One to One Functions | Definition & Tests

Let’s dive into Composite Functions! 🔗✨

Imagine you have two machines 🤖. One machine (g) takes an input and gives an output. Then, you take that output and feed it into a second machine (f), which then gives you a final output. That’s exactly what a composite function is! It’s when the output of one function becomes the input of another function.

What is a Composite Function? 🤔

A composite function is essentially a function inside another function. If you have two functions, say f and g, the composition of f and g is written as f ◦ g. This is defined by (f ◦ g)(x) = f(g(x)).

In simpler terms:

1. You start with an input x 🔢.
2. You apply the inner function g to x, getting g(x).
3. Then, you take that result g(x) and apply the outer function f to it, getting f(g(x)).

It’s like a chain reaction! ⛓️➡️

Determining the Domain of a Composite Function 🌍

When finding the domain of a composite function (f ◦ g)(x) = f(g(x)), there are two crucial rules to follow to ensure the function is well-defined:

1. Rule 1: Input must be in the domain of the inner function ⚠️ If your initial input x is not in the domain of g (x ∉ Domain(g)), then x cannot be in the domain of f ◦ g (x ∉ Domain(f ◦ g)). Simply put, the first machine must be able to process the input.

2. Rule 2: Output of inner function must be in the domain of the outer function ⚠️ You must exclude any x values for which the output of g(x) is not in the domain of f (g(x) ∉ Domain(f)). This means the output from the first machine must be something the second machine can actually use as its input.

Examples from the Sources 💡

Let’s look at some examples to make this clearer:

Example 1: Basic Composition If f(x) = 3x - 4 and g(x) = x²:

• (f ◦ g)(x) = f(g(x))

  1. First, we apply g(x) to x, which is x².
  2. Then, we substitute x² into f(x) wherever x appears: f(x²) = 3(x²) - 4 = **3x² - 4**.

• (g ◦ f)(x) = g(f(x))

  1. First, we apply f(x) to x, which is 3x - 4.
  2. Then, we substitute 3x - 4 into g(x) wherever x appears: g(3x - 4) = (3x - 4)² = **9x² - 24x + 16**.

Notice that (f ◦ g)(x) is not necessarily the same as (g ◦ f)(x). The order matters! 🔄

Example 2: Shop Discounts (Real-World Application) 🛍️💰 A shop offers two discounts on a product with a MRP of £14,000:

• First discount (f(x)): 15% off the MRP. So, f(x) = 0.85x (you pay 85% of the price).
• Second discount (g(x)): A flat £3,000 discount. So, g(x) = x - 3000.

If both discounts are applied, with the first discount applied then the second, this is a composition h(x) = g(f(x)):

1. f(x) (15% off) is applied first.
2. The result f(x) is then fed into g(x) (flat £3,000 discount). h(x) = g(f(x)) = 0.85x - 3000.

For an MRP of £14,000: h(14000) = 0.85(14000) - 3000 = 11900 - 3000 = **£8,900**. This shows how composition models sequential operations! 💡

Example 3: Composition with Domain Restrictions 🌍 Let f(x) = 2/(x-1) and g(x) = 3/x. Find (f ◦ g)(x) and its domain.

• (f ◦ g)(x) = f(g(x))

  1. Substitute g(x) into f(x): f(3/x) = 2 / ( (3/x) - 1 ).
  2. Simplify: 2 / ( (3-x)/x ) = **2x / (3-x)**.

• Domain of (f ◦ g)(x) 🌍:

  1. Rule 1 (Domain of g): g(x) = 3/x is undefined if x = 0. So, x ≠ 0.
  2. Rule 2 (Output of g in domain of f): f(x) = 2/(x-1) is undefined if x - 1 = 0, i.e., x = 1. This means g(x) cannot be 1. Set g(x) = 1: 3/x = 1 ➡️ x = 3. So, x ≠ 3.

  Therefore, the domain of (f ◦ g)(x) is all real numbers except 0 and 3 (R \ {0, 3}).

Practice Questions with Solutions 📝

Question 1: Given p(x) = x + 5 and q(x) = x² - 1. Find: (a) (p ◦ q)(x) 🤔 (b) (q ◦ p)(x) 🤔

Solution:

(a) (p ◦ q)(x) = p(q(x)) 💡 1. Substitute q(x) into p(x): p(x² - 1). 2. Replace x in p(x) with x² - 1: (x² - 1) + 5. 3. Simplify: **x² + 4** ✅

(b) (q ◦ p)(x) = q(p(x)) 💡 1. Substitute p(x) into q(x): q(x + 5). 2. Replace x in q(x) with x + 5: (x + 5)² - 1. 3. Simplify: (x² + 10x + 25) - 1 = **x² + 10x + 24** ✅

Question 2: Let f(x) = √(x - 2) and g(x) = 1/x. Find the domain of (f ◦ g)(x). 🌍🤔

Solution:

To find the domain of (f ◦ g)(x) = f(g(x)), we follow the two rules:

1. Rule 1: Domain of g(x) ⚠️ The function g(x) = 1/x is defined for all real numbers except where the denominator is zero. So, x ≠ 0.

2. Rule 2: Output of g(x) must be in the domain of f(x) ⚠️ The function f(x) = √(x - 2) is defined only when its input (x) is greater than or equal to 2. So, x - 2 ≥ 0 implies x ≥ 2. This means g(x) must be greater than or equal to 2: g(x) ≥ 2. Substitute g(x) = 1/x: 1/x ≥ 2.

   Now, we solve 1/x ≥ 2:

   • If x is positive, multiply both sides by x: 1 ≥ 2x ➡️ x ≤ 1/2. Combining x > 0 and x ≤ 1/2 gives 0 < x ≤ 1/2.
   • If x is negative, 1/x will always be negative. Since 2 is positive, 1/x ≥ 2 can never be true for negative x.

   So, the condition g(x) ∈ Domain(f) implies 0 < x ≤ 1/2.

Combining results from Rule 1 (x ≠ 0) and Rule 2 (0 < x ≤ 1/2): The only values that satisfy both conditions are **0 < x ≤ 1/2** ✅. Therefore, the domain of (f ◦ g)(x) is **(0, 1/2]** 🌍.

Question 3: A new online course offers a discount. First, a 20% discount is applied to the original price (P). Then, a service fee of £50 is added. If D(P) represents the discount function and F(P) represents the service fee function, write the composite function that represents the final cost of the course and calculate the final cost if the original price P is £500. 📚💸

Solution:

1. Discount Function (D(P)) 💡: A 20% discount means you pay 80% of the original price. D(P) = 0.80P.

2. Service Fee Function (F(P)) 💡: A £50 service fee is added. F(P) = P + 50 (Note: Here P represents the price after the discount, which will be the input to F).

3. Composite Function for Final Cost 🔗: Since the discount is applied first, then the service fee is added, the output of D(P) becomes the input for F(P). This is **(F ◦ D)(P) = F(D(P))**. Substitute D(P) into F(P): F(D(P)) = F(0.80P) = (0.80P) + 50 So, the final cost function is **C(P) = 0.80P + 50** ✅.

4. Calculate Final Cost for P = £500 💰: C(500) = 0.80(500) + 50 C(500) = 400 + 50 C(500) = **£450** ✅.