f(g(x)) from formulas, tables, and graphs — inside first, always(f∘g)(x) symbolically and simplifyA store discounts a $80 jacket by 25%, and you have a $10-off coupon. Two functions:
d(p) = 0.75p (discount) c(p) = p − 10 (coupon)
Coupon after discount: c(d(80)) = c(60) = $50. Discount after coupon: d(c(80)) = d(70) = $52.50.
Same two operations, different order, $2.50 apart (and the store will happily pick its favorite). Function composition is order-sensitive machinery — f∘g and g∘f are usually different functions, and the exam checks whether you respect that.
The composition (f∘g)(x) = f(g(x)) feeds x into g, then feeds g's output into f.
f∘g means "f after g" — the circle notation lists them in the opposite order from execution. If that trips you, always rewrite (f∘g)(x) as f(g(x)) before doing anything.Formulas: substitute the entire inner function for every x in the outer.
f(x) = x² + 1, g(x) = 3x − 2:
(f∘g)(x) = f(3x − 2) = (3x − 2)² + 1 = 9x² − 12x + 5
(g∘f)(x) = g(x² + 1) = 3(x² + 1) − 2 = 3x² + 1 ← different!
Tables: chase the values, one hop at a time.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 3 | 1 | 4 | 2 |
| g(x) | 2 | 4 | 1 | 3 |
f(g(3)): g(3) = 1, then f(1) = 3. Two lookups; no shortcuts exist.
Graphs: read the inner function's output on the y-axis, carry that number to the x-axis of the outer graph. (Same value-chasing as tables, with reading steps.)
x is in the domain of f∘g when: (1) x is in g's domain, and (2) g(x) is in f's domain. Condition (2) is the one students forget. Example: f(x) = 1/x, g(x) = x − 3: f∘g = 1/(x − 3) needs g(x) ≠ 0 → x ≠ 3.
Sneakier: f(x) = √x, g(x) = x − 3 → domain of f∘g is x ≥ 3, inherited from requiring g(x) ≥ 0 — even though the simplified formula √(x−3) shows it, always derive it from the pieces (simplification sometimes hides a restriction).
id(x) = x is composition's "do nothing" element: f∘id = f and id∘f = f. It matters next lesson: f and its inverse will be defined by composing to the identity: f(f⁻¹(x)) = x.
Writing h(x) = √(2x + 7) as f(g(x)) with g(x) = 2x + 7 (inner: what's done first) and f(x) = √x (outer: what's done last). Spot the inner function as "the expression that got wrapped." Decomposition is how you recognize transformations (Lesson 8 was compositions with linear functions all along: a·f(b(x−c))+d) and how you'll unravel equations in Unit 3.
Decompositions aren't unique — h(x) = (x+1)³ − 4(x+1) splits nicely as g(x) = x + 1 inner with f(u) = u³ − 4u outer — any split that recomposes correctly is valid (except the trivial f = h, g = id).
Problem: f(x) = x² + 1, g(x) = 3x − 2. Find f(g(2)) and g(f(2)).
Solution: f(g(2)): g(2) = 4, f(4) = 17. → 17 g(f(2)): f(2) = 5, g(5) = 13. → 13
Interpretation: 17 ≠ 13: order matters even at a single point. Inside first, every time.
Problem: With the same f and g, find and simplify (f∘g)(x), and verify at x = 2.
Solution:
(f∘g)(x) = (3x − 2)² + 1 = 9x² − 12x + 4 + 1 = 9x² − 12x + 5
Verify: 9(4) − 12(2) + 5 = 36 − 24 + 5 = 17 = f(g(2)) ✓
Interpretation: The whole inner expression replaces x — parentheses non-negotiable. The numeric check catches expansion slips (here, forgetting the −12x cross term is the classic one).
Problem: f(x) = √x, g(x) = 5 − x². Find (f∘g)(x) and its domain.
Solution: (f∘g)(x) = √(5 − x²). Need g(x) ≥ 0: 5 − x² ≥ 0 → x² ≤ 5 → −√5 ≤ x ≤ √5.
Interpretation: The domain requirement flows from the outer function's needs applied to the inner function's outputs. Write the inequality about g(x), then solve for x.
Problem: The radius of a circular oil spill grows with time: r(t) = 3t meters (t in hours). Area of a circle: A(r) = πr². Find (A∘r)(t), state what it computes, and evaluate at t = 4.
Solution: (A∘r)(t) = A(3t) = π(3t)² = 9πt² — the spill's area as a function of time directly, skipping the radius middleman. At t = 4: 9π(16) = 144π ≈ 452.4 m².
Interpretation: Composition chains models: time → radius → area becomes time → area. FRQ modeling scenarios use exactly this "units chain" — check that the inner function's output units match the outer function's input units (meters → meters ✓).
Problems 1–3 use f(x) = x² + 1 and g(x) = 3x − 2.
1. (A). g(2) = 4; f(4) = 17. (B) is the reversed order g(f(2)); (D) squares before applying g.
2. (B). f(2) = 5; g(5) = 13. (A) is f(g(2)).
3. (C). (3x−2)² + 1 = 9x² − 12x + 5. (B) drops the cross term; (A)/(D) are g∘f.
4. (D). Inner = what happens first = 2x + 7; outer = the wrap = √. (A) reverses the order.
Problems 5–6 use the table:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 3 | 1 | 4 | 2 |
| g(x) | 2 | 4 | 1 | 3 |
5. (D). g(3) = 1 → f(1) = 3. (A) stops after the first lookup.
6. (A). f(1) = 3 → g(3) = 1. (B) stops halfway.
7. (B). The identity passes x through untouched, so h∘id = h. (A) is id itself; (D) composes h twice.
8. (C). Need g(x) = x − 3 ≠ 0 → x ≠ 3. (A) applies f's restriction to x instead of to g(x).
9. (D). Composition is order-sensitive (warm-up: $50 vs. $52.50). (B) confuses composition with multiplication.
10. (B). Inner g(x) = x + 1 (the repeated block), outer f(u) = u³ − 4u. Recompose to check: (x+1)³ − 4(x+1) ✓. (C) reverses; recomposing gives (x³ − 4x) + 1 ✗.
11. (C). g(−3) = 9; f(9) = 2⁹ = 512. (D) is f(−3) then g — reversed order ((1/8)² = 1/64); (B) invents a sign.
12. (FRQ-style, 6 points) (i) [2 pts] (V∘r)(t) = (4/3)π(2 + 0.5t)³ — the balloon's volume in cm³ as a function of time in seconds. (ii) [2 pts] r(4) = 4; V = (4/3)π(64) = 256π/3 cm³. (iii) [2 pts] r expects a time input (seconds) but V outputs a volume (cm³); feeding cm³ into r(t) = 2 + 0.5t treats a volume as a time — the units chain breaks, so r∘V has no contextual meaning. (Compositions are only meaningful when the inner function's output units match the outer's input units.)
12. (FRQ-style) 🚫 A balloon's radius t seconds after inflation begins is r(t) = 2 + 0.5t cm, and its volume is V(r) = (4/3)πr³.
(i) Find (V∘r)(t) and state what it represents, including units.
(ii) Evaluate (V∘r)(4) exactly (leave π in the answer).
(iii) Explain why (r∘V)(t) is not meaningful in this context, referring to units.
The functions f and g are defined for all real numbers. Selected values:
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| f(x) | 5 | 2 | 1 | 2 | 5 |
| g(x) | 1 | 0 | −1 | −2 | 0 |
(a) (i) Evaluate (f∘g)(2) and (g∘f)(−1). (ii) A student concludes from these two values that f∘g = g∘f. Explain why the student's reasoning is invalid even if the two values had been equal.
(b) The table for f is consistent with f being an even function. (i) State what f being even would imply about f(g(x₁)) and f(g(x₂)) whenever g(x₁) and g(x₂) are opposite values. (ii) Verify this prediction with the table using x₁ = −2 and x₂ = 0.
(c) Let h(x) = f(x − 1) + 3. Express h as a composition involving f and linear functions, identifying the inner and outer pieces.
(a) [2 pts] (i) [1 pt] (f∘g)(2): g(2) = 0, f(0) = 1. (g∘f)(−1): f(−1) = 2, g(2) = 0. (ii) [1 pt] Equality of two functions requires agreement at every input; matching at one point (or any finite sample) proves nothing. A single counterexample disproves equality, but examples can never prove it.
(b) [2 pts] (i) [1 pt] If f is even, then f(−u) = f(u) for every u — so whenever two inputs give g-outputs that are opposites (g(x₁) = −g(x₂)), the compositions agree: f(g(x₁)) = f(g(x₂)). (ii) [1 pt] g(−2) = 1 and g(0) = −1 are opposites; f(g(−2)) = f(1) = 2 and f(g(0)) = f(−1) = 2 — equal ✓, as evenness predicts.
(c) [2 pts] h(x) = (v∘f∘u)(x) where u(x) = x − 1 (inner: shift input), then f, then v(x) = x + 3 (outer: shift output). [1 pt inner placement, 1 pt outer placement] (Lesson 8's transformations are exactly compositions with linear functions on either side of f.)
1. (A). g(2) = 4; f(4) = 17. (B) is the reversed order g(f(2)); (D) squares before applying g.
2. (B). f(2) = 5; g(5) = 13. (A) is f(g(2)).
3. (C). (3x−2)² + 1 = 9x² − 12x + 5. (B) drops the cross term; (A)/(D) are g∘f.
4. (D). Inner = what happens first = 2x + 7; outer = the wrap = √. (A) reverses the order.
5. (D). g(3) = 1 → f(1) = 3. (A) stops after the first lookup.
6. (A). f(1) = 3 → g(3) = 1. (B) stops halfway.
7. (B). The identity passes x through untouched, so h∘id = h. (A) is id itself; (D) composes h twice.
8. (C). Need g(x) = x − 3 ≠ 0 → x ≠ 3. (A) applies f's restriction to x instead of to g(x).
9. (D). Composition is order-sensitive (warm-up: $50 vs. $52.50). (B) confuses composition with multiplication.
10. (B). Inner g(x) = x + 1 (the repeated block), outer f(u) = u³ − 4u. Recompose to check: (x+1)³ − 4(x+1) ✓. (C) reverses; recomposing gives (x³ − 4x) + 1 ✗.
11. (C). g(−3) = 9; f(9) = 2⁹ = 512. (D) is f(−3) then g — reversed order ((1/8)² = 1/64); (B) invents a sign.
12. (FRQ-style, 6 points) (i) [2 pts] (V∘r)(t) = (4/3)π(2 + 0.5t)³ — the balloon's volume in cm³ as a function of time in seconds. (ii) [2 pts] r(4) = 4; V = (4/3)π(64) = 256π/3 cm³. (iii) [2 pts] r expects a time input (seconds) but V outputs a volume (cm³); feeding cm³ into r(t) = 2 + 0.5t treats a volume as a time — the units chain breaks, so r∘V has no contextual meaning. (Compositions are only meaningful when the inner function's output units match the outer's input units.)
🎯 Exam tip: On table-composition items, the traps are pre-built: the reversed order, the halfway stop, and the value from the wrong row all appear among the choices. Move your finger: input → inner row → result becomes new input → outer row. Mechanical beats clever here.