Activity: 1

"Double a number and add 3", do it for 5.

Calculation - 1

$$2 \times 5 = 10$$

Calculation - 2

$$10 + 3 = 13$$

In picture:

Observation:

Initial value of calcualtion - 1 is 5, where as initial value of calculation - 2 is 10, which is result of calculation - 1.

Conclusion:

Two calculations combines together to give a result.

Activity: 2 (Real Life Analogy)

Observation:

Bus 'A' takes Ram from Pokhara to Mugling.
Bus 'B' takes Ram from Mugling to Kathmandu.
Ram travel from Pokhara from Kathmandu by two buses.

Conclusion:

Ram enters Bus 'A' in Pokhara and exits in Mugling whereas he enters Bus 'B' in Mugling and exits in Kathmandu.
Passenger dropped by Bus 'A' is taken by Bus 'B'.

Activity: 3 (Function Machine Analogy)

Observation:

Machine - 1 takes '4' and gives '16'.
Machine - 2 takes '16' and gives '32'.
From '4' to '32' is work of two machines.

Conclusion:

Output of one function is input of another function, when two functions act/work in this way we call them composite functions.

Definition:

Composite Function:

Let f: A $\rightarrow$ B be a function defined from set A to B and g: B $\rightarrow$ C be a function defined from set B to C. The function which is defined from set A to C is called COMPOSITE FUNCTION of f and g.

It is denoted by $gof$(x).

Mathematically,
$(g \circ f)(x)$ = $g(f(x))$

Good Question:

Q: Why it is written $(g \circ f)(x)$ instead of $(f \circ g)(x)$?

Ans:

In mathematics, function notation is read from right to left regarding the order of operations.

In the expression g(f(x)), the f is physically closer to the x.
Because f acts on x first, it is written on the inside.
The notation $(g \circ f)$ is designed to mirror the nested notation g(f(x)).

The Logic: If we wrote it as $f \circ g$ it meant "apply $g$ and then apply $f$'' which is logically wrong as $f$ is the function which takes the input first, then $g$ takes the output of $f$ as its input. To understand it more in a simple way, the function which is close of $x$ takes the input first or the function which takes the input first must be written close to $x$.

Composite Function in Mapping Diagram:

Composite Function: (Functions written as ordered pairs)

Let $f$ = {(a, p), (b, q), (c, r)} and $g$ = {(p, 1), (q, 2), (r, 3)}

For: $g \circ f$:

To understand:

$a, b, c$ are inputs of function $f$

$p, q, r$ are outputs of function $f$

$p, q, r$ are inputs of function $g$

$1, 2, 3$ are outputs of function of $g$

Solution:

$g \circ f$ = $g(f(a))$ = $g(p)$ = 1

$g \circ f$ = $g(f(b))$ = $g(q)$ = 2

$g \circ f$ = $g(f(c))$ = $g(r)$ = 3

$\therefore$ $g \circ f$ = {(a, 1), (b, 2), (c, 3)}

In mapping diagram:

Composite Function: (Functions written as formula)

Let, $f(x)$ = $2x - 1$ and $g(x)$ = 3x + 5, $x \in \mathbb{R}$

Solution:

Given,

$f(x)$ = $2x - 1$

$g(x)$ = $3x + 5$

Now,

$(g \circ f)(x)$ = g(f(x)) = g(2x - 1) = 3(2x - 1) + 5 = 6x - 3 + 5 = 6x + 2

[Understand: like $x$, (2x - 1) is input value of function $g$, so it replaces input variable of function $g$ i.e. $x$.]

Again,

$(f \circ g)(x)$ = f(g(x)) = f(3x + 5) = 2 (3x + 5) - 1 = 6x + 10 - 1 = 6x + 9

$\therefore$ $(g \circ f)(x)$ $\neq$ $(f \circ g)(x)$

Conclusion: The composite of any two functions is not commutative.

Remember:

$f^2$ = $f \circ f$

$g^2$ = $g \circ g$

In composite function $g \circ f$:

Domain = Domain of $f$

Range = Range of $g$