Composite Functions

Grade 12 · algebra · 86 practice problems · read aloud

🔊 Listen to this explanation

🔗 Composite Functions: The Big Idea

A composite function is created when one function is applied to the result of another function. If you have two functions, f(x) and g(x), the composite function f(g(x)) (read as "f of g of x") means you plug g(x) into f(x). This is like a "function chain" and is fundamental for modeling complex, multi-step processes in advanced mathematics, physics, and engineering.

🧩 Step-by-Step Guide to Solving

  1. Identify the Inner and Outer Functions: For f(g(x)), g(x) is the inner function and f(x) is the outer function.
  2. Substitute the Inner Function: Replace every 'x' in the outer function with the entire inner function. Use parentheses! f(g(x)) = f( (g(x)) )
  3. Simplify: Expand and simplify the resulting expression using algebra.
  4. Check the Domain: Ensure the input x is valid for g(x), and that the output g(x) is valid for f(x).

📚 Worked Examples

Example 1: Basic Composition

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

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

Example 2: With a Rational Function

Let f(x) = 1/x and g(x) = x + 4. Find g(f(x)).

  1. Substitute: g(f(x)) = g(1/x) = (1/x) + 4
  2. Simplify: = 1/x + 4 or (1 + 4x)/x
  3. Domain: x ≠ 0 (because of division in f(x)).

🚨 Common Mistakes to Avoid

  • Misreading the Order: f(g(x)) is NOT the same as g(f(x)). Composition is generally not commutative.
  • Forgetting Parentheses: Writing f(g(x)) = f(2x-3) as 2x-3² + 1 is wrong. It must be (2x-3)² + 1.
  • Ignoring the Domain: Always consider if the output of the inner function creates an issue for the outer function (e.g., division by zero, square root of a negative).

💡 Tips & Tricks

  • Think "Inside Out": Always work from the inside function to the outside.
  • Notation Reminder: (f ∘ g)(x) is just another way to write f(g(x)).
  • Check Your Work: Pick a simple number for x and evaluate g(x) first, then plug that result into f(x). See if it matches your composite function.

🎯 How to Practice

  • Start by finding f(g(x)) and g(f(x)) for the same function pair to see how the order changes the result.
  • Practice with various function types: polynomials, rational, and radical functions.
  • Move on to word problems where one process depends on the outcome of another.
  • Challenge yourself by finding one of the original functions given the composite and the other function.

Practice problems

6 of the 86, worked through step by step — try them before opening the answer.

1 f(x) = √(x - 2), g(x) = 1/(x - 4), find the domain of (g∘f)(x)

Hint: For composite functions, consider restrictions from both the inner and outer functions. The inner function's output must be valid input for the outer function.

Show the answer

Answer: (2, 6) ∪ (6, ∞)

When finding domains of composite functions, you need to ensure the inner function's output falls within the domain of the outer function. For example, if h(x) = √x and k(x) = 1/(x-1), then (k∘h)(x) requires that √x is defined (x ≥ 0) AND that √x ≠ 1 (to avoid division by zero).

2 f(x) = √(x + 3), g(x) = 1/(x - 1), find the domain of (g∘f)(x)

Hint: For composite functions, consider restrictions from both the inner and outer functions. The inner function's output must be valid input for the outer function.

Show the answer

Answer: [ -3, 1 ) ∪ ( 1, ∞ )

  1. The composite function is (g∘f)(x) = g(f(x)) = 1/(√(x + 3) - 1)
  2. For f(x) = √(x + 3), the domain requires x + 3 ≥ 0, so x ≥ -3
  3. For g(f(x)) = 1/(√(x + 3) - 1), the denominator cannot be zero, so √(x + 3) - 1 ≠ 0
  4. Solve √(x + 3) - 1 ≠ 0: √(x + 3) ≠ 1, x + 3 ≠ 1, x ≠ -2
  5. Combine restrictions: x ≥ -3 (from
  6. AND x ≠ -2 (from
  7. Final domain: [-3, -2) ∪ (-2, ∞)

3 f(x) = √(x - 2) and g(x) = 1/(x - 4), find the domain of (g∘f)(x)

Hint: Consider the restrictions from both the inner function and the composition, including square root domains and denominator exclusions.

Show the answer

Answer: (2, 6) ∪ (6, ∞)

When finding domains of composite functions, you must ensure the input works for the inner function, and the output of the inner function works for the outer function. For square root functions, the radicand must be non-negative. For rational functions, denominators cannot be zero.

4 f(x) = √(x + 3) and g(x) = 1/(x - 1), find the domain of (g∘f)(x)

Hint: Consider the restrictions from both functions: the inner function's domain and the outer function's domain applied to the output of the inner function.

Show the answer

Answer: [ -3, 1 ) ∪ ( 1, ∞ )

  1. Find (g∘f)(x) = g(f(x)) = 1/(√(x + 3) - 1)
  2. Domain of f(x): x + 3 ≥ 0 → x ≥ -3
  3. Domain of g∘f: f(x) ≠ 1 → √(x + 3) ≠ 1 → x + 3 ≠ 1 → x ≠ -2
  4. Combine restrictions: x ≥ -3 AND x ≠ -2
  5. Write in interval notation: [-3, -2) ∪ (-2, ∞)

The answer is [-3, -2) ∪ (-2, ∞).

5 f(x) = √(x + 3) and g(x) = 1/(x - 2), find the domain of (g∘f)(x)

Hint: For composite functions, the output of the inner function must be valid input for the outer function. Consider restrictions from both square root and rational functions.

Show the answer

Answer: [ -3, 11 ) ∪ ( 11, ∞ )

  1. Write the composite function (g∘f)(x) = g(f(x)) = 1/(√(x + 3) - 2)
  2. For f(x) = √(x + 3) to be defined: x + 3 ≥ 0 → x ≥ -3
  3. For g(f(x)) to be defined: √(x + 3) - 2 ≠ 0 → √(x + 3) ≠ 2 → x + 3 ≠ 4 → x ≠ 1
  4. Combine both conditions: x ≥ -3 AND x ≠ 1
  5. Write in interval notation: [-3, 1) ∪ (1, ∞) The domain is [-3, 1) ∪ (1, ∞).

6 f(x) = √(x + 3) and g(x) = 1/(x² - 9), find the domain of (g∘f)(x)

Hint: For composite functions, the output of the inner function must be valid input for the outer function. Consider restrictions from square roots and denominators.

Show the answer

Answer: (-3, 3) ∪ (3, ∞)

  1. Find (g∘f)(x) = g(f(x)) = g(√(x + 3)) = 1/((√(x + 3))² - 9) = 1/((x + 3) - 9) = 1/(x - 6)
  2. Domain restrictions come from both functions: - From f(x) = √(x + 3): x + 3 ≥ 0 → x ≥ -3 - From g(f(x)) = 1/(x - 6): x - 6 ≠ 0 → x ≠ 6
  3. Combine restrictions: x ≥ -3 AND x ≠ 6
  4. Write in interval notation: [-3, 6) ∪ (6, ∞) The domain is [-3, 6) ∪ (6, ∞).
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