Composition of Functions by vLqjRHz

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									                   Composition of Functions



Composite functions are frequently encountered in calculus and sometimes in physics.
At first sight this is often perceived as a difficult topic but once you look at this
material a bit closer it will turn out to be quite simple and logical.



In this document and for the following examples we will use the two functions:
f ( x )  x 2  2 x  5 and g(x)  3x  1



It is of course easy to find f(3). Just replace the x by a 3 and you will get
f (3)  3 2  2  3  5  20 .

Likewise, it should be simple to calculate f(B). Replace the x by a B and get
f (B)  B 2  2B  5 .

Now let us calculate f(A-1). Replace the x by (A-1) and get
f (A  1)  (A  1) 2  2(A  1)  5 .
The only thing you can do here is to distribute and collect the like terms.
This gives us:
f (A  1)  A 2  2A  1  2A  2  5  A 2  4




Example 1: Find f g( x )

Following the above procedure we can try find f g ( x ) . Compare this with f(B), now
the quantity B  g( x ) so we get:

 f g ( x)  g ( x)  2  g ( x)  5 replace g x  by its value:
                     2


 f g ( x)  (3x  1) 2  2(3x  1)  5
 f g ( x)  9 x 2  6 x  1  6 x  2  5
 f g ( x)  9 x 2  4




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Example 2: Find gf ( x )

Using the same procedure as in the previous example we get:

g  f ( x)  3   f ( x)  1 replace f x  by its value :
                           
g  f ( x)  3 x 2  2 x  5  1
g  f ( x)  3x 2  6 x  15  1
g  f ( x)  3x 2  6 x  14



Example 3: Find f g(1)

This is easy. First find g(1):
g(1)  3  1  1  2 .

Now replace g(1) with 2 to get:
f g (1)  f 2

Now evaluate f(2)
f 2  2 2  2  2  5  13

So we now have:
 f g (1)  f 2  13




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