Understand (f ∘ g)(x) = f(g(x)), compose functions step by step, find domains, decompose, and evaluate from tables — with fully worked examples.
(f ∘ g)(x) = f(g(x))
Read: "f composed with g of x" or "f of g of x"
When you compose two functions, you apply them in sequence. In (f ∘ g)(x), g is applied first — you plug x into g and get g(x). Then you feed that result into f — giving f(g(x)).
The ∘ symbol (a small circle) means composition. It is not multiplication. The notation (f ∘ g) defines a new function whose rule is "do g, then do f."
Input x
↓ apply g
g(x) [intermediate value]
↓ apply f
f(g(x)) [final output]
Key distinction: (f ∘ g)(x) and (g ∘ f)(x) are different operations. Order matters — composition is not commutative in general.
Substitute g(x) wherever x appears in f
Take the formula for f(x) and replace every x with the entire expression g(x). Use parentheses — this is where errors happen.
Simplify the result
Expand, combine like terms, and simplify the expression. Don't leave unsimplified products or nested powers.
State the domain
Find all x-values for which the composition is defined. x must be in the domain of g, and g(x) must be in the domain of f.
f(x) = 2x + 1, g(x) = x²
Find (f ∘ g)(x):
(f ∘ g)(x) = f(g(x)) = f(x²)
= 2(x²) + 1
= 2x² + 1
Find (g ∘ f)(x):
(g ∘ f)(x) = g(f(x)) = g(2x + 1)
= (2x + 1)²
= 4x² + 4x + 1
= 4x² + 4x + 1
Conclusion: (f ∘ g)(x) ≠ (g ∘ f)(x)
2x² + 1 is not the same function as 4x² + 4x + 1. Composition is NOT commutative in general — order always matters.
f(x) = √x, g(x) = x − 4
Find (f ∘ g)(x):
(f ∘ g)(x) = f(g(x)) = f(x − 4)
= √(x − 4)
= √(x − 4)
Domain:
g(x) = x − 4 has domain: all real numbers
f(x) = √x requires input ≥ 0, so g(x) ≥ 0:
x − 4 ≥ 0 → x ≥ 4
Domain of (f ∘ g): [4, ∞)
f(x) = 1/x, g(x) = x + 2
Find (f ∘ g)(x):
(f ∘ g)(x) = f(g(x)) = f(x + 2)
= 1/(x + 2)
= 1/(x + 2)
Domain:
g(x) = x + 2: domain all reals
f(x) = 1/x: undefined when denominator = 0
g(x) = 0 when x + 2 = 0 → x = −2
Domain: all real x, x ≠ −2
f(x) = x², g(x) = sin(x)
Find (f ∘ g)(x):
(f ∘ g)(x) = f(g(x)) = f(sin x)
= (sin x)²
= sin²(x)
Find (g ∘ f)(x):
(g ∘ f)(x) = g(f(x)) = g(x²)
= sin(x²)
= sin(x²)
sin²(x) ≠ sin(x²)
sin²(x) = (sin x)² — square the output of sine. | sin(x²) — square the input first, then take sine. These are fundamentally different functions.
The Full Rule
The domain of (f ∘ g)(x) is the set of all x such that:
Both conditions must hold simultaneously. Even if x is a perfectly fine input for g, if g(x) cannot be fed into f, that x is excluded.
Find the domain of (f ∘ g)(x) where f(x) = √x and g(x) = x − 4.
Find the domain of g
g(x) = x − 4 is a polynomial → domain is all real numbers: (−∞, ∞)
Find the domain of f
f(x) = √x requires x ≥ 0 → domain of f is [0, ∞)
Require g(x) to be in the domain of f
Need g(x) ≥ 0:
x − 4 ≥ 0
x ≥ 4
Domain of (f ∘ g): [4, ∞)
Decomposition is the reverse: given a composite function h(x), find f and g so that h = f ∘ g. This is an important skill for calculus (chain rule).
Find f and g such that h(x) = (f ∘ g)(x).
Strategy: identify the "inner" and "outer" operations.
Most common decomposition:
g(x) = 2x + 3 (inner — the expression inside the power)
f(x) = x⁵ (outer — raise to the 5th power)
Check: f(g(x)) = f(2x + 3) = (2x + 3)⁵ = h(x) ✓
Alternative valid decomposition:
g(x) = 2x + 3
f(x) = x⁵ ← same as above (most natural)
Note: there are infinitely many valid decompositions (e.g., g(x) = 2x, f(x) = (x+3)⁵), but exams typically expect the most natural split where the outer function operates on the full inner expression.
When functions are defined by tables rather than formulas, use the table values directly — no formula needed.
| x | f(x) | g(x) |
|---|---|---|
| 1 | 4 | 2 |
| 2 | 1 | 5 |
| 3 | 6 | 1 |
| 4 | 3 | 6 |
| 5 | 2 | 3 |
| 6 | 5 | 4 |
Find (f ∘ g)(2) = f(g(2)):
Step 1: g(2) = 5 (from table, row x = 2)
Step 2: f(5) = 2 (from table, row x = 5)
(f ∘ g)(2) = 2
Find (g ∘ f)(3) = g(f(3)):
Step 1: f(3) = 6 (from table, row x = 3)
Step 2: g(6) = 4 (from table, row x = 6)
(g ∘ f)(3) = 4
Find (f ∘ f)(1) = f(f(1)):
Step 1: f(1) = 4 (from table, row x = 1)
Step 2: f(4) = 3 (from table, row x = 4)
(f ∘ f)(1) = 3
Composition extends naturally to three or more functions. Apply from right to left:
(f ∘ g ∘ h)(x) = f(g(h(x)))
Apply h first, then g, then f
Find (f ∘ g ∘ h)(x).
Step 1: h(x) = x²
Step 2: g(h(x)) = g(x²) = 2x²
Step 3: f(g(h(x))) = f(2x²) = 2x² + 1
(f ∘ g ∘ h)(x) = 2x² + 1
Verify with a specific value: let x = 3.
h(3) = 9 → g(9) = 18 → f(18) = 19
Formula: 2(3)² + 1 = 18 + 1 = 19 ✓
In (f ∘ g)(x), g goes first. The symbol closest to x is applied first. Writing it out as f(g(x)) makes the order obvious. When in doubt, expand the notation.
When substituting g(x) into f, always wrap g(x) in parentheses. f(x) = 2x + 1 and g(x) = x − 3: f(g(x)) = 2(x − 3) + 1, NOT 2x − 3 + 1. Skipping parentheses is the #1 arithmetic error.
The domain of f ∘ g requires BOTH that x is in domain(g) AND g(x) is in domain(f). On free-response questions, show both conditions — examiners look for the two-step justification.
(f ∘ g)(x) means f(g(x)) — you apply g first, then apply f to that result. Read it as 'f composed with g of x' or 'f of g of x'. The right-hand function is always applied first.
No. Function composition is generally NOT commutative. In most cases (f ∘ g)(x) ≠ (g ∘ f)(x). For example, with f(x) = 2x + 1 and g(x) = x², (f ∘ g)(x) = 2x² + 1 but (g ∘ f)(x) = (2x+1)² = 4x² + 4x + 1. These are different functions.
The domain of (f ∘ g)(x) = f(g(x)) requires two conditions: (1) x must be in the domain of g, and (2) g(x) must be in the domain of f. Start with the domain of g, then exclude any x-values where g(x) falls outside the domain of f. Example: for f(x) = √x and g(x) = x − 4, the domain of g is all reals, but f requires g(x) ≥ 0, so x − 4 ≥ 0, giving domain x ≥ 4.
Interactive problems with step-by-step solutions and private tutoring — work through composition, domains, and decomposition with instant feedback.
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