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Functions Prepared by: Richard Mitchell Humber College 4.
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Transcript of Functions Prepared by: Richard Mitchell Humber College 4.
Functions
Prepared by: Richard Mitchell Humber College
4
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
Domain (All of the x values)
Range (All of the y values)
{( , ) ( , ) ( , ) ( , ) (
-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
Domain (All of the x values)
Range (All of the y values)
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.
4.1-DEFINITIONS-Pages 117 to 118
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).
4.1-DEFINITIONS-Pages 118 to 120
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.
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.
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
4.2-EXAMPLE 29-Page 123
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
4.2-EXAMPLE 30-Page 123
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
4.2-EXAMPLE 31-Page 124
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)
4.2-EXAMPLE 32-Page 124
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
4.2-EXAMPLE 32-Page 124
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
4.2-EXAMPLE 33-Page 124
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
4.2-EXAMPLE 36-Page 125
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)
4.2-EXAMPLE 38-Page 125
= 2∙(9) + 4∙(3) (15)
= 18 + 12 15
= 30 15
ANS: 2
4.2-EXAMPLE 38-Page 1252
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
4.2-EXAMPLE 39-Page 125
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
4.3-EXAMPLE 45-Page 130
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
4.3-EXAMPLE 46-Page 130
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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