Worksheet 2: Composition of functions
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Let there be two functions defined as:
f:A→B by f(x) for all x∈A
g: B→C by g(x) for all x∈B
Then, the new function, “gof” read
as "g circle f" or "g composed with
f", is defined as:
g
Domain of f(x)
Domain of h(x)
Range of f(x)
Domain of g(f(x))
f  ( x)  g ( f ( x)) , for all x∈A
Range of g(x)
Range of h(x) = g(f(x))
Example 1:
Let two functions be defined as:
f={(1,2) ,(2,3) ,(3,4) ,(4,5) } and g={(2,4) ,(3,2) ,(4,3) ,(5,1) }
Check whether “gof” and “fog” exist for the given functions.
Solution:
Hence,
Domain Range
⇒Range of “f”⊂Domain of “g” “gof”exists.
f {1,2,3,4} {2,3,4,5}
g {2,3,4,5} {4,2,3,1}={1,2,3,4} ⇒Range of “g”⊂Domain of “f” “fog” exists.
It means that both compositions “gof” and “fog” exist for the given sets.
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Example 2:
Given f ( x)  2 x  1 and g ( x)   x 2  3 , find:
a. ( f o g)(x).
b. ( g o f)(x).
c. ( f o f)(x).
d. ( g o g)(x).
Solution:
a.  f g  x   f  g  x    f   x 2  3  2   x 2  3  1  2 x 2  7 .
b.
c.
d.
g
f  x   g  f  x    g  2 x  1    2 x  1  3
f
g
f  x   f  f  x    f  2x  1  2  2 x  1  1  4x  3
2
   4 x 2  4 x  1  3  4 x 2  4 x  2
g  x   g  g  x    g   x 2  3     x 2  3  3
2
   x4  6x2  9  3   x4  6 x2  6
Note that:
 f g  x   2x2  7   g f  x   4x2  4x  2
That is, ( f o g)(x) is not the same as (g o f )(x). The open dot "o" is not the same
as a multiplication dot "•", nor does it mean the same thing.
f(x) • g(x) = g(x) • f(x)
[always true for multiplication]
...you cannot say that:
( f o g)(x) = (g o f )(x)
[generally false for composition]
Domain and range of the composition of functions
Consider the function:
f ( x) 
1
, when x  1  Domain of f is   1 , i.e. all real numbers but 1.
1 x
2
Let us now see the expression of composition of function with itself,
f
1
 1 
f  x   f  f  x    f 

 1 x  1 1
1 x
1
1
1  x x 1
 valid for real values of x≠0.




1 x
1
x

x
x

1 x 1 x 1 x
Since f is undefined for x = 1, and f
composition  f
f is undefined for x = 0, thus the domai n of the
f  x  is :   0,1 ; i.e. all real numbers except 0 and 1.
Sometimes you have to be careful with the domain and range of the composite function.
General rule to determine the domain:
f ( x)
Polynomial
f ( x) 

x0
Domain
1
x
f ( x)  x
f ( x)  log( x)
f ( x)  a x
x0
x0
 , for a  0 ,
  0 ,for a  0
Example:
Given f ( x)  x and g ( x)  x  3 , find the domains of ( f o g)(x) and (g o f )(x).
Solution:
f ( x)  x  x  0
So:
f
g  x   f  g  x    x  3  x  3  0  x  3
Hence, the domain of ( f o g)(x) is "all x > 3".
 Now do the other composition:
g
f  x   g  g  x    ...
Hence, the domain of (gof)(x) is …
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Going backward: given composed function, find original functions
Usually composition is used to combine two functions. But sometimes you are asked to
go backwards. That is, they will give you a function, and they'll ask you to come up with
the two original functions that they composed.
Example 1:
Given h  x    x  5   3  x  5   7 , determine two functions f (x) and g(x) which, when
composed, generate h(x).
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Solution:
This is asking you to notice patterns and to figure out what is "inside" something else.
In this case, this looks similar to the quadratic x 2  3 x  7 , except that, instead of squaring
x, they're squaring x + 5.
So let's make g(x) = x + 5, and then plug this function into f  x   x2  3x  7 :
f
g  x   f  g  x    f  x  5    x  5   3  x  5   7
2
Then h(x) may be stated as the composition of f  x   x2  3x  7 and g(x) = x + 5.
Example 2:
Given h  x   3x  4 , determine two functions f (x) and g(x) which, when composed,
generate h(x).
Solution:
Since the square root is "on" (or "around") the "3x + 4", then the 3x + 4 is put inside the
square root, that is:
h ( x )  f ( g ( x ))
 
x 
  3x  4  

g x
f x
 3x  4 
Thus, g(x) = 3x + 4, f  x   x , and h(x) = ( f o g)(x).
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Exercise
For the given functions:
a. f(x) = x + 1 , g(x) = 3x
b. f ( x)  x 2  1, g ( x)  2 x
c. f ( x)   x  1, g ( x)  x 2  5
d. f(x) = 2x + 1 , g(x) = x2
Find:
1. Domain and range of each f(x) and g(x)
f  x
g  x
a. Domain =
Range =
b. Domain =
Range =
b. Domain =
Range =
Domain =
Range =
Domain =
Range =
Domain =
Range =
c. Domain =
Range =
Domain =
Range =
2. Determine  f g  x  and its domain
a.
f
g  x 
Domain =
b.
f
g  x 
Domain =
c.
f
g  x 
5
Domain =
d.
f
g  x 
Domain =
3. Determine  g f  x  and its domain
a.
g
f  x 
Domain =
b.
g
f  x 
Domain =
c.
g
f  x 
Domain =
d.
g
f  x 
Domain =
4. Determine  f
a.
f
f  x  and its domain
f  x 
Domain =
b.
f
f  x 
6
Domain =
c.
f
f  x 
Domain =
d.
f
f  x 
Domain =
5. Determine  g g  x  and its domain
a.
g
g  x 
Domain =
b.
g
g  x 
Domain =
c.
g
g  x 
Domain =
d.
g
g  x 
Domain =
6. A function is defined for real values by : f ( x) 


1
for all real values except x
1 x
=1 . Determine f f  f  x   and draw the graph of resulting composition!
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7. Given f(2) = 3, g(3) = 2, f(3) = 4 and g(2) = 5, evaluate (f o g)(3)!
8. Functions f and g are as sets of ordered pairs
f = {(-2,1),(0,3),(4,5)} and g = {(1,1),(3,3),(7,9)}
Find the composite function defined by g o f and describe its domain and range.
9. Write function F given below as the composition of two functions f and g, where
1
1
g ( x )  and F ( x)  x
1 x
x
8
10. Evaluate f(g(h(1))), if possible, given that h( x)   x , g ( x)  x  1, and
f ( x) 
1
.
x2
11. For the composite function  f g  ( x) and f ( x) , find g  x  !
a.
f
g  ( x)  x , f ( x)  x 2  2
b.
f
g  ( x)  2 x 6  x 2  1 , f ( x )  2 x 3  x  1
c.
f
g  ( x)   x  1  4 , f  x   x 2  4
d.
f
g  ( x)  x, f  x   x 2  5
2
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12. For the composite function  f g  ( x) and g ( x ) , find f  x  !
e.
f
g  ( x)  sin  x 2  1 , g ( x)  x 2  1
f.
f
g  ( x)  x , g ( x)  1  x 2
g.
f
g  ( x)  4  x, g ( x)  x
1
x , g ( x)  1  1
h.  f g  ( x) 
x
 1
1    1
 x
1
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