Composite Functions | Domain

Alright, let’s dive into composite functions again, this time focusing on their domain with some friendly emojis! 🤖✨

What are Composite Functions? 🤔

As we discussed, imagine two magical machines 🤖⚙️:

• Machine G (the inner function g): Takes your initial idea (x) and transforms it into something new (g(x)).
• Machine F (the outer function f): Takes what Machine G made (g(x)) and transforms that into a final product (f(g(x))).

A composite function is essentially a “function of a function” [Conversation History]. If you have two functions, f and g, their composition is typically written as f ◦ g (read as “f of g”), and it’s formally defined by the equation:

(f ◦ g)(x) = f(g(x)) [Conversation History, 5.9]

This means you apply g first, then f to the result [Conversation History, 5.9].

Understanding the Domain of a Composite Function 🌍

Determining the domain of a composite function (f ◦ g)(x) = f(g(x)) is super important because it tells you which input values x are allowed for the entire process to work seamlessly. Think of it as ensuring both magical machines can handle the materials at each step! ⚙️🔗

There are two crucial rules to follow to find the domain of (f ◦ g)(x) [Conversation History, 5.9]:

1. Rule 1: The initial input x must be acceptable to the inner function g 🚫.

   • If your x is not in the domain of g (x ∉ Domain(g)), then g(x) simply won’t exist. If g(x) doesn’t exist, f can’t even start its work, so (f ◦ g)(x) cannot exist either [Conversation History, 5.9].
   • Emoji Analogy: You can’t put square pegs 🟩 into round holes ⭕! The first machine (g) needs to be able to process the raw material (x) you feed it.

2. Rule 2: The output of the inner function g(x) must be acceptable to the outer function f ✅.

   • You must exclude any x values for which the output g(x) falls outside the domain of f (g(x) ∉ Domain(f)) [Conversation History, 5.9]. Even if g(x) produces a value, that value might be something f cannot handle (e.g., trying to take the square root of a negative number, or dividing by zero).
   • Emoji Analogy: The output from the first machine ⚙️ must be compatible raw material for the second machine 🤖. If machine G makes something too hot 🔥 or too cold ❄️ for machine F to handle, the whole operation breaks down!

Combining these two rules, the domain of (f ◦ g)(x) consists of all x such that:

• x is in the domain of g [5.9].
• g(x) is in the domain of f [5.9].

Examples from the Sources 💡

Let’s look at some examples to illustrate these concepts:

Example 1: Basic Algebraic Composition If f(x) = 3x - 4 and g(x) = x² [Conversation History, 5.9]:

• (f ◦ g)(x) = f(g(x)):

  1. g(x) transforms x into x² (input for f) [Conversation History].
  2. f(x²) = 3(x²) - 4 = **3x² - 4** (final result) [Conversation History, 5.9].
  • Domain: Both f(x) and g(x) have domains of all real numbers (ℝ) [Conversation History]. So, x can be any real number, and g(x) will always be a real number that f(x) can handle. Thus, the domain of (f ◦ g)(x) is ℝ.

• (g ◦ f)(x) = g(f(x)):

  1. f(x) transforms x into 3x - 4 (input for g) [Conversation History].
  2. g(3x - 4) = (3x - 4)² = **9x² - 24x + 16** (final result) [Conversation History, 5.9].
  • Domain: Similar to above, both f(x) and g(x) have domains of ℝ. So, the domain of (g ◦ f)(x) is also ℝ.

Notice how the order of composition produces different results [Conversation History].

Example 2: Shop Discounts (Real-World Application) 🛍️💰 A shop offers two discounts on a product with a Manufacturer’s Recommended Price (MRP) of £14,000 [Conversation History, 5.9].

• First discount (f(x)): 15% off the MRP. This means you pay 85% of the price. So, f(x) = 0.85x [Conversation History, 5.9].
• Second discount (g(x)): A flat £3,000 discount. So, g(x) = x - 3000 [Conversation History, 5.9].

If the first discount is applied then the second, this is h(x) = g(f(x)) [Conversation History, 5.9]:

1. f(x) (15% off) is applied first.
2. The result, f(x), is then fed into g(x) (the flat £3,000 discount). h(x) = g(f(x)) = g(0.85x) = 0.85x - 3000 [Conversation History, 5.9].

For an MRP of £14,000: h(14000) = 0.85(14000) - 3000 = 11900 - 3000 = **£8,900** [Conversation History, 5.9].

• Domain: In this real-world scenario, x (price) must be positive, so Domain(f) is (0, ∞). The output f(x) will also be positive, which is valid for g(x). Thus, the domain of h(x) is (0, ∞).

Example 3: Composition with Domain Restrictions 🌍 Let f(x) = 2/(x-1) and g(x) = 3/x [Conversation History, 5.9]. Let’s 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 ) [Conversation History, 5.9].
  2. Simplify: 2 / ( (3-x)/x ) = **2x / (3-x)** [Conversation History, 5.9].

• Domain of (f ◦ g)(x) 🌍:

  1. Rule 1 (Domain of g): g(x) = 3/x. The denominator cannot be zero. So, x ≠ 0 🚫 [Conversation History, 5.9].
  2. Rule 2 (Output of g in domain of f): f(x) = 2/(x-1). Its input (g(x)) cannot make its denominator zero. So, g(x) cannot be 1 [Conversation History, 5.9]. Set g(x) = 1: 3/x = 1 ➡️ x = 3 [Conversation History, 5.9]. Therefore, x ≠ 3 🚫.

  Combining these, the domain of (f ◦ g)(x) is all real numbers except 0 and 3. This can be written as ℝ \ {0, 3} or (-∞, 0) ∪ (0, 3) ∪ (3, ∞) [Conversation History, 5.9].

Practice Questions with Solutions 📝

Let’s test your understanding with some fresh examples!

Question 1: Given the functions f(x) = x² - 4 and g(x) = x + 2. Find: (a) (f ◦ g)(x) 🤔 (b) (g ◦ f)(x) 🤔

Solution:

(a) (f ◦ g)(x) = f(g(x)) 💡 1. First, the inner function g(x) gives us x + 2 [5.9]. 2. Then, we substitute x + 2 into f(x) wherever x appears: f(x + 2) = (x + 2)² - 4 [5.9]. 3. Expand and simplify: (x² + 4x + 4) - 4 = **x² + 4x** ✅

(b) (g ◦ f)(x) = g(f(x)) 💡 1. First, the inner function f(x) gives us x² - 4 [5.9]. 2. Then, we substitute x² - 4 into g(x) wherever x appears: g(x² - 4) = (x² - 4) + 2 [5.9]. 3. Simplify: **x² - 2** ✅

Question 2: A bakery has a process for making special cookies 🍪.

• Step 1 (P(x)): For every x kilograms of dough, they add 0.5x + 1 kilograms of chocolate chips 🍫. So, P(x) = 0.5x + 1.
• Step 2 (C(y)): The total mixture y (dough + chips) is then cooked, and 0.1y kilograms evaporate 🌬️, so the final cookie weight is y - 0.1y = 0.9y. So, C(y) = 0.9y.

Write the composite function that represents the final weight of cookies for x kilograms of dough, and calculate the final cookie weight if they start with 10 kilograms of dough. 🏭⚖️

Solution:

1. Chocolate Chip Function (P(x)) 💡: P(x) = 0.5x + 1. This function represents the amount of chocolate chips added based on the dough. Note: The problem states “total mixture y (dough + chips)”. So, P(x) should represent the total mixture. Let’s adjust P(x) to be the total mass after adding chips: P(x) = x + (0.5x + 1) = 1.5x + 1.

2. Cooking Function (C(y)) 💡: C(y) = 0.9y. This function represents the final weight after cooking.

3. Composite Function for Final Cookie Weight 🔗: Since P(x) determines the mixture before cooking, and C(y) takes that mixture as input, the composite function is **(C ◦ P)(x) = C(P(x))** [5.9]. Substitute P(x) into C(y): C(P(x)) = C(1.5x + 1) = 0.9(1.5x + 1) Simplify: **W(x) = 1.35x + 0.9** (where W is the final weight).

4. Calculate Final Weight for x = 10 kg of dough 📈: W(10) = 1.35(10) + 0.9 W(10) = 13.5 + 0.9 W(10) = **14.4 kilograms** ✅.

Question 3: Let f(x) = 1/x and g(x) = x - 4. Find the domain of: (a) (f ◦ g)(x) 🌍 (b) (g ◦ f)(x) 🌍

Solution:

(a) Domain of (f ◦ g)(x) = f(g(x)) 💡: * First, find the expression for (f ◦ g)(x): f(g(x)) = f(x - 4) = **1/(x - 4)** * Rule 1 (Domain of g): g(x) = x - 4 is defined for all real numbers. So, x ∈ ℝ ✅. * Rule 2 (Output of g in domain of f): f(x) = 1/x. Its input (g(x)) cannot be zero. So, g(x) ≠ 0. Set g(x) = 0: x - 4 = 0 ➡️ x = 4 🚫. * Therefore, the domain of (f ◦ g)(x) is all real numbers except 4. In interval notation: **(-∞, 4) ∪ (4, ∞)** 🌍.

(b) Domain of (g ◦ f)(x) = g(f(x)) 💡: * First, find the expression for (g ◦ f)(x): g(f(x)) = g(1/x) = **(1/x) - 4** * Rule 1 (Domain of f): f(x) = 1/x is defined for all real numbers except where its denominator is zero. So, x ≠ 0 🚫. * Rule 2 (Output of f in domain of g): g(x) = x - 4. Its input (f(x)) can be any real number because there are no restrictions (e.g., square roots of negatives, division by zero) [Conversation History, 5.9]. So, f(x) can be any real number, and g can always process it ✅. * Therefore, the domain of (g ◦ f)(x) is all real numbers except 0. In interval notation: **(-∞, 0) ∪ (0, ∞)** 🌍.