Learning Objectives
Evaluate a composition in the correct order and determine its domain.
Find the inverse of a one-to-one function.
Restrict a quadratic domain so its inverse is a function.
Prerequisites: function notation, domain and range, and solving equations. A composition applies one rule after another. An inverse reverses a one-to-one input-output pairing.
1. Composition Works from the Inside Out
The notation (f ∘ g)(x) means f(g(x)). First evaluate g using x, then use that result as the input to f. The order generally matters: f(g(x)) and g(f(x)) can be different.
Worked Example: Compare Orders
Let f(x) = 2x + 1 and g(x) = x². Then f(g(x)) = 2x² + 1.
In the other order, g(f(x)) = (2x + 1)² = 4x² + 4x + 1.
At x = 2, the first composition equals 9 and the second equals 25. The two operations do not commute.
2. Composition Has Two Domain Checks
For f(g(x)) to be defined, x must lie in the domain of g, and g(x) must lie in the domain of f. An algebraic simplification does not erase either original restriction.
Worked Example: A Composite Domain
Let f(u) = √u and g(x) = 3 − x. The inner function accepts every real x, but the outer function requires its input to be nonnegative.
Therefore 3 − x ≥ 0, so the composition f(g(x)) = √(3 − x) has domain x ≤ 3.
In contrast, g(f(x)) = 3 − √x has domain x ≥ 0. Both the formulas and the domains depend on order.
3. An Inverse Undoes a One-to-One Rule
A one-to-one function never sends two different inputs to the same output. Its inverse f⁻¹ reverses each pairing. The inverse's domain is the original function's range, and the inverse's range is the original domain. The superscript −1 here does not mean reciprocal.
Worked Example: Invert a Linear Function
For f(x) = 3x − 5, write y = 3x − 5 and solve for x: x = (y + 5)/3.
Interchange the variable names to obtain f⁻¹(x) = (x + 5)/3.
Check: f(f⁻¹(x)) = 3[(x + 5)/3] − 5 = x. The reverse composition also equals x for every real x.
4. Restrict a Domain before Inverting
The function f(x) = x² on all real inputs is not one-to-one, because x and −x give the same square. If its domain is restricted to x ≥ 0, it becomes one-to-one and its inverse is f⁻¹(x) = √x, with inverse domain x ≥ 0.
If the original domain is instead x ≤ 0, the inverse is −√x. The sign depends on which original branch must be recovered. Writing ±√x would assign two outputs to most positive inputs and would not define an inverse function.
Common Mistakes
Read a composition from the inside outward.
Do not confuse f⁻¹(x) with 1/f(x).
Choose a domain restriction before deciding which square-root branch is the inverse.
5. Practice
1. For f(x) = x − 4 and g(x) = 3x, find f(g(2)) and g(f(2)).
2. Find the inverse of f(x) = (x + 2)/5.
3. Find the domain of f(g(x)) when f(u) = 1/u and g(x) = x² − 9.
4. Find the inverse of f(x) = (x − 1)² on x ≥ 1, including its domain.
Worked Solutions
1. f(g(2)) = f(6) = 2. In the other order, g(f(2)) = g(−2) = −6.
2. From y = (x + 2)/5, solve x = 5y − 2. Hence f⁻¹(x) = 5x − 2.
3. The inner function is defined for all real x, but its value cannot be zero. Exclude x = −3 and x = 3.
4. Since the original input is at least 1, x − 1 is nonnegative. The inverse is f⁻¹(x) = 1 + √x, with domain x ≥ 0 and range x ≥ 1.
6. Summary
- Composition feeds an inner output into an outer function.
- Both component domains constrain a composition.
- A function must be one-to-one on its chosen domain to have an inverse function.
- Check an inverse by composing and by comparing domains and ranges.