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Subject: MAT060.Calculus with Analytic Geometry I
Chapter: 1
Chapter Title: Elementary Functions
Time Frame: 4 hrs
Chapter Objectives:
- To define a function and identify different kinds of functions
- To evaluate and perform operations on functions
- To determine the properties of linear, quadratic, rational, exponential, logarithmic, trigonometric
and inverse trigonometric functions algebraically and graphically
- To sketch the graph of linear, quadratic, rational, exponential, logarithmic, trigonometric and inverse trigonometric functions manually and using graphing calculators
Introduction:
The first chapter of this course serves as a review of your senior high school algebra and
trigonometry. In this chapter we study definition and examples of functions, their properties using
algebraic and graphical methods that include plotting points on the rectangular coordinate system.
However, modern day methods employ sophisticated computer software applications, which are
now widely available, to generate extremely accurate graphical representations of functions. These
generated graphs help us understand the properties of functions - vertex, domain, range, and
asymptotes. The student is expected to run through this chapter focusing greatly on how to
determine the domain and range of the different kinds of functions, and is encouraged to make this
a reference as the course progresses.
1 Definition and Examples of Functions
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 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. 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. 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.
Example 1.2 The following sets are examples of functions in R × R.
- f = {(1, 0), (2, 0), (3, 0), (4, 0)}.
- f = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}.
- f = {(x, y) ∈ R × R : 2x − y = 4}.
- f = {(x, y) ∈ R × R : x^2 − y = 1}.
- f = {(x, y) ∈ R × R : y =
x − 1 }.
- f = {(x, y) ∈ R × R : 2x − xy + y = 0}.
- f = {(x, y) ∈ R × R : y = 3
x − 1 }.
- f = {(x, y) ∈ R × R : y = x^3 }.
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.
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 1.3 The functions in Example 1.2 can be written in the notation y = f (x).
- y = f (x) = 0, x ∈ { 1 , 2 , 3 , 4 }.
- y = f (x) = x^2 , x ∈ { 1 , 2 , 3 , 4 , 5 , 6 }.
- y = f (x) = 2x − 4.
- y = f (x) = x^2 + 1.
- y = f (x) =
x − 1 }.
- y = f (x) = (^) x^2 −x 1.
- y = f (x) = 3
x − 1 }.
- y = f (x) = x^3.
Example 1.4 Let f be a function defined by f (x) = x^2 − 3 x + 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 = 9a^2 − 9 a + 4;
(e) f (2x − 1) = (2x − 1)^2 − 3(2x − 1) + 4 = 4x^2 − 4 x + 1 − 6 x + 3 + 4 = 4x^2 − 10 x + 8;
(f) f (x + h) = (x + h)^2 − 3(x + h) + 4 = x^2 − 2 xh + h^2 − 3 x − 3 h + 4.
Example 1.5 Let f be a function 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) =
(b) f (5) =
(c) f (9) =
(d) f (3a) =
3 a + 5 − 1 =
3 a + 4;
(e) f (2x − 1) =
2 x − 1 − 1 =
2 x − 2;
(f) f (x + h) =
x + h − 1.
Example 1.6 Let f be a function defined by f (x) =
x + 4, if x ≤ − 4
4 − x, if − 4 < x
; find (a) f (−6); (b)
f (−4); (c) f (0); (d) f (4).
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.
Example 1.7 Let f be a function defined by f (x) =
x + 3, if x < 2
4 , if x = 2
2 x − 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.
Exercises 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).
- Given f (x) = 3x^2 + 2x − 4, find (a) f (1); (b) f (−1); (c) f ( 12 ); (d) f (3a + 5); (e) f (2x − 1);
Solution: (a) (f + g)(x) = x x+4− 3 +
x − 1;
(b) (f − g)(x) =
x + 4
x − 3
x − 1;
(c) (f · g)(x) =
x + 4
x − 3
x − 1 =
(x + 4)
x − 1
x − 3
(d)
f
g
(x) =
x+ √^ x−^3 x − 1
x + 4
(x − 3)
x − 1
Definition 2.4 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 2.5 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.
Example 2.6 Let f and g be functions defined by f (x) = x^2 + 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(x^2 + 4) =
x^2 + 4 − 2 =
x^2 + 2.
(c) (f ◦ f )(x) = f (f (x)) = f (x^2 + 4) = (x^2 + 4)^2 + 4 = x^4 + 8x^2 + 20;
(b) (g ◦ g)(x) = g(g(x)) = g(
x − 2) =
x − 2 − 2.
Exercises 2.7 Perform the following operations on functions.
- Let f and g be functions defined by f (x) =
x^2 + 1 and g(x) = 2x + 3. Define the following
functions: (a) f + g; (b) f − g; (c) f · g; and (d)
f
g
- Let f and g be functions defined by f (x) =
x + 2
x − 5
and g(x) = 3
x. Define the following
functions: (a) f + g; (b) f − g; (c) f · g; and (d)
f
g
- Let f and g be functions defined by f (x) = x^2 − 3 and g(x) =
x + 2. Define the following
functions: (a) f ◦ g; (b) g ◦ f ; (c) f ◦ f and; (d) g ◦ g.
- Let f and g be functions defined by f (x) =
x^2 + 1 and g(x) = 3x − 4. Define the following
functions: (a) f ◦ g; (b) g ◦ f ; (c) f ◦ f and; (d) g ◦ g.
3 Linear Functions
Definition 3.1 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 3.2 Let f be a function defined by 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 3.3 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 decreasing. 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.
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Example 3.4 Let f be a function defined by f (x) = − 2 x + 2. Find D(f ) and R(f ). Identify the
graph of f and determine whether it is increasing or decreasing. 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.
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Exercises 3.5 Given the following linear functions, find D(f ) and R(f ); and determine whether
the graph of f is increasing or decreasing. Sketch the graph of f.
- f (x) = 3x + 6 2. f (x) = 3x − 6 3. f (x) = − 2 x + 10
- f (x) = − 4 x + 6 5. f (x) = 2x − 5 6. f (x) = − 5 x − 10
Example 4.4 Let f be a function defined by f (x) = −x^2 + 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
b
2 a
= 1 and
4 ac − b^2
4 a
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 0 1 2
y 2 3 2
Using these three points, sketch a graph of the parabola.
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V = (1, 3)
Exercises 4.5. 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.
- f (x) = x^2 − 2 x + 3 2. f (x) = x^2 + 4x + 5 3. f (x) = −x^2 − 2 x − 3
- f (x) = − 2 x^2 + 4x 2. f (x) = x^2 − 4 x + 1 3. f (x) = −x^2 + 3x − 3
5 Rational Functions
Definition 5.1 A function f in R × R defined by f (x) =
p(x) q(x) , where^ p(x) and^ q(x) are polynomial
functions and q(x) 6 = 0, is called a rational function.
Theorem 5.2 Let f be a function defined by f (x) =
p(x) q(x) , where^ p(x)^ and^ q(x)^ are polynomial
functions 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 5.3 Let f be a function defined by f (x) =
x − 2
. Find D(f ) and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x ∈ R :
x − 2
∈ R
= {x ∈ R : x − 2 6 = 0}
= {x ∈ R : x 6 = 2}
= R{ 2 }
= (−∞, 2) ∪ (2, +∞).
f (x) =
x − 2
⇒ y =
x − 2
⇒ xy − 2 y = 1
⇒ xy = 2y + 1
⇒ x =
2 y + 1
y
⇒ f −^1 (y) =
2 y + 1
y
R(f ) =
y ∈ R : f −^1 (y) ∈ R
y ∈ R :
2 y + 1
y
∈ R
= {y ∈ R : y 6 = 0}
= (−∞, 0) ∪ (0, +∞).
Example 5.4 Let f be a function defined by f (x) =
x + 1
x + 3
. Find D(f ) and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x ∈ R :
x + 1
x + 3
∈ R
= {x ∈ R : x + 3 6 = 0}
= {x ∈ R : x 6 = − 3 }
= (−∞, −3) ∪ (− 3 , +∞).
f (x) =
x + 1
x + 3
⇒ y =
x + 1
x + 3
⇒ xy + 3y = x + 1
⇒ xy − x = 1 − 3 y
⇒ x(y − 1) = 1 − 3 y
⇒ x =
1 − 3 y
y − 1
⇒ f −^1 (y) =
1 − 3 y
y − 1
f (x) =
4 − 3 x
2 x
⇒ y =
4 − 3 x
2 x ⇒ 2 xy = 4 − 3 x
⇒ 2 xy + 3x = 4
⇒ x(2y + 3) = 4
⇒ x =
2 y + 3
⇒ f
− 1 (y) =
2 y + 3
R(f ) =
y ∈ R : f
− 1 (y) ∈ R
y ∈ R :
2 y + 3
∈ R
= {y ∈ R : 2y + 3 6 = 0}
y ∈ R : y 6 = −
Definition 5.7 Let f be a function in R × R. Then the line x = a is a vertical asymptote of the
graph of f if f (x) increases or decreases without bounds as x approaches a.
Theorem 5.8 Let f be a function defined by f (x) =
p(x)
q(x)
, where p(x) and q(x) have no common
factors. If a is a zero of q(x), then x = a is a vertical asymptote of the graph of f.
Example 5.9 Let f be a function defined by f (x) =
x − 2
. Find the vertical asymptote/s, if any.
Solution: Set x − 2 = 0. Then x = 2.
Therefore, the vertical asymptote of the graph of f is x = 2.
Example 5.10 Let f be a function defined by f (x) =
2 x + 1
x + 3
. Find the vertical asymptote/s, if
any.
Solution: Set x + 3 = 0. Then x = −3.
Therefore, the vertical asymptote of the graph of f is x = −3.
Example 5.11 Let f be a function defined by f (x) =
2 x
(x + 1)(x − 2)
. Find the vertical asymptote/s,
if any.
Solution: Set (x + 1)(x − 2) = 0. Then x + 1 = 0 or x − 2 = 0. Thus, x = −1 or x = 2.
Therefore, the vertical asymptotes of the graph of f are x = −1 and x = 2.
Definition 5.12 Let A ⊆ R and f : A → R be a real function. Then the line y = b is a horizontal
asymptote of the graph of f if f (x) approaches b as x increases or decreases without bounds.
Theorem 5.13 Let A ⊆ R and f : A → R be a a rational function given by
f (x) =
anxn^ + an− 1 xn−^1 + ... + a 1 x + a 0
bmxm^ + bm− 1 xm−^1 + ... + b 1 x + b 0
, where an 6 = 0 and bm 6 = 0.
(i) If n < m, then the line y = 0 is the horizontal asymptote of the graph of f.
(ii) If n = m, then the line y =
an bm is the horizontal asymptote of the graph of^ f^. (iii) if n > m, then the graph of f has no horizontal asymptote.
Example 5.14 Let f be a function defined by f (x) =
x − 2
. Find the horizontal asymptote, if
any.
Solution: n = 0 and m = 1. Then n < m.
Therefore, the horizontal asymptote of the graph of f is y = 0.
Example 5.15 Let f be a function defined by f (x) =
2 x + 1
x + 3
. Find the horizontal asymptote, if
any.
Solution: n = 1 and m = 1. Then n = m.
Therefore, the horizontal asymptote of the graph of f is y = 2.
Example 5.16 Let f be a function defined by f (x) =
2 x^2 + 3
x + 1
. Find the horizontal asymptote, if
any.
Solution: n = 2 and m = 1. Then n > m.
Therefore, the graph of f has no horizontal asymptote.
Example 5.17 Let f be a function defined by f (x) =
2 x
x − 1
. Find D(f ), R(f ), the vertical and
horizontal asymptotes, if any. Sketch the graph of f.
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x ∈ R :
2 x
x − 1
∈ R
= {x ∈ R : x − 1 6 = 0}
= {x ∈ R : x 6 = 1}
= (−∞, 1) ∪ (1, +∞).
f (x) =
2 x
x − 1
⇒ y =
2 x
x − 1
⇒ xy − y = 2x
⇒ xy − 2 x = y
⇒ x(y − 2) = y
⇒ x =
y
y − 2
⇒ f −^1 (y) =
y
y − 2
R(f ) =
y ∈ R : f
− 1 (y) ∈ R
y ∈ R :
y
y − 2
∈ R
= {y ∈ R : y − 2 6 = 0}
= {y ∈ R : y 6 = 2}
= (−∞, 2) ∪ (2, +∞).
R(f ) =
y ∈ R : f
− 1 (y) ∈ R
y ∈ R :
− 3 y − 2
2 y − 1
∈ R
= {y ∈ R : 2y − 1 6 = 0}
y ∈ R : y 6 =
= R\
Set 2x + 3 = 0. Then x = − 32. The vertical asymptote of the graph of f is x = − 32.
We have n = 1 and m = 1. Then n = m. The horizontal asymptote of the graph of f is y = 12.
Exercises 5.19. Given the following rational functions, find D(f ), R(f ), the vertical and horizontal
asymptotes, if any. Skecth the graph of f.
- f (x) =
3 x
x − 1
- f (x) =
x + 5
- f (x) =
x − 2
x + 2
- f (x) =
4 − x
- f (x) =
x + 2
x − 2
- f (x) =
2 x − 5
4 x + 3
6 Functions of the Form f (x) =
g(x)
Theorem 6.1 Let f be a function in R × R defined by f (x) =
g(x), where 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 }.
Example 6.2 Let f be a function defined by f (x) =
x − 1. Find D(f ) and R(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
⇒ y^2 = x − 1
⇒ x = y^2 + 1
⇒ f
− 1 (y) = y
2
R(f ) =
y ∈ R : f
− 1 (y) ∈ R and y ≥ 0
y ∈ R : y
2
= {y ∈ R : y ∈ R and y ≥ 0 }
= {y ∈ R : y ≥ 0 }
= [0, +∞).
Example 6.3 Let f be a function defined by f (x) =
x^2 − 4. Find D(f ) and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x ∈ R :
x^2 − 4 ∈ R
x ∈ R : x
2 − 4 ≥ 0
= {x ∈ R : (x + 2)(x − 2) ≥ 0 }
= {x ∈ R : x + 2 ≤ 0 and x − 2 ≤ 0 } ∪ {x ∈ R : x + 2 ≥ 0 and x − 2 ≥ 0 }
= {x ∈ R : x ≤ −2 and x ≤ 2 } ∪ {x ∈ R : x ≥ −2 and x ≥ 2 }
= {x ∈ R : x ≤ − 2 } ∪ {x ∈ R : x ≥ 2 }
= (−∞, −2] ∪ [2, +∞).
f (x) =
x^2 − 4 ⇒ y =
x^2 − 4
⇒ y^2 = x^2 − 4
⇒ x
2 = y
2
⇒ x = ±
y^2 + 4
⇒ f −^1 (y) = ±
y^2 + 4.
R(f ) =
y ∈ R : f −^1 (y) ∈ R and y ≥ 0
y ∈ R : ±
y^2 + 4 ∈ R and y ≥ 0
y ∈ R : y
2
= {y ∈ R : y ∈ R and y ≥ 0 }
= {y ∈ R : y ≥ 0 }
= [0, +∞).
f (x) =
x + 1
x − 2
⇒ y =
x + 1
x − 2
⇒ y
2
x + 1
x − 2
⇒ xy
2 − 2 y
2 = x + 1
⇒ xy
2 − x = 2y
2
⇒ x(y^2 − 1) = 2y^2 + 1
⇒ x =
2 y^2 + 1
y^2 − 1
⇒ f
− 1 (y) =
2 y^2 + 1
y^2 − 1
R(f ) =
y ∈ R : f −^1 (y) ∈ R and y ≥ 0
y ∈ R :
2 y^2 + 1
y^2 − 1
∈ R and y ≥ 0
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, +∞).
Example 6.6 Let f be a function defined by f (x) =
2 x
4 − x
. Find D(f ) and R(f ).
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
x ∈ R :
2 x
4 − x
∈ R
x ∈ R :
2 x
4 − x
= {x ∈ R : 2x ≤ 0 and 4 − x < 0 } ∪ {x ∈ R : 2x ≥ 0 and 4 − x > 0 }
= {x ∈ R : x ≤ 0 and 4 < x} ∪ {x ∈ R : x ≥ 0 and 4 > x}
= { } ∪ {x ∈ R : 0 ≤ x and x < 4 }
= {x ∈ R : 0 ≤ x and x < 4 }
= [0, 4).
f (x) =
2 x
4 − x
⇒ y =
2 x
4 − x
⇒ y
2
2 x
4 − x
⇒ 4 y
2 − xy
2 = 2x
⇒ xy
2
2
⇒ x(y^2 + 2) = 4y^2
⇒ x =
4 y^2
y^2 + 2
⇒ f
− 1 (y) =
4 y^2
y^2 + 2
R(f ) =
y ∈ R : f −^1 (y) ∈ R and y ≥ 0
y ∈ R :
4 y^2
y^2 + 2
∈ R and y ≥ 0
y ∈ R : y
2
= {y ∈ R : y ∈ R and y ≥ 0 }
= {y ∈ R : y ≥ 0 }
= [0, +∞).
Exercises 6.7 Find the domain and range of the following functions.
- f (x) =
3 x + 5 2. f (x) =
x^2 − x − 12 3.f (x) =
9 − x^2
- f (x) =
x^2 − 6 x + 8 5. f (x) =
x + 5
x + 3
6.f (x) =
4 x
x − 6
- f (x) =
x^2
x^2 − 4
- f (x) =
2 x + 5
4 − 3 x
9.f (x) =
4 x √ x − 6
7 Exponential and Logarithmic Functions
Definition 7.1 A function f in R × R defined by f (x) = bx, where b > 0 and b 6 = 1, is called an
exponential function.
Example 7.2 The following are examples of exponential function.
- f (x) = 2x^ 2. f (x) = 3x^ 3. f (x) = (−2)x
- f (x) = ex^ 5. f (x) = ( 12 )x^ 3. f (x) = (− 23 )x.
Theorem 7.3 Let A ⊆ R and let f : A → R be defined by f (x) = bu(x), where b > 0 , b 6 = 1, and
u(x) is a real function. Then
(i) D(f ) = {x ∈ R : f (x) = bu(x)^ ∈ R} = {x ∈ R : u(x) ∈ R}.
(ii) R(f ) = {y ∈ R : f −^1 (y) ∈ R}.
Definition 7.4 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 7.10 Let f be a function defined by f (x) = ln
4 − x. Find D(f ) and R(f ). Sketch
the graph of f.
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
⇒ e
2 y = 4 − x
⇒ x = 4 − e
2 y
⇒ f −^1 (y) = 4 − e^2 y.
R(f ) = {y ∈ R : f
− 1 (y) ∈ R}
= {y ∈ R : 4 − e
2 y ∈ R}
= {x ∈ R : 2y ∈ R}
= {x ∈ R : y ∈ R}
= R.
We may construct a table of values using the relation
y = ln
4 − x ⇔ x = 4 − e^2 y.
x 4 − e−^4 4 − e−^2 3 4 − e^2 4 − e^4
y = f (x) -2 -1 0 1 2
Using these points, sketch the graph of f.
Example 7.11 Let f be a function defined by f (x) = 2x. Find D(f ) and R(f ). Sketch the graph
of f.
Solution:
D(f ) = {x ∈ R : f (x) ∈ R}
= {x ∈ R : 2x^ ∈ R}
= {x ∈ R : x ∈ R}
= R.
f (x) = 2
x ⇒ y = 2
x
⇒ x = log 2 y
⇒ f −^1 (y) = log 2 y.
R(f ) =
y ∈ R : f
− 1 (y) ∈ R
= {y ∈ R : log 2 y ∈ R}
= {y ∈ R : y > 0 }
= (0, +∞).
We may construct a table of values:
x -2 -1 0 1 2
y = f (x) 14 12 1 2 4
Using these points, sketch the graph of f.
