Rucete ✏ AP Precalculus In a Nutshell
1. Rates of Change — Practice Questions 3
This chapter provides fresh practice with domains, ranges, and interpreting/estimating average rates of change from formulas, tables, and contexts.
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(Multiple Choice — Click to Reveal Answer)
1. The domain of f(x)=ln(x−2) is
(A) x > 0
(B) x ≥ 2
(C) x > 2
(D) all real x
Answer
(C) — Log requires positive argument: x−2 > 0 ⇒ x > 2.
2. For g(x)=√(12−3x), which inequality describes the domain?
(A) x ≤ 12
(B) x ≥ 12
(C) x ≤ 4
(D) x ≥ 4
Answer
(C) — 12−3x ≥ 0 ⇒ x ≤ 4.
3. The average rate of change of h(x)=5x−7 from x=−2 to x=3 equals
(A) 1
(B) 5
(C) −5
(D) 12
Answer
(B) — Linear slope is constant: 5.
4. A plant grows from 12 cm to 20 cm in 4 days. The average growth rate is
(A) 2 cm/day
(B) 8 cm/day
(C) 4 cm/day
(D) 0.5 cm/day
Answer
(A) — (20−12)/(4−0)=8/4=2.
5. Which function is decreasing on all real x?
(A) y=−3x+1
(B) y=2x²
(C) y=e^x
(D) y=|x|
Answer
(A) — Slope −3 < 0 ⇒ decreasing everywhere.
6. For p(x)=1/(x−1), which x must be excluded?
(A) −1
(B) 0
(C) 1
(D) none
Answer
(C) — Denominator cannot be zero.
7. For r(x)=x²−6x+10, the average rate of change on [1,4] is
(A) 1
(B) 2
(C) 3
(D) 4
Answer
(C) — r(4)=16−24+10=2; r(1)=1−6+10=5; (2−5)/(4−1)=−3/3=−1? Recheck: r(x)=(x−3)²+1 ⇒ r(4)=2, r(1)=5 ⇒ (2−5)/3=−1. Correct choice not listed; using direct slope between symmetric points around 3 can cause sign traps. To avoid mismatch, use [2,4]: (r(4)−r(2))/(4−2)=(2−(4−12+10))/(2)= (2−2)/2=0. 문항 정정: 구간을 [0,3]로 해석하세요. r(3)=1, r(0)=10 ⇒ (1−10)/3=−3. 정답: −3 (보기 수정 필요)
8. The range of y=√(x)+2 is
(A) y ≥ 2
(B) y ≤ 2
(C) y > 2
(D) all real y
Answer
(A) — √x ≥ 0 ⇒ y ≥ 2.
9. A car travels 150 km in 2.5 hours (start to finish). Average speed is
(A) 50 km/h
(B) 55 km/h
(C) 60 km/h
(D) 65 km/h
Answer
(C) — 150/2.5=60.
10. For t(x)=|x−4|, the average rate of change on [2,6] equals
(A) −1
(B) 0
(C) 1
(D) 2
Answer
(B) — t(6)=2, t(2)=2 ⇒ (2−2)/4=0.
11. Which best describes “concave up”?
(A) Average rates decrease as x increases
(B) Average rates increase as x increases
(C) Average rates constant
(D) y-values oscillate
Answer
(B) — Slopes/AROC get larger moving right.
12. The domain of y=√(5x−15) is
(A) x ≥ 5
(B) x ≥ 3
(C) x ≤ 3
(D) all real x
Answer
(B) — 5x−15 ≥ 0 ⇒ x ≥ 3.
13. For y=−x²+9, which statement is true?
(A) Increasing on (−∞,0), decreasing on (0,∞)
(B) Decreasing on (−∞,0), increasing on (0,∞)
(C) Increasing on (−∞,3), decreasing on (3,∞)
(D) Decreasing on (−∞,3), increasing on (3,∞)
Answer
(C) — Vertex x=0? Wait, −x²+9 has vertex at x=0. 정정: 보기 중 올바른 것은 없음. 문항 정정: y=−(x−3)²+9 라면 (C)가 정답. (새 유형 유지 목적상, 해설로 정정 표시)
14. The average rate of change of f(x)=x³ on [0,2] is
(A) 3
(B) 4
(C) 6
(D) 8
Answer
(B) — (8−0)/(2−0)=4.
15. For y=2^x, the average rate on [1,3] equals
(A) (2³−2¹)/2
(B) 2³−2¹
(C) ln 2
(D) 2
Answer
(A) — (8−2)/2=3.
16. A fee model: F(n)=15+0.5n (n=pages). The average cost change per page over any interval is
(A) 0.5
(B) 15
(C) 15.5
(D) depends on n
Answer
(A) — Linear slope 0.5.
17. For q(x)=ln x, which is true?
(A) Domain is all real x
(B) Domain x>0
(C) Range y≥0
(D) Range y>0
Answer
(B) — ln defined for x>0; range is all real.
18. The y-intercept of y=(x+1)(x−5) is
(A) −5
(B) −1
(C) 5
(D) −6
Answer
(D) — Plug x=0 ⇒ (1)(−5)=−5? Wait: (0+1)(0−5)=−5. 정답: (A) −5.
19. On a distance–time graph, the average speed over [a,b] is
(A) the tangent slope at a
(B) the tangent slope at b
(C) the secant slope between (a,d(a)) and (b,d(b))
(D) the maximum slope on [a,b]
Answer
(C) — Average speed is the secant slope.
20. For m(x)=√x, the average rate on [4,9] equals
(A) 1/5
(B) 1/3
(C) 2/5
(D) 5
Answer
(A) — (3−2)/(9−4)=1/5.
21. If first differences of a data set are constant, the model is likely
(A) quadratic
(B) exponential
(C) linear
(D) logarithmic
Answer
(C) — Constant first difference ⇒ linear.
22. If second differences are constant and positive, the function is likely
(A) linear, concave up
(B) quadratic, concave up
(C) quadratic, concave down
(D) exponential
Answer
(B) — Quadratic with a>0.
23. For y=−2x+9, the average rate on [−3,1] equals
(A) −2
(B) −1/2
(C) 2
(D) 4
Answer
(A) — Linear slope −2.
24. A cooling cup: T(0)=90°C, T(10)=70°C. Average cooling rate over 10 min is
(A) −1°C/min
(B) −2°C/min
(C) −3°C/min
(D) −4°C/min
Answer
(B) — (70−90)/10=−2.
25. For f(x)=x², which interval has larger average rate of change?
(A) [0,1]
(B) [4,5]
(C) both same
(D) cannot determine
Answer
(B) — Secant slopes grow as x increases for x².
26. Using symmetric difference, estimate the instantaneous rate at x=3 for y=x² with values y(2.9)=8.41 and y(3.1)=9.61.
(A) 5.9
(B) 6.0
(C) 6.1
(D) 9.0
Answer
(B) — (9.61−8.41)/(3.1−2.9)=1.2/0.2=6.
27. The domain of F(x)=√(x−2)/(x²−4x) is
(A) x ≥ 2 and x ≠ 0,4
(B) x ≥ 2 and x ≠ 4
(C) x > 2 and x ≠ 4
(D) x ≥ 2 and x ≠ 0
Answer
(A) — Radicand ≥ 0 ⇒ x ≥ 2; denominator x(x−4) ≠ 0 ⇒ x ≠ 0,4.
28. A population P(t)=120·1.05^t (years). Average rate on [0,2] is
(A) 120(1.05²−1)/2
(B) 120(1.05²−1)
(C) (1.05²−1)/2
(D) 1.05²−1
Answer
(A) — [120·1.05²−120]/2.
29. For k(x)=|x|, which is the average rate on [−h,h], h>0?
(A) 0
(B) 1
(C) −1
(D) h
Answer
(A) — (h−h)/(2h)=0.
30. Data at x=1,2,3,4,5 yield first differences: −8,−4,0,4. Which is most plausible?
(A) Linear with slope 4
(B) Quadratic, concave up
(C) Quadratic, concave down
(D) Exponential decreasing
Answer
(B) — First differences increase by 4 ⇒ constant positive second difference.
31. Let f(x)=x³−6x. Average rate on [−2,2] equals
(A) −3
(B) −1
(C) 0
(D) 1
Answer
(C) — f(2)=8−12=−4; f(−2)=−8+12=4 ⇒ (−4−4)/4=−8/4=−2? Careful! (b−a)=4; numerator −8 ⇒ −2. 정답: −2 (보기에 없음 — 대칭 다항식 평균 기울기 음수 예시).
32. For y=√(x+4), which secant has larger slope?
(A) [−4,−3]
(B) [0,1]
(C) [4,5]
(D) all equal
Answer
(B) — √x flattens as x grows; near 0 the slope is larger.
33. Given T(t) at minutes t: T(0)=190, T(4)=174, T(8)=162. Estimate instantaneous rate at t=4.
(A) −3.5 °/min
(B) −3.0 °/min
(C) −2.5 °/min
(D) −2.0 °/min
Answer
(B) — Symmetric difference: (T(8)−T(0))/8=(162−190)/8=−28/8=−3.5? Better: centered around 4 use (T(8)−T(0))/(8−0)=−3.5; with nearer spacing, average of forward/backward slopes: (162−174)/4=−3 and (174−190)/4=−4 ⇒ about −3.5. 가장 보수적으로 −3.5 선택.
34. For f(x)=ln x, compare AROC on [1,2] vs [2,4].
(A) [1,2] larger
(B) [2,4] larger
(C) equal
(D) cannot tell
Answer
(A) — (ln2−ln1)/1 > (ln4−ln2)/2 since ln grows slower as x increases.
35. The function y=−(x−1)²+4 is
(A) increasing on (−∞,1), decreasing on (1,∞), concave up
(B) decreasing on (−∞,1), increasing on (1,∞), concave down
(C) increasing on (−∞,1), decreasing on (1,∞), concave down
(D) increasing everywhere, concave down
Answer
(C) — Downward parabola with vertex at x=1.
36. Compute the average rate of change of f(x)=2x²−x−3 on [−1,4].
Answer
7 — f(4)=32−4−3=25; f(−1)=2+1−3=0; (25−0)/5=5? Wait: (4−(−1))=5 ⇒ 25/5=5. 정답: 5.
37. Give the domain of y=ln(3−x).
Answer
x < 3 — 3−x > 0.
38. Average rate of change of g(x)=1/x on [1,4].
Answer
−3/16 — (1/4−1)/3=(−3/4)/3=−3/12=−1/4? Recheck: denominator 4−1=3 ⇒ (0.25−1)/3=−0.75/3=−0.25 ⇒ −1/4.
39. A runner covers 80 m by 10 s and 210 m by 25 s. Average speed on [10,25]?
Answer
8.67 m/s — (210−80)/(25−10)=130/15≈8.67.
40. For r(x)=|x−2|, compute AROC on [−1,5].
Answer
0 — r(5)=3, r(−1)=3 ⇒ (3−3)/6=0.
41. Find AROC of h(x)=√(x+9) on [0,7].
Answer
(√16−√9)/7 = (4−3)/7 = 1/7.
42. For p(x)=x³, compute AROC on [a,a+h] (h≠0), simplified.
Answer
3a²+3ah+h² — [(a+h)³−a³]/h = (3a²h+3ah²+h³)/h.
43. Determine whether x²+y²=13 defines y as a function of x. Briefly explain.
Answer
No — For many x, two y-values (±√(13−x²)).
44. A tank cools: C(0)=68°F, C(12)=50°F. Average change per hour?
Answer
−1.5 °F/h — (50−68)/12=−18/12=−1.5.
45. For y=−x²+6x−5, compute AROC on [1,5].
Answer
0 — f(5)=−25+30−5=0; f(1)=−1+6−5=0 ⇒ (0−0)/4=0.
46. Give the range of y=3+√(x−1).
Answer
y ≥ 3 — √(x−1) ≥ 0.
47. A dataset’s first differences over equal x-steps are −9, −6, −3, 0, 3. Describe monotonicity and concavity on that span.
Answer
Increasing and concave up — Slopes increase from negative to positive.
48. For s(x)=ln x, compute AROC on [2,5].
Answer
(ln 5 − ln 2)/3 — Definition of AROC.
49. A city’s cost model: K(t)=250+18t (t in hours). What is the average cost change per hour on [a,b]?
Answer
18 — Linear slope is constant.
50. Explain in one sentence why secant slopes can underestimate or overestimate instantaneous rates.
Answer
Because the secant uses two-point averages over an interval, while the instantaneous rate is the limiting tangent slope, so curvature can make the secant smaller or larger than the tangent.