Composition of Functions Questions with Solutions

Questions on the Composition of Functions along with their detailed solutions are presented. There is also a tutorial on compositions of functions that may be used. Some of the questions are about finding the composition of two functions and its domain. An answer to one of these questions is correct when both the formula of the function and the domain are correct. Also examples of Applications of Composition of Functions are included in this site.

Questions

Question 1

Find the composition function \( (f \circ g)(x) \) and its domain given
\( f(x) = x + 1 \), \( g(x) = 3x \)

Question 2

Find the composition function \( (f \circ g)(x) \) and its domain given
\( f(x) = x^2 + 1 \), \( g(x) = \sqrt{2x} \)

Question 3

Find the composition function \( (f \circ g)(x) \) and its domain given
\( f(x) = \sqrt{-x + 1} \), \( g(x) = x^2 - 8 \)

Question 4

Find, if possible, \( (f \circ g)(-2) \) given
\( f(x) = 2x + 1 \), \( g(x) = x^2 \)

Question 5

Find, if possible, \( (f \circ g)(0) \) given
\( f(x) = \dfrac{1}{x + 1} \), \( g(x) = \dfrac{1}{x - 1} \)

Solutions to the Above Questions

Solution to Question 1

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

Solution to Question 2

The composition \( (f \circ g)(x) \) of
\( f(x) = x^2 + 1 \), \( g(x) = \sqrt{2x} \)
is given by
\( (f \circ g)(x) = f(g(x)) \)
\( = g(x)^2 + 1 \)
\( = [\sqrt{2x}]^2 + 1 \)
\( = 2x + 1 \)
To find the domain of the composition of the two functions, we proceed as follows: \( x \) must be in the domain of \( g \).
\( 2x \geq 0 \) which is equivalent to \( x \geq 0 \)
\( g(x) \) must be in the domain of \( f(x) \). The domain of \( f \) is the set of all real numbers. The condition \( x \geq 0 \) 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
\[ [0 , + \infty) \]

Solution to Question 3

The composition \( (f \circ g)(x) \) of
\( f(x) = \sqrt{- x + 1} \), \( g(x) = x^2 - 8 \)
is given by
\( (f \circ g)(x) = f(g(x)) \)
\( = \sqrt{- g(x) + 1} \)
\( = \sqrt{-(x^2 - 8) + 1} \)
\( = [\sqrt{9 - x^2}] \)
The domain of \( g \) is the set of real numbers. Hence to find the domain of the composition, \( x \) must satisfy the condition
\( 9 - x^2 \geq 0 \)
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 \( g(x) = x^2 \),
\( (f \circ g)(-2) \) is calculated as follows
\( (f \circ g)(-2) = f(g(-2)) \)
\( = f(4) = 9 \)

Solution to Question 5

Given
\( f(x) = \dfrac{1}{x + 1} \) and \( g(x) = \dfrac{1}{x - 1} \),
\( (f \circ g)(0) \) is calculated as follows
\( (f \circ g)(0) = f(g(0)) \)
\( = f(-1) = \text{undefined} \) since \( x = -1 \) makes the denominator of \( f \) equal to zero.