2.1 the basic language of functions x

The Basic Language of Functions

The Basic Language of FunctionsA 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.

A function is a procedure that assigns each input




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.

A function is a procedure that assigns each input




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}



The Basic Language of FunctionsFunctions may be given graphically:

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

The output



input box

For example,

f(X) = X2 – 2X + 3 = y


the function



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) = (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 + 2



a. Evaluate f(–2)

f(x) = –3x + 2

f(–2)



a. Evaluate f(–2)

f(x) = –3x + 2

f(–2)

copy the input



a. Evaluate f(–2)

f(x) = –3x + 2

f(–2)

copy the input then paste the input

where the x is



a. Evaluate f(–2)

f(x) = –3x + 2

f(–2) = –3(–2) + 2


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}


The Basic Language of FunctionsGraphs 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).





function y = f(x).

For example, let the function

f(x) = x + 1,





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).





function y = f(x).






x 0 1 2 3

y = f(x) 1 2 3 4





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.





function y = f(x).






x 0 1 2 3

y = f(x) 1 2 3 4

y = x + 1


a line.





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.





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.

The Basic Language of FunctionsSince 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

Linear Equations and LinesC. Estimate the slope by eyeballing two points then find an

equation of each line below.

1. 2. 3. 4.

5. 6. 7. 8.

Linear Equations and LinesD. Draw each line that passes through the given two points.

Find the slope and an equation of the line.

Again Identify the vertical lines and the horizontal lines by

inspection and solve for them first.

(Fraction Review: slide 99 of 1.2)

1. (0, –1), (–2, 1) 2. (3, –1), (3, 1)

4. (1, –2), (–2, 3)

3. (2, –1), (3, –1)

6. (4, –2), (4, 0)5. (7, –2), (–2, –6)

7. (3/2, –1), (3/2, 1) 8. (3/4, –1/3), (1/3, 3/2)

9. (–1/4, –3/2), (2/3, –3/2) 10. (–1/3, –1/6), (–3/4, 1/2)

11. (2/5, –3/10), (–1/2, –3/5) 12. (–3/4, 5/6), (–3/4, –4/3)

Linear Equations and Lines

2. It’s perpendicular to 2x – 4y = 1 and passes through (–2, 1)

6. It’s perpendicular to 3y = x with x–intercept at x = –3.

12. It has y–intercept at y = 3 and is parallel to 3y + 4x = 1.

8. It’s perpendicular to the y–axis with y–intercept at 4.

9. It has y–intercept at y = 3 and is parallel to the x axis.

10. It’s perpendicular to the x– axis containing the point (4, –3).

11. It is parallel to the y axis has x–intercept at x = –7.

5. It is parallel to the x axis has y–intercept at y = 7.

E. Find the equations of the following lines.

1. The line that passes through (0, 1) and has slope 3.

7. The line that passes through (–2 ,1) and has slope –1/2.

3. The line that passes through (5, 2) and is parallel to y = x.

4. The line that passes through (–3, 2) and is perpendicular

to –x = 2y.

Linear Equations and Lines

The cost y of renting a tour boat consists of a base–cost plus

the number of tourists x. With 4 tourists the total cost is $65,

with 11 tourists the total is $86.

1. What is the base cost and what is the charge per tourist?

2. Find the equation of y in terms of x.

3. What is the total cost if there are 28 tourists?

The temperature y of water in a glass is rising slowly.

After 4 min. the temperature is 30 Co, and after 11 min. the

temperature is up to 65 Co. Answer 42–44 assuming the

temperature is rising linearly.

4. What is the temperature at time 0 and what is the rate of

the temperature rise?

5. Find the equation of y in terms of time.

6. How long will it take to bring the water to a boil at 100 Co?

F. Find the equations of the following lines.

Linear Equations and LinesThe cost of gas y on May 3 is $3.58 and on May 9 is $4.00.

Answer 45–47 assuming the price is rising linearly.

7. Let x be the date in May, what is the rate of increase in

price in terms of x?

8. Find the equation of the price in term of the date x in May.

9. What is the projected price on May 20?

G. Find the coordinates of the following points assuming

all points are evenly spaced.

1. 1 4

2. –1 5

1 3 11

3. a. Find x and y.

x zy

The number z is a “weighted average” of {1, 3, 11}

whose average is 5. In this case z is the average of

{1, 3, 3,11} instead because “3” is used both for x and y.

1 3 11

b. Find z the mid-point of x and y.

x y







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.

The Basic Language of FunctionsExercise 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

The Basic Language of FunctionsExercise 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( )a1 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 + 2x – 3

2.1 the basic language of functions x

Education