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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