Functions Prepared by: Richard Mitchell Humber College 4

Functions

Prepared by: Richard Mitchell Humber College

4

http://en.wikipedia.org/wiki/Function_(mathematics)

CASE STUDY

4.1 – FUNCTIONS AND RELATIONS

Relations

f(x) = 3x – 5 (Functional Notation)

y = 3x3 (Power)

y = 3x2 + 4 (Quadratic)

y = 3 sin 2x (Trigonometric)

y = 52x (Exponential)

y = log(x + 2) (Logarithmic)

Relations

4.1-DEFINITIONS-Pages 111 to 112

Each value of x results in only one value of y.In these relations, y is a function of x.

y x2 2 4y x

Each value of x results in two values of y. In these relations, y is not a function of x.

ordered pairs{(2,3)(4,3)(6,3)(8,3)( 10,3)}

THESE ARE NOT FUNCTIONS

THESE ARE FUNCTIONS

ordered pairs{(2,3)(2, 3)(4,5)(4, 5)}

4.1-EXAMPLE 3-Page 112Relations

Domain (All of the x values)

Range (All of the y values)

{( , ) ( , ) ( , - ) ( , ) -3 6 12 4 3 6 2(6 , 3)}

THIS IS NOT A FUNCTION

(2, -3)

( )x components ( )y componentsDomain Range

{ , , , an2 3 4 d 6}

-3 -1 3{ , , , and 6}

Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are two points associated with the same x value: (2, -3) and (2, 3). This one value, x = 2, gives two possible y values.

4.1-EXAMPLE 4-Page 113Relations



{( , ) ( , ) ( , ) ( , ) (

-4 -3 -2 -10

-3 -3 -3 -3-3, ) ( ,1 -3)}

THIS IS A FUNCTION

( , -4 -3)

( )x components ( )y componentsDomain Range

-4 -3 -2 -{ , , , , and1 0 1}

{-3}

Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are no duplicates of x values in the data set so this relation is a function.

4.1-DEFINITIONS-Page 114 Graphing



Graph 3y x

THIS IS A FUNCTION(no duplicates){ , , , , , ..0 1 2 3 4 5 .}

0 3 6 9 { , , , , , .12 15 ..}

Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are no duplicates of x values in the data set so this relation is a function.

x-values (Domain)

y-values (Range)

0 0

+1 3

+2 6

+3 9

+4 12

+5 15

0 0

1 3

2 6

3 9

4 12

5 15

4.1-EXAMPLE 5-Page 116Vertical Line Test

THIS IS A FUNCTION

A vertical line drawn anywhere on the curve intersects only at one point therefore it is a function.

THIS IS NOT A FUNCTION

A vertical line drawn anywhere on the curve intersects at two points therefore it is not a function.


The equation is a function rule that associates exactly one y with one x.(Domain x and Range y are Real Numbers).

The ordered pair is a function since each value of x (load) is associated with one value of y (stretch).(Domain x and Range y are Real Numbers).

Load (Kg) Domain 0 1 2 3 4 5 6 7 8

Stretch (cm) Range 0 0.5 0.9 1.4 2.1 2.4 3.0 3.6 4.0

The table of ordered pairs is a function since each valueof x (load) is associated with one value of y (stretch).(Domain x and Range y are Real Numbers).

I. Equations

6y + 2x = 6

Types of Functions

II. Ordered Pairs

A load of 6 kg causes a spring to stretch 3 cm. The ordered pair that results is (6, 3) where load is first and distance is second.

III. Table of Values

IV. Verbal Statement

Write an equation for the volume of a cone

in terms of its base (75 units) and altitude.

V = 1/3 x (base area) x (altitude)

V = 1/3 x (75) x H

V = 25 H

The formula is a function since each value of H (altitude)is associated with one value of V (volume).(Domain H and Range V are Real Numbers).

V. Graphs

f(x) = x2 - 4x - 3

The graph is a function since each value on thex (axis) is associated with one value on the y (axis).(Domain x and Range y are Real Numbers).


Test what values of x and y do not work. Domain (x) and Range (y) will be those values that do work.

Domain (x) and Range (y)

2y x

: Domain x

:Range 0y

2y x

Domain: 2x

:Range 0y

9

4y

x

Domain: <4x

:Range 0y

Example 15

Example 16

Example 18

Every value of x gives a real value of y.

No value of x will make y negative.



Any value of x less than 2 will make the quantity under the radical sign negative.

An x value of 2 gives a y value of zero. An x value larger than 2 gives a y value greater than zero.

An x value of 4 will make the quantity in the denominator equal to zero. Any value of x greater than 4 will make the quantity under the radical sign negative.

All values of y will be positive.

4.2 – FUNCTIONAL NOTATION

Explicit Form (One variable is isolated on one side) y = 2x3 + 5

z = ay + b

x = 3z2 + 2z – 5

Implicit Form (One variable is not isolated to one side)

y = x2 + 4y x2 + y2 = 25w + x = y + z + x

Dependent and Independent Variables (Value of Dependent Variable y depends upon value of Independent Variable x)

y = x + 5 y = 2x2 – 3

4.2-DEFINITIONS-Page 121

Manipulating Functions (Re-arranging)

If 2x + y = 5, then

y = f(x) x = f(y) f(x,y) = 0 y = 5 – 2x x = 2x + y – 5 = 0

If y – 4x = 5 – z, then

y = f(x,z) x = f(y,z) z = f(x,y) f(x,y,z) = 0 y = 4x – z + 5 x = z = 4x – y + 5 4x-y-z+5 = 0

5

2

y

5

4

y z

4.2-STRATEGY extra

Write the equation y = 2x – 3 in the form x = f(y)

x = f(y)

2x = y + 3

x = 3

2

y

3ANS: =

2

yx

4.2-EXAMPLE 27-Page 123

Write the equation y = 3x2 – 2x in the form f(x,y) = 0

f(x,y) = 0

3x2 – 2x – y = 0

ANS: 3x2 – 2x – y = 0


Substitution into Functions

Given f(x) = (x)3 – 5(x), find f(2)

f(2) = (2)3 – 5(2)

f(2) = 8 – 10

f(2) = -2

ANS: f(2) = -2


f(x) = (x)3 – 5(x)

Given y(x) = 3(x)2 – 2(x), find y(5)

y(x) = 3(x)2 – 2(x)

y(5) = 3(5)2 – 2(5)

y(5) = 3(25) – 10

y(5) = 75 – 10

y(5) = 65

ANS: y(5) = 65


Given f(x) = (x)2 – 3(x) + 4, find

f(x) = (x)2 – 3(x) + 4 f(x) = (x)2 – 3(x) + 4 f(x) = (x)2 – 3(x) + 4

f(5) = (5)2 – 3(5) + 4 f(2) = (2)2 – 3(2) + 4 f(3) = (3)2 – 3(3) + 4

f(5) = 25 – 15 + 4 f(2) = 4 – 6 + 4 f(3) = 9 – 9 + 4

f(5) = 10 + 4 f(2) = -2 + 4 f(3) = 0 + 4

f(5) = 14 f(2) = 2 f(3) = 4

(con’t)

f(5) – 3∙f(2) 2∙f(3)


Given f(x) = (x)2 – 3(x) + 4, find (where f(5) = 14 f(2) = 2 and f(3) = 4)

f(5) – 3∙f(2) 2∙f(3)

= (14) – 3∙(2)2∙(4)

= 14 – 6 8

= 8 8

f(5) – 3∙f(2) 2∙f(3)

ANS: 1


Given f(x) = 3(x)2 – 2(x) + 3, find f(5a)

f(x) = 3(x)2 – 2(x) + 3

f(5a) = 3(5a)2 – 2(5a) + 3

f(5a) = 3(25a2) – 10a + 3

f(5a) = 75a2 – 10a + 3

ANS: f(5a) = 75a2 – 10a +3


Given f(x) = 5(x) – 2, find f(x + a)

f(x) = 5(x) – 2

f(x + a) = 5(x + a) – 2

f(x + a) = 5x + 5a – 2

ANS: f(x+ a) = 5x + 5a – 2


2∙g(3) + 4∙h(9) f(5)

2andGiven ( ) 3( ) ( ) ( ) ( ) ( ) evalua e t x xf g x x hx x

2( ) 3( ) ( ) ( ) ( ) ( )x x xf hxx x g

2( ) 3( ) ( ) ( ) 3 5 ( )935 ( )9f g h

3( ) 15 ( ) 9 9( 35 )f g h (con’t)


= 2∙(9) + 4∙(3) (15)

= 18 + 12 15

= 30 15

ANS: 2


where ( ) 15, (3) 9 an5 9d ( ) 3)(

andGiven ( ) 3( ) ( ) ( ( ) ( ) evaluate) f g h

f g x x xxx hx

2∙g(3) + 4∙h(9) f(5)

Given f(x,y,z) = 2(y) – 3(z) + x, find f(3,1,2)

f(3,1,2) = 2(1) – 3(2) + 3

f(3,1,2) = 2 – 6 + 3

f(3,1,2) = -4 + 3

ANS: f(3,1,2) = -1


4.3 – COMPOSITE FUNCTIONS AND

INVERSE FUNCTIONS

Given g(x) = (x) + 1, find g(2), g(z2) and g[f(x)]

g(x) = (x) + 1 g(x) = (x) + 1 g(x) = (x) + 1

g(2) = (2) + 1 g(z2) = (z2) + 1 g[f(x)] = (f(x)) + 1

g(2) = 3 g(z2) = z2 + 1 g[f(x)] = f(x) + 1

ANS: g(2) = 3 ANS: g(z2) = z2 + 1 ANS: g[f(x)] = f(x) +

1

4.3-EXAMPLE 40 (a), (b) and (c)-Page 128

Given g(x) = (x)2 and f(x) = (x) + 1, find the following g[f(x)] f[g(2)] g[f(2)]

g(x) = (x)2 f(x) = (x) + 1 g(x) = (x)2

g[f(x)] = (f(x))2 f[g(2)] = (g(2)) + 1 g[f(2)] = (f(2))2

= (x + 1)2 = (22) + 1 = (2 + 1)2

= x2 + 2x + 1 = 5 = 9

ANS: x2 + 2x + 1 ANS: 5 ANS: 9

4.3-EXAMPLE 42 (b), (c) and (d)-Page 129

Find the inverse f -1(x) of the function y = f(x) = (x)3

y = f(x) = (x)3


3x yStep 1: Solve the given equation for x.

1 3( ) xfy x Step 2: Interchange the x and y variables.

1 3ANS: ( )f x x

Find the inverse f -1(x) of the function y = f(x) = 2x + 5

y = f(x) = 2x + 5


5

2x

y Step 1: Solve the given equation for x.

1 5( )

2f x

xy

Step 2: Interchange the x and y variables.

1 5ANS: ( )

2

xf x

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Functions Prepared by: Richard Mitchell Humber College 4