Composition of Functions | Mathematics (Maths) Class 12 - JEE PDF Download

Composite Functions

Let f:  X → Y₁ and g: Y₂ → Z be two functions and the set D = {x ∈ X: f(x) ∈ Y₂}. If D ≡ Ø, then the function h defined on D by h(x) = g{f(x)} is called composite function of g and f and is denoted by gof. It is also called function of a function.

Remark: Domain of gof is D which is a subset of X (the domain of f). Range of gof is a subset of the range of g. If D = X, then f(X) ⊆ Y₂.

Properties of composite functions

1. The composite of functions is not commutative i.e. gof ≠ fog.
2. The composite of functions is associative i.e. if f, g, h are three functions such that fo (goh) & (fog) oh are defined, then fo (goh) = (fog) oh.

Example 1. Let f(x) = e^x ; R⁺ → R and g(x) = sin^-1 x; [-1, 1] →[- π/2 ,  π/2]. Find domain and range of fog (x)
Solution. Domain of f(x): (0, ∞), Range of g(x): [- π/2 ,  π/2]
The values in range of g(x) which are accepted by f(x) are (0, π/2]
⇒ 0 < g(x) ≤ π/2   0 < sin^-1 x ≤ π/2 , 0 < x ≤ 1
Hence domain of fog(x) is x ∈ (0, 1]

Example 2. Let f(x) = (x-1)/(x+1) ,  f²(x) = f{f(x)}, f³ (x) = f{f²(x)},.....fk + 1 (x) = f{fk(x)}. for k = 1, 2, 3,...., Find f1998 (x).
Solution.

Thus, we can see that fk(x) repeats itself at intervals of k = 4.
Hence, we have f1998(x) = f2(x) = -1/x, [∴ 1998 = 499 × 4 + 2]

Example 3. Let g : R → R be given by g(x) = 3 + 4x. If gⁿ(x) = gogo....og(x), show that fⁿ(x) = (4ⁿ - 1) + 4ⁿx if g^-n (x) denotes the inverse of gⁿ (x).
Solution. Since g(x) = 3 + 4x
g²(x) = (gog) (x) = g {g (x)} = g (3 + 4x) = 3 + 4 (3 + 4x) or g²(x) = 15 + 4²x = (4² – 1) + 4²x
Now g³(x) = (gogog) x = g {g² (x)} = g (15 + 42 x) = 3 + 4 (15 + 42 x) = 63 + 43 x = (43 –1) + 43x
Similarly we get gⁿ(x) = (4ⁿ – 1) + 4ⁿx
Now let gⁿ (x) = y  ⇒ x = g^–n(y) ....(1)
∴ y = (4ⁿ – 1) + 4ⁿx or x = (y + 1 – 4ⁿ)4^–n ...(2)
From (1) and (2) we get g^–n (y) = (y + 1 – 4ⁿ) 4^–n.
Hence g^–n (x) = (x + 1 – 4ⁿ) 4^–n

Example 4. If f(x) = If f(x)  = | |x – 3| – 2 | ; 0 ≤ x ≤  4 and g(x) = 4 – |2 – x| ; –1 ≤  x ≤  3 then find fog(x).
Solution.
 

Example 5. Prove that f(n) = 1 - n is the only integer valued function defined on integers such that
(i) f(f(n)) = n for all n ε Z and 
(ii) f(f(n + 2) + 2) = n for all n ε Z and 
(iii) f(0) = 1.
Solution. The function f(n) = 1 - n clearly satisfies conditions (i), (ii) and (iii). Conversely, suppose a function
f: Z → Z satisfies (i), (ii) and (iii). Applying f to (ii) we get, f(f(f(n + 2) + 2))) = f(n) and this gives because of (i), f(n + 2) + 2 = f(n), ........(1)
for all n ε Z. Now using (1) it is easy to prove by induction on n that for all n ε Z,

Also by (iii), f(0) = 1. Hence by (i), f(1) = 0. Hence f(n) = 1 - n for all n ε Z.

General Definition

• Identity function: A function f ; A → defined by f(x) = x ∀ x ∈ A is called the identity of A & denoted by I_A.
  Ex: f : R⁺ → R⁺ ; f(x) = e^ℓnx and f : R → R ; f(x) = ℓn e^x
  Every Identity function is a bijection.
• Constant function: A function f: A → B is said to be constant function. If every element of set A has the same functional image in set B i.e. f: A → B ; f(x) = c ∀ x ε A & c ε B is called constant function.
• Homogeneous function: A function is said to be homogeneous w.r.t. any set of variables when each of its term is of the same degree w.r.t. those variables.
• Bounded Function: A function y = f(x) is said to be bounded if it can be express is the form of
  a ≤ f(x) ≤ b where a and b are finite quantities.
  Example: -1 ≤ sin x ≤ 1 ; 0 ≤ {x} < 1 ; -1 ≤ sgn (x) ≤ 1 but e^x is not bounded.
  Any function having singleton range like constant function.
• Implicit function & Explicit function: If y has been expressed entirely in terms of `x' then it is called an explicit function.
  If x & y are written together in the form of an equation then it is known as implicit equation corresponding to each implicit equation there can be one, two or more explicit function satisfying it
  Example: 
  y = x<sup>3</sup> + 4x<sup>2</sup> + 5x → Explicit function
  x + y = 1 → Implicit equation
  y = 1 - x → Explicit function
• Even & Odd Functions
  Function must be defined in symmetric interval [-x, x]
  If f (-x) = f (x) for all x in the domain of `f' then f is said to be an even function.
  e.g. f (x) = cos x; g (x) = x² + 3.
  If f (-x) = -f (x) for all x in the domain of `f' then f is said to be an odd function.
  e.g. f (x) = sin x ; g (x) = x³ + x.

Remark

• f (x) - f (-x) = 0 ⇒ f (x) is even & f (x) + f (-x) = 0 ⇒ f (x) is odd.
• A function may be neither even nor odd.
• Inverse of an even function is not defined.
• Every even function is symmetric about the y-axis & every odd function is symmetric about the origin.
• A function (whose domain is symmetric about origin) can be expressed as a sum of an even & an odd function. e.g.
   
• The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.
• If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd and other even then f.g will be odd.

Example 6. Which of the following functions is odd ?
(a) sgn x + x²⁰⁰⁰  
(b) | x | - tan x 
(c) x³ cot x 
(d) cosec x⁵⁵
Solution.
Let’s name the function of the parts (A), (B), (C) and (D) as f(x), g(x), h(x) & f(x) respectively. Now
(a) f(–x) = sgn (–x) + (–x)²⁰⁰⁰ = –sgn x + x²⁰⁰⁰ ≠ f(x) & ≠ –f(x) ∴  f is neither even nor odd.
(b) g(–x) = |–x| – tan (–x) = |x| + tan x ∴ g is neither even nor odd.
(c) h(–x) = (–x)³ cot (–x) = –x³ (–cot x) = x³ cot x = h(x) ∴ h is an even function
(d) f(–x) = cosec (–x)⁵⁵ = cosec (–x⁵⁵) = –cosec x⁵⁵ = – f(x) ∴ f is an odd function.
Alternatively
(a) f(x) = sgn (x) + x²⁰⁰⁰ = O + E = neither E nor O
(b) g(x) = E – O = Neither E nor O
(c) h(x) = O × O = E (D) f(–x) = O o O = O
(d) is the correct option

Example 7. 

(a) An even function 
(b) An odd function
(c) Neither even nor odd function 
(d) None of these
Solution.

= O × e^{O × O} = O × e^E = O × E = O

Example 8. Let f: [-2, 2] → R be a function if f(x) = 

 so that
(i) f is an odd function 
(ii) f is an even function (where [*] denotes the greatest integer function)
Solution.



(i) If f is an odd function then f(x) = –f (–x)

(ii) If f is an even function  


Example 9. Let f(x) = e^x + sin x be defined on the interval [-4, 0]. Find the odd and even extension of f(x) in the interval [-4, 4].
Solution.
Odd Extension:  Let g₀ be the odd extension of f(x), then

Even Extension:  Let g_e be the odd extension of f(x), then