Worksheet A Topic 2.7 Composition Of Functions

The composition of functions is a fundamental concept in mathematics that allows us to combine two or more functions to create a new function. This process involves applying one function to the result of another, effectively chaining them together. Understanding composition of functions is crucial for various areas of mathematics, including calculus, analysis, and computer science. It provides a powerful tool for modeling complex relationships and solving problems involving multiple steps.

Introduction to Composition of Functions

Composition of functions can be visualized as a "function machine" where the output of one machine becomes the input of the next. Imagine two machines, f and g. When you input a value x into machine g, it produces an output g(x). Now, if you take that output and feed it into machine f, the final output is f(g(x)). This combined process is the composition of f with g, denoted as f ∘ g.

Mathematically, the composition of function f with function g is defined as:

(f ∘ g)(x) = f(g(x))

This means that to evaluate the composite function (f ∘ g)(x), first evaluate g(x), and then substitute that result into f(x). It's important to note that the order matters; f(g(x)) is generally not the same as g(f(x)).

Key Concepts:

• Inner Function: The function that is applied first (in this case, g(x)).
• Outer Function: The function that is applied second (in this case, f(x)).
• Domain: The set of all possible input values for which the function is defined.
• Range: The set of all possible output values of the function.

Why is Composition of Functions Important?

• Modeling Complex Systems: It allows us to break down complex relationships into smaller, manageable parts. For example, in physics, we can model the trajectory of a projectile by composing functions that describe its horizontal and vertical motion.
• Simplifying Expressions: Composition can be used to simplify complex expressions by identifying and substituting intermediate functions.
• Computer Science: It is fundamental in programming, where functions are often composed to create complex algorithms.
• Calculus: Understanding composition is essential for the chain rule, which is used to differentiate composite functions.

Step-by-Step Guide to Finding the Composition of Functions

Let's go through a step-by-step guide on how to find the composition of two functions, f(x) and g(x), to obtain (f ∘ g)(x) and (g ∘ f)(x).

Step 1: Understand the Notation

Remember that (f ∘ g)(x) means f(g(x)), and (g ∘ f)(x) means g(f(x)). The order of operations is crucial here.

Step 2: Find f(g(x))

1. Identify g(x): Determine the expression for the inner function g(x).
2. Substitute g(x) into f(x): Replace every instance of x in the function f(x) with the entire expression for g(x).
3. Simplify the resulting expression: Expand and simplify the expression obtained in the previous step, if possible.

Step 3: Find g(f(x))

1. Identify f(x): Determine the expression for the inner function f(x).
2. Substitute f(x) into g(x): Replace every instance of x in the function g(x) with the entire expression for f(x).
3. Simplify the resulting expression: Expand and simplify the expression obtained in the previous step, if possible.

Step 4: Determine the Domain of the Composite Functions

The domain of a composite function is the set of all x values that satisfy two conditions:

1. x must be in the domain of the inner function (either g(x) for (f ∘ g)(x) or f(x) for (g ∘ f)(x)).
2. The output of the inner function must be in the domain of the outer function.

This often requires a bit of careful analysis, especially when dealing with functions that have restrictions on their domains, such as square roots or rational functions.

Example 1: Simple Polynomials

Let's say f(x) = x² + 1 and g(x) = 2x - 3. Find (f ∘ g)(x) and (g ∘ f)(x).

• (f ∘ g)(x) = f(g(x)) = f(2x - 3)

  • Substitute (2x - 3) into f(x): (2x - 3)² + 1
  • Simplify: 4x² - 12x + 9 + 1 = 4x² - 12x + 10

• (g ∘ f)(x) = g(f(x)) = g(x² + 1)

  • Substitute (x² + 1) into g(x): 2(x² + 1) - 3
  • Simplify: 2x² + 2 - 3 = 2x² - 1

In this example, both f(x) and g(x) are polynomials, so their domains are all real numbers. Therefore, the domains of (f ∘ g)(x) and (g ∘ f)(x) are also all real numbers.

Example 2: Functions with Restrictions

Let f(x) = √(x) and g(x) = x - 2. Find (f ∘ g)(x) and (g ∘ f)(x), and determine their domains.

• (f ∘ g)(x) = f(g(x)) = f(x - 2)

  • Substitute (x - 2) into f(x): √(x - 2)

• (g ∘ f)(x) = g(f(x)) = g(√(x))

  • Substitute √(x) into g(x): √(x) - 2

Now let's consider the domains:

• For (f ∘ g)(x) = √(x - 2), the expression under the square root must be non-negative. So, x - 2 ≥ 0, which means x ≥ 2. Therefore, the domain of (f ∘ g)(x) is [2, ∞).

• For (g ∘ f)(x) = √(x) - 2, the expression under the square root must be non-negative. So, x ≥ 0. Therefore, the domain of (g ∘ f)(x) is [0, ∞).

Example 3: Rational Functions

Let f(x) = 1/x and g(x) = x + 3. Find (f ∘ g)(x) and (g ∘ f)(x), and determine their domains.

• (f ∘ g)(x) = f(g(x)) = f(x + 3)

  • Substitute (x + 3) into f(x): 1/(x + 3)

• (g ∘ f)(x) = g(f(x)) = g(1/x)

  • Substitute (1/x) into g(x): (1/x) + 3

Now let's consider the domains:

• For (f ∘ g)(x) = 1/(x + 3), the denominator cannot be zero. So, x + 3 ≠ 0, which means x ≠ -3. Therefore, the domain of (f ∘ g)(x) is all real numbers except -3, which can be written as (-∞, -3) ∪ (-3, ∞).

• For (g ∘ f)(x) = (1/x) + 3, the denominator cannot be zero. So, x ≠ 0. Therefore, the domain of (g ∘ f)(x) is all real numbers except 0, which can be written as (-∞, 0) ∪ (0, ∞).

Common Mistakes to Avoid:

• Forgetting to Substitute Correctly: Make sure you replace every instance of x in the outer function with the entire expression for the inner function.
• Incorrect Order of Operations: Always evaluate the inner function first.
• Ignoring Domain Restrictions: Always check for domain restrictions on both the inner and outer functions. The domain of the composite function must satisfy the restrictions of both.
• Assuming Commutativity: Remember that (f ∘ g)(x) is generally not equal to (g ∘ f)(x). Composition of functions is not commutative.

Worksheet Examples and Solutions

Let's work through some more examples that you might encounter on a worksheet:

Worksheet Problem 1:

Given f(x) = 3x - 2 and g(x) = x² + 1, find:

a) (f ∘ g)(x) b) (g ∘ f)(x) c) (f ∘ f)(x) d) (g ∘ g)(x)

Solution:

a) (f ∘ g)(x) = f(g(x)) = f(x² + 1) = 3(x² + 1) - 2 = 3x² + 3 - 2 = 3x² + 1

b) (g ∘ f)(x) = g(f(x)) = g(3x - 2) = (3x - 2)² + 1 = 9x² - 12x + 4 + 1 = 9x² - 12x + 5

c) (f ∘ f)(x) = f(f(x)) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8

d) (g ∘ g)(x) = g(g(x)) = g(x² + 1) = (x² + 1)² + 1 = x⁴ + 2x² + 1 + 1 = x⁴ + 2x² + 2

Worksheet Problem 2:

Given f(x) = √(x + 4) and g(x) = x - 5, find:

a) (f ∘ g)(x) and its domain b) (g ∘ f)(x) and its domain

Solution:

a) (f ∘ g)(x) = f(g(x)) = f(x - 5) = √((x - 5) + 4) = √(x - 1)

Domain: For √(x - 1) to be defined, x - 1 ≥ 0, so x ≥ 1. Therefore, the domain is [1, ∞).

b) (g ∘ f)(x) = g(f(x)) = g(√(x + 4)) = √(x + 4) - 5

Domain: For √(x + 4) to be defined, x + 4 ≥ 0, so x ≥ -4. Therefore, the domain is [-4, ∞).

Worksheet Problem 3:

Given f(x) = 1/(x - 2) and g(x) = 3/x, find:

a) (f ∘ g)(x) and its domain b) (g ∘ f)(x) and its domain

Solution:

a) (f ∘ g)(x) = f(g(x)) = f(3/x) = 1/((3/x) - 2) = 1/((3 - 2x)/x) = x/(3 - 2x)

Domain: We need to consider two restrictions: 1. The denominator of g(x) cannot be zero, so x ≠ 0. 2. The denominator of (f ∘ g)(x) cannot be zero, so 3 - 2x ≠ 0, which means x ≠ 3/2.

Therefore, the domain is all real numbers except 0 and 3/2, which can be written as (-∞, 0) ∪ (0, 3/2) ∪ (3/2, ∞).

b) (g ∘ f)(x) = g(f(x)) = g(1/(x - 2)) = 3/(1/(x - 2)) = 3(x - 2) = 3x - 6

Domain: We need to consider the restriction on the inner function, f(x). The denominator of f(x) cannot be zero, so x - 2 ≠ 0, which means x ≠ 2.

Therefore, the domain is all real numbers except 2, which can be written as (-∞, 2) ∪ (2, ∞).

Worksheet Problem 4: Evaluating Composite Functions at Specific Points

Given f(x) = x² - 3 and g(x) = 2x + 1, find:

a) (f ∘ g)(2) b) (g ∘ f)(-1)

Solution:

a) (f ∘ g)(2) = f(g(2))

• First, find g(2): g(2) = 2(2) + 1 = 5
• Then, find f(5): f(5) = 5² - 3 = 25 - 3 = 22
• Therefore, (f ∘ g)(2) = 22

b) (g ∘ f)(-1) = g(f(-1))

• First, find f(-1): f(-1) = (-1)² - 3 = 1 - 3 = -2
• Then, find g(-2): g(-2) = 2(-2) + 1 = -4 + 1 = -3
• Therefore, (g ∘ f)(-1) = -3

Advanced Topics and Applications

While the basic concept of composition of functions is straightforward, it opens the door to more advanced topics:

• Decomposition of Functions: This is the reverse process of composition. Given a composite function, the goal is to find the individual functions f(x) and g(x) that, when composed, produce the given function. This can be useful for simplifying complex expressions or understanding the underlying structure of a mathematical model.

• Iterated Functions: This involves composing a function with itself multiple times. For example, f(f(x)) is the second iterate of f(x), and f(f(f(x))) is the third iterate. Iterated functions are used in chaos theory and dynamical systems.

• Applications in Calculus: Composition is crucial for understanding and applying the chain rule in differentiation. The chain rule states that the derivative of (f ∘ g)(x) is f'(g(x)) * g'(x).

• Applications in Computer Science: Function composition is a fundamental concept in functional programming. It allows programmers to build complex programs by combining smaller, reusable functions.

Conclusion

Composition of functions is a powerful and versatile tool in mathematics. Mastering this concept will not only improve your understanding of functions but also provide a solid foundation for more advanced mathematical topics. By understanding the steps involved in finding the composition of functions, paying attention to domain restrictions, and practicing with various examples, you can confidently tackle any worksheet problem or real-world application involving composite functions. Remember to focus on understanding the underlying principles and the order of operations, and you'll be well on your way to mastering this important mathematical concept.