Let f: X → Y_{1} and g: Y_{2 }→ Z be two functions and the set D = {x ∈ X: f(x) ∈ Y_{2}}. 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) ⊆ Y_{2}.
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) = (x1)/(x+1) , f^{2}(x) = f{f(x)}, f^{3} (x) = f{f^{2}(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 g^{n}(x) = gogo....og(x), show that f^{n}(x) = (4^{n}  1) + 4^{n}x if g^{n} (x) denotes the inverse of g^{n} (x).
Solution. Since g(x) = 3 + 4x
g^{2}(x) = (gog) (x) = g {g (x)} = g (3 + 4x) = 3 + 4 (3 + 4x) or g^{2}(x) = 15 + 4^{2}x = (4^{2} – 1) + 4^{2}x
Now g^{3}(x) = (gogog) x = g {g^{2} (x)} = g (15 + 42 x) = 3 + 4 (15 + 42 x) = 63 + 43 x = (43 –1) + 43x
Similarly we get g^{n}(x) = (4^{n} – 1) + 4^{n}x
Now let g^{n }(x) = y ⇒ x = g^{–n}(y) ....(1)
∴ y = (4^{n} – 1) + 4^{n}x or x = (y + 1 – 4^{n})4^{–n} ...(2)
From (1) and (2) we get g^{–n} (y) = (y + 1 – 4^{n}) 4^{–n}.
Hence g^{–n} (x) = (x + 1 – 4^{n}) 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.
Example 6. Which of the following functions is odd ?
(a) sgn x + x^{2000 }
(b)  x   tan x
(c) x^{3} cot x
(d) cosec x^{55}
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 + x^{2000} ≠ 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) = –x^{3} (–cot x) = x^{3} cot x = h(x) ∴ h is an even function
(d) f(–x) = cosec (–x)^{55} = cosec (–x^{55}) = –cosec x^{55} = – f(x) ∴ f is an odd function.
Alternatively
(a) f(x) = sgn (x) + x^{2000} = 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 g_{0} be the odd extension of f(x), then
Even Extension: Let g_{e} be the odd extension of f(x), then
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1. What is a composite function? 
2. How do you find the composite of two functions? 
3. What is the notation for composite functions? 
4. How can composite functions be used in reallife applications? 
5. Can you have a composite function of more than two functions? 

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