Calculus I
Chapter 1
Functions and Graphs
Chapter 1
Functions and Graphs
Introduction
The different instructional approaches in this module allow you to explore concepts on your
own or by working with others. Some lessons may be led by your teacher, while others may be
done through the use of hands-on activities. For some lessons, you may be asked to read solved
examples on your own to build an understanding of a concept.
Specific Objectives
At the end of the lesson, the students should be able to:
-
Interpret and use function notation properly, including evaluating functions given as
tables, graphs, or formulas.
Distinguish between the sketches of the graphs of basic functions: linear, x 2, x3,
square root, |x|
Determine the domain and range of a function from a graph or formula
Find the composition of two functions
Graph functions using technology on an appropriate viewing window
Duration
Chapter 1:
Functions and Graphs
= 9 hours
(7 hours discussion; 2 hours
assessment)
Lesson Proper
RELATION
-
A relation is any set of ordered pairs, x and y. The domain of a relation is the set of
all first coordinates (x-coordinate) from the ordered pairs. The range of a relation is
the set of all second coordinates (y-coordinate) from the ordered pairs.
-
Each ordered pair has two parts, an x-value and a y-value. The x-values of a given
relation are called the domain. The y-values of the relation are called the range. When
you list down the domain and range of a relation, you place each (the domain and
range) in a separate set of brackets.
-
The graph of a relation is the set of points in the plane that correspond to the ordered
pairs of that relation.
Example 1
Determine the domain and range and plot the set of points defined by the relation
A = {(1, -1), (1, 1), (2, -2), (2, 2), (3, 2), (3, -2)}
Solution:
The domain is the set of all the first coordinates of the ordered pairs while the range is
the set of all the second coordinates of the ordered pairs (see Figure 1.1). Thus,
Domain: {1, 2, 3}
Range: {-1, 1, -2, 2}
Figure 1.1
Example 2
Sketch the graph and determine the domain and range of the relation defined by
𝑥2 + 𝑦2 = 1
Solution:
The graph of the relation 𝑥 2 + 𝑦 2 = 1 is a circle with center at the origin and whose
radius is 1 unit (see Figure 1.2). It is the set of points lying on the circle.
The domain is the set of all real numbers x such that 𝑥 ≥ −1 and 𝑥 ≤ 1. Using set notation, we
get
Domain: {𝑥|𝑥∈ℝ, −1 ≤ 𝑥 ≤ 1}
The range is the set of all real numbers y such that 𝑦 ≥ −1 and 𝑦 ≤ 1. Thus,
Range: {𝑦|𝑦∈ℝ, −1 ≤ 𝑦 ≤ 1}
Figure 1.2
FUNCTION
-
A function f is special type of a relation such that no two ordered pairs of the set have
different second coordinates corresponding to the same first coordinate The set of all
the first coordinates of the ordered pairs is the domain of the function. The set of all
second coordinates of the ordered pairs is the range of the function (see figure 1.3).
-
A rule for a relationship between an input, or independent, quantity and an output, or
dependent, quantity in which each input value uniquely determines one output value.
We say “the output is a function of the input.”
-
In symbols, 𝑦=𝑓(𝑥) is read as “y is a function of x” where y is the dependent variable
and x is the independent variable.
Figure 1.3
Example 1
Plot the points that represent the relation
{(1, -1), (1, 1), (2, -2), (2, 2), (3, 2), (3, -2)}
State the domain and range of the relation.
Is the relation a function?
Solution
Domain: {1, 2, 3}
Range: {-1, 1, -2, 2}
The domain is paired with more
than one member of the range,
so this relation is not a
function.
Figure 1. 4
Example 2
Is the set of the ordered pairs that satisfies y = x2 a function?
Solution
To determine this, we prepare a table to find ordered pairs that satisfy the equation by assigning
values for x. Then we find the corresponding values of y.
x
y = x2
-3
9
-2
4
-1
1
0
0
1
1
2
4
3
9
Now we plot these ordered pairs and sketch the graph of y = x2. Since x can be any real number,
there is an infinite number of ordered pairs that can be plotted. If all of them were plotted, they
would take on the shape of a curve called parabola.
Since for every x there
corresponds exactly one y,
the relation is a function.
Figure 1.5
Example 3
Graph the relation y2 = x + 1 and determine if it is a function or not.
Solution
We need to prepare a table to find ordered pairs that satisfy the equation. In this case it is easier
to assign values for y and find the corresponding values of x.
x
y =x+1
y=√𝑥 + 1
3
2, -2
2
0
1, -1
-1
0
or
y
x = y2 - 1
-2
3
-1
0
0
-1
1
0
2
3
Figure 1.6
You can observe from the graph that there can be two y values for almost all x values on the
graph. This relation is not a function. If a vertical line can be drawn that intersects the graph of
a relation at more than one point, then that relation is not a function. This method is often
called vertical line test for a function.
CLASSIFICATIONS OF FUNCTION
-
All functions are classed as either algebraic or transcendental. The algebraic functions
are rational integral functions, or polynomials; rational fractions, or quotient of
polynomials; and irrational functions, of which the simplest are those formed from
rational functions by extraction of roots. The elementary transcendental functions are
trigonometric and inverse trigonometric functions; exponential functions, in which the
variable occurs as an exponent; ad logarithms.
TYPES OF FUNCTION
 Linear Function
-
The simplest type of function is the linear function, where the graph is a line.
To graph a linear function, we need only two points in keeping with Euclid’s Axiom:
“Two points determine a line” or “The shortest distance between two points is a line.”
Notice that when we say line in Mathematics, that means it is straight otherwise, it is
a curve.
Any equation of the form
𝑦=𝑚𝑥+𝑏
(where m is the slope of the line and b is the y-intercept) is called a linear function in
x.
Remarks: Notice that if m = 0, then 𝑓(𝑥)=𝑏, which is called a constant function. If the
domain of a constant function is the set of real numbers. Then the graph of 𝑓(𝑥)=𝑏 is
horizontal line. The line parallel to the y-axis is called a vertical line. Notice that
vertical lines are not function.
Example
Sketch the graph, and determine the domain and range of
𝑦=𝑓(𝑥)=2𝑥−3, 𝑥 ∈ [−1,3)
Solution
The domain of the linear function is restricted. Notice that when x = 3, there is a hole in the
graph (see Figure 1.4). Thus,
when x = -1
𝑦 =𝑓 (−1) = 2 (−1) – 3 = − 2 − 3 = −5
We assume that x = 3 so that 𝑦=𝑓 (3) = 2 (3) – 3 = 6 – 3 = 3.
Thus, the domain is the set of real numbers greater than or equal to -1 but less than 3. Using
set notation, we write
Domain: {𝑥 | −1 ≤ 𝑥 < 3, 𝑥 ∈ ℝ}
The range is the set of real numbers greater than or equal to -5, but less than 3. That is,
Range: {𝑦 |−5 ≤ 𝑦 <3, 𝑦∈ ℝ}
Figure 1.7
 QUADRATIC FUNCTION
- Any equation in the form of 𝑦=𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐 where 𝑎≠0 is called a quadratic
function. The restriction that 𝑎≠0 is necessary because if a = 0, then the equation
becomes a linear function. The graph of a quadratic function is called a parabola.
-
In a quadratic function 𝑦=𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐, 𝑎≠0
i. If 𝑎>0, the parabola opens upward and has a minimum value at the vertex.
ii. If 𝑎<0, the parabola opens downward and has a maximum value at the vertex.
The vertex of a parabola is given by the equation
Example
Sketch the graph and determine the vertex, domain, and range of
𝑦 = 𝑥2 and 𝑦 = − 𝑥2
Solution
The vertices of these two parabola lie at origin. The first parabola opens upward (since a = 1 >
0) while the second parabola opens downward (since a = -1 < 0).
Figure 1.8
For 𝑦 = 𝑥2
For 𝑦 = −𝑥 2
Domain:
{𝑥|𝑥 ∈ ℝ}
{𝑥|𝑥 ∈ℝ}
Range:
{𝑦|𝑦 ≥ 0, 𝑦∈ℝ}
{𝑦|𝑦 ≤ 0, 𝑦∈ℝ}
Vertex:
(0, 0)
(0, 0)
SPECIAL TYPESS OF FUNCTION
 The Square Root Function
-
The domain and the range of the square root function are the set of nonnegative real
numbers. The intercept of the graph is at (0, 0). The square root function is neither
even nor odd and is increasing on the interval (0, ∞).
𝑓(𝑥) = √𝑥
Domain: {𝑥|𝑥 ≥ 0, 𝑥 ∈ℝ }
Range: {𝑦|𝑦 ≥ 0}
Figure 1.9
 The Absolute Value Function
-
The domain of the absolute value function is the set of all real numbers; its range is the
set of nonnegative real numbers. The intercept of the graph is at (0, 0). If x ≥ 0, then
f(x) = x and the graph of f is part of the line y = x; if x < o then f(x) = -x and the graph
of f is part of the line y = -x. the absolute value function is an even function; it is
decreasing on the interval (−∞, 0) and increasing on the interval (0, ∞).
f(x) =|x|
Domain: {𝑥|𝑥 ∈ℝ}
Range: {y | y ≥ 0, y ∈ ℝ}
Figure 1.10
 The Signum Function
-
Signum is a latin word for sign. In practice, signum is usually abbreviated to sgn.
Except at zero, the value of the signum function is determined by the algebraic sign of
its argument. When the argument x is positive, 𝑠𝑔𝑛𝑥 = +1; when x is negative, 𝑠𝑔𝑛 𝑥
= −1. Thus,
sgn x = -1, for x < 0
= 0, for x= 0
=+1, for x > 0
Figure 1.11
EVALUATION OF FUNCTION
To evaluate a function, substitute the input for function’s variable. Replace x with the
given number or expression.
-
Example 1
f(x) = 3x-5, find f(4)
Solution:
f(4) = 3(4) - 5
= 12 - 5
f(4) = 7
Example 2
f(x) = x2 -2x+1 , find f(2x)
Solution:
f(2x) = (2x)2 -2(2x)+1
f(2x) = 4x2 – 4x +1
Example 3
f(x) = 3x-3 +1 , find f(2)
Solution:
f(2) = 3(2)-3 +1
𝟑
= 𝟐𝟑 + 𝟏
𝟑
=𝟖+𝟏
f(2) =
𝟏𝟏
𝟖
-
COMPOSITION O F FUNCTION
Another way to combine two functions to create a new function is called the
composition of functions.
-
Given 2 functions f and g, the composite function denoted by f o g (read as “ f of g”, “f
composition g”, or “f circle g”) is defined by (f o g)(x) = f(g(x))
-
In the composition of functions, we substitute an entire function into another function.
-
The domain of f ∘ g is the set of all numbers x in the domain of g such that g(x) is in the
domain of f.
Let A = {1,2,3} and B = {a,b,c,d}
g: A→ A
1→3
2→ 1
3→2
f: A→B
1→ b
2→ a
3→ d
Find (f o g): A→B
Answer
1→ d
2→ b
3→ a
Example 1
f(x)= 2x and g(x) = x2 , find (f o g)(x).
Solution:
(f o g)(x) = 𝑓 (𝑥 2 )
= 2 (𝑥 2 )
(f o g)(x) = 𝟐𝒙𝟐
Example 2
f(x)= 2x+1 and g(x) = 6x + 4, find (g o f)(x).
Solution:
(g o f)(x) = 𝑓 (2𝑥 + 1)
= 6(2𝑥 + 1) + 4
= 12𝑥 + 6 + 4
(f o g)(x) =1𝟐𝒙 + 𝟏𝟎
The next example is finding components of a composite function.
Example 3
Given f(x) = 2x + 3 and f(g(x)) = 2x – 3, find g(x).
Solution:
Since 𝑓(𝑥) = 2x + 3 and 𝑓(𝑔(𝑥)) = 2x – 3 , then
f(x) = 2x + 3
f(g(x)) = 2(g(x)) + 3
2x – 3 = 2(g(x)) + 3
Solve for g(x),
2x – 3 = 2(g(x)) + 3
2x – 3 – 3 = 2(g(x)) + 3 – 3
2x – 6 = 2(g(x))
2𝑥−6
2
= g(x)
x – 3 = g(x)
Therefore, g(x) = x – 3