﻿ Composition of Functions in Math-interactive lesson with pictures , examples and several practice problems

# Composition of Functions in Math-interactive

#### What is a composition of functions?

Answer: It is the application of one function to the output of the first function

In the following flow chart, our second function $$g(x))$$ is applied to the result of $$f(x)$$.

As you can see the range of f(x) is the domain of g(x) .

### Step By Step Flow Chart

The flow chart below shows a step by step walk through of $$(f \cdot g)(x)$$

As you can see, you always go right to left !

Try some practice problems down below

#### Are compositions commutative?

Does $$f(g(x)) = g(f ( x))$$?

Remember that the commutative property states that "order does not matter".

Multiplication is commutative because $$3 \cdot 2 = 2 \cdot 3$$.

On the other hand, subtraction is not commutative because $$3 - 2 \ne 2 - 3$$

We want to know if this property applies to composition of functions.

In other words, does it make if difference whether we write $$\bf{ f(g(x)) }$$ vs $$\bf{ g(f(x)) } \text{ ?}$$

To answer that question, let's explore a specific example

Let $$\bf{ f(x) = x + 5}$$ and $$\bf{ g(x) = 4x}$$

Let's evaluate $$\red {f}( \blue{g}( 3))$$ and $$\blue{g}( \red {f}( 3))$$ and see if we get the same answer.

$$\red {f}( \blue{g}( 3)) \\ \blue{g}( 3) =4 \cdot 3 = 12 \\ \red {f}( 12 ) = 12 + 5 = 17$$

$$\blue{g}( \red {f}( 3)) \\ \red {f}( 3) = 3+ 5 = 8 \\ \blue{g}(8) = 4 \cdot 8 = 32$$

Conclusion: Composition of functions are not commutative because $$f(g( 3)) \red { \ne} g(f(3))$$

#### What about the composition of inverse function

In other words, what if $$f(x)$$ and $$g(x)$$ are inverses?

Again, the best way to understand this is to try some examples, and see what happens.

$$\red {f(}\blue{g(x)} \red {)} \\ = \red {f(}\blue{ x-3} \red {)} \\ = \red {(}\blue{ x-3} \red {) +3}= \bf{ x}$$

$$\red {f(}\blue{g(x)} \red {)} \\ =\red {f(}\blue{ \sqrt{x}} \red {)} \\ = \red {(} \blue{ \sqrt{x}} \red { )^2} = \bf{ x}$$

Conclusion:

$$f( f^{-1}(x)) = \bf {x}$$

Therefore, you know that:

$$g( g^{-1}(5)) = 5 \\ a( a^{-1}(232232)) = 232232 \\ k( k^{-1}(-90210)) = -90210 \\ k(k ^{-1}(\bigstar)) = \bigstar \\ f( f^{-1}(stuff)) = stuff$$

If you think of the inverse function as 'undoing' the original function, this rule might make sense. Read more about inverse functions here.

### Practice Problems Part I

Directions: Evaluate each composition of functions listed below.

Step 1

Evaluate function on right $$f( \red {g(2) })$$

$$g(2) =$$
$$4(2) = \blue{8}$$
Step 2

Substitute that value into left side function $$\red {f}( g(2) )$$

$$f( \blue{ 8 })$$
Step 3
$$f( \blue{ 8 })$$ =
$$8 + 3 = 11$$
Step 1

Evaluate function on right $$f( \red {g(8) })$$

$$g(8) =$$
$$8+1 = \blue{9}$$
Step 2

Substitute that value into left side function $$\red {f}( g(8) )$$

$$f( \blue{ 9 })$$
Step 3
$$f( \blue{ 9 })$$ =
$$9^2 = 81$$
Step 1

Evaluate function on right $$f( \red {h(18) })$$

$$h(18) =$$
$$\sqrt{2\cdot 18} =\sqrt{36}= \blue{6}$$
Step 2

Substitute that value into left side function $$\red {f}( h(18) )$$

$$f( \blue{ 6 })$$
Step 3
$$f( \blue{ 6 })$$ =
$$| 7\cdot 6 + 4 | = 46$$
Step 1

Evaluate function on right $$f( \red {b( 5 ) })$$

$$b(5) =$$
$$5 ^3= \blue{125}$$
Step 2

Substitute that value into left side function $$\red {f}( b(5) )$$

$$f( \blue{ 125 })$$
Step 3
$$f( \blue{ 125 })$$ =
$$3(\blue{125})-1 = 374$$
Step 1

Evaluate function on right $$f( \red {m(11) })$$

$$m(11) =$$
$$5 ^3= \blue{125}$$
Step 2

Substitute that value into left side function $$\red {f}(m(11) )$$

$$f( \blue{ 27 })$$
Step 3
$$f( \blue{ 27 })$$ =
$$\frac{1}{3} \blue{27} + 2 = 9 + 2=11$$

Did you notice that the input of '11' is the same as the output for this composition.

#### Why do you think that is?

Because these two functions are inverses of each other!
Remember that the rule about the composition of inverse functions.

Advanced problems - solving algebraically.

### Practice Problems Part II

Step 1

Substitute right side function into left side

$$\red { k(} \blue{ h(x) } \red {)} = \red { k(} \blue{ x + 3 } \red {)} \\ \red { k(} \blue{ x + 3 } \red {)} = \red {(} \blue{ x + 3 } \red {)^2 + (}\blue{ x + 3 } \red {)}$$

Step 2

Simplify

$$\red {(}x^2 + 6x + 9 \red {) + ( } x + 3 \red {)} \\ x^2 + 6x + 9 + x + 3 \\ x^2 + 7x + 12$$

Step 1

Substitute right side function into left side

$$\red { g(} \blue{ h(x) } \red {)} = \red { g(} \blue{ x + 4 } \red {)} \\ \red { g(} \blue{ x + 4 } \red {)} = \red {2(} \blue{ x + 4 } \red {)^2 - 3(}\blue{ x + 4 } \red {)}$$

Step 2

Simplify

$$\red {2(} \blue{ x + 4 } \red {)^2 - 3(}\blue{ x + 4 } \red {)} \\ 2 (x^2 + 8x + 16) - 3 \cdot x - 3 \cdot 4 \\ 2x^2 + 2\cdot 8x + 2\cdot 16 - 3 x - 12 \\ 2x^2 + 16x + 32 - 3x - 12 \\ 2x^2 + 13x +20$$

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