functions – composite functions rules :. functions – composite functions rules : read as f at g...
TRANSCRIPT
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Read as “ f ” at “g” of “x”
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Read as “ f ” at “g” of “x”
Read as “ g ” at “f” of “x”
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Read as “ f ” at “g” of “x”
Read as “ g ” at “f” of “x”
Symbol DOES NOT mean multiply !!!
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Read as “ f ” at “g” of “x”
Read as “ g ” at “f” of “x”
It is the symbol used to show composite…
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( ƒ ◦ g )(3) if ƒ(x) = 2x – 6 and g(x) = x2
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( ƒ ◦ g )(3) if ƒ(x) = 2x – 6 and g(x) = x2
Using the Rule )3()3(
)()(
gfgf
xgfxgf
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( ƒ ◦ g )(3) if ƒ(x) = 2x – 6 and g(x) = x2
Using the Rule )3()3(
)()(
gfgf
xgfxgf
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( ƒ ◦ g )(3) if ƒ(x) = 2x – 6 and g(x) = x2
Using the Rule )3()3(
)()(
gfgf
xgfxgf
9)3()3( 2 g We need to find g(3) first
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( ƒ ◦ g )(3) if ƒ(x) = 2x – 6 and g(x) = x2
Using the Rule )3()3(
)()(
gfgf
xgfxgf
9)3()3( 2 g
12618)9(
6)9(2)9(
f
f Now we place 9 into f(x)
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( g ◦ ƒ )(-2) if ƒ(x) = x2 – x – 12 and g(x) = 0.5x + 4
Using the Rule : )2()2(
)()(
fgfg
xfgxfg
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( g ◦ ƒ )(-2) if ƒ(x) = x2 – x – 12 and g(x) = 0.5x + 4
Using the Rule : )2()2(
)()(
fgfg
xfgxfg
First find ƒ( -2 )
61224)2(
12)2()2()2( 2
f
f
FUNCTIONS – Composite Functions
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating numeric composites : work “ inside out “
EXAMPLE : Find ( g ◦ ƒ )(-2) if ƒ(x) = x2 – x – 12 and g(x) = 0.5x + 4
Using the Rule : )2()2(
)()(
fgfg
xfgxfg
Now substitute -6 in g(x)
61224)2(
12)2()2()2( 2
f
f
143)6(
4)6(5.0)6(
g
g
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( ƒ ◦ g )(x) if ƒ(x) = x – 1 and g(x) = 0.5x + 4
45.0)()( xfxgfxgf
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( ƒ ◦ g )(x) if ƒ(x) = x – 1 and g(x) = 0.5x + 4
45.0)()( xfxgfxgf
1)45.0( xxf Substitute into “x”
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( ƒ ◦ g )(x) if ƒ(x) = x – 1 and g(x) = 0.5x + 4
45.0)()( xfxgfxgf
1)45.0( xxf Substitute into “x”
145.0)45.0( xxf
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( ƒ ◦ g )(x) if ƒ(x) = x – 1 and g(x) = 0.5x + 4
45.0)()( xfxgfxgf
1)45.0( xxf
145.)45.0( xxf
35. x Combined like terms
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( g ◦ ƒ )(x) if ƒ(x) = x + 3 and g(x) = 3x2 – 2x + 8
3)()( xgxfgxfg
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( g ◦ ƒ )(x) if ƒ(x) = x + 3 and g(x) = 3x2 – 2x + 8
3)()( xgxfgxfg
8)3(2)3(33 2 xxxg
823)( 2 xxxg
Substituted ( x + 3 ) for all x’s
RULES :
)()(
)(
xfgxfg
xgfxgf
Calculating algebraic composites : substitute the rule of the inside function into the outside function wherever there is a variable
EXAMPLE : Find ( g ◦ ƒ )(x) if ƒ(x) = x + 3 and g(x) = 3x2 – 2x + 8
3)()( xgxfgxfg
8)3(2)3(33 2 xxxg
823)( 2 xxxg
291633
862271833
8629633
2
2
2
xxxg
xxxxg
xxxxg
Composite Functions : going backwards
Suppose we were given a function h(x) that was the result of a composite operation. How could we determine the two functions that were combined to get that function ?
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses or under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Composite Functions : going backwards
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses of under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Example : If h(x) = ( x + 2 )2 was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
Composite Functions : going backwards
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses of under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Example : If h(x) = ( x + 2 )2 was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
h(x) = ( x + 2 )2 ** you can see, that x + 2 is raised to the 2nd power
** so x + 2 is our something raised to a power
Composite Functions : going backwards
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses of under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Example : If h(x) = ( x + 2 )2 was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
h(x) = ( x + 2 )2 ** you can see, that x + 2 is raised to the 2nd power
** so x + 2 is our something raised to a power
Therefore : g ( x ) = x + 2
Composite Functions : going backwards
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses of under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Example : If h(x) = ( x + 2 )2 was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
h(x) = ( x + 2 )2 ** you can see, that x + 2 is raised to the 2nd power
** so x + 2 is our something raised to a power
Therefore : g ( x ) = x + 2
Removing the x + 2 from the parentheses and replacing it with just x
creates our f(x)
Composite Functions : going backwards
We will always use h(x) = ( f ◦ g )(x) or f[g(x)]
Generally, look for things inside parentheses of under roots.
What you’ll see is “something” raised to a power or under a root.
That something becomes our g(x), and then f(x) becomes a simple
equation with x replacing the “something”.
Example : If h(x) = ( x + 2 )2 was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
h(x) = ( x + 2 )2 ** you can see, that x + 2 is raised to the 2nd power
** so x + 2 is our something raised to a power
Therefore : g ( x ) = x + 2 AND f(x) = ( x )2
Removing the x + 2 from the parentheses and replacing it with just x
creates our f(x)
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
Can you see the “something” ??
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
Can you see the “something” ?? It’s x – 5
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
Can you see the “something” ?? It’s x – 5
Therefore : g ( x ) = x – 5
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
g ( x ) = x – 5
Now replace the x – 5 inside each parentheses with just “x”
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
g ( x ) = x – 5
Now replace the x – 5 inside each parentheses with just “x”
f ( x ) = 4( x )3 + 2( x )
Composite Functions : going backwards
Example : If h(x) = 4( x – 5 )3 + 2( x – 5 ) was created by ( f ◦ g )(x)
find the functions f(x) and g(x) that created h(x)
g ( x ) = x – 5
f ( x ) = 4( x )3 + 2( x )
Composite Functions : going backwards
Example : If h(x) = find the f(x) and g(x) that created h(x)
What is the “something” ??
103 x
Composite Functions : going backwards
Example : If h(x) = find the f(x) and g(x) that created h(x)
What is the “something” ?? 3x – 10 is under a root
103 a
Composite Functions : going backwards
Example : If h(x) = find the f(x) and g(x) that created h(x)
What is the “something” ?? 3x – 10 is under a root
So : g ( a ) = 3x - 10
103 a
Composite Functions : going backwards
Example : If h(x) = find the f(x) and g(x) that created h(x)
g ( x ) = 3x – 10
Now replace the 3x – 10 under the root with just “x”
103 a
Composite Functions : going backwards
Example : If h(x) = find the f(x) and g(x) that created h(x)
g ( x ) = 3x – 10
f ( x ) =
Now replace the 3x – 10 under the root with just “x”
103 a
x