ± ln y − 1
p
⇒ f −1 (y) = ± ln y − 1.
2 +1
. Find D(f )
37
R(f ) = y ∈ R : f −1 (y) ∈ R
o
n
p
= y ∈ R : ± ln y − 1 ∈ R
= {y ∈ R : ln y − 1 ≥ 0}
= {y ∈ R : ln y ≥ 1}
= {y ∈ R : ln y ≥ ln e}
= {y ∈ R : y ≥ e}
= [e, +∞).
Sample Problems 1.10.16 Given the following functions, find the domain
the range.
1. f (x) = log5 (x + 4)
2. f (x) = log3 (x + 6)
3. f (x) = log10 (2 − x)
2
4. f (x) = log10 (5 − x)
5. f (x) = 2x+2
6. f (x) = ex +4
x−4
7. f (x) = 12
8. f (x) = 2x+2
9. f (x) = 102x−3
1.11
Trigonometric Functions
Definition 1.11.1 A function f in R × R defined by f (x) = sin x is called
the sine function.
Theorem 1.11.2 Let f be a function defined by f (x) = sin x. Then
D(f ) = R and R(f ) = [−1, 1].
Theorem 1.11.3 Let f be a function defined by f (x) = sin u(x), where u(x)
is a function in R × R. Then
D(f ) = {x ∈ R : u(x) ∈ R}.
Example 1.11.4 Let f be a function defined by f (x) = sin(x2 − 2x + 5).
Find D(f ).
Solution:
D(f ) = x ∈ R : x2 − 2x + 5 ∈ R
= {x ∈ R : x ∈ R}
= R.
38
Example 1.11.5 Let f be a function defined by f (x) = sin
x+4
. Find
x−4
D(f ).
Solution:
x+4
D(f ) = x ∈ R :
∈R
x−4
= {x ∈ R : x − 4 6= 0}
= {x ∈ R : x 6= 4}
= (−∞, 4) ∪ (4, +∞).
Definition 1.11.6 A function f in R × R defined by f (x) = cos x is called
the cosine function.
Theorem 1.11.7 Let f (x) = cos x. Then
D(f ) = R and R(f ) = [−1, 1].
Theorem 1.11.8 Let f be a function defined by f (x) = cos u(x), where u(x)
is a function in R × R. Then
D(f ) = {x ∈ R : u(x) ∈ R}.
Example 1.11.9 Let f be a function defined by f (x) = cos
D(f ).
√
x − 1. Find
Solution:
√
D(f ) = x ∈ R : x − 1 ∈ R
= {x ∈ R : x − 1 ≥ 0}
= {x ∈ R : x ≥ 1}
= [1, +∞).
Example 1.11.10 Let f be a function defined by f (x) = cos ln(x − 4). Find
D(f ).
Solution:
D(f ) = {x ∈ R : ln(x − 4) ∈ R}
= {x ∈ R : x − 4 > 0}
= {x ∈ R : x > 4}
= (4, +∞).
39
Definition 1.11.11 A function f in R × R defined by f (x) = tan x is called
the tangent function.
Theorem 1.11.12n Let f be a function definedoby f (x) = tan x. Then
π
(i) D(f ) = x ∈ R : x 6= + πk, k ∈ Z .
2
(ii) R(f ) = R.
π
(iii) The lines x = + πk, k ∈ Z are the vertical asymptotes of the
2
graph of f .
Theorem 1.11.13 Let f be a function defined by f (x) = tan u(x), where u(x)
is a function in R × R. Then
o
n
π
D(f ) = x ∈ R : u(x) 6= + πk, k ∈ Z .
2
π
Example 1.11.14 Let f be a function defined by f (x) = tan x +
. Find
2
D(f ).
Solution:
n
o
π
π
D(f ) = x ∈ R : x + 6= + πk, k ∈ Z
2
2
n
o
π π
= x ∈ R : x 6= − + + πk, k ∈ Z
2
2
= {x ∈ R : x 6= πk, k ∈ Z} . 1
Example 1.11.15 Let f be a function defined by f (x) = tan x. Find D(f ).
2
Solution:
1
π
D(f ) = x ∈ R : x 6= + πk, k ∈ Z
2
2
n
π
o
= x ∈ R : x 6= 2
+ πk , k ∈ Z
2
= {x ∈ R : x 6= π + 2πk, k ∈ Z} . Definition 1.11.16 A function f in R × R defined by f (x) = cot x is called
the cotangent function.
Theorem 1.11.17 Let f be a function defined by f (x) = cot x. Then
(i) D(f ) = {x ∈ R : x 6= πk, k ∈ Z}.
(ii) R(f ) = R.
(iii) The lines x = πk, k ∈ Z are the vertical asymptotes of the graph
of f .
40
Theorem 1.11.18 Let f be a function defined by f (x) = cot u(x), where u(x)
is a function in R × R. Then
D(f ) = {x ∈ R : u(x) 6= πk, k ∈ Z}.
π
Example 1.11.19 Let f be a function defined by f (x) = cot x −
. Find
2
D(f ).
Solution:
o
n
π
6 πk, k ∈ Z
D(f ) = x ∈ R : x − =
2
n
o
π
= x ∈ R : x 6= + πk, k ∈ Z . 2
1
Example 1.11.20 Let f be a function defined by f (x) = cot x. Find D(f ).
2
Solution:
1
D(f ) = x ∈ R : x 6= πk, k ∈ Z
2
= {x ∈ R : x 6= 2πk, k ∈ Z} . Definition 1.11.21 A function f in R × R defined by f (x) = sec x is called
the secant function.
Theorem 1.11.22n Let f be a function definedoby f (x) = sec x. Then
π
(i) D(f ) = x ∈ R : x 6= + πk, k ∈ Z .
2
(ii) R(f ) = (−∞, −1] ∪ [1, +∞).
π
(iii) The lines x = + πk, k ∈ Z are the vertical asymptotes of the
2
graph of f .
Theorem 1.11.23 Let f be a function defined by f (x) = sec u(x), where u(x)
is a function in R × R. Then
n
o
π
D(f ) = x ∈ R : u(x) 6= + πk, k ∈ Z .
2
π
Example 1.11.24 Let f be a function defined by f (x) = sec x −
. Find
2
D(f ).
41
Solution:
o
n
π
π
D(f ) = x ∈ R : x − 6= + πk, k ∈ Z
2
2
n
o
π π
= x ∈ R : x 6= + + πk, k ∈ Z
2
2
= {x ∈ R : x 6= π + πk, k ∈ Z}
= {x ∈ R : x 6= π(k + 1), k ∈ Z} . Example 1.11.25 Let f be a function defined by f (x) = sec 3x. Find D(f ).
Solution:
o
n
π
D(f ) = x ∈ R : 3x 6= + πk, k ∈ Z
2
1 π
+ πk , k ∈ Z
= x ∈ R : x 6=
3 2
n
o
π π
= x ∈ R : x 6= + k, k ∈ Z . 6
3
Definition 1.11.26 A function f in R × R defined by f (x) = csc x is called
the cosecant function.
Theorem 1.11.27 Let f be a function defined by f (x) = csc x. Then
(i) D(f ) = {x ∈ R : x 6= πk, k ∈ Z}.
(ii) R(f ) = (−∞, −1] ∪ [1, +∞).
(iii) The lines x = πk, k ∈ Z are the vertical asymptotes of the graph
of f .
Theorem 1.11.28 Let f be a function defined by f (x) = csc u(x), where u(x)
is a function in R × R. Then
D(f ) = {x ∈ R : u(x) 6= πk, k ∈ Z}.
Example 1.11.29 Let f be a function defined by f (x) = csc (x + π). Find
D(f ).
Solution:
D(f ) = {x ∈ R : x + π 6= πk, k ∈ Z}
= {x ∈ R : x 6= −π + πk, k ∈ Z}
= {x ∈ R : x 6= π(k − 1), k ∈ Z} . 42
Example 1.11.30 Let f be a function defined by f (x) = csc 2x. Find D(f ).
Solution:
D(f ) = {x ∈ R : 2x 6= πk, k ∈ Z}
o
n
π
= x ∈ R : x 6= k, k ∈ Z . 2
Sample Problems 1.11.31 Given the following trigonometric functions, find
the domain and the vertical asymptotes, if any.
√
x
1. f (x) = sin(3x3 −4x+1)
2. f (x) = sin x+1
3. f (x) = cos x + 3
√
1
4. f (x) = cos 1 − x
5. f (x) = tan x
6. f (x) = tan 2x
3
9. f (x) = csc 3x
7. f (x) = cot 2x
8. f (x) = sec 13 x
π
10. f (x) = cot x − 3
11. f (x) = sec x − 3π
12. f (x) =
2
π
csc x − 4
1.12
Inverse Trigonometric Functions
Definition 1.12.1 A function f in R × R defined by f (x) = arcsin x is called
the inverse sine function.
Definition 1.12.2 The inverse sine function is the inverse of the restricted
sine function
y = sin x,
− π2 ≤ x ≤ π2 .
Thus,
y = arcsin x ⇔ x = sin y and − π2 ≤ y ≤ π2 .
Example 1.12.3 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using calculator.
√
1. arcsin 0;
2. arcsin 1;
3. arcsin(− 12 );
4. arcsin( 23 )
Solution:
1. Let y = arcsin 0. Then
y = arcsin 0 ⇒ sin y = 0 and −
⇒ y = 0.
π
π
≤y≤
2
2
43
Therefore, arcsin 0 = 0.
2. Let y = arcsin 1. Then
π
π
y = arcsin 1 ⇒ sin y = 1 and − ≤ y ≤
2
2
π
⇒y= .
2
π
Therefore, arcsin 1 = .
2
3. Let y = arcsin(− 12 ). Then
1
π
π
1
y = arcsin −
⇒ sin y = − and − ≤ y ≤
2
2
2
2
π
⇒y=− .
6
1
π
Therefore, arcsin −
=− .
2
6
√
4. Let y = arcsin(
3
).
2
y = arcsin
Therefore, arcsin
Then
√ !
√
3
3
π
π
⇒ sin y =
and − ≤ y ≤
2
2
2
2
π
⇒y= .
3
√ !
3
π
= .
2
3
Theorem 1.12.4 Let by f (x) = arcsin x. Then
h π πi
D(f ) = [−1, 1] and R(f ) = − , .
2 2
Definition 1.12.5 A function f in R × R defined by f (x) = arccos x is called
the inverse cosine function.
Definition 1.12.6 The inverse cosine function is the inverse of the restricted
cosine function
y = cos x,
0 ≤ x ≤ π.
44
Thus,
y = arccos x ⇔ x = cos y and 0 ≤ y ≤ π.
Example 1.12.7 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using calculator.
√
4. arccos( 23 )
1. arccos 0;
2. arccos 1;
3. arccos(− 21 );
Solution:
1. Let y = arccos 0. Then
y = arccos 0 ⇒ cos y = 0 and 0 ≤ y ≤ π.
π
⇒y= .
2
π
Therefore, arccos 0 = .
2
2. Let y = arccos 1. Then
y = arccos 1 ⇒ cos y = 1 and 0 ≤ y ≤ π.
⇒ y = 0.
Therefore, arccos 1 = 0.
3. Let y = arccos(− 21 ). Then
1
1
⇒ cos y = − and 0 ≤ y ≤ π
y = arccos −
2
2
2π
⇒y=
.
3
2π
1
Therefore, arccos −
=
.
2
3
√
4. Let y = arccos(
3
).
2
Then
y = arccos
Therefore, arccos
√
√ !
3
3
⇒ cos y =
and 0 ≤ y ≤ π
2
2
π
⇒y= .
6
√ !
3
π
= .
2
6
45
Theorem 1.12.8 Let f (x) = arccos x. Then
D(f ) = [−1, 1] and R(f ) = [0, π].
Definition 1.12.9 A function f in R × R defined by f (x) = arctan x is called
the inverse tangent function.
Definition 1.12.10 The inverse tangent function is the inverse of the restricted
tangent function
π
π
y = tan x, − < x < .
2
2
Thus,
y = arctan x ⇔ x = tan y and −
π
π
<y< .
2
2
Example 1.12.11 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using calculator.
√
4. arctan 3
1. arctan 0;
2. arctan 1;
3. arctan(− √13 );
Solution:
1. Let y = arctan 0. Then
y = arctan 0 ⇒ tan y = 0 and −
π
π
<y< .
2
2
⇒ y = 0.
Therefore, arctan 0 = 0.
2. Let y = arctan 1. Then
π
π
y = arctan 1 ⇒ tan y = 1 and − < y < .
2
2
π
⇒y= .
4
π
Therefore, arctan 1 = .
4
3. Let y = arctan(− √13 ). Then
1
1
π
π
y = arctan − √
⇒ tan y = − √ and − < y < .
2
2
3
3
π
⇒y=− .
6
46
π
1
=− .
Therefore, arctan − √
6
3
√
4. Let y = arctan 3. Then
√
√
π
π
y = arctan 3 ⇒ tan y = 3 and − < y < .
2
2
π
⇒y= .
3
√
π
Therefore, arctan 3 = . 3
Theorem 1.12.12 Let f (x) = arctan x. Then
(i) D(f ) = R.
π π
(ii) R(f ) = − ,
.
2 2
π
π
(iii) The lines y = and y = − are the horizontal asymptotes of the
2
2
graph of f .
Definition 1.12.13 A function f in R × R defined by f (x) = arccot x is
called the inverse cotangent function.
Definition 1.12.14 The inverse cotangent function is the inverse of the restricted
cotangent function
y = cot x,
0 < x < π.
Thus,
y = arccot x ⇔ x = cot y and 0 < y < π.
Example 1.12.15 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using calculator.
√
1. arccot 0;
2. arccot 1;
3. arccot (− √13 );
4. arccot 3
Solution:
1. Let y = arccot 0. Then
y = arccot 0 ⇒ cot y = 0 and 0 < y < π.
π
⇒y= .
2
π
Therefore, arccot 0 = .
2
47
2. Let y = arccot 1. Then
y = arccot 1 ⇒ cot y = 1 and 0 < y < π.
π
⇒y= .
4
π
Therefore, arccot 1 = .
4
3. Let y = arccot (− √13 ). Then
1
1
y = arccot − √
⇒ cot y = − √ and 0 < y < π.
3
3
2π
⇒y=
.
3
2π
1
=
.
Therefore, arccot − √
3
3
√
4. Let y = arccot 3. Then
√
√
y = arccot 3 ⇒ cot y = 3 and 0 < y < π.
π
⇒y= .
6
√
π
Therefore, arccot 3 = . 6
Theorem 1.12.16 Let f (x) = arccot x. Then
(i) D(f ) = R.
(ii) R(f ) = (0, π).
(iii) The lines y = 0 and y = π are the horizontal asymptotes of the
graph of f .
Definition 1.12.17 A function f in R × R defined by f (x) = arcsec x is
called the inverse secant function.
Definition 1.12.18 The inverse secant function is the inverse of the restricted
secant function
π
π
y = sec x, 0 ≤ x < or −π ≤ x < − .
2
2
Thus,
y = arcsec x ⇔ x = sec y and, 0 ≤ y <
π
π
or −π ≤ y < − .
2
2
48
Example 1.12.19 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using
√ calculator.
4. arcsec √23
1. arcsec 2;
2. arcsec 1;
3. arcsec (− 2);
Solution:
1. Let y = arcsec 2. Then
π
π
y = arcsec 2 ⇒ sec y = 2 and, 0 ≤ y < or − π ≤ y < − .
2
2
π
⇒y= .
3
Therefore, arcsec 2 =
π
.
3
2. Let y = arcsec 1. Then
y = arcsec 1 ⇒ sec y = 1 and, 0 ≤ y <
π
π
or − π ≤ y < − .
2
2
⇒ y = 0.
Therefore, arcsec 1 = 0.
√
3. Let y = arcsec (− 2). Then
√ √
π
π
y = arcsec − 2 ⇒ sec y = − 2 and, 0 ≤ y < or − π ≤ y < − .
2
2
3π
⇒y=− .
4
√ 3π
Therefore, arcsec − 2 = − .
4
4. Let y = arcsec
√2 .
3
Then
2
2
π
π
y = arcsec √ ⇒ sec y = √ and, 0 ≤ y < or − π ≤ y < − .
2
2
3
3
π
⇒y= .
6
2
π
Therefore, arcsec √ = . 6
3
Theorem 1.12.20 Let f (x) = arcsec x. Then
(i) D(f ) = (−∞, −1] ∪ [1, +∞).
49
h
π h π
∪ 0,
.
(ii) R(f ) = −π, −
2
2
π
π
(iii) The lines y = and y = − are the horizontal asymptotes of the
2
2
graph of f .
Definition 1.12.21 A function f in R × R defined by f (x) = arccsc x is
called the inverse cosecant function.
Definition 1.12.22 The inverse cosecant function is the inverse of the restricted
cosecant function
π
π
y = csc x, 0 < x ≤ or −π < x ≤ − .
2
2
Thus,
π
y = arccsc x ⇔ x = csc y and 0 < y ≤ π2 or −π < y ≤ − .
2
Example 1.12.23 Use the above definition to find the value (in radians) of
the following expressions. Verify your answer using
√ calculator.
1. arccsc 2;
2. arccsc 1;
3. arccsc (− 2);
4. arccsc √23
Solution:
1. Let y = arccsc 2. Then
π
π
y = arccsc 2 ⇒ csc y = 2 and, 0 < y ≤ or − π < y ≤ − .
2
2
π
⇒y= .
6
π
Therefore, arccsc 2 = .
6
2. Let y = arccsc 1. Then
π
π
y = arccsc 1 ⇒ csc y = 1 and, 0 < y ≤ or − π < y ≤ − .
2
2
π
⇒y= .
2
π
Therefore, arccsc 1 = .
2
√
3. Let y = arccsc (− 2). Then
√ √
π
π
y = arccsc − 2 ⇒ csc y = − 2 and, 0 < y ≤ or − π < y ≤ − .
2
2
3π
⇒y=− .
4
50
√ 3π
Therefore, arccsc − 2 = − .
4
4. Let y = arccsc
√2 .
3
Then
2
2
π
π
y = arccsc √ ⇒ csc y = √ and, 0 < y ≤ or − π < y ≤ − .
2
2
3
3
π
⇒y= .
3
2
π
Therefore, arccsc √ = . 3
3
Theorem 1.12.24 Let f (x) = arccsc x. Then
(i) D(f ) = (−∞,
π i−1]∪ [1, +∞).
πi
(ii) R(f ) = 0,
∪ −π, − .
2
2
(iii) The lines y = 0 and y = −π are the horizontal asymptotes of the
graph of f .
Sample Problems 1.12.25 Use the definition to find the value (in radians)
of the following expressions. Verify
√ your answer using calculator.
1. arcsin(−1);
2. arcsin(− 23 );
3. arccos(−1);
4. arccos 21 ;
√
5. arctan(−1);
6. arctan(− 3);
7. arccot (−1);
8. arccsc √13 ;
√
√
9. arcsec (−1);
10. arcsec 2;
11. arccsc (−1);
12. arccsc 2