Given two functions f and g, the composition function, denoted by f ∘ g (read as f composed with g), is defined by
(f ∘ g)(x) = f[g(x)]
We know that composition of two functions f and g is
(f ∘ g)(x) = f[g(x)]
Let g(x) be undefined when x = a.
Then, 'a' has to be excluded from the domain of composition of two functions f and g, (f ∘ g)(x).
And also, let (f ∘ g)(x) be undefined when x = b.
Then, 'b' has to be excluded from the domain of composition of two functions f and g, (f ∘ g)(x).
Therefore, the domain of (f ∘ g)(x) is all real values except 'a' and 'b'.
Domain of (f ∘ g)(x) : R - {a, b}
Case (i) :
If both f(x) and g(x) are defined for all real values of x, then the domain of (f ∘ g)(x) is all real values of x.
Domain of (f ∘ g)(x) = R
Case (ii) :
Let g(x) be undefined when x = a and f(x) be defined for all real values of x. Then the domain of (f ∘ g)(x) is all real values of x except 'a'.
Domain of (f ∘ g)(x) : R - {a}
Case (iii) :
Let g(x) be undefined when x = a and f(x) be undefined when x = b.
Since g(x) is undefined when x = a, exclude 'a' from the domain of (f ∘ g)(x).
(f ∘ g)(x) = f[g(x)]
Since f(x) is undefined when x = b, g(x) can not be equal to 'b' in f[g(x)].
To find the additional value/s to be excluded from the domain of (f ∘ g)(x), equate g(x) to b and solve for x.
Let x = k be the solution to the equation g(x) = b
g(x) = b ----> x = k
Now, you have to exclude k from the domain of (f ∘ g)(x)
Therefore,
Domain of (f ∘ g)(x) : R - {a, k}
Example 1 :
Let f(x) = 2x2 - 3 and g(x) = 4x.
Find (a) f o g and (b) g o f.
Then find the domain of each composite function.
Solution :
In f(x) = 2x2 - 3 and g(x) = 4x, both f(x) and g(x) are defined for all real values of x.
So, the domain of f and the domain of g are the set of all real numbers.
f o g :
f o g = f[g(x)]
= f(4x)
= 2(4x)2 - 3
= 2(16x2) - 3
= 32x2 - 3
Since the domains of both f and g are the set of all real numbers, the domain of f o g is the set of all real numbers.
g o f :
g o f = g[f(x)]
= g(2x2 - 3)
= 4(2x2 - 3)
= 8x2 - 12
Since the domains of both f and g are the set of all real numbers, the domain of g o f is the set of all real numbers.
Example 2 :
Let f(x) = x2 + 2x - 1 and g(x) = 2x + 3.
Find (a) f o g and (b) g o f.
Then find the domain of each composite function.
Solution :
In f(x) = x2 + 2x - 1 and g(x) = 2x + 3, both f(x) and g(x) are defined for all real values of x.
So, the domain of f and the domain of g are the set of all real numbers.
f o g :
f o g = f[g(x)]
= f(2x + 3)
= (2x + 3)2 + 2(2x + 3) - 1
= (2x)2 + 2(2x)(3) + 32 + 4x + 6 - 1
= 4x2 + 12x + 9 + 4x + 6 - 1
= 4x2 + 16x + 14
Since the domains of both f and g are the set of all real numbers, the domain of f o g is the set of all real numbers.
g o f :
g o f = g[f(x)]
= g(x2 + 2x - 1)
= 2(x2 + 2x - 1) + 3
= 2x2 + 4x - 2 + 3
= 2x2 + 4x - 1
Since the domains of both f and g are the set of all real numbers, the domain of g o f is the set of all real numbers.
Example 3 :
Let f(x) = 1/(x + 2) and g(x) = 4/(x - 1).
Find the domain of f o g.
Solution :
f o g = f[g(x)]
In g(x) = 4/(x - 1), g(x) is undefined when x = 1.
So, the domain of g(x) is all real values except 1.
Domain of g = R - {1}
So, exclude 1 from the domain of f o g.
In f(x) = 1/(x + 2), f(x) is undefined when x = -2.
So, the domain of f(x) is all real values except -2.
Domain of f = R - {-2}
Since -2 is excluded from the domain of f(x), in f[g(x)], g(x) can not take the value -2.
Equate g(x) to -2 to find the additional value(s) of x to be excluded from the domain of f o g.
g(x) = -2
4/(x - 1) = -2
Take reciprocal on both sides.
(x - 1)/4 = -1/2
Multiply both sides by 4.
x - 1 = -2
Add 1 to both sides.
x = -1
So, exclude -1 from the domain of f o g.
Therefore, the domain of f o g = R - {-1, 1}.
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