Problem 1: A function f(x) is given as :

f x = { a − x n } 1 n f x = { a − x n } 1 n

where a > 0, x > 0 and "n" is a positive integer. Find f{f(x)}

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to find, the composition, we evaluate f(y), where y = f(x).

⇒ f { f x } = f y = a − y n 1 n = [ a − { a − x n 1 / n } n ] 1 / n ⇒ f { f x } = f y = a − y n 1 n = [ a − { a − x n 1 / n } n ] 1 / n

⇒ f { f x } = a − a − x n 1 n = x n 1 n = x ; x > 0 ⇒ f { f x } = a − a − x n 1 n = x n 1 n = x ; x > 0

Here, composition is that of function with itself. As such, domain of composition is equal to intersection of domain of the given function with itself. But, the intersection of an interval with itself is same interval. Hence, we have retained the domain interval of the composition same as that of given function.

Problem 2: A function f(x) is given as :

f x = { 2 − x n } 1 n f x = { 2 − x n } 1 n

where x > 0 and "n" is a positive integer. Prove that :

f { f x } + f { f 1 x } ≥ 2 f { f x } + f { f 1 x } ≥ 2

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to prove the inequality, we need to determine each composition on the left hand side of the given inequality.

We have seen in earlier example that if f x = { a − x n } 1 ∕ n f x = { a − x n } 1 ∕ n , then f{f(x) = x. Hence, if f x = { 2 − x n } 1 ∕ n f x = { 2 − x n } 1 ∕ n , then f{f(x) = x. Similarly, we determine f{f(1/x)}. Here,

⇒ f { f 1 x } = f y = [ a − { a − 1 x n 1 / n } n ] 1 / n = a − a − 1 x n 1 / n ⇒ f { f 1 x } = f y = [ a − { a − 1 x n 1 / n } n ] 1 / n = a − a − 1 x n 1 / n

⇒ f { f 1 x } = 1 x n 1 n = 1 x ⇒ f { f 1 x } = 1 x n 1 n = 1 x

Substituting these values in the LHS of the inequality, we have :

⇒ LHS = f { f x } + f { f 1 x } = x + 1 x ⇒ LHS = f { f x } + f { f 1 x } = x + 1 x

Using algebraic identity a 2 + b 2 = a − b 2 + 2 a b a 2 + b 2 = a − b 2 + 2 a b , we have :

⇒ LHS = x − 1 x 2 + 2 ⇒ LHS = x − 1 x 2 + 2

But, the square term is a non-negative number. Hence,

⇒ LHS ≥ 2 ⇒ LHS ≥ 2

⇒ f { f x } + f { f 1 / x } ≥ 2 ⇒ f { f x } + f { f 1 / x } ≥ 2

Problem 3: A function is defined as :

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

Find composition f(|x|) and its domain.

Solution :

Statement of the problem : The function is defined by different rules in two intervals.

The composition consists of two functions “f(x)” and “|x|”. We know that modulus is defined for all values of “x”. However, domain of “f(x)” is [-2,2]. Hence, domain of composition is intersection of two domains, which is [-2,2]. Here,

         | -1	     ; -2≤|x|≤0  and R
f(|x|) = | 
         | |x|-1     ;  0≤|x|≤2  and R

The interval “-2≤[|x|≤0” can be interpreted in parts. The left part is “|x|≥-2”, which is always true. The right part “|x|≤0” is meaningless, which yields no solution for “x”. Therefore, upper interval of the function is not a valid interval. On the other hand, interval “0≤|x|≤2” has two parts. The left part “|x|≥0” is true for all values of “x”. The right part is “|x|≤2”. This expands to :

- 2 ≤ x ≤ 2 - 2 ≤ x ≤ 2

The intersection of “-2≤x≤2” and “R” is “-2≤x≤2”. Hence, composition is :

f | x | = | x | - 1 ; - 2 ≤ x ≤ 2 f | x | = | x | - 1 ; - 2 ≤ x ≤ 2

Problem 4: Two functions are given as :

f x = - 1 + | x − 1 | ; - 1 ≤ x ≤ 3 f x = - 1 + | x − 1 | ; - 1 ≤ x ≤ 3

g x = 2 − | x + 1 | ; - 2 ≤ x ≤ 2 g x = 2 − | x + 1 | ; - 2 ≤ x ≤ 2

Find fog(x).

Solution :

Statement of the problem : The domains of the given functions are different. We need to determine composition and interpret domain of the composition.

Let y = g x = 2 − | x + 1 | ; - 2 ≤ x ≤ 2 Let y = g x = 2 − | x + 1 | ; - 2 ≤ x ≤ 2

Now, let us first determine the composition,

⇒ f o g x = f y = f 2 − | x + 1 | | = - 1 + | 2 − | x + 1 | − 1 | ⇒ f o g x = f y = f 2 − | x + 1 | | = - 1 + | 2 − | x + 1 | − 1 |

This is the rule of the composition function. In order to find the domain of the composition, we write domains of two functions as an intersection :

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

We interpret the interval “-1 ≤ y ≤ 3 “ as :

- 1 ≤ 2 − | x + 1 | ≤ 3 ⇒ - 3 ≤ - | x + 1 | ≤ 1 - 1 ≤ 2 − | x + 1 | ≤ 3 ⇒ - 3 ≤ - | x + 1 | ≤ 1

Multiplying each term with “-1” and reversing inequality, we have :

3 ≥ | x + 1 | ≥ - 1 3 ≥ | x + 1 | ≥ - 1

But, "|x+1| ≥ -1" is true for all values of “x”. Hence, above inequality is equal to interval given by first part "|x+1| ≤ 3" as :

⇒ - 3 ≤ x + 1 ≤ 3 ⇒ - 3 ≤ x + 1 ≤ 3

⇒ - 4 ≤ x ≤ 2 ⇒ - 4 ≤ x ≤ 2

Hence, interval of the composition is intersection of intervals “-4≤ x ≤2“ and “-2 ≤ x ≤ 2”.

Domain = - 2 ≤ x ≤ 2 Domain = - 2 ≤ x ≤ 2

The composition, therefore, is :

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

Problem 5: Two functions are defined as :

g x = 1 − x 1 + x ; 0 ≤ x ≤ 1 g x = 1 − x 1 + x ; 0 ≤ x ≤ 1

h x = 4 x 1 − x ; 0 ≤ x ≤ 1 h x = 4 x 1 − x ; 0 ≤ x ≤ 1

Determine the composition goh(x) and hog(x).

Solution :

Statement of the problem : The domains of the given functions are same. We need to interpret domain of the composition as we compose the required function.

Let

y = h x = 4 x 1 − x = 4 x − 4 x 2 ; 0 < = x < = 1 y = h x = 4 x 1 − x = 4 x − 4 x 2 ; 0 < = x < = 1

Then,

⇒ g o h x = g y ; 0 ≤ y ≤ 1 and 0 ≤ x ≤ 1 ⇒ g o h x = g y ; 0 ≤ y ≤ 1 and 0 ≤ x ≤ 1

⇒ g o h x = g ( 4 x − 4 x 2 ) ; 0 ≤ 4 x − 4 x 2 ≤ 1 and 0 ≤ x ≤ 1 ⇒ g o h x = g ( 4 x − 4 x 2 ) ; 0 ≤ 4 x − 4 x 2 ≤ 1 and 0 ≤ x ≤ 1

Let us first interpret rule of the function :

⇒ g o h x = 1 − y 1 + y = 1 − 4 x − 4 x 2 1 + 4 x − 4 x 2 = 1 − 4 x + 4 x 2 1 + 4 x − 4 x 2 ⇒ g o h x = 1 − y 1 + y = 1 − 4 x − 4 x 2 1 + 4 x − 4 x 2 = 1 − 4 x + 4 x 2 1 + 4 x − 4 x 2

Now, we interpret the interval “ 0 ≤ 4 x − 4 x 2 ≤ 1 0 ≤ 4 x − 4 x 2 ≤ 1 ” in parts. The left part is :

4 x − 4 x 2 ≥ 0 ⇒ x 1 − x ≥ 0 ⇒ x ≥ 0 , x ≤ 1 ⇒ 0 ≤ x ≤ 1 4 x − 4 x 2 ≥ 0 ⇒ x 1 − x ≥ 0 ⇒ x ≥ 0 , x ≤ 1 ⇒ 0 ≤ x ≤ 1

The right part of the interval is :

4 x − 4 x 2 ≤ 1 ⇒ 4 x 2 − 4 x ≥ - 1 ⇒ 4 x 2 − 4 x + 1 ≥ 0 4 x − 4 x 2 ≤ 1 ⇒ 4 x 2 − 4 x ≥ - 1 ⇒ 4 x 2 − 4 x + 1 ≥ 0

Applying sign scheme, this inequality is valid for all values of “x” :

⇒ - ∞ ≤ x ≤ ∞ ⇒ - ∞ ≤ x ≤ ∞

The intersection of two parts is “0≤x≤1”. Thus, interval of composition is intersection of intervals “0≤x≤1” and “0≤x≤1”, which is “0≤x≤1”. Therefore, “goh(x)” is :

⇒ g o h x = 1 − 4 x + 4 x 2 1 + 4 x − 4 x 2 ; 0 ≤ x ≤ 1 ⇒ g o h x = 1 − 4 x + 4 x 2 1 + 4 x − 4 x 2 ; 0 ≤ x ≤ 1

For determining hog(x), Let

y = g x = 1 − x 1 + x ; 0 ≤ x ≤ 1 y = g x = 1 − x 1 + x ; 0 ≤ x ≤ 1

Then,

⇒ h o g x = g y ; 0 ≤ y ≤ 1 and 0 ≤ x ≤ 1 ⇒ h o g x = g y ; 0 ≤ y ≤ 1 and 0 ≤ x ≤ 1

⇒ h o g x = g 1 − x 1 + x ; 0 ≤ 1 − x 1 + x ≤ 1 and 0 ≤ x ≤ 1 ⇒ h o g x = g 1 − x 1 + x ; 0 ≤ 1 − x 1 + x ≤ 1 and 0 ≤ x ≤ 1

Let us first interpret rule of the function :

⇒ h o g x = 4 y 1 − y = 4 1 - x 1 + x 1 - 1 - x 1 + x ⇒ h o g x = 4 y 1 − y = 4 1 - x 1 + x 1 - 1 - x 1 + x

⇒ h o g x = 4 1 - x 1 + x 1 + x - 1 + x 1 + x = 8 x 1 − x 1 + x 2 ⇒ h o g x = 4 1 - x 1 + x 1 + x - 1 + x 1 + x = 8 x 1 − x 1 + x 2

Now, we interpret the interval “ 0 ≤ 1 − x 1 + x ≤ 1 0 ≤ 1 − x 1 + x ≤ 1 ” in parts. The left part is :

1 - x 1 + x ≥ 0 ⇒ 1 − x ≥ 0 1 - x 1 + x ≥ 0 ⇒ 1 − x ≥ 0

Applying sign scheme, this inequality is valid for all values of “x” :

⇒ 0 ≤ x ≤ 1 ⇒ 0 ≤ x ≤ 1

The right part of the interval is :

1 − x 1 + x ≤ 1 ⇒ 1 − x ≤ 1 + x ⇒ 0 ≤ 2 x ⇒ x ≥ 0 1 − x 1 + x ≤ 1 ⇒ 1 − x ≤ 1 + x ⇒ 0 ≤ 2 x ⇒ x ≥ 0

The intersection of two parts is “0≤x≤1”. Thus, interval of composition is intersection of intervals “0≤x≤1” and “0≤x≤1”, which is “0≤x≤1”. Therefore, “goh(x)” is :

⇒ h o g x = 8 x 1 − x 1 + x 2 ; 0 ≤ x ≤ 1 ⇒ h o g x = 8 x 1 − x 1 + x 2 ; 0 ≤ x ≤ 1

Problem 6: A function is defined as :

       | 1 + x	   ; x≥0
f(x) = | 
       | 1- x      ; x < 0

Determine composition f{f(x)}.

Solution :

Statement of the problem : The function is defined by different rules in two intervals.

For f(1+x) in the interval x>=0

         | 1 + 1+ x	 ; 1+x ≥0 and x≥0
f(1+x) = | 
         | 1- 1 - x      ; 1+x < 0 and x≥0  

         | 2+ x	       ; x ≥-1 and x≥0
f(1+x) = | 
         |  - x        ; x < -1 and x≥0 

The intersection of upper intervals “x ≥-1 and x≥0” is equal to “x≥0”. There is no common interval for the intersection of lower intervals. Hence,

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

For f(1-x) in the interval x<0

         | 1 + 1- x	; 1-x ≥0 and x≤0
f(1-x) = | 
         | 1- 1 + x     ; 1- x < 0 and x<0  

         | 2- x       ; x ≤1 and x<0
f(1-x) = | 
         |  x         ; x > 1 and x<0  

The intersection of upper intervals “x ≤1 and x<0” is equal to “x<0”. There is no common interval for the intersection of lower intervals. Hence,

f 1 − x = 2 - x ; x < 0 f 1 − x = 2 - x ; x < 0

Therefore, the composition is :

          | 2+x       ; x>0
f{f(x)}=  |
          | 2- x      ; x < 0