Function Composition

Applying One Function After Another

A store applies the same discount and fixed price reduction in two different orders:

1. Promo A: $20\%$ discount, then another deduction of $\text{Rp}25{,}000.00$.
2. Promo B: Price reduction of $\text{Rp}25{,}000.00$, then a $20\%$ discount.

Do both promotions produce the same final price? We can answer this by writing each operation as a function and composing the two functions in the stated order.

Definition of Function Composition

Function composition is the sequential combination of two or more functions to produce a new function.

If we have a function $g: A \to B$ and a function $f: B \to C$, then their composition, written as $(f \circ g)(x)$, is a new function that maps directly from the domain $A$ to the codomain $C$.

This means we apply function $g$ first, and then we input its result into function $f$.

Mathematically, this is written as:

The composite is defined only for inputs $x$ in the domain of $g$ whose output $g(x)$ lies in the domain of $f$:

Promo Calculation

Calculate the final price for an item worth $\text{Rp}200{,}000.00$. This simplified example ignores taxes, minimum-purchase rules, and price floors so that only the order of the two operations changes.

Let $x$ be the initial price of the item.

• $20\%$ discount function: $d(x) = x - 0.20x = 0.80x$
• $\text{Rp}25{,}000.00$ reduction function: $p(x) = x - 25000$

Now compose these two functions according to the promo sequence:

1. Promo A (Discount first, then price reduction): We are looking for $(p \circ d)(x)$

   $$\begin{aligned} (p \circ d)(x) &= p(d(x)) \\ &= p(0.80x) \\ &= 0.80x - 25000 \end{aligned}$$

   For $x = 200000$:

   $$\begin{aligned} &(p \circ d)(200000) \\ &= 0.80(200000) - 25000 \\ &= 160000 - 25000 \\ &= 135000 \end{aligned}$$

   So, the final price with Promo A is $\text{Rp}135{,}000.00$.

2. Promo B (Price reduction first, then discount): We are looking for $(d \circ p)(x)$

   $$\begin{aligned}(d \circ p)(x) &= d(p(x)) \\ &= d(x - 25000) \\ &= 0.80(x - 25000) \\ &= 0.80x - 20000\end{aligned}$$

   For $x = 200000$:

   $$\begin{aligned} &(d \circ p)(200000) \\ &= 0.80(200000) - 20000 \\ &= 160000 - 20000 \\ &= 140000 \end{aligned}$$

   So, the final price with Promo B is $\text{Rp}140{,}000.00$.

For this item, Promo A ($p \circ d$) gives a final price of $\text{Rp}135{,}000.00$, while Promo B ($d \circ p$) gives $\text{Rp}140{,}000.00$. Reversing the order changes the result, so in general $(f \circ g)(x) \neq (g \circ f)(x)$.

Comparing Both Orders of Composition

Suppose we have two functions:

Determine $(f \circ g)(x)$ and $(g \circ f)(x)$.

Solution:

1. Finding $(f \circ g)(x)$:

   $$(f \circ g)(x) = f(g(x)) = f(x^2 - 3)$$

   Substitute $x$ in $f(x)$ with $g(x)$:

   $$\begin{aligned} f(x^2 - 3) &= 2(x^2 - 3) + 1 \\ &= 2x^2 - 6 + 1 \\ &= 2x^2 - 5 \end{aligned}$$

   So, $(f \circ g)(x) = 2x^2 - 5$.

2. Finding $(g \circ f)(x)$:

   $$(g \circ f)(x) = g(f(x)) = g(2x + 1)$$

   Substitute $x$ in $g(x)$ with $f(x)$:

   $$\begin{aligned} g(2x + 1) &= (2x + 1)^2 - 3 \\ &= (4x^2 + 4x + 1) - 3 \\ &= 4x^2 + 4x - 2 \end{aligned}$$

   So, $(g \circ f)(x) = 4x^2 + 4x - 2$.

Because $(f \circ g)(x) \neq (g \circ f)(x)$, function composition is not commutative.

The Domain Must Follow the Order

Composition is not the same as multiplying two functions, and its domain cannot be found by looking only at the final formula. Let

Then

The input must satisfy $x\in D_g=\mathbb{R}$ and $g(x)=x^2\in D_f=[1,\infty)$. Therefore,

This check explains why values between $-1$ and $1$ are excluded: they make the expression under the square root negative.