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Functions Prepared by: Richard Mitchell Humber College 4

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Chapter Four

FunctionsPrepared by: Richard Mitchell Humber College

4

1Click on the computer image at the bottom right for a direct web link to an interesting Wikipedia Math Site.

CASE STUDY

4.1 Functions and relations3Relations 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 112Each value of x results in only one value of y.In these relations, y is a function of x.

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

THESE ARE NOT FUNCTIONSTHESE ARE FUNCTIONS

4.1-EXAMPLE 3-Page 112Relations

Domain (All of the x values)

Range (All of the y values)

THIS IS NOT A FUNCTION

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

Domain (All of the x values)

Range (All of the y values)

THIS IS A FUNCTION

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

Domain (All of the x values)

Range (All of the y values)

THIS IS A FUNCTION

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)00+13+26+39+412+51500132639412515

4.1-EXAMPLE 5-Page 116Vertical Line Test

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

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

4.1-DEFINITIONS-Pages 117 to 118The 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) Domain012345678Stretch (cm) Range00.50.91.42.12.43.03.64.0The 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. Equations6y + 2x = 6Types of Functions II. Ordered PairsA 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 ValuesIV. Verbal StatementWrite 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 HThe 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 - 3The 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).9

4.1-DEFINITIONS-Pages 118 to 120Test 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)

Example 15Example 16Example 18Every value of x gives a real value of y.No value of x will make y negative.Test what values of x and y do not work. Domain (x) and Range (y) will be those values that do work.Test what values of x and y do not work. Domain (x) and Range (y) will be those values that do work.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 notation11Explicit Form (One variable is isolated on one side) y = 2x3 + 5 z = ay + bx = 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 + 5y = 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

4.2-STRATEGY extraWrite the equation y = 2x 3 in the form x = f(y)x = f(y)

2x = y + 3

x =

4.2-EXAMPLE 27-Page 123Write the equation y = 3x2 2x in the form f(x,y) = 0f(x,y) = 0

3x2 2x y = 0

ANS: 3x2 2x y = 0

4.2-EXAMPLE 29-Page 123Substitution 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) = -24.2-EXAMPLE 30-Page 123f(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) = 654.2-EXAMPLE 31-Page 124Given 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

(cont)f(5) 3f(2) 2f(3)4.2-EXAMPLE 32-Page 124Given f(x) = (x)2 3(x) + 4, find (where f(5) = 14 f(2) = 2 and f(3) = 4)

f(5) 3f(2) 2f(3)

= (14) 3(2)2(4)

= 14 6 8

= 8 8f(5) 3f(2) 2f(3)ANS: 14.2-EXAMPLE 32-Page 124Given 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 +34.2-EXAMPLE 33-Page 124Given 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 24.2-EXAMPLE 36-Page 125

2g(3) + 4h(9) f(5)

(cont)4.2-EXAMPLE 38-Page 125

= 2(9) + 4(3) (15)= 18 + 12 15= 30 15ANS: 24.2-EXAMPLE 38-Page 125

2g(3) + 4h(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) = -14.2-EXAMPLE 39-Page 1254.3 composite functions and inverse functions25Given 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 128Given 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 129Find the inverse f -1(x) of the function y = f(x) = (x)3y = f(x) = (x)3

4.3-EXAMPLE 45-Page 130

Step 1: Solve the given equation for x.

Step 2: Interchange the x and y variables.

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

4.3-EXAMPLE 46-Page 130

Step 1: Solve the given equation for x.

Step 2: Interchange the x and y variables.

CopyrightCopyright 2012 John Wiley & Sons Canada, Ltd. All rights reserved. Reproduction or translation of this work beyond that permitted by Access Copyright (The Canadian Copyright Licensing Agency) is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons Canada, Ltd. The purchaser may make back-up copies for his or her own use only and not for distribution or resale. The author and the publisher assume no responsibility for errors, omissions, or damages caused by the use of these programs or from the use of the information contained herein.30