#### What kind of notation ?

Notation | Meaning |

$$ g \cdot f (x)$$ | "g of f of x" |

$$ g ( f (x) ) $$ | "g of f of x" |

Debug

Notation | Meaning |

$$ g \cdot f (x)$$ | "g of f of x" |

$$ g ( f (x) ) $$ | "g of f of x" |

A composition involves 2 (or more) functions. In a composition, you use the output of one function as the input of a second function.

In the following flow chart, The output of $$ f(x) $$ is used as the input of our second function $$ g(x)) $$

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

We have 2 functions that we will use for our composition:

$ f(x) = 2x $

$ g(x) = x- 1 $

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

Step 1

Perform right side function $$ g(x)$$ .

Step 2

Apply the left side function $$ f(x) $$

to the output of Step 1 .

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

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

Remember that the commutative property states that "*order does not matter*". ( link )

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

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

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

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

Let

$$ \red{ f(x) = x + 5} $$

$$ \blue{ g(x) = 4x} $$

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

$$ \red {f}(\blue{g}(3)) $$

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

$$ \blue{g}(\red {f}(3)) $$

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

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

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.

Example 1

Let $$ \bf{ \red { f(x) = x + 3}} \\ \bf{ \blue{ g(x) = f^{-1}(x) = x -3}} $$

What is $$ f(g(x)) = ? $$

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

Example 2

Let $$ \bf{ \red { f(x) = x^2}} \\ \bf{ \blue{ g(x) = f^{-1}(x) = \sqrt{x}}} $$

What is $$ f(g(x)) = ? $$

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

Composition of inverse functions always evaluate to $$ x $$ (ie the input itself) .

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

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.

**Directions:** Evaluate each composition of functions listed below.

Step 1

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

$$ g(2) = $$

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

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 $$ f(\red {g(8) }) $$.

$$ g(8) = $$

$$ 8+1 = \blue{9} $$

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 $$ f(\red {h(18) }) $$.

$$ h(18) = $$

$$ \sqrt{2\cdot 18} =\sqrt{36}= \blue{6} $$

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 $$ f(\red {b(5) }) $$.

$$ b(5) = $$

$$ 5 ^3= \blue{125} $$

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 $$ f(\red {m(11) }) $$.

$$m(11) = $$

$$ 5 ^3= \blue{125} $$

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.

Because these two functions are inverses of each other!

**Remember** that the rule about the composition of inverse functions.

Advanced problems - solving algebraically.

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.

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