Example

Problem 2: Let two functions be defined as :

f = { 1,2 , 2,3 , 3,4 , 4,5 } f = { 1,2 , 2,3 , 3,4 , 4,5 }

g = { 2,4 , 3,2 , 4,3 , 5,1 } g = { 2,4 , 3,2 , 4,3 , 5,1 }

Check whether “gof” and “fog” exit for the given functions?

Solution : Here,

⇒ Domain of “f” = { 1,2,3,4 } ⇒ Domain of “f” = { 1,2,3,4 }

⇒ Range of “f” = { 2,3,4,5 } ⇒ Range of “f” = { 2,3,4,5 }

⇒ Domain of “g” = { 2,3,4,5 } ⇒ Domain of “g” = { 2,3,4,5 }

⇒ Range of “g” = { 4,2,3,1 } = { 1,2,3,4 } ⇒ Range of “g” = { 4,2,3,1 } = { 1,2,3,4 }

Hence,

⇒ Range of “f” ⊂ Domain of “g” ⇒ Range of “f” ⊂ Domain of “g”

and

⇒ Range of “g” ⊂ Domain of “f” ⇒ Range of “g” ⊂ Domain of “f”

It means that both compositions “gof” and “fog” exist for the given sets.

Algorithm for finding interval of composition

The most important aspect of working with composition of function is to combine intervals i.e. domains of two functions. Consider for example the function given by :

f x = x + 1 ; 0 ≤ x ≤ 2 f x = x + 1 ; 0 ≤ x ≤ 2

g x = | x | ; 0 ≤ x ≤ 3 g x = | x | ; 0 ≤ x ≤ 3

Here, "|x|" is modulus function, which returns non-negative number for all real values of "x". We are required to find fog(x). What would be the domain of the resulting composition? Let us have a closer look at the definition of composition,

f o g x = f g x = f | x | f o g x = f g x = f | x |

In plain words, it means that argument of the function “f” is the function “g” itself. The function “f” is defined in the domain "0 ≤ x ≤ 2", whereas function “g” is defined in "0≤x≤3". From the expression of composition as above, it is clear that we need to ensure that value of “x” should lie in the domain interval of function “g”. Hence,

f o g x = f g x = f | x | ; 0 ≤ x ≤ 3 f o g x = f g x = f | x | ; 0 ≤ x ≤ 3

But function “f” is defined for values of “x”, which lie in its interval “0≤x≤2”. So when we expand the composition in accordance with the rule of function “f”, we should ensure that value of its argument – note that it is not the value of independent variable “x” - lies in the interval specified by its domain.

⇒ f o g x = | x | + 1 ; 0 ≤ | x | ≤ 2 and 0 ≤ x ≤ 3 ⇒ f o g x = | x | + 1 ; 0 ≤ | x | ≤ 2 and 0 ≤ x ≤ 3

Note that interval of function “f” is written with respect to function “g” i.e. “|x|” – not “x”. Since requirements of both functions are required to be met simultaneously, the domain of the resulting composition is intersection of two domains. It is this reason that we use either “and” or a comma “,” to combine two intervals.

Now, we interpret the interval of modulus function “0≤ |x| ≤2”. We see that part of the interval, "|x| ≥0", is always true for all values of “x”. Whereas part of the interval, "|x| ≤2", means (we shall learn about interpreting modulus inequality in a separate module) :

- 2 ≤ x ≤ 2 - 2 ≤ x ≤ 2

Combining intervals of two parts of the interval “0≤ |x| ≤2”, we conclude that it is equal to :

- 2 ≤ x ≤ 2 - 2 ≤ x ≤ 2

Hence,

⇒ f o g x = | x | + 1 ; - 2 ≤ x ≤ 2 and 0 ≤ x ≤ 3 ⇒ f o g x = | x | + 1 ; - 2 ≤ x ≤ 2 and 0 ≤ x ≤ 3

⇒ f o g x = | x | + 1 ; 0 ≤ x ≤ 2 ⇒ f o g x = | x | + 1 ; 0 ≤ x ≤ 2

Problem 3: Consider the function as given here :

f x = x + 1 ; 0 ≤ x ≤ 2 f x = x + 1 ; 0 ≤ x ≤ 2

Determine fof(x).

Solution : The composition of the function with itself is :

f o f x = f f x = f x + 1 ; 0 ≤ x ≤ 2 f o f x = f f x = f x + 1 ; 0 ≤ x ≤ 2

⇒ f o f x = x + 1 + 1 ; 0 ≤ x + 1 ≤ 2 and 0 ≤ x ≤ 2 ⇒ f o f x = x + 1 + 1 ; 0 ≤ x + 1 ≤ 2 and 0 ≤ x ≤ 2

f o f x = x + 2 ; - 1 ≤ x ≤ 1 and 0 ≤ x ≤ 2 f o f x = x + 2 ; - 1 ≤ x ≤ 1 and 0 ≤ x ≤ 2

f o f x = x + 2 ; 0 ≤ x ≤ 1 f o f x = x + 2 ; 0 ≤ x ≤ 1

Problem 4: A function is defined for real values by :

f x = 1 1 - x for all real values except x = 1 f x = 1 1 - x for all real values except x = 1

Determine f[f{f(x)}] and draw the graph of resulting composition.

Solution : This is composition triplet. We have already seen that :

⇒ f o f = f { f x } = x - 1 x when x ≠ 0 , 1 ⇒ f o f = f { f x } = x - 1 x when x ≠ 0 , 1

Compositing again with f(x), we have :

⇒ f [ f { f x } ] = 1 1 − x − 1 x = x x − x + 1 when x ≠ 0 , 1 ⇒ f [ f { f x } ] = 1 1 − x − 1 x = x x − x + 1 when x ≠ 0 , 1

y = f [ f { f x } ] = x when x ≠ 0 , 1 y = f [ f { f x } ] = x when x ≠ 0 , 1

The graph of the composition is as shown here :

Figure 2: The plot is a straight line with two undefined points.

Plot of compostion

Problem 5: A function is defined for real values by :

       | 1 + x ;  0 ≤ x ≤ 2    
f(x) = |  
       | 3 – x ;  2 < x ≤ 3  

Determine f(f(x)).

Solution : The function “f(x)” is combined with itself. Here,

          | f(1 + x) ;  0 ≤ x ≤ 2    
f(f(x)) = |
          | f(3 - x) ;  2 < x ≤ 3    

We need to evaluate function for each of the above two intervals :

f(1 + x) ;  0 ≤ x ≤ 2 

  | 1 + (1+x) ; 0 ≤ 1+ x ≤ 2 and 0 ≤ x ≤ 2    
= |
  | 3 – (1+x); 2 < 1+ x ≤ 3 and 0 ≤ x ≤ 2  

  | 2 + x ; -1 ≤ x ≤ 1 and 0 ≤ x ≤ 2    
= |
  | 2 – x ; 1 < x ≤ 2 and 0 ≤ x ≤ 2 

  | 2 + x ; 0 ≤ x ≤ 1 
= |
  | 2 – x ; 1 < x ≤ 2 

Similarly,

f(3 - x) ;  2 < x ≤ 3

  | 1 + (3-x) ; 0 ≤ 3- x ≤ 2 and 2 < x ≤ 3    
= |
  | 3 – (3-x) ; 2 < 3- x ≤3 and 2 < x ≤3  

  | 4 - x ;  -3 ≤ x ≤ -1 and 2 < x ≤ 3    
= |
  | x     ; -1 < x ≤ 0 and 2 < x ≤ 3  

  | 4 - x ;  1 ≤ x ≤ 3 and 2 < x ≤ 3    
= |
  | x     ; 0 < x ≤ 1 and 2 < x ≤ 3 

  | 4 - x ;  2 < x ≤ 3    
= |
  |   x   ; No common interval  

= | 4 - x ;  2 < x ≤ 3    

Putting the results in the expression of “fof”, we have :

          | 2 + x ; 0 ≤ x ≤1
f(f(x)) = | 2 – x ; 1 < x ≤ 2
          | 4 - x ; 2 < x ≤ 3