DOMAIN OF COMPOSITE FUNCTIONS WORKSHEET

Problem 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.

Problem 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.

Problem 3 :

Let f(x) = 1/(x + 2) and g(x) = 4/(x - 1).

Find the domain of f o g.

Problem 4 :

Let f(x) = 1/(x - 3) and g(x) = 6/(x + 1).

Find the domain of f o g.

1. Solution :

f(x) = 2x² - 3 and g(x) = 4x

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.

2. Solution :

f(x) = x² + 2x - 1 and g(x) = 2x + 3

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.

3. Solution :

f(x) = 1/(x + 2) and g(x) = 4/(x - 1)

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}.

4. Solution :

f(x) = 1/(x - 3) and g(x) = 6/(x + 1)

f o g = f[g(x)]

In g(x) = 6/(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 - 3), f(x) is undefined when x = 3.

So, the domain of f(x) is all real values except 3.

Domain of f = R - {3}

Since 3 is excluded from the domain of f(x), in f[g(x)], g(x) can not take the value 3.

Equate g(x) to 3 to find the additional value(s) of x to be excluded from the domain of f o g.

g(x) = 3

6/(x + 1) = 3

Take reciprocal on both sides.

(x + 1)/6 = 1/3

Multiply both sides by 6.

x + 1 = 2

Subtract 1 from 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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