# Solutions to Composition of Functions Questions

Detailed solutions to Composition of Functions Questions are presented.

### Solution to Question 1

• The composition (f o g) (x) of
f(x) = x + 1 , g(x) = 3x
• is given by
(f o g) (x) = f(g(x)) = g(x) + 1 = 3x + 1
• Since the domain of both functions is the set of all real numbers, the composition (f o g) (x) also have the set of all real numbers as its domain.

### Solution to Question 2

• The composition (f o g) (x) of
f(x) = x 2 + 1 , g(x) = √(2 x)
• is given by
(f o g) (x) = f(g(x))
= g(x) 2 + 1
= [ √(2 x) ] 2 + 1
= 2 x + 1
• To find the domain of the composition of the two functions, we proceed as follows: x must be in the domain of g.
2 x ≥ 0 which is equivalent to x ≥ 0
• g(x) must be in the domain of f(x). The domain of f is the set of all real numbers. The condition x ≥ 0 makes g(x) real and is therefore in the domain of f. Hence the domain of the composition is the set of all values defined by the interval
[0 , + ∞)

### Solution to Question 3

• The composition (f o g) (x) of
f(x) = √(- x + 1) , g(x) = x 2 - 8
• is given by
(f o g) (x) = f(g(x))
= √(- g(x) + 1)
= √(- (x 2 - 8) + 1)
= [ √(9 - x 2) ]
• The domain of g is the set of real numbers. Hence to find the domain of the composition, x must satisfy the condition
9 - x 2 ≥ 0
• The solution set of the above inequality is also the domain given by the interval
[-3 , 3]

### Solution to Question 4

• Given
f(x) = 2x + 1 and g(x) = x 2 ,
• (f o g) (-2) is calculated as follows
(f o g) (-2) = f(g(-2))
= f(4) = 9

### Solution to Question 5

• Given
f(x) = 1 / (x + 1) and g(x) = 1 / (x - 1) ,
• (f o g) (0) is calculated as follows
(f o g) (0) = f(g(0))
= f(-1) = undefined since x = -1 makes the denominator of f equal to zero.

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