2. The Basic Language of Functions
A function is a procedure that assigns each input
exactly one output.
3. The Basic Language of Functions
A function is a procedure that assigns each input
exactly one output.
Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.
N
4. The Basic Language of Functions
A function is a procedure that assigns each input
exactly one output.
Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.
This is a function because for each license number,
there is exactly one owner of that license.
N
5. Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.
This is a function because for each license number,
there is exactly one owner of that license.
The Basic Language of Functions
A function is a procedure that assigns each input
exactly one output. In mathematics, usually we name
functions as f, g, h…and by default we let x represent
the input and y represent the output.
N
6. Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.
This is a function because for each license number,
there is exactly one owner of that license.
The Basic Language of Functions
b. The inputs are driver-license numbers x,
the outputs y are the names of all the siblings of the
license-holders.
A function is a procedure that assigns each input
exactly one output. In mathematics, usually we name
functions as f, g, h…and by default we let x represent
the input and y represent the output.
7. Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.
This is a function because for each license number,
there is exactly one owner of that license.
The Basic Language of Functions
b. The inputs are driver-license numbers x,
the outputs y are the names of all the siblings of the
license-holders. This is not a function because for
some inputs of license numbers, there might be many
outputs of names of the siblings of the license holders.
A function is a procedure that assigns each input
exactly one output. In mathematics, usually we name
functions as f, g, h…and by default we let x represent
the input and y represent the output.
8. Given a function, the set D of all the inputs is called
the domain of the function.
The Basic Language of Functions
9. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers.
The Basic Language of Functions
N
10. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers. The set R of all
the outputs is called the range and the range R of the
license-number-to-name function are all the names
of drivers with a valid license.
The Basic Language of Functions
N
11. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers. The set R of all
the outputs is called the range and the range R of the
license-number-to-name function are all the names
of drivers with a valid license.
A function may be given by a chart where
the input and outputs are listed explicitly.
The Basic Language of Functions
x y
–1 4
2 3
5 –3
6 4
7 2
12. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers. The set R of all
the outputs is called the range and the range R of the
license-number-to-name function are all the names
of drivers with a valid license.
A function may be given by a chart where
the input and outputs are listed explicitly.
From the table we see that 3 is the output for
the input 2.
The Basic Language of Functions
x y
–1 4
2 3
5 –3
6 4
7 2
13. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers. The set R of all
the outputs is called the range and the range R of the
license-number-to-name function are all the names
of drivers with a valid license.
A function may be given by a chart where
the input and outputs are listed explicitly.
From the table we see that 3 is the output for
the input 2. The domain D = {–1, 2, 5, 6, 7}
and the range is R = {4, 3, –3, 2}.
The Basic Language of Functions
x y
–1 4
2 3
5 –3
6 4
7 2
14. Given a function, the set D of all the inputs is called
the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers. The set R of all
the outputs is called the range and the range R of the
license-number-to-name function are all the names
of drivers with a valid license.
A function may be given by a chart where
the input and outputs are listed explicitly.
From the table we see that 3 is the output for
the input 2. The domain D = {–1, 2, 5, 6, 7}
and the range is R = {4, 3, –3, 2}.
Note that we may have the same output 4
for two different inputs –1 and 6.
The Basic Language of Functions
x y
–1 4
2 3
5 –3
6 4
7 2
15. Functions may be given graphically:
The Basic Language of Functions
16. Functions may be given graphically:
For instance, Nominal Price(1975) $0.50
The Basic Language of Functions
17. Functions may be given graphically:
Domain (Nominal Price) = {year 1918 2005}
For instance, Nominal Price(1975) $0.50
The Basic Language of Functions
18. Functions may be given graphically:
Domain (Nominal Price) = {year 1918 2005}
Range (Nominal Price) = {$0.20$2.51}
For instance, Nominal Price(1975) $1.00
The Basic Language of Functions
19. The Basic Language of Functions
Functions may be given graphically:
Inflation Adjusted Price(1975) $1.85
20. The Basic Language of Functions
Functions may be given graphically:
Inflation Adjusted Price(1975) $1.85
Domain (Inflation Adjusted Price) = {1918 2005}
21. The Basic Language of Functions
Functions may be given graphically:
Domain (Inflation Adjusted Price) = {1918 2005}
Range (Inflation Adjusted Price) = {$1.25$3.50}
Inflation Adjusted Price(1975) $1.85
22. The Basic Language of Functions
Most functions are given by mathematics formulas.
23. For example,
f(X) = X2 – 2X + 3 = y
The Basic Language of Functions
Most functions are given by mathematics formulas.
24. For example,
f(X) = X2 – 2X + 3 = y
name of
the function
The Basic Language of Functions
Most functions are given by mathematics formulas.
25. For example,
f(X) = X2 – 2X + 3 = y
name of
the function
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
26. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
27. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
28. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
The input box holds the input for the formula.
29. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
The input box holds the input for the formula.
Hence f (2) means to replace x by (2) in the formula,
so f(2) =
30. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
The input box holds the input for the formula.
Hence f (2) means to replace x by (2) in the formula,
so f(2) = (2)2 – 2(2) + 3 = 3 = y.
31. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
The input box holds the input for the formula.
Hence f (2) means to replace x by (2) in the formula,
so f(2) = (2)2 – 2(2) + 3 = 3 = y.
The domain of this f(x) is the set of all real numbers.
32. For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
The output
The Basic Language of Functions
Most functions are given by mathematics formulas.
input box
The input box holds the input for the formula.
Hence f (2) means to replace x by (2) in the formula,
so f(2) = (2)2 – 2(2) + 3 = 3 = y.
Function names are used with the +, –, /, and *
with the obvious interpretation.
The domain of this f(x) is the set of all real numbers.
33. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
34. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
f(x) = –3x – 3
35. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)
36. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)
copy the input
37. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)
copy the input then paste the input
where the x is
38. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)
f(x) = –3x – 3
f(–2) = –3(–2) – 3
39. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
40. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
41. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(x) = –2x2 – 3x + 1
42. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(x) = –2x2 – 3x + 1
g(–2)
43. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(x) = –2x2 – 3x + 1
g(–2)
copy the input
44. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(x) = –2x2 – 3x + 1
g(–2)
copy the input
then paste the input
where the x’s are
45. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(x) = –2x2 – 3x + 1
g(–2) = –2(–2)2 – 3(–2) + 1
copy the input
46. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
47. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
48. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
49. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
=
50. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
= 3 – (–1) = 4
51. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
= 3 – (–1) = 4
d. Expand and simplify g(a + b).
52. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
= 3 – (–1) = 4
d. Expand and simplify g(a + b).
g(a + b) = –2(a + b)2 – 3(a + b) + 1
53. The Basic Language of Functions
Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
= 3 – (–1) = 4
d. Expand and simplify g(a + b).
g(a + b) = –2(a + b)2 – 3(a + b) + 1
= –2a2 – 4ab – 2b2 – 3a – 3b + 1
54. The Basic Language of Functions
e. Find the x where f(x) = g(x). f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
55. The Basic Language of Functions
e. Find the x where f(x) = g(x).
Setting f(x) = g(x) we’ve a 2nd degree equation:
–3x – 3 = –2x2 –3x + 1
f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
56. The Basic Language of Functions
e. Find the x where f(x) = g(x).
Setting f(x) = g(x) we’ve a 2nd degree equation:
–3x – 3 = –2x2 –3x + 1 so
2x2 = 4 → x = ±√2
f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
57. The Basic Language of Functions
The function f(x) = c where c is a number is called
a constant function, its output is the same.
e. Find the x where f(x) = g(x).
Setting f(x) = g(x) we’ve a 2nd degree equation:
–3x – 3 = –2x2 –3x + 1 so
2x2 = 4 → x = ±√2
f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
58. The Basic Language of Functions
The function f(x) = c where c is a number is called
a constant function, its output is the same.
The function P(x) = anxn + an-1xn-1 + .. + a1x + a0,
with a’s as numbers, is called a polynomial function.
The 1st degree polynomials L(x) = ax + b and
the 2nd degree polynomials Q(x) = ax2 + bx + c
are the most basic and important functions.
e. Find the x where f(x) = g(x).
Setting f(x) = g(x) we’ve a 2nd degree equation:
–3x – 3 = –2x2 –3x + 1 so
2x2 = 4 → x = ±√2
f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
59. The Basic Language of Functions
The function f(x) = c where c is a number is called
a constant function, its output is the same.
The function P(x) = anxn + an-1xn-1 + .. + a1x + a0,
with a’s as numbers, is called a polynomial function.
The 1st degree polynomials L(x) = ax + b and
the 2nd degree polynomials Q(x) = ax2 + bx + c
are the most basic and important functions.
A rational or fractional function is a function of the form
R(x) = N(x)/D(x), where N and D are polynomials.
e. Find the x where f(x) = g(x).
Setting f(x) = g(x) we’ve a 2nd degree equation:
–3x – 3 = –2x2 –3x + 1 so
2x2 = 4 → x = ±√2
f(x) = –3x – 3,
g(x) = –2x2 – 3x + 1
60. There are two main things to consider when
determining the domains of functions of real numbers.
The Basic Language of Functions
61. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
The Basic Language of Functions
62. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
The Basic Language of Functions
63. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
b. f (X) = 2x + 6
The Basic Language of Functions
64. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0
b. f (X) = 2x + 6
The Basic Language of Functions
65. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0 x = -3
b. f (X) = 2x + 6
The Basic Language of Functions
66. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0 x = -3
So the domain = {all numbers except x = -3}.
b. f (X) = 2x + 6
The Basic Language of Functions
67. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0 x = -3
So the domain = {all numbers except x = -3}.
b. f (X) = 2x + 6
We must have square root of nonnegative numbers.
The Basic Language of Functions
68. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0 x = -3
So the domain = {all numbers except x = -3}.
b. f (X) = 2x + 6
We must have square root of nonnegative numbers.
Hence 2x + 6 > 0 x > -3
The Basic Language of Functions
69. There are two main things to consider when
determining the domains of functions of real numbers.
1. The denominators can't be 0.
2. The radicand of square root (or any even root)
can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0 x = -3
So the domain = {all numbers except x = -3}.
b. f (X) = 2x + 6
We must have square root of nonnegative numbers.
Hence 2x + 6 > 0 x > -3
So the domain = {all numbers x > -3}
The Basic Language of Functions
71. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
72. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1,
73. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
74. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
75. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
Plot the (x, y)’s and we have the graph of f(x) = x + 1,
a line.
76. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
y = x + 1
Plot the (x, y)’s and we have the graph of f(x) = x + 1,
a line.
77. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
y = x + 1
Plot the (x, y)’s and we have the graph of f(x) = x + 1,
a line. Note that the graph of a function may cross any
vertical line at most at one point because for each x
there is only one corresponding output y.
78. The Basic Language of Functions
Graphs of Functions
The plot of all points (x, y)’s
that satisfy a given function
y = f(x) is the graph of the
function y = f(x).
For example, let the function
f(x) = x + 1, set y = f(x) and
make a table of few of the
ordered pairs (x, y)’s that
satisfy the equation y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
y = x + 1
Plot the (x, y)’s and we have the graph of f(x) = x + 1,
a line. Note that the graph of a function may cross any
vertical line at most at one point because for each x
there is only one corresponding output y.
79. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
80. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2,
81. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
82. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
So x = y2 is not a function.
83. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
So x = y2 is not a function.
Plot the graph by the table
shown.
x 0 1 1 4 4
y 0 1 -1 2 -2
84. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
So x = y2 is not a function.
Plot the graph by the table
shown.
x 0 1 1 4 4
y 0 1 -1 2 -2
x 0 1 1 4 4
y 0 1 -1 2 -2
x = y2
85. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
So x = y2 is not a function.
Plot the graph by the table
shown. In particular, if we
draw the vertical line x = 4,
x 0 1 1 4 4
y 0 1 -1 2 -2
x 0 1 1 4 4
y 0 1 -1 2 -2
x = y2
it intersects the graph at two points (4, 2) and (4, –2).
86. The Basic Language of Functions
The equation x = y2, treating x
as the input, is not a function.
For instance, for the input
x = 4, there are two outputs,
y’s that satisfy 4 = y2, namely
y = 2 and y = –2.
So x = y2 is not a function.
Plot the graph by the table
shown. In particular, if we
draw the vertical line x = 4,
x 0 1 1 4 4
y 0 1 -1 2 -2
x 0 1 1 4 4
y 0 1 -1 2 -2
x = y2
it intersects the graph at two points (4, 2) and (4, –2).
In general, if any vertical line crosses a graph at two
or more points, then the graph does not represent
any function.
87. The Basic Language of Functions
Since for functions each
input x has exactly one
output, therefore each
vertical line can only
intersect it’s graph at exactly
one location (e.g. y = x + 1).
88. The Basic Language of Functions
Since for functions each
input x has exactly one
output, therefore each
vertical line can only
intersect it’s graph at exactly
one location (e.g. y = x + 1).
y = x + 1
89. The Basic Language of Functions
Since for functions each
input x has exactly one
output, therefore each
vertical line can only
intersect it’s graph at exactly
one location (e.g. y = x + 1).
y = x + 1
However, if any vertical line
intersects a graph at two or
more points, i.e. there are two
or more outputs y associated
to one input x (eg. x = y2),
90. The Basic Language of Functions
Since for functions each
input x has exactly one
output, therefore each
vertical line can only
intersect it’s graph at exactly
one location (e.g. y = x + 1).
y = x + 1
However, if any vertical line
intersects a graph at two or
more points, i.e. there are two
or more outputs y associated
to one input x (eg. x = y2),
then the graph must not be
the graph of a function.
91. The Basic Language of Functions
Since for functions each
input x has exactly one
output, therefore each
vertical line can only
intersect it’s graph at exactly
one location (e.g. y = x + 1).
x 0 1 1 4 4
y 0 1 -1 2 -2
y = x + 1
However, if any vertical line
intersects a graph at two or
more points, i.e. there are two
or more outputs y associated
to one input x (eg. x = y2),
then the graph must not be
the graph of a function.
92. The Basic Language of Functions
Exercise A. For problems 1 – 6, determine if the
given represents a function. If it’s not a function,
give a reason why it’s not.
x y
2 4
2 3
4 3
1. x y
2 4
3 4
4 4
2. 3. 4.
x
y y
6. For any real number input x that is a rational
number, the output is 0, otherwise the output is 1
5. For any input x that is a positive integer, the
outputs are it’s factors.
x
All the (x, y)’s on the curve
93. The Basic Language of Functions
Exercise B.
Given the functions
f, g and h, find the
outcomes of the
following expressions.
If it’s not defined,
state so.
x y = g(x)
–1 4
2 3
5 –3
6 4
7 2
y = h(x)
f(x) = –3x + 7
1. f(–1) 2. g(–1) 3.h(–1)
4. –f(3) 5. –g(3) 6. –h(3)
7. 3g(6) 8. 2f(2) 9. h(3) + h(0)
10. 2f(4) + 3g(2) 11. –f(4) + f(–4) 12. h(6)*[f(2)]2
94. The Basic Language of Functions
1. f(x) =
1
2x – 6 2. f (x) = 2x – 6
C. Find the domain of the following functions.
5. f(x) =
1
(x – 2)(x + 6)
7. f (x) = (x – 2)(x + 6)
3. f(x) =
1
3 – 2x 4. f (x) = 3 – 2x
6. f(x) = 1
x2 – 1
8. f (x) = 1 – x2
9. f (x) = (x – 2)/(x + 6) 10. f (x) = 4x2 – 9
For problem 7-10, draw sign charts (section 1.6)
to solve for the domains.
95. The Basic Language of Functions
1. –f(3) 2. –g(3) 3. –h(3)
4. 3g(2) 5. 2f(2) 6. h(3) + h(0)
7. 2f(4) + 3g(2) 8. f(–3/2) 9. g(1/2)
13. f(3 + a) 14. g(3 + a) 15. g(3 + a)
16. f(a – b) 17. g(a – b)
18. f( )
a
1 19. f(a)
1
11. h(–3/2) 12. g(–1/2)
10. 1/g(2)
Exercise D. Given the functions f, g and h, find the
outcomes of the following expressions. If it’s not
defined, state so.
f(x) = –2x + 3 g(x) = –x2 + 3x – 2 h(x) =
x + 2
x – 3
96. The Basic Language of Functions
1. where f(x) = 5.
Exercise E. Given the functions f, g and h,
f(x) = 3x – 4 g(x) = –x2 + 3x – 2 h(x) = 4
x + 1
2. where f(x) = –2.
Find the x or x’s
3. where g(x) = 0.
5. where g(x) = f(x).
4. where g(x) = –3.
6. where h(x) = 1.
8. where h(x) = f(x).
7. where h(x) = x.
97. The Basic Language of Functions
(Answers to odd problems) Exercise A.
1. It’s not a function 3. Not a function
5. Not a function
Exercise B.
1. f(–1) = 10 3. h(–1) = 2 5. Not defined
7. 3g(6) = 12 9. h(3) + h(0) = 2 11. –f(4) + f(–4) = 24
1. x ≠ 3
Exercise C.
5. x ≠ 2 and x ≠ -6
3. x ≠ 3/2
98. The Basic Language of Functions
7. (𝑥 – 2)(𝑥 + 6)
x=-6 x=2
+
+
The domain of f(x) = (x – 2)(x + 6) is −∞, −6 ∪ [2, ∞)
9. (𝑥 – 2)/(𝑥 + 6)
x=-6 x=2
+
+
The domain of f(x) = (x – 2)/(x + 6) is (−∞, −6) ∪ [2, ∞)
99. The Basic Language of Functions
1. –f(3) = 3 3. Not defined 5. 2f(2) = –2
7. 2f(4) + 3g(2) = –10 9. g(1/2) = –3/4
13. f(3 + a) = –2a – 3
15. g(3 + a) = –a2 – 3a – 2
17. g(a – b) = –a2 + 2ab – b2 + 3a – 3b – 2
19. f(a)
1 = 1
11. h(–3/2) = –1/9
Exercise D.
–2a+3
1. x = 3
Exercise E.
3. x = 1, 2 5. x = ±√2 7. x = (–1±√17)/2