1 composition and inverse composition functions standards 4, 24, 25 problem 2 inverse of functions...
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1
COMPOSITION AND INVERSE
COMPOSITION FUNCTIONS
Standards 4, 24, 25
PROBLEM 2
INVERSE OF FUNCTIONS AND RELATIONS
PROBLEM 3
PROBLEM 1
PROBLEM 4
PROBLEM 5
END SHOW
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2
STANDARD 4:
Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes.
STANDARD 24:
Students solve problems involving functional concepts, such as composition, defining the inverse function and performing arithmetic operations on functions.
STANDARD 25:
Students use properties from number systems to justify steps in combining and simplifying functions.
ALGEBRA II STANDARDS THIS LESSON AIMS:
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3
ESTÁNDAR 4:
Los estudiantes factorizan diferencias de cuadrados, trinomios cuadrados perfectos, y la suma y diferencia de dos cubos.
ESTÁNDAR 24:
Los estudinates resuelven problemas que involucran conceptos como composición de funciones, definición de inversa de funciones y efectuan operaciones aritméticas en funciones.
ESTÁNDAR 25:
Los estudiantes usan propiedades de los sistemas numéricos para combinar y simplificar funciones.
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4
Standards 4, 24, 25
COMPOSITION OF FUNCTIONS
Suppose f and g are functions such that the range of g is a subset of the domain of f. Then the composite function of f g can be described by the equation [f g](x)=f[g(x)]
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5
Standards 4, 24, 25Suppose f(x)=3x+2 and g(x)=4x +1 find 2 [g f](x)[f g](x) and
[f g](x)=f[g(x)]
=f( )4x + 12
=3( )+24x + 12
=12x + 3 +22
=12x + 52
[g f](x)=g[f(x)]
=g( )3x+2
= 4( ) +123x+2
= 4( ) +19x + 12x +42
= 36x + 48x + 16 + 12
= 36x + 48x + 172
Now find [g f](3)and[f g](1)
[f g](x) =12x + 52
[f g](1) =12( ) + 521
=12(1)+5
=12 + 5
= 17
[g f](x)= 36x + 48x + 172
= 36( ) + 48( ) + 172[g f](3) 3 3
= 36(9) + 144 + 17
= 324 + 144 + 17
= 485
then evaluate them for 1 and 3 respectively.
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6
Standards 4, 24, 25Suppose f(x)=2x-1 and g(x)=6x +2 find 2 [g f](x)[f g](x) and
[f g](x)=f[g(x)]
=f( )6x + 22
=2( )-16x + 22
=12x + 4 -12
=12x + 32
[g f](x)=g[f(x)]
=g( )2x-1
= 6( ) +222x-1
= 6( ) +24x - 4 x + 12
= 24x - 24x + 6 + 22
= 24x - 24x + 82
Now find [g f](5)and[f g](-2)
[f g](x) =12x + 32
[f g](-2)=12( ) + 32-2
=12(4)+3
=48 + 3
= 51
[g f](x)= 24x - 24x + 82
= 24( ) - 24( ) + 82[g f](5) 5 5
= 24(25)-120 +8
= 600 - 120 + 8
= 488
then evaluate them for -2 and 5 respectively.
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7
Standards 4, 24, 25
INVERSE FUNCTIONS AND RELATIONS
• Two functions f and g are inverse functions if and only if both of their compositions are the identity function. That is, and [f g](x)=x [g f](x)=x
• Two relations are inverse relations if and only if whenever one relation contains the element (a,b), the other contains the element (b,a).
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8
Standards 4, 24, 25
Find the inverse for the following relation and graph it:
(6,-2), (4,2), (7,5), (9,5), (10,-7)f (x) =-1
(-2,6), (2,4), (5,7), (5,9), (-7,10)f(x)=
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
yf(x)=x Identity
function
We can observe the symmetry respect the identity function.
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9
Standards 4, 24, 25
Find the inverse of f(x)= 3x -9, graph both functions and then get the compositions from one to the other.
f(x)= 3x - 9 y = 3x -9
f (x)-1
x = 3y - 9 Solving for y+9 +9
x + 9 = 3y
3 3
y = x + 93 3
y = x + 313
y
84 12-4-8-12
4
8
12
-4
-8
-12
16 20-16-20
16
-16
20
x
f(x)= 3x - 9 f(x)=x
f (x)-1
= x + 313
f (x)-1
= x + 313
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10
Standards 4, 24, 25
=f( )
[f f ](x)=f[f (x)]-1 -1 [f f ](x)=f [f(x)]-1 -1
x + 313
f (x)-1
= x + 313
f(x)= 3x - 9
= 3( ) - 9x + 313
=(3) x + 9 - 913
= x
=g( )3x-9
= ( ) + 313
3x-9
13
= (3x) –( )(9) +313
= x – 3 +3
= x
Now getting the composition from each other
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11
Standards 4, 24, 25
Find the inverse of f(x)= 2x +8, graph both functions and then get the compositions from one to the other.
f(x)= 2x + 8 y = 2x +8
f (x)-1
x = 2y + 8 Solving for y-8 -8
x – 8 = 2y
2 2
y = x - 82 2
y = x - 412
y
84 12-4-8-12
4
8
12
-4
-8
-12
16 20-16-20
16
-16
20
x
f(x)= 2x+8 f(x)=x
f (x)-1
= x - 412
f (x)-1
= x - 412
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12
Standards 4, 24, 25
=f( )
[f f ](x)=f[f (x)]-1 -1 [f f ](x)=f [f(x)]-1 -1
x - 412
f (x)-1
= x - 412
f(x)= 2x + 8
= 2( ) + 8
x - 412
=(2) x - 8 + 812
= x
=g( )2x+8
= ( ) - 412
2x+8
12
= (2x) –( )(8) -412
= x – 4 +4
= x
Now getting the composition from each other
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