DOMAIN OF A COMPOSITE FUNCTION

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)]

How to Find the Domain of a Composite Function

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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