1. 4.3 Notes.notebook October 25, 2012
4.3 Introduction to Composite Functions
So far we have learned how to do the 4 arithmetic operations
on functions.
Now we switch to a very different operation.
To look at composite functions we look at the idea of placing
one function INSIDE another function. The inside function
is evaluated first and its resulting value is placed inside the
second function which is then evaluated.
The key concept is to work from the INSIDE OUT.
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3. 4.3 Notes.notebook October 25, 2012
If f(x) = 2x 4 and g(x) = (x1)2
Then f(g(x)) means to substitute g(x) into the x of f(x).
Therefore f(g(x)) = 2(g(x))4 = 2(x1)2 4.
This can then be evaluated into:
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This process can also work in the reverse direction.
Find g(f(x)). In other words substitue f(x) into the x of g(x).
Therefore g(f(x) = ((2x4) 1)2
This can then be evaluated into:
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4. 4.3 Notes.notebook October 25, 2012
Notice the second way to say f(g(x))
Use the above diagrams to find:
f(g(1)) __________
If f(g(x)) = 4 then solve for x: _____________
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5. 4.3 Notes.notebook October 25, 2012
Find the y value for g(1).
This value then becomes the x value for f(x).
Now find g(f(4)) _________________
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9. 4.3 Notes.notebook October 25, 2012
If you find g(1) and place this value into f,
the result MUST be the same as h(1) if
h(x)=f(g(x))
In other words h(1) must equal f(g(1)). CHECK TO SEE.
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10. 4.3 Notes.notebook October 25, 2012
Let h(x)=f(g(x))
Prove that h(1)=f(g(1))
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11. 4.3 Notes.notebook October 25, 2012
Caution: Do not confuse the composition instruction
(f o g)(x) with the multiplication instruction (f . g)(x)
as used in some of the supplementary questions.
Homework:
Page 298 #4,6,7,9, (10,11 Do not find
domain, range), 1215
Multiple Choice #1,2
Supplementary 2 Handout
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