1. COMPOSITION AND INVERSE
Standards 4, 24, 25
COMPOSITION AND INVERSE
COMPOSITION FUNCTIONS
PROBLEM 1
PROBLEM 2
INVERSE OF FUNCTIONS AND RELATIONS
PROBLEM 3
PROBLEM 4
PROBLEM 5
END SHOW
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2. ALGEBRA II STANDARDS THIS LESSON AIMS:
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.
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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. COMPOSITION OF FUNCTIONS
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. Suppose f(x)=3x+2 and g(x)=4x +1 find [g f](x) [f g](x) and
Standards 4, 24, 25
Suppose f(x)=3x+2 and g(x)=4x +1 find 2
[g f](x)
[f g](x)
and
then evaluate them for 1 and 3 respectively.
[f g](x)=f[g(x)]
[g f](x)=g[f(x)]
=f( )
4x + 1 2
=g( )
3x+2
= 4( ) +1 2
=3( )+2
4x + 1 2
3x+2
=12x 2
= 4( ) +1
9x + 12x +4 2
=12x + 5 2
= 36x + 48x 2
= 36x + 48x + 17 2
Now find
[g f](3)
and
[f g](1)
[f g](x)
=12x + 5 2
[g f](x)
= 36x + 48x + 17 2
[f g](1)
=12( ) + 5 2 1
= 36( ) + 48( ) + 17 2
[g f](3) 3
=12(1)+5
= 36(9)
=12 + 5 =
= 17
= 485
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6. Suppose f(x)=2x-1 and g(x)=6x +2 find [g f](x) [f g](x) and
Standards 4, 24, 25
Suppose f(x)=2x-1 and g(x)=6x +2 find 2
[g f](x)
[f g](x)
and
then evaluate them for -2 and 5 respectively.
[f g](x)=f[g(x)]
[g f](x)=g[f(x)]
=f( )
6x + 2 2
=g( )
2x-1
= 6( ) +2 2
=2( )-1
6x + 2 2
2x-1
=12x 2
= 6( ) +2
4x - 4 x + 1 2
=12x + 3 2
= 24x x 2
= 24x x + 8 2
Now find
[g f](5)
and
[f g](-2)
[f g](x)
=12x + 3 2
[g f](x)
= 24x - 24x + 8 2
[f g](-2)
=12( ) + 3 2 -2
= 24( ) - 24( ) + 8 2
[g f](5) 5
=12(4)+3
= 24(25)
=48 + 3 =
= 51
= 488
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7. Standards 4, 24, 25 [f g](x) [g f](x)
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)
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. Find the inverse for the following relation and graph it:
Standards 4, 24, 25
Find the inverse for the following relation and graph it:
(-2,6), (2,4), (5,7), (5,9), (-7,10)
f(x)=
f (x) = -1
(6,-2), (4,2), (7,5), (9,5), (10,-7)
Identity function
f(x)=x 4 2 6 -2 -4 -6 8 10 -8
-10 x y
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 8 4 12 -4 -8
-12 16 20
-16
-20 x
f(x)=x
f(x)= 3x - 9
y = 3x -9
f (x) -1
x = 3y - 9
Solving for y
f (x) -1
= x + 3 1 3
x + 9 = 3y
y = x + 9
y = x + 3 1 3
f (x) -1
= x + 3 1 3
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10. Now getting the composition from each other
Standards 4, 24, 25
Now getting the composition from each other
f (x) -1
= x + 3 1 3
f(x)= 3x - 9
[f f ](x)=f[f (x)] -1
[f f ](x)=f [f(x)] -1
x + 3 1 3
=f( )
=g( )
3x-9
x + 3 1 3
= ( ) + 3 1 3
= 3( ) - 9
3x-9
=(3) x 1 3 1 3
= (3x) –( )(9) +3
= x
= x – 3 +3
= x
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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 8 4 12 -4 -8
-12 16 20
-16
-20 x
f(x)=x
f(x)= 2x + 8
y = 2x +8
f (x) -1
x = 2y + 8
Solving for y
x – 8 = 2y
f (x) -1
= x - 4 1 2
y = x - 8
y = x - 4 1 2
f (x) -1
= x - 4 1 2
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12. Now getting the composition from each other
Standards 4, 24, 25
Now getting the composition from each other
f (x) -1
= x - 4 1 2
f(x)= 2x + 8
[f f ](x)=f[f (x)] -1
[f f ](x)=f [f(x)] -1
x - 4 1 2
=f( )
=g( )
2x+8
x - 4 1 2
= ( ) - 4 1 2
= 2( ) + 8
2x+8
=(2) x 1 2 1 2
= (2x) –( )(8) -4
= x
= x – 4 +4
= x
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