The CLEP exam's largest category, Representations of Functions (30%), plus the Functions concept category (15%), both run directly on this lesson. Nearly half the exam asks you to move between a function's formula, its graph, and its table — this lesson builds the vocabulary everything else uses.
A function f assigns each input x exactly one output f(x). "Exactly one" is the entire definition — one input can never produce two outputs. The set of allowed inputs is the domain; the set of outputs actually produced is the range.
Vertical line test: a graph in the xy-plane represents y as a function of x exactly when every vertical line intersects it at most once. A circle like x² + y² = 25 fails (the line x = 0 hits it twice: at y = 5 and y = −5), so a circle's equation does not define y as a function of x. The CLEP exam asks this directly with stems like "Which of the following defines y as a function of x?" — scan each option for any x-value that yields two y-values.
f(a) means "the output of f at input a" — substitute a for every x and compute:
f(x) = 2x² − 3x + 1
f(−2) = 2(−2)² − 3(−2) + 1 = 8 + 6 + 1 = 15
Two slips cause most wrong answers: squaring before applying the coefficient incorrectly (2(−2)² is 2·4 = 8, not (−4)² = 16) and mishandling the sign on the middle term (−3(−2) = +6, not −6). Substitute in parentheses, every time.
Evaluation also happens from tables (find the row, read the output) and from graphs (find x on the horizontal axis, read the height). "What is the value of g(4)?" is one of the most common CLEP stems — it can arrive in any of the three representations.
A piecewise function uses different formulas on different parts of the domain:
f(x) = { x² + 1, x < 2
{ 5x − 3, x ≥ 2
The only skill: check which condition the input satisfies before choosing the formula. f(2) uses the second branch (2 ≥ 2), giving 5(2) − 3 = 7. f(−1) uses the first branch (−1 < 2), giving (−1)² + 1 = 2. Boundary inputs (here x = 2) are the exam's favorite — the inequality signs decide which branch owns the boundary.
For the algebraic functions on this exam, only two things break:
Apply both when both appear. For f(x) = √(x − 4)/(x − 7): the root demands x ≥ 4, the denominator demands x ≠ 7, so the domain is x ≥ 4 with x ≠ 7. Missing the second condition (7 sits inside x ≥ 4, so it genuinely must be carved out) is the standard error.
[GRAPH: A continuous curve on the interval −3 ≤ x ≤ 5. It starts at its highest point (−3, 4), descends to its lowest point (1, −2), then rises to end at (5, 3). Domain shown as the segment [−3, 5] on the x-axis; range shown as the segment [−2, 4] on the y-axis.]
For this graph: domain [−3, 5], range [−2, 4]. Keep the two axes straight — mixing a domain endpoint into the range is the trap the answer choices are built around.
Over an interval of the domain:
Always report these as intervals of x-values, never y-values. "f is increasing on (−1, 2)" refers to inputs between −1 and 2.
A local maximum is a point where the graph turns from rising to falling; a local minimum turns falling to rising. The absolute maximum/minimum on an interval is the largest/smallest value attained anywhere on it — and on a closed interval, an endpoint can beat every interior turning point. When a question asks for "the maximum value of g on [−5, 5]," compare the local peaks and the endpoint values before answering. (In Section 1 of the exam, the online graphing calculator will locate maxima and minima for you; in Section 2 you read them from the given figure.)
| Question asked | What to report |
|---|---|
| Where is f increasing? | interval(s) of x |
| What is the maximum value? | the y-value attained |
| Where does the maximum occur? | the x-value |
The last two rows are different questions with different answers — read the stem's noun carefully.
Two symmetry types, tested algebraically:
Most functions are neither: x³ + 1 fails both tests (the constant term is even-type, the cube is odd-type). Don't eyeball — compute f(−x) and compare.
Even/odd symmetry is also a computation shortcut: if f is even, then f(−3) = f(3) automatically, no formula needed. If f is odd and defined at 0, then f(0) = 0.
A function is periodic with period p if f(x + p) = f(x) for every x — the graph repeats every p units horizontally. Consequence: f(a) = f(a + p) = f(a + 2p) = f(a − p) = … — any two inputs that differ by a multiple of the period share an output.
If f has period 6 and f(2) = −3, then f(20) = −3 because 20 − 2 = 18 = 3·6. But f(12) is unknown: 12 − 2 = 10 is not a multiple of 6. The test is always "do the inputs differ by a multiple of p?" — not whether the input is itself a multiple of p. Periodicity is the defining property of the trigonometric functions later in the course, and the fact sheet's warning applies: trig thinking permeates far more of this exam than its 15% category suggests.
1. B — Correct: f(−2) = 2(−2)² − 3(−2) + 1 = 8 + 6 + 1 = 15. A) computes −3(−2) as −6 (sign slip on the middle term: 8 − 6 + 1 = 3). C) applies the negative to the squared term, −8 + 6 + 1 = −1. D) reads 2x² as (2x)², giving 16 + 6 + 1 = 23. E) drops the constant term (8 + 6 = 14). Fix: substitute inside parentheses — write f(−2) = 2(−2)² − 3(−2) + 1 before computing anything.
2. B — Correct: y = x² gives exactly one y for each x. A) x = y² gives two y-values for x = 4 (y = ±2). C) the circle fails the vertical line test at x = 0. D) x = |y| pairs x = 3 with y = 3 and y = −3. E) y² = x + 1 gives y = ±√(x+1), two outputs. Fix: test one x-value in each relation — if any x produces two y's, it's not a function of x.
3. E — Correct: the square root requires x − 4 ≥ 0 (x ≥ 4) and the denominator requires x ≠ 7; since 7 lies inside x ≥ 4, both conditions bind. A) applies only the root condition and forgets the denominator. B) wrongly excludes 4 (the root of 0 is fine — only the denominator can't be 0). C) applies only the denominator condition. D) confuses which number comes from which restriction. Fix: list every restriction (roots ≥ 0, denominators ≠ 0) separately, then intersect them.
4. A — Correct: f(2) uses the second branch since 2 ≥ 2, giving 5(2) − 3 = 7; f(−1) uses the first branch since −1 < 2, giving (−1)² + 1 = 2; sum = 9. B) evaluates f(2) with the wrong branch (2² + 1 = 5) and gets 5 + 2 = 7. C) evaluates f(−1) with the wrong branch (5(−1) − 3 = −8): 7 + (−8) = −1. D) uses the wrong branch both times: 5 + (−8) = −3. E) evaluates f(2) with both branches (7 and 5) and adds those, never reaching f(−1) — the "boundary input belongs to both pieces" error. Fix: before substituting, write down which inequality the input satisfies; the boundary value belongs to the branch whose inequality includes equality.
5. E — Correct: the range is the set of heights attained — from the lowest point y = −2 up to the highest point y = 4: [−2, 4]. A) is the domain. B) mixes the x-coordinate of the minimum with the maximum height. C) and D) each mix one domain endpoint with one range endpoint. Fix: range questions are answered entirely on the y-axis — project the graph sideways and read the interval of heights.
6. D — Correct: f rises exactly on the piece from (−1, −2) to (2, 5), so f is increasing on (−1, 2). A) is where f is decreasing (the first falling piece). B) reads the y-coordinates (−2 to 5) instead of x-coordinates. C) is the second decreasing piece. E) merges a falling piece with the rising piece. Fix: increasing/decreasing intervals are always reported in x-values — trace left to right and record where the graph goes uphill.
7. C — Correct: for f(x) = x³ − 5x, f(−x) = −x³ + 5x = −f(x); all exponents are odd. A) x³ + 1 is neither — the constant term (even-type) breaks odd symmetry. B) x² − 4 is even, not odd. D) |x| is even (|−x| = |x|). E) x³ + x² mixes parities, so it is neither. Fix: compute f(−x) in full and compare to f(x) and −f(x); for polynomials, "all exponents odd" = odd, "all even (including constants)" = even, mixed = neither.
8. D — Correct: the maximum value on a closed interval is the largest value attained anywhere, and the endpoint value g(5) = 6 exceeds the local maximum value 3. A) stops at the local maximum without checking endpoints. B) is the local minimum value. C) is the x-coordinate of the local maximum, not a value of g. E) is the x-coordinate where the maximum occurs, not the maximum value. Fix: on a closed interval, compare all turning-point values and both endpoint values — and answer with the y-value when asked for "the maximum value."
9. A — Correct: evenness gives f(−3) = f(3) = 7 and f(5) = f(−5) = 2, so the sum is 9. B) applies the odd-function rule f(−x) = −f(x), getting −7 + (−2). C) subtracts instead of adding (7 − 2). D) assumes even functions vanish at reflected points. E) applies the odd rule to only one term. Fix: even means outputs match at ±x (f(−a) = f(a)); odd means they flip sign — identify which symmetry you have before moving any signs.
10. C — Correct: f(20) = f(2) because 20 − 2 = 18 = 3·6, a multiple of the period. A) f(6): 6 − 2 = 4 is not a multiple of 6 — the input being (near) the period doesn't matter; the difference of inputs does. B) f(12): 12 − 2 = 10, not a multiple of 6 — "multiples of the period" applies to input differences, not to inputs. D) f(−2) would need f to be even, which isn't given. E) f(3): 3 − 2 = 1, unrelated. Fix: for period p, f(a) = f(b) is guaranteed exactly when b − a is a multiple of p — always subtract the inputs and divide by the period.
11. B — Correct: the denominator x² − 9 = (x − 3)(x + 3) is zero at x = ±3; those are the only exclusions. A) excludes the numerator's zero — a zero numerator is allowed (it just makes g(x) = 0). C) solves x² = 9 as x = 9. D) wrongly adds the numerator's zero to the genuine exclusions. E) ignores the denominator entirely. Fix: domain exclusions come only from denominators (and even roots) — set each denominator equal to zero, solve fully, and exclude exactly those values.
12. C — Correct: a graph fails to define y as a function of x exactly when some vertical line hits it more than once, and x = 0 meets the circle at (0, 4) and (0, −4). A) cites the horizontal line test, which detects one-to-one functions (invertibility), not function-ness. B) functions need not be lines — parabolas are functions. D) passing through the origin is irrelevant to the definition. E) solvability for x is about x as a function of y, the reverse question. Fix: "is y a function of x?" is always answered by the vertical line test — one x, at most one y.
1. B — Correct: f(−2) = 2(−2)² − 3(−2) + 1 = 8 + 6 + 1 = 15. A) computes −3(−2) as −6 (sign slip on the middle term: 8 − 6 + 1 = 3). C) applies the negative to the squared term, −8 + 6 + 1 = −1. D) reads 2x² as (2x)², giving 16 + 6 + 1 = 23. E) drops the constant term (8 + 6 = 14). Fix: substitute inside parentheses — write f(−2) = 2(−2)² − 3(−2) + 1 before computing anything.
2. B — Correct: y = x² gives exactly one y for each x. A) x = y² gives two y-values for x = 4 (y = ±2). C) the circle fails the vertical line test at x = 0. D) x = |y| pairs x = 3 with y = 3 and y = −3. E) y² = x + 1 gives y = ±√(x+1), two outputs. Fix: test one x-value in each relation — if any x produces two y's, it's not a function of x.
3. E — Correct: the square root requires x − 4 ≥ 0 (x ≥ 4) and the denominator requires x ≠ 7; since 7 lies inside x ≥ 4, both conditions bind. A) applies only the root condition and forgets the denominator. B) wrongly excludes 4 (the root of 0 is fine — only the denominator can't be 0). C) applies only the denominator condition. D) confuses which number comes from which restriction. Fix: list every restriction (roots ≥ 0, denominators ≠ 0) separately, then intersect them.
4. A — Correct: f(2) uses the second branch since 2 ≥ 2, giving 5(2) − 3 = 7; f(−1) uses the first branch since −1 < 2, giving (−1)² + 1 = 2; sum = 9. B) evaluates f(2) with the wrong branch (2² + 1 = 5) and gets 5 + 2 = 7. C) evaluates f(−1) with the wrong branch (5(−1) − 3 = −8): 7 + (−8) = −1. D) uses the wrong branch both times: 5 + (−8) = −3. E) evaluates f(2) with both branches (7 and 5) and adds those, never reaching f(−1) — the "boundary input belongs to both pieces" error. Fix: before substituting, write down which inequality the input satisfies; the boundary value belongs to the branch whose inequality includes equality.
5. E — Correct: the range is the set of heights attained — from the lowest point y = −2 up to the highest point y = 4: [−2, 4]. A) is the domain. B) mixes the x-coordinate of the minimum with the maximum height. C) and D) each mix one domain endpoint with one range endpoint. Fix: range questions are answered entirely on the y-axis — project the graph sideways and read the interval of heights.
6. D — Correct: f rises exactly on the piece from (−1, −2) to (2, 5), so f is increasing on (−1, 2). A) is where f is decreasing (the first falling piece). B) reads the y-coordinates (−2 to 5) instead of x-coordinates. C) is the second decreasing piece. E) merges a falling piece with the rising piece. Fix: increasing/decreasing intervals are always reported in x-values — trace left to right and record where the graph goes uphill.
7. C — Correct: for f(x) = x³ − 5x, f(−x) = −x³ + 5x = −f(x); all exponents are odd. A) x³ + 1 is neither — the constant term (even-type) breaks odd symmetry. B) x² − 4 is even, not odd. D) |x| is even (|−x| = |x|). E) x³ + x² mixes parities, so it is neither. Fix: compute f(−x) in full and compare to f(x) and −f(x); for polynomials, "all exponents odd" = odd, "all even (including constants)" = even, mixed = neither.
8. D — Correct: the maximum value on a closed interval is the largest value attained anywhere, and the endpoint value g(5) = 6 exceeds the local maximum value 3. A) stops at the local maximum without checking endpoints. B) is the local minimum value. C) is the x-coordinate of the local maximum, not a value of g. E) is the x-coordinate where the maximum occurs, not the maximum value. Fix: on a closed interval, compare all turning-point values and both endpoint values — and answer with the y-value when asked for "the maximum value."
9. A — Correct: evenness gives f(−3) = f(3) = 7 and f(5) = f(−5) = 2, so the sum is 9. B) applies the odd-function rule f(−x) = −f(x), getting −7 + (−2). C) subtracts instead of adding (7 − 2). D) assumes even functions vanish at reflected points. E) applies the odd rule to only one term. Fix: even means outputs match at ±x (f(−a) = f(a)); odd means they flip sign — identify which symmetry you have before moving any signs.
10. C — Correct: f(20) = f(2) because 20 − 2 = 18 = 3·6, a multiple of the period. A) f(6): 6 − 2 = 4 is not a multiple of 6 — the input being (near) the period doesn't matter; the difference of inputs does. B) f(12): 12 − 2 = 10, not a multiple of 6 — "multiples of the period" applies to input differences, not to inputs. D) f(−2) would need f to be even, which isn't given. E) f(3): 3 − 2 = 1, unrelated. Fix: for period p, f(a) = f(b) is guaranteed exactly when b − a is a multiple of p — always subtract the inputs and divide by the period.
11. B — Correct: the denominator x² − 9 = (x − 3)(x + 3) is zero at x = ±3; those are the only exclusions. A) excludes the numerator's zero — a zero numerator is allowed (it just makes g(x) = 0). C) solves x² = 9 as x = 9. D) wrongly adds the numerator's zero to the genuine exclusions. E) ignores the denominator entirely. Fix: domain exclusions come only from denominators (and even roots) — set each denominator equal to zero, solve fully, and exclude exactly those values.
12. C — Correct: a graph fails to define y as a function of x exactly when some vertical line hits it more than once, and x = 0 meets the circle at (0, 4) and (0, −4). A) cites the horizontal line test, which detects one-to-one functions (invertibility), not function-ness. B) functions need not be lines — parabolas are functions. D) passing through the origin is irrelevant to the definition. E) solvability for x is about x as a function of y, the reverse question. Fix: "is y a function of x?" is always answered by the vertical line test — one x, at most one y.