Domain and Range of Composite Functions- Practice Guide
What Are Composite Functions?
A composite function is one function applied to the result of another. If you have f(g(x)), you first calculate g(x), then plug that result into f.
Mathematically, if f: A → B and g: B → C, then the composite fog: A → C maps elements from A through B to C.
The notation matters. (f ∘ g)(x) means f(g(x)). The function on the right gets evaluated first.
Finding the Domain of Composite Functions
The domain isn't just "all x values." You have restrictions from both functions involved.
Step-by-Step Process
- Identify the inner function g(x) and its domain restrictions
- Find what x-values make g(x) produce valid outputs
- Take those outputs and check if they satisfy f(x)'s domain
- The final domain is the intersection of these restrictions
Example 1: f(g(x)) where g(x) = √x and f(x) = 1/x
Working through (f ∘ g)(x) = f(√x) = 1/√x:
- g(x) = √x requires x ≥ 0
- f(x) = 1/x requires the input ≠ 0
- Since √x = 0 only when x = 0, and 1/0 is undefined, exclude x = 0
- Domain: (0, ∞)
Example 2: f(g(x)) where g(x) = x² - 4 and f(x) = √x
Working through (f ∘ g)(x) = f(x² - 4) = √(x² - 4):
- The inner function x² - 4 has no restrictions on x
- The outer function √x requires x ≥ 0
- So we need x² - 4 ≥ 0
- Solving: x² ≥ 4 means x ≤ -2 or x ≥ 2
- Domain: (-∞, -2] ∪ [2, ∞)
Finding the Range of Composite Functions
Range is trickier. You need to track what outputs the inner function can produce, then see how the outer function transforms those outputs.
The Process
- First find the range of g(x) over its domain
- That range becomes the input set for f(x)
- Apply f to every value in g's range
- The resulting set is your composite's range
Example: f(g(x)) where g(x) = 2x + 1 and f(x) = x²
Working through (f ∘ g)(x) = (2x + 1)²:
- g(x) = 2x + 1 produces all real numbers (linear function)
- Range of g: (-∞, ∞)
- f(x) = x² takes all real inputs and outputs only non-negative values
- Since g's range covers all reals, f(x²) covers all x² ≥ 0
- Range: [0, ∞)
Practice Problems with Solutions
Problem 1
Find the domain and range of h(x) = √(x² - 9) where h = f ∘ g, g(x) = x² - 9, f(x) = √x.
Solution:
- Domain: x² - 9 ≥ 0 → (x - 3)(x + 3) ≥ 0 → x ≤ -3 or x ≥ 3
- Range: x² - 9 produces values ≥ 0, and √x produces √(≥ 0) ≥ 0
- Domain: (-∞, -3] ∪ [3, ∞), Range: [0, ∞)
Problem 2
Find the domain and range of h(x) = 1/(x² - 4) where h = f ∘ g, g(x) = x² - 4, f(x) = 1/x.
Solution:
- g(x) = x² - 4: all real numbers, but output = 0 when x = ±2
- f(x) = 1/x: input cannot be 0
- So x cannot be ±2
- Domain: (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)
- Range: x² - 4 produces (-4, ∞), excluding 0
- f(1/y) where y ∈ (-4, ∞), y ≠ 0
- As y → -4⁺, 1/y → -0.25; as y → 0⁻, 1/y → -∞; as y → 0⁺, 1/y → +∞
- Range: (-∞, -0.25] ∪ (0, ∞)
Common Mistakes to Avoid
- Ignoring inner function restrictions: Just because g(x) works for all x doesn't mean the composite does
- Forgetting to check outputs of g: f might reject values that g produces
- Algebra errors when solving inequalities: Test points to verify
- Assuming range equals domain: They are usually different for composites
Domain and Range of Common Function Combinations
| Composite Type | Domain Condition | Range Notes |
|---|---|---|
| f(g(x)) where f(x) = 1/x | g(x) ≠ 0 | Never produces 0 |
| f(g(x)) where f(x) = √x | g(x) ≥ 0 | Always ≥ 0 |
| f(g(x)) where f(x) = ln(x) | g(x) > 0 | All real numbers possible |
| f(g(x)) where f(x) = x² | Same as g(x) | ≥ 0 if g's range ⊆ ℝ |
Getting Started: Quick Checklist
Before solving any composite function problem:
- Identify which function is inner (g) and which is outer (f)
- Write out the composite explicitly: f(g(x)) = ?
- List domain restrictions for g(x) — solve if needed
- List domain restrictions for f(x)
- Combine: find x-values satisfying both
- For range, find g's range first, then apply f's transformation to that set
When One Function Inverts the Other
If you have f(g(x)) where f and g are inverses, the domain and range simplify significantly.
For (f ∘ f⁻¹)(x) = x:
- Domain equals f⁻¹'s domain
- Range equals f⁻¹'s range
- The composite identity function x has no restrictions
For (f⁻¹ ∘ f)(x) = x:
- Domain equals f's domain
- Range equals f's range
- The composite identity function x has no restrictions
This only works when the inner function's outputs stay within the outer function's domain. If g's range exceeds f's domain, the composite isn't defined for those x-values.