How to find the composition of functions and its domain? A tutorial including detailed explanations is presented. Questions with answers are also included at the end of this page. Also examples of Applications of Composition of Functions are included in this website.

Solution to question 1

note that

\( (f \circ g)(-2) = f( g(-2) ) \)

evaluate g(-2).

\( g(-2) = |-2 - 4| = 6 \)

evaluate \( f( g(-2) ) \).

\( f( g(-2) ) = f(6) = -3 \times 6 + 2 = -16 \)

conclusion:

\( (f \circ g)(-2) = -16 \)

Solution to question 2

First find \( (f \circ g)(x) \)

\( (f \circ g)(x) = f( g(x) ) = \dfrac{{g(x) - 1}}{{g(x) + 2}} \)

=\( \dfrac{{(x + 1)/(x - 2) - 1}}{{(x + 1)/(x - 2) + 2}} \)

= \( \dfrac{{3}}{{3x - 3}} \)

First find domain of \( f \) and \( g \)

domain of \( f \) : \( x \) not equal to -2

domain of \( g \) : \( x \) not equal to 2

\( g(x) \) has to be in the domain of \( f \)

\( g(x) \) not equal to -2

solve for \( x \) the equation \( g(x) = -2 \)

\( \dfrac{{x + 1}}{{x - 2}} = -2 \)

\( x + 1 = -2x + 4 \)

\( 3x = 3 \)

\( x = 1 \)

for \( g(x) \) to be different from - 2, \( x \) has to be different from 1.

conclusion:
The domain of \( f \circ g \) is: \( (- \infty , 1) \cup (1 , 2) \cup (2 , \infty) \)

Solution to question 3

First find \( (f \circ g)(x) \)

\( (f \circ g)(x) = f( g(x) ) = g(x)^2 + 2\)

\( = ( \sqrt{x - 2} )^2 + 2 \)

\( = x \)

First find domain of \( f \) and \( g \)

domain of \( f \) : all real numbers

domain of \( g \) : \( x - 2 \geq 0 \) ; \( x \geq 2 \)

Since the domain of \( f \) is all real numbers, we have to make sure that \( x \) is in the domain of \( g \) so that \( g \) has a real value.

conclusion:
The domain of \( f \circ g \) is: \( [2 , \infty \))

a) \( f(x) = 2x^3 + x - 1 \) and \( g(x) = x^2 \)

b) \( f(x) = | x^2 - 4 | \) and \( g(x) = x - 1 \)

c) \( f(x) = x^2 - 5 \) and \( g(x) = \sqrt{x + 5} \)

d) \( f(x) = \ln x \) and \( g(x) = \sqrt{x + 5} \)

e) \( f(x) = \sin x \) and \( g(x) = x - 2 \)

Answers to Above Questions.

Find the composition \( (f \circ g)(x) \) and its domain given \( f \) and \( g \) below:

a) \( (f \circ g)(x) = 2 x^6 + x^2 -1 \) , domain: \( (- \infty , + \infty ) \)

b) \( (f \circ g)(x) = | x^2 - 2x - 3 | \) , domain:\( (- \infty , + \infty ) \)

c) \( (f \circ g)(x) = x \) , domain: \( [ -5 , + \infty ) \)

d) \( (f \circ g)(x) = \dfrac{1}{2} \ln (x + 5) \) , domain: \( [ -5 , + \infty ) \)

e) \( (f \circ g)(x) = \sin (x - 2) \) , domain: \( (- \infty , + \infty ) \)

A video with more on composite functions is available.

Composition of Functions Questions a

Questions on Composite Functions with Solutions.