9 the basic language of functions

The Basic Language of Functions

A function is a procedure that assigns each input
exactly one output.

exactly one output. In mathematics, usually we name
functions as f, g, h…

functions as f, g, h…and by default we let x represent
the input and y represent the output.

Example A.
a. The inputs are driver-license numbers x,
the outputs y are the names of the license-holders.

Example A.
This is a function because for each license number,
there is exactly one owner of that license.

Example A.
b. The inputs are driver-license numbers x,
the outputs y are the names of all the siblings of the
license-holders.

Example A.
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.

Given a function, the set D of all the inputs is called
the domain of the function.

the domain of the function. The domain D of the
license-number-to-name function in example A
are all valid driver license numbers.

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.
x y
–1 4
2 3
5 –3
6 4
7 2

From the table we see that 3 is the output for
the input 2.
x y
–1 4
2 3
5 –3
6 4
7 2

the input 2. The domain D = {–1, 2, 5, 6, 7}
and the range is R = {4, 3, –3, 2}.
x y
–1 4
2 3
5 –3
6 4
7 2

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.
x y
–1 4
2 3
5 –3
6 4
7 2

Functions may be given graphically:

For instance, Nominal Price(1975)  $0.50

Domain (Nominal Price) = {year 1918  2005}

Domain (Nominal Price) = {year 1918  2005}
Range (Nominal Price) = {$0.20$2.51}

Inflation Adjusted Price(1975)  $1.85

Domain (Inflation Adjusted Price) = {1918  2005}

Domain (Inflation Adjusted Price) = {1918  2005}
Range (Inflation Adjusted Price) = {$1.25$3.50}

Most functions are given by mathematics formulas.

For example,
f(X) = X2 – 2X + 3 = y

For example,
f(X) = X2 – 2X + 3 = y
name of
the function

For example,
f(X) = X2 – 2X + 3 = y
name of
the function
input box

For example,
f(X) = X2 – 2X + 3 = y
name of actual formula
the function
input box

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box
The input box holds the input for the formula.

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box
Hence f (2) means to replace x by (2) in the formula,
so f(2) =

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box
so f(2) = (2)2 – 2(2) + 3 = 3 = y.

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box
so f(2) = (2)2 – 2(2) + 3 = 3 = y.
The domain of this f(x) is the set of all real numbers.

For example,
f(X) = X2 – 2X + 3 = y
the function
The output
input box
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.

Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3.
a. Evaluate f(–2)

a. Evaluate f(–2)
f(x) = –3x – 3

a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)

a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)
copy the input

a. Evaluate f(–2)
f(x) = –3x – 3
f(–2)
copy the input then paste the input
where the x is

a. Evaluate f(–2)
f(x) = –3x – 3
f(–2) = –3(–2) – 3

Example B. Let f(x) = –3x – 3, g(x) = –2x2 – 3x + 1.
a. Evaluate f(–2).
f(–2) = –3(–2) – 3 = 3

f(–2) = –3(–2) – 3 = 3
b. Evaluate g(–2).

f(–2) = –3(–2) – 3 = 3
g(x) = –2x2 – 3x + 1

f(–2) = –3(–2) – 3 = 3
g(x) = –2x2 – 3x + 1
g(–2)

f(–2) = –3(–2) – 3 = 3
g(x) = –2x2 – 3x + 1
g(–2)
copy the input

f(–2) = –3(–2) – 3 = 3
g(x) = –2x2 – 3x + 1
g(–2)
copy the input
then paste the input
where the x’s are

f(–2) = –3(–2) – 3 = 3
g(x) = –2x2 – 3x + 1
g(–2) = –2(–2)2 – 3(–2) + 1
copy the input

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
c. Evaluate f(–2) – g(–2).

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
Using the outputs of parts a and b we’ve
f(–2) – g(–2)
=

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
f(–2) – g(–2)
= 3 – (–1) = 4

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
f(–2) – g(–2)
= 3 – (–1) = 4
The function f(x) = c where c is a number is called
a constant function.

f(–2) = –3(–2) – 3 = 3
g(–2) = –2(–2)2 – 3(–2) + 1
= –8 + 6 + 1 = –1
f(–2) – g(–2)
= 3 – (–1) = 4
The function f(x) = c where c is a number is called
a constant function. The outputs of such functions
do not change.

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.

can't be negative.
Example C. Find the domain of the following functions.
a. f(x) = 1/(2x + 6)
b. f (X) =  2x + 6

can't be negative.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0
b. f (X) =  2x + 6

can't be negative.
a. f(x) = 1/(2x + 6)
The denominator can’t be 0 i.e. 2x + 6 = 0  x = -3
b. f (X) =  2x + 6

can't be negative.
a. f(x) = 1/(2x + 6)
So the domain = {all numbers except x = -3}.
b. f (X) =  2x + 6

can't be negative.
a. f(x) = 1/(2x + 6)
b. f (X) =  2x + 6
We must have square root of nonnegative numbers.

can't be negative.
a. f(x) = 1/(2x + 6)
b. f (X) =  2x + 6
Hence 2x + 6 > 0  x > -3

can't be negative.
a. f(x) = 1/(2x + 6)
b. f (X) =  2x + 6
Hence 2x + 6 > 0  x > -3
So the domain = {all numbers x > -3}

Graphs 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).

Graphs of Functions
function y = f(x).
For example, let the function
f(x) = x + 1,

Graphs of Functions
function y = f(x).
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).

Graphs of Functions
function y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4

Graphs of Functions
function 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.

Graphs of Functions
function y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
y = x + 1
a line.

Graphs of Functions
function y = f(x).
x 0 1 2 3
y = f(x) 1 2 3 4
y = 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.

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,

y’s that satisfy 4 = y2, namely
y = 2 and y = –2.

y = 2 and y = –2. So x = y2 is
not a function.

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

y = 2 and y = –2. So x = y2 is
not a function.
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

y = 2 and y = –2. So x = y2 is
not a function.
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).

y = 2 and y = –2. So x = y2 is
not a function.
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.

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

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),

y = x + 1
then the graph must not be
the graph of a function.

x 0 1 1 4 4
y 0 1 -1 2 -2
y = x + 1
then the graph must not be
the graph of a function.

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

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

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.

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

(Answers to odd problems) Exercise A.
1. It’s 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 ≠ -63. x ≠ 3/2

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, ∞)

1. –f(3) = 3 3. Not defined 5. 2f(2) = –2
7. 2f(4) + 3g(2) = –10 9. g(1/2) = –1/4
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

9 the basic language of functions

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9 the basic language of functions