Composition of Functions

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

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!

The Commutative Law and Compositions

Compositions of Functions are not commutative.
In other words, f(g(x)) ≠ g(f(x)). To see for yourself, look at the following example:

f(x) = x² +1
g(x) = 3x

Does f(g(1)) = g(f(1))?

f(g(1)) = f(3) = 10
g(f(1)) = g(2) = 6

Since these compositions are not equal, compositions of functions are not commutative.

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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