Composition of Functions

Compositions of Functions are a well known example of a process that is not commutative. In other words, f(g(x)) ≠ g(f(x)). The examples below illustrate that although sometimes composition of functions will preserve commuativity (problems 1 and 2), this commutativity is not always true (see problems 3 and 4)

Practice Problems: Composition of Functions

The two functions t(x) and v(x) are defined below.

t(x) = 3x – 1

v(x) = x²+1

Practice problem 1)

Evaulate the composition of functions v(t(1))

Answer

Step 1) Evaluate inner function : t(1) = 3•1 -1 = 3-1 = 2
Step 2) Insert answer from step 1 into outer function and evaulate v(2) = 2² + 1 = 4 +1 = 5
v(t(1)) = 5

Practice problem 2)

Evaulate the composition of functions t(v(1))

Answer

Step 1) Evaluate inner function : v(1) = 1² +1 = 1 + 1 = 2
Step 2) Insert answer from step 1 into outer function and evaulate
t(2) = 3 • 2 – 1= 6 -1 = 5
t(v(1)) = 5

t(x) = 3x – 1

v(x) = x²+1

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Are compositions of functions commutative?

From practice problems 1 and 2, it appears that compositions of functions are commutative, but that is not always true. Compositions of functions are not , as a rule, commutative. In fact, they very rarely happen to be commutative. For example look at the next two compositions and you will see for yourself!

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Practice problem 3)

Evaulate the composition of functions v(t(2))

Answer

Step 1) Evaluate inner function : t(2) = 3•2 -1 = 6-1 = 5
Step 2) Insert answer from step 1 into outer function and evaulate v(5) = 5² + 1 = 25 +1 = 26
v(t(2)) = 26

Practice problem 4)

Evaulate the composition of functions t(v(2))

Answer

Step 1) Evaluate inner function : v(2) = 2² +1 = 4 + 1 = 5
Step 2) Insert answer from step 1 into outer function and evaulate t(5) = 3 • 5 – 1= 15 - 1 = 14
t(v(2)) = 14