A cup of coffee comes off the machine at 190°F and sits in a 70°F room. Two true statements:
Statement 1 is about the direction of change. Statement 2 is about the rate of change — and how that rate itself changes. AP Precalculus lives in the gap between those two sentences. Most students can say "it goes down." The exam pays you for saying "it decreases at a decreasing rate." This lesson builds that language.
A function f maps each input x to exactly one output f(x). When we say two quantities "change in tandem," we're describing how the output responds as the input increases. Four representations of the same relationship:
T(t) = 70 + 120·(0.93)^t (you'll build models like this in Unit 2)The exam constantly asks you to translate between these. If you learn one habit from Lesson 1, make it this: when you see a graph, say a sentence; when you read a sentence, sketch a graph.
Over an interval of its domain:
f is increasing if, whenever a < b, f(a) < f(b) — outputs rise as inputs rise.f is decreasing if, whenever a < b, f(a) > f(b).Report these as intervals of the input (x-values), never y-values. "f is increasing on (−2, 1)" means for x between −2 and 1.
This is the concept that separates AP Precalc from Algebra 2.
Direction and concavity are independent, giving four behaviors:
| Behavior | Rate of change | Shape | Example context |
|---|---|---|---|
| Increasing, concave up | positive, increasing | rises, steepening | viral video views taking off |
| Increasing, concave down | positive, decreasing | rises, leveling off | height of a filling bathtub as it widens |
| Decreasing, concave up | negative, increasing (toward 0) | falls, flattening | cooling coffee |
| Decreasing, concave down | negative, decreasing | falls, steepening | ball's height just after its peak |
[GRAPH: Four mini-panels in a 2×2 grid, each showing one behavior. Top-left: increasing & concave up (curve rising steeply like the right half of a parabola). Top-right: increasing & concave down (curve rising but flattening, like √x). Bottom-left: decreasing & concave up (curve falling and flattening, like the coffee-cooling curve). Bottom-right: decreasing & concave down (curve falling and steepening, like the left of a downward parabola's descent). Each panel labeled with its behavior pair.]
The graders' phrasing — memorize these sentence frames:
f is an input c with f(c) = 0. Graphically: an x-intercept at (c, 0).(0, f(0)) — a function has at most one (vertical line test).The average rate of change of f over the interval [a, b] is
AROC = ( f(b) − f(a) ) / ( b − a )
It is the slope of the secant line through (a, f(a)) and (b, f(b)). Three things the exam wants you to know cold:
Rate of change at a point. A curve doesn't have a single slope, but we can estimate the rate of change at a point by computing the AROC over a tiny interval containing that point. (Next year, calculus makes this exact and calls it the derivative. This year, we estimate.)
Connecting AROC to concavity: on a concave-up stretch, AROCs over successive equal-width intervals increase; on a concave-down stretch, they decrease. This is how you diagnose concavity from a table — no graph needed.
Problem: The graph of g falls from (−3, 4) to a low point at (0, −1), then rises through (2, 3). The left piece falls and flattens; the right piece rises and steepens. Describe where g is increasing/decreasing and its concavity.
Strategy: Direction first, then shape.
Solution: g is decreasing on (−3, 0) and increasing on (0, 2). "Falls and flattens" = concave up; "rises and steepens" = also concave up. So g is concave up on the whole interval (−3, 2), with a minimum at x = 0.
Interpretation: One concavity can contain both decreasing and increasing behavior — the bottom of a cup does exactly that.
Problem: The height of a model rocket is h(t) = −5t² + 40t meters, t in seconds. Find the average rate of change of height over [1, 3] and interpret it.
Strategy: Evaluate at both endpoints, then slope.
Solution:
h(1) = −5(1) + 40(1) = 35
h(3) = −5(9) + 40(3) = −45 + 120 = 75
AROC = (75 − 35) / (3 − 1) = 40 / 2 = 20
Interpretation: Between 1 and 3 seconds, the rocket's height increases by an average of 20 meters per second. (It's not moving at 20 m/s the whole time — faster near t = 1, slower near t = 3.)
Problem: Selected values of f:
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| f(x) | 3 | 11 | 15 | 17 |
Is f increasing or decreasing? What does the table suggest about concavity?
Strategy: Compute AROC on each successive interval.
Solution:
[0,2]: (11−3)/2 = 4
[2,4]: (15−11)/2 = 2
[4,6]: (17−15)/2 = 1
Outputs rise, so f is increasing. The rates 4 → 2 → 1 are decreasing, so the table is consistent with f being concave down: increasing at a decreasing rate.
Interpretation: "Consistent with" is the honest phrasing — a table can't prove concavity between sample points, and the AP rewards that precision.
Problem: For f(x) = x², estimate the rate of change of f at x = 3 using the interval [2.9, 3.1].
Solution:
f(3.1) = 9.61 f(2.9) = 8.41
AROC = (9.61 − 8.41)/(3.1 − 2.9) = 1.20 / 0.2 = 6
The rate of change of f at x = 3 is approximately 6.
Interpretation: A symmetric interval around the point gives a good estimate (here it's exactly the calculus answer, a happy accident of parabolas). The skill: rate at a point ≈ AROC over a small surrounding interval.
(f(b)−f(a))/(b−a) — endpoints only, one subtraction over one subtraction. Don't average f(a) and f(b), and don't average several table rates unless asked.(Answers in section (g). Letters checked for A–D spread.)
f is decreasing and concave down on an interval. Which description fits?1. (B). Decreasing = falling. Concave down = rate of change decreasing = the fall steepens. (A) is decreasing/concave up (coffee!); (C) increasing/concave down; (D) increasing/concave up.
f(x) = x² − 3x over [1, 4] is2. (C). f(4) = 16 − 12 = 4; f(1) = 1 − 3 = −2. AROC = (4 − (−2))/(4 − 1) = 6/3 = 2. (A) uses (f(4)−f(1))/(4+1)... a made-up formula; (B) is f(4)−f(1) ÷ wrong width; (D) is f(4) alone.
g has zeros at x = −2 and x = 5 only, with g(0) > 0 and g continuous. Which must be true?3. (A). Continuous with zeros only at −2 and 5, and positive at 0 (inside the interval) → it can't cross zero inside (−2, 5), so it stays positive there. (B) confuses positive with increasing. (C) unknown — sign outside could be anything without more info... careful: for x < −2 there are no zeros, so the sign is constant there, but we don't know which sign. (D) nothing forces the max at 0.
h: h(1)=2, h(3)=8, h(5)=18, h(7)=32. The table is consistent with h being4. (D). Rates: (8−2)/2 = 3, (18−8)/2 = 5, (32−18)/2 = 7. Increasing outputs, increasing rates → increasing and concave up. (C) fails since rates aren't constant; (A) reverses the rate pattern.
5. (B). Volume falls (draining) → decreasing. "Quickly, then slower" → rate of change is negative but increasing toward 0 → concave up. Same shape family as the cooling coffee. (C) would drain slowly then faster.
f over [2, 6] equals 5, and f(2) = −3. Then f(6) =6. (A). 5 = (f(6) − (−3))/4 → f(6) + 3 = 20 → f(6) = 17. (B) forgets the negative sign on −3; (D) is the rise 5·4 without adjusting for f(2); (C) solves 5 = (f(6)+3)/(6+2).
f is true?7. (C). A function assigns one output to x = 0 (if 0 is in the domain), so at most one y-intercept. (A) violates the definition of function; (B) reverses coordinates — it's (0, f(0)); (D) fails when 0 isn't in the domain, e.g. f(x) = 1/x.
p is increasing at a decreasing rate. The secant slopes over [0,1], [1,2], [2,3], [3,4] are, in order, possibly8. (B). Increasing → positive slopes; decreasing rate → slopes shrink: 8, 5, 3, 2. (A) is increasing at an increasing rate; (C) is decreasing (negative slopes); (D) is linear.
T(m) of an oven, in °F after m minutes, satisfies T(4) = 180 and T(10) = 330. The AROC over [4, 10] is9. (A). (330 − 180)/(10 − 4) = 150/6 = 25 °F per minute. (B) forgets to divide; (C) averages the temperatures ((180+330)/2 ÷ something) — nonsense here; (D) divides by 10 instead of 6.
f(x) = x³. The rate of change of f at x = 1, estimated with the interval [0.9, 1.1], is closest to10. (A). f(1.1) = 1.331, f(0.9) = 0.729. AROC = (1.331 − 0.729)/0.2 = 0.602/0.2 = 3.01. (D) is (1.331 − 0.729)/... miscomputed with width 0.1 minus rounding; (B) and (C) are f(1) and e — plausible-looking numbers, wrong idea.
11. (B). Concave up = rate of change increasing; concave down = rate decreasing. At the switch (x = 2), the rate of change stops increasing and starts decreasing. The function stays increasing, so its rate stays positive — (C) is the classic trap conflating the sign of the rate with the trend of the rate. (This switch point is called a point of inflection — Lesson 3.)
12. (FRQ-style, scored as 4 points) (i) [1 pt value + 1 pt interpretation] AROC = (S(9) − S(3))/(9 − 3) = (41 − 14)/6 = 27/6 = 4.5 thousand subscribers per month. Interpretation: from month 3 to month 9, the channel gained subscribers at an average rate of 4,500 per month. (ii) [1 pt] Successive rates: (14−5)/3 = 3, (26−14)/3 = 4, (41−26)/3 = 5, (59−41)/3 = 6. The rates increase, so the data are consistent with S being concave up. (Not linear — rates aren't constant.) (iii) [1 pt] Concave up with increasing outputs means S is increasing at an increasing rate: growth is speeding up.
12. (FRQ-style) 🚫 The number of subscribers to a streaming channel is modeled by S(t), in thousands, t months after launch. Selected values:
| t | 0 | 3 | 6 | 9 | 12 |
|---|---|---|---|---|---|
| S(t) | 5 | 14 | 26 | 41 | 59 |
(i) Compute the average rate of change of S over [3, 9] and interpret it with units. (ii) Are the data consistent with S being concave up, concave down, or linear? Justify using rates of change. (iii) Based on your answer to (ii), is the channel's growth speeding up or slowing down?
FRQ practice begins in Lesson 3, once we have enough machinery for a full 6-point task. Problem 12 above is your FRQ warm-up.
1. (B). Decreasing = falling. Concave down = rate of change decreasing = the fall steepens. (A) is decreasing/concave up (coffee!); (C) increasing/concave down; (D) increasing/concave up.
2. (C). f(4) = 16 − 12 = 4; f(1) = 1 − 3 = −2. AROC = (4 − (−2))/(4 − 1) = 6/3 = 2. (A) uses (f(4)−f(1))/(4+1)... a made-up formula; (B) is f(4)−f(1) ÷ wrong width; (D) is f(4) alone.
3. (A). Continuous with zeros only at −2 and 5, and positive at 0 (inside the interval) → it can't cross zero inside (−2, 5), so it stays positive there. (B) confuses positive with increasing. (C) unknown — sign outside could be anything without more info... careful: for x < −2 there are no zeros, so the sign is constant there, but we don't know which sign. (D) nothing forces the max at 0.
4. (D). Rates: (8−2)/2 = 3, (18−8)/2 = 5, (32−18)/2 = 7. Increasing outputs, increasing rates → increasing and concave up. (C) fails since rates aren't constant; (A) reverses the rate pattern.
5. (B). Volume falls (draining) → decreasing. "Quickly, then slower" → rate of change is negative but increasing toward 0 → concave up. Same shape family as the cooling coffee. (C) would drain slowly then faster.
6. (A). 5 = (f(6) − (−3))/4 → f(6) + 3 = 20 → f(6) = 17. (B) forgets the negative sign on −3; (D) is the rise 5·4 without adjusting for f(2); (C) solves 5 = (f(6)+3)/(6+2).
7. (C). A function assigns one output to x = 0 (if 0 is in the domain), so at most one y-intercept. (A) violates the definition of function; (B) reverses coordinates — it's (0, f(0)); (D) fails when 0 isn't in the domain, e.g. f(x) = 1/x.
8. (B). Increasing → positive slopes; decreasing rate → slopes shrink: 8, 5, 3, 2. (A) is increasing at an increasing rate; (C) is decreasing (negative slopes); (D) is linear.
9. (A). (330 − 180)/(10 − 4) = 150/6 = 25 °F per minute. (B) forgets to divide; (C) averages the temperatures ((180+330)/2 ÷ something) — nonsense here; (D) divides by 10 instead of 6.
10. (A). f(1.1) = 1.331, f(0.9) = 0.729. AROC = (1.331 − 0.729)/0.2 = 0.602/0.2 = 3.01. (D) is (1.331 − 0.729)/... miscomputed with width 0.1 minus rounding; (B) and (C) are f(1) and e — plausible-looking numbers, wrong idea.
11. (B). Concave up = rate of change increasing; concave down = rate decreasing. At the switch (x = 2), the rate of change stops increasing and starts decreasing. The function stays increasing, so its rate stays positive — (C) is the classic trap conflating the sign of the rate with the trend of the rate. (This switch point is called a point of inflection — Lesson 3.)
12. (FRQ-style, scored as 4 points) (i) [1 pt value + 1 pt interpretation] AROC = (S(9) − S(3))/(9 − 3) = (41 − 14)/6 = 27/6 = 4.5 thousand subscribers per month. Interpretation: from month 3 to month 9, the channel gained subscribers at an average rate of 4,500 per month. (ii) [1 pt] Successive rates: (14−5)/3 = 3, (26−14)/3 = 4, (41−26)/3 = 5, (59−41)/3 = 6. The rates increase, so the data are consistent with S being concave up. (Not linear — rates aren't constant.) (iii) [1 pt] Concave up with increasing outputs means S is increasing at an increasing rate: growth is speeding up.
🎯 Exam tip: FRQ 4 ("Communicating about Functions") hands out points for exactly the sentences drilled here. Practice writing "f is increasing at a decreasing rate on the interval (a, b) because the average rates of change over successive equal-length intervals are positive and decreasing" until it's automatic. It reads like boilerplate because it is — points-bearing boilerplate.