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PreCalculus
Section 1-2
Functions
Objectives
• Determine whether a relation between two
variables is a function.
• Use function notation and evaluate
functions.
• Find the domain of functions.
• Use functions to model and solve real-life
problems.
Definition of Relations
A relation is a rule of correspondence that
relates two sets.
For instance, the formula I = 500r describes a
relation between the amount of interest I
earned in one year and the interest rate r.
In mathematics, relations are represented by
sets of ordered pairs (x, y) .
Definition of Function
A function from a set A to a set B is a relation
that assigns to each element x in the set A
exactly one element y in the set B (for each x
value there is one and only one y value).
The set A (x values) is called the domain of the
function.
The set B (y values) is called the range of the
function.
Ways to Write a Domain/Range
Let’s use the function y = 2x, for example, the domain and
range of this function is any real number. The domain and
range of a function can be expressed using set, interval,
and inequality notation.
• Set Notation - Domain: {x| x∈ℝ} and Range: {y| y∈ℝ} ,
where ∈ means “element of” and ℝ means “real numbers”.
• Interval Notation - Domain: (-∞, ∞) and Range: (-∞, ∞),
where -∞ represents “negative infinity” and ∞ represents
“positive infinity”. Open parentheses are used to indicate
the interval is infinite, does not include the endpoints. A
square bracket ‘[ ]’ indicates that the interval includes the
endpoints and must never be used with -∞ or ∞.
• Inequality Notation - Domain: -∞< x <∞ and
Range: -∞< y <∞.
Set Builder Notation
• A method of representing points on the real
number line.
“such that”
: or |
“set of”
•
{x: 2x + 3 > 5, xєR}
Variable
“all x”
Rule(s) that defines
the variable
Read “the set of all x such
that 2x + 3 > 5 and x is an
element of real numbers” π
Set Builder Notation
• Other useful set notation
– U Union (or statement)
– ∩ Intersection (and statement)
– \ excludes whatever follows
– Ø empty set or null set
π
Graphing a set on the real number line
• If the end point of a set is included, then a
“closed circle” is used.
– {x: x ≥ 3}
• If the end point of a set is excluded, then an
“open circle” is used.
– {x: x > 3}
π
Examples
• Write the following intervals on the number
line in set builder notation.
1) |
|
|
|
|
|
|
|
–4
–3
–2
–1
0
1
2
3
 { x : x >-2, x𝛜ℝ }
 If the variable is on the left side of the
inequality, then the inequality sign is the same
direction as the arrow.
|
4
Continued
2)
|
|
|
|
|
|
|
|
|
–4
–3
–2
–1
0
1
2
3
4
 { y | y ≤ 1, y𝛜ℤ }
Another
3)
|
|
|
|
|
|
|
|
|
–4
–3
–2
–1
0
1
2
3
4




Conjunction “and statement”
x ≥-3 and x < 2
As compound inequality -3 ≤ x < 2
Set builder { x | -3 ≤ x < 2, x𝛜ℝ }
One more
4)
|
|
|
|
|
|
|
|
|
–4
–3
–2
–1
0
1
2
3
4
 Disjunction “or statement”
 x ≤-3 or x > 0
 Disjunction must be written with the word or
(or symbol ∪).
 Set builder { x : x ≤-3 ∪ x > o, x𝛜ℝ }
Last one
5)
|
|
|
|
|
|
|
|
|
–4
–3
–2
–1
0
1
2
3
4
 x equals all real numbers except 0
 { x : x𝛜ℝ\0 }
Interval Notation
• Uses an interval, lower endpoint coma
upper endpoint, and “square brackets or
square brackets & round brackets” to
indicate if the endpoint is included or
excluded.
– Endpoint included: lower [ or upper ]
– Endpoint excluded: lower (, ] or upper ),[
– Note, ∞ is never included.
π
Notation Dilemma
],( & ),[ mean the same as
Both notations mean that the number IS NOT a solution!
[ & ] mean the same as
Both notations mean that the number IS a solution.
Intervals on the real number line.
Interval type Notation Inequality
Closed
Open
[ a, b]
( a, b)
[ a, b)
( a, b]
a xb
a xb
Graph
[
]
a
b
(
)
a
b
[
)
a
b
a xb
a xb
(
]
a
b
Intervals on the real number line.
Interval Type
Notation Inequality
[ a, )
xa
Graph
[
a
Open
( a,  )
(, b]
xa (
xb
a
]
b
Open
Entire real line
(, b)
(, )
xb
)
b
Interval
Notation
Summary:
Example
1) Write an inequality to represent the
interval (2, 10].
 This notation represents all real numbers x
between 2 and 10, including 10.
 Inequality notation
2  x  10
2) Write an inequality to represent the
interval [-5, ∞).
 This notation represents all real numbers x
greater than or equal to –5.
 Inequality notation
x  5
Function Diagram
Set B
Function
Set A
y
y
x
y
x
y
x
x
Domain
y
Range
Example: Determine whether the relation
represents y as a function of x.
a)
{(-2, 3), (0, 0), (2, 3), (4, -1)}
Function
b)
{(-1, 1), (-1, -1), (0, 3), (2, 4)}
Not a Function
Your Turn:
Determine whether each relation is a function?
{(2, -2), (3, 0), (4, 1), (5, -1)}
Function
{(4, -3), (2, 0), (5, -2), (3, 1), (2, -1)}
Not Function
{(2, -1), (3, -1), (4, -1), (5, -1)}
Function
{(0, 0), (3, -2), (5, 0), (2, -3), (0, 5)}
Not Function
Four Ways to Represent a Function
• Verbally (description in words)
• Numerically (by a table of values)
• Visually (by a graph)
• Algebraically (by an equation or
formula)
A Function Verbally
The amount of sales tax depends on the
amount of the purchase.
OR
Divide x seconds by 5 to obtain y miles.
A Function Numerically
Use a table of values:
x (seconds) 1
2
3
4
y (miles) 0.2 0.4 0.6 0.8
5
1
6
7
1.2 1.4
A Function Visually
Any function can be visually represented
by a graph.
Dependent variable
Independent variable
A Function Algebraically
Any function can be algebraically
represented by a equation in two
variables.
y = 3x + 10
OR
f(x) = 3x + 10
Testing for Functions
1. Set of ordered pairs or points on a graph.
•
Use definition of a function and determine
whether each input value (x) is matched with
exactly one output value (y).
2. Function represented algebraically
(equation).
•
•
Solve for y in terms of x.
Determine whether each value of x
corresponds to exactly one value of y.
Testing for Functions
Example:
Determine whether the relation represents y as a function of x.
y
4
3
2
1
x
4
3 2 1
2
3
4
1
2
3
4
The ordered pairs in the graph are:
(– 3, 0), (– 2, 2), (– 1, 4), (0, 1),
(1 – 2), and (2, 1).
y is a function of x.
Your Turn
Determine whether the relation represents y as a function of x.
y
5
4
3
2
1
0
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
y is not a function of x.
x
Example
Determine whether y is a function of x.
1. x2+y=3
y=-x2+3
solve for y
each value of x corresponds to exactly one value
of y, so y is a function of x.
2. x2+y2=8
y2=8-x2
subtract x2 from both sides
y   8  x2
solve for y
each value of x corresponds to 2 values of y, so y
is not a function of x.
Your Turn
1. y2-x=0
Not Function
2. y=√x
Function
3. y3+x=1
Function
Function Notation - f(x) : “f of x”
Functions represented by equations are often
named using a letter such as f or g.
The symbol f (x), read as “the value of f at x” or
simply as “f of x”, is the element in the range of f
that corresponds with the domain element x. That
is,
y = f (x)
Inputs & Outputs
The domain elements, x, can be thought of as
the inputs and the range elements, f (x), can
be thought of as the outputs.
Function
Input
Output
f
x
f (x)
Independent
variable
Dependent
variable
Evaluating Functions
To evaluate a function f (x) at x = a, substitute
the specified value a for x into the given
function.
2
Example: Let f (x) = x – 3x – 1. Find f (–2).
2
f (x) = x – 3x – 1
2
f (–2) = (–2) – 3(–2) – 1
Substitute –2 for x.
f (–2) = 4 + 6 – 1
Simplify.
f (–2) = 9
The value of f at –2 is 9.
Your Turn
2
Let f (x) = 4x – x . Find f (x + 2).
2
f (x) = 4x – x
f (x + 2) = 4(x + 2) – (x + 2)2
Substitute x + 2 for x.
f (x + 2) = 4x + 8 – (x2 + 4x + 4)
Expand (x + 2)2.
2
f (x + 2) = 4x + 8 – x – 4x – 4
f (x + 2) = 4 – x2
Distribute –1.
The value of f at x + 2
2
is 4 – x .
Piecewise Function
• A piecewise function is an equation with
multiple parts in which each part matches up
with a different part of the domain.
Piecewise Function
A piecewise function is an equation with multiple parts or
pieces. It is made up of a series of different equations.
Each different equation covers a different set of input
values (or numbers on the x-axis) over the domain.
A different piece of the domain is paired with each
equation. The domain determines if the end points on
the line are included or not. If the value is stated as
equal to, then the value is represented with a closed
circle on the line. If the value in the equation is stated
as less than or greater than, but not equal too, then it is
represented by an open circle.
Piecewise Example:
x-1 if x<-1
f(x)= -1 if x=-1
x+1 if x>-1
A closed dot indicates that the graph does not extend
beyond this point, and the point belongs to the graph.
An open dot indicates that the graph does not
extend beyond this point and the point
does not belong to the graph.
An arrow indicates that the graph extends
indefinitely in the direction in
which the arrow points.
Piecewise Functions
 2 x  1,if x  1
f  x  
 3x  1,if x  1
•One equation gives the value of f(x)
when x ≤ 1
•And the other when x>1
Example:
Evaluate f(x) when x=0, x=2, x=4
 x  2,if x  2
f ( x)  
 2 x  1,if x  2
•First you have to figure out which equation to use
•You NEVER use both
X=0
So:
This one fits
0+2=2
into the top
f(0)=2
equation
X=2
This
So: one fits here
2(2) + 1 = 5
f(2) = 5
X=4
So: one fits here
This
2(4) + 1 = 9
f(4) = 9
Your Turn
Evaluate the function when x=-2, 0, and 2.
3x  4, x  0
f ( x)  
 3x  1, x  0
f(-2)=-10
f(0)=1
f(2)=7
The Domain
• When defining a function you MUST
specify the set of input values to be used
(the domain).
– Example: y=3x2-5, x≥1
• A function is only defined over it’s domain,
outside of it’s domain the function does not
exist.
• If the domain is omitted we assume an
implied domain.
Implied Domain
• When no domain or restrictions on the xvalues have been provided, we assume
the largest possible set of x-values for
which the relation is defined as the
implied domain.
Determining Implied Domain
• Look for restrictions on the input or values
of x where the relation is undefined.
• These restrictions on x are caused by;
– Dividing by zero
– Taking the square root of a negative number
Example
Find the domain of the function
3x  10 x  8
g ( x)  2
x  3x  28
2
Solution:
We can substitute any real number in the
numerator, but we must avoid inputs that make the
denominator 0.
Solve x2  3x  28 = 0.
(x  7)(x + 4) = 0
x  7 = 0 or x + 4 = 0
x = 7 or
x = 4
The domain consists of the set of all real numbers
except 4 and 7 or {x|x  4 and x  7}.
Example
Find the domain of the function
f ( x)  x  3
The function is defined only for x-values for
which x – 3  0. Solving the inequality yields
x–30
x3
Domain: {x| x  3}
Your Turn
x2
Find the domain of the function g ( x)  2
x 1
The x values for which the function is undefined
are excluded from the domain. The function is
2
undefined when x – 1 = 0.
2
x –1=0
(x + 1)(x – 1) = 0
x=1
Domain: {x| x   1}
Your Turn
Find the domain of the function
f ( x)  x  1
The function is defined only for x-values for
which x – 1  0. Solving the inequality yields
x–10
x1
Domain: {x| x  1}
Applications
Example:
The total volume V (in billions of dollars) of farm real estate in the
United States from 1980 through 1997 can be modeled by
 13.826t 2  59.96t  775.4, 0  t  6
V (t )  
7  t  17
382.4  28.7t ,
where t=0 represents 1980. Use this model to approximate the
value of farm real estate 1n 1982, 1987, and 1996.
1982 : t  2  use 13.826t 2  59.96t  775.4
V (2)  13.826(2) 2  59.96(2)  775.4
V (2)  840  1982 : $840 billion
1987 : t  7  use 382.4+28.7t
V (7)  382.4  28.7(7)
V (7)  583  1987 : $583 billion
1996 : t  16  use 382.4+28.7t
V (16)  382.4  28.7(16)
V (16)  842  1996 : $842 billion
Your Turn
A baseball is hit at a point 3 feet above ground at a velocity of
100 feet per second and an angle of 45˚. The path of the
baseball is modeled by y = -0.0032x2 + x + 3, where x and
y are measured in feet. Will the baseball clear a 20-foot
fence located 280 feet from home plate?
Yes, the ball will be 32 feet off the ground when it
passes over the fence.
The Difference Quotient
The difference quotient of a function f is an
expression of the form
where h ≠ 0.
Where does
it come
from?
f (x  h)  f (x)
h
Example: Calculating a Difference
Quotient
f(x+h)-f(x)
2
Find
for f(x)=x  2 x  5
h
First find f(x+h)
f(x+h)=(x+h)  2(x+h)-5
2
x  2hx  h  2 x  2h  5
2
2
f(x+h)-f(x)
Find
for f(x)=x 2  2 x  5
h
Use f(x+h) from the previous slide
f(x+h)-f(x)
Second find
h
2
2
2
x

2
hx

h

2
x

2
h

5

x
 2 x  5

f(x+h)-f(x)

h
h
x 2  2hx  h 2  2 x  2h  5  x 2  2 x  5
h
2hx  h 2  2h
h
h  2x  h  2
h
2x+h-2
Your Turn:
Find and simplify the expressions if f ( x)  2 x  1
Find f(x+h)
f(x+h)-f(x)
Find
, h0
h
f ( x  h)  2 x  2h  1
f ( x  h)  f ( x )
2
h
Example: Calculating a
Difference Quotient
Let the distance d in feet that a racehorse runs in
t seconds be d(t) = 2t2 for 0 ≤ t ≤ 10.
(a) Find d(t +h).
(b) Find the difference quotient of d and simplify
the result.
(c) Evaluate the difference quotient for t =7 and
h=0.1. Interpret your results.
(d) Evaluate the difference quotient for t =4 and
h=1. Then sketch a graph that illustrates the
result.
(a) Find d(t +h).
(b) Find the difference quotient of d and simplify the result.
Solution
(a) d t  h   2 t  h 
 2 t 2  2th  h 2
2

 2t  4th  2h
2
(b) d t  h   d t 
h
2
4th  h

h
h 4t  h 

h
 4t  h
2
2t  4ht  2h  2t

h
2

2
2
(c) Evaluate the difference quotient for t =7 and
h=0.1. Interpret your results.
Solution
(c) If t = 7 and h = 0.1, then the difference
quotient becomes
4t + 2h = 4(7) +2(0.1) = 28.2.
The average rate of change, or average
velocity, of the horse from 7 seconds to
7 + 0.1 = 7.1 seconds is 28.2 feet per
second.
(d) Evaluate the difference quotient for t =4 and h=1.
Solution
(d) If t = 4 and h = 1, then
4t + 2h = 4(4) + 2(1) = 18
If t = 4 and d = 2(4)2 = 32, the first
point is (4, 32). If t = 4 and h = 1, it
follows that t + h = 4 + 1 = 5 so d =
2(5)2 = 50, the second point is (5, 50).
Thus the slope of the line passing
through (4, 32) and (5, 50) is m = 18.
REVIEW PROBLEMS
Problems
1. Let A = {2, 3, 4, 5} and B = {-3, -2, -1, 0, 1}. Which of
the following sets of ordered pairs represent functions
from set A to set B?
a)
–
b)
–
c)
–
d)
–
{(2, -2), (3, 0), (4, 1), (5, -1)}
Each element of A is matched with exactly one element of B, Function.
{(4, -3), (2, 0), (5, -2), (3, 1), (2, -1)}
The element 2 in A is matched with two elements, 0 and -1, in B, Not a
Function.
{(2, -1), (3,-1), (4, -1), (5, -1)}
Each element of A is matched with exactly one element of B, Function.
{(3, -2), (5, 0), (2, -3)}
The element 4 in A is not matched with an element of B, Not a Function.
Problems
2. Determine whether the equation x2 + y2 = 8
represents y as a function of x.
• y is not a function of x.
3. Let f(x) = 10 – 3x2 and find the following:
a) f(2)
– f(2) = -2
b) f(-4)
– f(-4) = -38
c) f(x – 1)
– f(x – 1) = -3x2 + 6x +7
Problems
4. Evaluate the function below at x = -2, 0,
and 2.
3x  4, x  0
f ( x)  
 3x  1, x  0
• f(-2) = -10, f(0) = 1, f(2) = 7
5. Find the domain of g ( x)  x  16
• x ≥ 16 or [16, ∞)
.
Summary of Function Terminology
• Function: A function is a relationship between two variables such that
to each value of the independent variable there corresponds exactly
one value of the dependent variable.
• Function Notation: y = f(x)
–
–
–
–
f is the name of the function.
y is the dependent variable, or the output value.
x is the independent variable, or the input value.
f(x) is the value of the function at x, or the output value.
• Domain: The domain of a function is the set of all values (inputs) of
the independent variable for which the function is defined.
• Range: The range of a function is the set of all values (outputs)
assumed by the dependent variable.
• Implied Domain: If f is defined by an algebraic expression and the
domain is not specified, the implied domain consists of all real
numbers for which the expression is defined.
Homework
• Section 1.2, pg. 24 – 29:
Vocabulary Check #1 – 5 all
Exercises: #1-23 odd, 27-39 odd, 47-59odd,
83, 87, 89
• Read Section 1.3, pg. 30-37