This material sits in the Functions: Concept, Properties and Operations category (15% of the exam) and feeds directly into the 30% Representations category — table-based composition and inverse items are exam regulars in both the calculator and no-calculator sections.
Functions combine pointwise: (f + g)(x) = f(x) + g(x), (f·g)(x) = f(x)·g(x), (f/g)(x) = f(x)/g(x), and so on. The domain rule is the same for all of them:
The domain of a combination is the intersection of the two domains — with the extra exclusion, for quotients, of any x where the denominator function is zero.
Example: f(x) = √x (domain x ≥ 0) and g(x) = √(4 − x) (domain x ≤ 4). Then (f + g)(x) = √x + √(4 − x) has domain 0 ≤ x ≤ 4 — both conditions at once, endpoints included, since the square root of 0 is legal.
The composition (f∘g)(x) = f(g(x)) feeds x into g first, then feeds g's output into f. Read from the inside out — g acts first, even though the notation f∘g lists f first. If the circle notation trips you, rewrite it as nested parentheses before doing anything.
From formulas — substitute the entire inner expression for every x in the outer formula:
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!
Order matters: f∘g and g∘f are generally different functions. And when expanding (3x − 2)², keep the cross term: 9x² − 12x + 4, not 9x² + 4.
From tables — chase values one hop at a time:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 3 | 4 | 2 | 1 |
| g(x) | 4 | 1 | 3 | 2 |
f(g(1)): g(1) = 4, then f(4) = 1. Two lookups, no shortcuts. The wrong answers on the exam are pre-built: the reversed order, the halfway stop, and the wrong row all appear among the choices.
From graphs — same value-chasing: read the inner function's output as a height, then use that number as an input position on the outer function's graph.
Numeric check habit: after any symbolic composition, verify at one input. (f∘g)(2) from the formula above: 9(4) − 24 + 5 = 17; directly: g(2) = 4, f(4) = 17 ✓.
Writing h(x) = (2x − 3)⁵ as f(g(x)): the inner function is what happens first — the expression that got wrapped — so g(x) = 2x − 3, and the outer function is the wrapper, f(x) = x⁵. Recompose to check: f(g(x)) = (2x − 3)⁵ ✓. Decompositions aren't unique, but the recompose-and-check test settles any candidate instantly.
A function is one-to-one if every output comes from exactly one input. Graphically: every horizontal line crosses the graph at most once. Practical test: a function that is strictly increasing (or strictly decreasing) on its whole domain is one-to-one.
Parabolas, |x|, and even functions generally fail — two inputs share an output. Why it matters: only one-to-one functions have inverses. (A non-one-to-one function can be rescued by restricting its domain to a piece where it's monotonic — for example, x² restricted to x ≥ 0.)
The inverse f⁻¹ runs f backwards: if f(a) = b, then f⁻¹(b) = a. Three signature facts follow:
[GRAPH: The line y = x drawn dashed. A rising curve f through (0, 1) and (2, 5); its inverse through (1, 0) and (5, 2) — each curve the mirror image of the other across the dashed line, with the point pairs (0,1)/(1,0) and (2,5)/(5,2) marked.]
Finding f⁻¹ analytically: write y = f(x), solve for x in terms of y (undo operations in reverse order), then swap letters:
f(x) = 2x − 5: y = 2x − 5 → x = (y + 5)/2 → f⁻¹(x) = (x + 5)/2
f does "double, then subtract 5"; f⁻¹ does "add 5, then halve" — inverse operations, reverse order.
Verifying by composition: f and g are inverses exactly when f(g(x)) = x on the domain of g and g(f(x)) = x on the domain of f. Check the linear example: 2·((x + 5)/2) − 5 = x ✓. The two-sided requirement has teeth: with f(x) = x² and g(x) = √x, f(g(x)) = x holds for x ≥ 0, but g(f(−2)) = √4 = 2 ≠ −2 — so they are inverses only after restricting f to x ≥ 0.
⚠️ Notation trap: f⁻¹(x) is NOT 1/f(x). The −1 means "inverse function," not an exponent. The reciprocal would be written (f(x))⁻¹.
Inverse values without formulas: "f is one-to-one and f(6) = −2; find f⁻¹(−2)" is a pair-reversal question — the answer is 6, no algebra required. If you're solving equations for items like this, you're working too hard.
A standard CLEP setup: "The functions f and g are defined above. If f(a) = g(a), what is the value of a?" Set the formulas equal and solve:
f(x) = x² − 3, g(x) = 2x + 5, a > 0:
x² − 3 = 2x + 5 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2
With the constraint a > 0, the answer is a = 4. Two habits: factor carefully (sign errors in the factor pair are the planted distractors), and apply any stated constraint to discard the extra root — the discarded root is always among the answer choices. In Section 1 you could also solve this by graphing both functions on the built-in calculator and finding the intersection; in Section 2 the algebra must be by hand.
Questions 1 and 3 refer to f(x) = x² + 1 and g(x) = 3x − 2.
1. B — Correct: inside first — g(2) = 3(2) − 2 = 4, then f(4) = 4² + 1 = 17. A) is the reversed order g(f(2)) = g(5) = 13. C) multiplies f(2)·g(2) = 5·4 = 20, confusing composition with the product. D) adds f(2) + g(2) = 9, confusing composition with the sum. E) stops halfway at f(2) = 5. Fix: rewrite f(g(2)) as nested parentheses and evaluate the innermost first — composition is substitution, not arithmetic on outputs.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 3 | 4 | 2 | 1 |
| g(x) | 4 | 1 | 3 | 2 |
What is the value of f(g(1))?
2. B — Correct: g(1) = 4, then f(4) = 1. A) stops after the first lookup (g(1) = 4). C) reverses the order: g(f(1)) = g(3) = 3. D) reads f(4) from the wrong row (g(4) = 2). E) the table gives every value needed — nothing is undetermined. Fix: finger on the table, two hops: input → inner function's row → that output becomes the input → outer function's row.
3. C — Correct: (f∘g)(x) = (3x − 2)² + 1 = 9x² − 12x + 4 + 1 = 9x² − 12x + 5; check at x = 2: 36 − 24 + 5 = 17 = f(g(2)) ✓. A) is the other order, g(f(x)) = 3x² + 1. B) drops the cross term, expanding (3x − 2)² as 9x² + 4. D) is g∘f with the constant mangled. E) computes the cross term as −4x instead of −2·3x·2 = −12x. Fix: substitute the whole inner expression in parentheses, expand with the cross term, then numeric-check at one point.
4. E — Correct: f needs x ≥ 0, g needs 4 − x ≥ 0 (x ≤ 4); the sum exists where both do: 0 ≤ x ≤ 4. A) uses only f's restriction. B) uses only g's. C) ignores both root conditions. D) wrongly excludes the endpoints — √0 = 0 is defined, so 0 and 4 belong to the domain. Fix: the domain of f + g is the intersection of the separate domains; square roots allow their argument to equal 0.
5. D — Correct: x³ + 2 is strictly increasing everywhere, so it passes the horizontal line test and is one-to-one. A) x² fails: f(−1) = f(1). B) |x − 1| fails: f(0) = f(2). C) x⁴ − x² is even, so f(−x) = f(x) always. E) x² − 6x is a parabola: f(0) = f(6) = 0. Fix: monotonic on the whole domain ⇒ one-to-one; any turnaround (parabola, absolute value, even function) creates two inputs with one output.
6. A — Correct: y = 2x − 5 → x = (y + 5)/2, so f⁻¹(x) = (x + 5)/2 — undo "subtract 5" before undoing "double." Verify: 2·((x+5)/2) − 5 = x ✓. B) undoes the operations in the wrong order (halves before adding 5). C) is the reciprocal 1/f(x) — the notation trap. D) rearranges signs without inverting. E) repeats the forward operations instead of undoing them. Fix: invert by undoing operations in reverse order, then confirm f(f⁻¹(x)) simplifies to x.
7. B — Correct: g(6) = −2 puts (6, −2) on g, so the reversed pair (−2, 6) is on g⁻¹. A) is the point on g itself, unreversed. C) negates both coordinates instead of swapping them. D) swaps and negates. E) takes reciprocals — misreading the −1 exponent. Fix: every inverse-value question is pair reversal: (a, b) on f ⇔ (b, a) on f⁻¹; swap coordinates, change nothing else.
8. D — Correct: inverting swaps the roles — domain of f⁻¹ = range of f = [−3, 2]; range of f⁻¹ = domain of f = [1, 7]. A) leaves domain and range unswapped. B) negates both intervals, confusing inversion with reflection through the origin. C) swaps but then negates the new range. E) flips signs on the new domain — the swap moves the intervals intact, with no sign changes. Fix: for inverses, copy the range down as the new domain and the domain as the new range — verbatim, no arithmetic.
9. A — Correct: the two compositions equaling x mean f and g are inverses, so f(7) = 10 forces g(10) = 7. B) returns the output of f instead of running it backwards. C) treats g as the reciprocal of f's output. D) negates instead of inverting. E) the value is fully determined — inverses reverse every pair. Fix: f(g(x)) = g(f(x)) = x is the definition of inverse functions; from then on, every known pair of f reads backwards as a pair of g.
10. C — Correct: the inner function is what happens first — g(x) = 2x − 3 — and the outer is the wrapper, f(x) = x⁵; recomposing gives (2x − 3)⁵ ✓. A) reverses inner and outer: recomposing gives 2x⁵ − 3 ✗. B) recomposes to 2(x − 3)⁵ ✗. D) recomposes to (2x)⁵ − 3 ✗. E) recomposes to (2(x − 3))⁵ = (2x − 6)⁵ ✗. Fix: always recompose your candidate pair and compare with the original expression — the check takes ten seconds and is decisive.
11. E — Correct: x² − 3 = 2x + 5 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2; the constraint a > 0 selects a = 4. A) is the rejected root — it solves the equation but violates a > 0. B) and C) come from misfactoring as (x + 4)(x − 2), which gives roots 2 and −4 (check: that expands to x² + 2x − 8, wrong middle sign). D) treats the constant −8 as the answer after a sign flip, skipping the factoring. Fix: after solving f(a) = g(a), expand your factors to confirm the middle term, then apply the stated constraint to choose among the roots.
12. A — Correct: inverse functions require both compositions to return x on the appropriate domains, and g(f(−2)) = √((−2)²) = √4 = 2 ≠ −2 shows the pair fails for negative inputs; restricting f to x ≥ 0 repairs it. B) states the one-composition shortcut, which is exactly the flaw in the student's reasoning. C) is false — x³ and ∛x are a polynomial/root inverse pair. D) sharing a point proves nothing about being inverses. E) is false — √x is itself invertible (it's one-to-one). Fix: to certify inverses, check f(g(x)) = x and g(f(x)) = x, watching the domains; one direction alone can silently hide a failed horizontal line test.
1. B — Correct: inside first — g(2) = 3(2) − 2 = 4, then f(4) = 4² + 1 = 17. A) is the reversed order g(f(2)) = g(5) = 13. C) multiplies f(2)·g(2) = 5·4 = 20, confusing composition with the product. D) adds f(2) + g(2) = 9, confusing composition with the sum. E) stops halfway at f(2) = 5. Fix: rewrite f(g(2)) as nested parentheses and evaluate the innermost first — composition is substitution, not arithmetic on outputs.
2. B — Correct: g(1) = 4, then f(4) = 1. A) stops after the first lookup (g(1) = 4). C) reverses the order: g(f(1)) = g(3) = 3. D) reads f(4) from the wrong row (g(4) = 2). E) the table gives every value needed — nothing is undetermined. Fix: finger on the table, two hops: input → inner function's row → that output becomes the input → outer function's row.
3. C — Correct: (f∘g)(x) = (3x − 2)² + 1 = 9x² − 12x + 4 + 1 = 9x² − 12x + 5; check at x = 2: 36 − 24 + 5 = 17 = f(g(2)) ✓. A) is the other order, g(f(x)) = 3x² + 1. B) drops the cross term, expanding (3x − 2)² as 9x² + 4. D) is g∘f with the constant mangled. E) computes the cross term as −4x instead of −2·3x·2 = −12x. Fix: substitute the whole inner expression in parentheses, expand with the cross term, then numeric-check at one point.
4. E — Correct: f needs x ≥ 0, g needs 4 − x ≥ 0 (x ≤ 4); the sum exists where both do: 0 ≤ x ≤ 4. A) uses only f's restriction. B) uses only g's. C) ignores both root conditions. D) wrongly excludes the endpoints — √0 = 0 is defined, so 0 and 4 belong to the domain. Fix: the domain of f + g is the intersection of the separate domains; square roots allow their argument to equal 0.
5. D — Correct: x³ + 2 is strictly increasing everywhere, so it passes the horizontal line test and is one-to-one. A) x² fails: f(−1) = f(1). B) |x − 1| fails: f(0) = f(2). C) x⁴ − x² is even, so f(−x) = f(x) always. E) x² − 6x is a parabola: f(0) = f(6) = 0. Fix: monotonic on the whole domain ⇒ one-to-one; any turnaround (parabola, absolute value, even function) creates two inputs with one output.
6. A — Correct: y = 2x − 5 → x = (y + 5)/2, so f⁻¹(x) = (x + 5)/2 — undo "subtract 5" before undoing "double." Verify: 2·((x+5)/2) − 5 = x ✓. B) undoes the operations in the wrong order (halves before adding 5). C) is the reciprocal 1/f(x) — the notation trap. D) rearranges signs without inverting. E) repeats the forward operations instead of undoing them. Fix: invert by undoing operations in reverse order, then confirm f(f⁻¹(x)) simplifies to x.
7. B — Correct: g(6) = −2 puts (6, −2) on g, so the reversed pair (−2, 6) is on g⁻¹. A) is the point on g itself, unreversed. C) negates both coordinates instead of swapping them. D) swaps and negates. E) takes reciprocals — misreading the −1 exponent. Fix: every inverse-value question is pair reversal: (a, b) on f ⇔ (b, a) on f⁻¹; swap coordinates, change nothing else.
8. D — Correct: inverting swaps the roles — domain of f⁻¹ = range of f = [−3, 2]; range of f⁻¹ = domain of f = [1, 7]. A) leaves domain and range unswapped. B) negates both intervals, confusing inversion with reflection through the origin. C) swaps but then negates the new range. E) flips signs on the new domain — the swap moves the intervals intact, with no sign changes. Fix: for inverses, copy the range down as the new domain and the domain as the new range — verbatim, no arithmetic.
9. A — Correct: the two compositions equaling x mean f and g are inverses, so f(7) = 10 forces g(10) = 7. B) returns the output of f instead of running it backwards. C) treats g as the reciprocal of f's output. D) negates instead of inverting. E) the value is fully determined — inverses reverse every pair. Fix: f(g(x)) = g(f(x)) = x is the definition of inverse functions; from then on, every known pair of f reads backwards as a pair of g.
10. C — Correct: the inner function is what happens first — g(x) = 2x − 3 — and the outer is the wrapper, f(x) = x⁵; recomposing gives (2x − 3)⁵ ✓. A) reverses inner and outer: recomposing gives 2x⁵ − 3 ✗. B) recomposes to 2(x − 3)⁵ ✗. D) recomposes to (2x)⁵ − 3 ✗. E) recomposes to (2(x − 3))⁵ = (2x − 6)⁵ ✗. Fix: always recompose your candidate pair and compare with the original expression — the check takes ten seconds and is decisive.
11. E — Correct: x² − 3 = 2x + 5 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2; the constraint a > 0 selects a = 4. A) is the rejected root — it solves the equation but violates a > 0. B) and C) come from misfactoring as (x + 4)(x − 2), which gives roots 2 and −4 (check: that expands to x² + 2x − 8, wrong middle sign). D) treats the constant −8 as the answer after a sign flip, skipping the factoring. Fix: after solving f(a) = g(a), expand your factors to confirm the middle term, then apply the stated constraint to choose among the roots.
12. A — Correct: inverse functions require both compositions to return x on the appropriate domains, and g(f(−2)) = √((−2)²) = √4 = 2 ≠ −2 shows the pair fails for negative inputs; restricting f to x ≥ 0 repairs it. B) states the one-composition shortcut, which is exactly the flaw in the student's reasoning. C) is false — x³ and ∛x are a polynomial/root inverse pair. D) sharing a point proves nothing about being inverses. E) is false — √x is itself invertible (it's one-to-one). Fix: to certify inverses, check f(g(x)) = x and g(f(x)) = x, watching the domains; one direction alone can silently hide a failed horizontal line test.