Let f(x) = √x. Using the tangent line to f at x = 16, find the linear approximation of √16.5.
A4.125
B4.5
C4.0625
D4.03125
Explanation
The linearization is L(x) = f(a) + f′(a)(x − a) with a = 16. Here f(16) = 4 and f′(x) = 1/(2√x), so f′(16) = 1/8. Then L(16.5) = 4 + (1/8)(0.5) = 4.0625. Using slope 1/32 (mistakenly writing f′(x) = 1/(2x) instead of 1/(2√x)) gives 4.03125; using slope 1/4 = 1/√16 forgets the factor of 2 in the derivative of √x.
Question 2 of 10
MCQU7Topic 7.1Medium No calc Word
A population P grows so that its rate of change with respect to time t is proportional to the square root of the population present. Which differential equation models this situation, where k is a positive constant?
AdP/dt = kP
BP = k√t
CdP/dt = k√P
DdP/dt = k/√P
Explanation
‘Rate of change of P with respect to t’ is dP/dt, and ‘proportional to’ means it equals a constant k times the quantity described — here the square root of P — giving dP/dt = k√P. Writing dP/dt = kP makes the rate proportional to P itself, not to √P. Writing dP/dt = k/√P makes the rate proportional to the reciprocal of √P, the opposite dependence. Setting P = k√t is a candidate solution formula, not the differential equation the phrase describes.
Question 3 of 10
MCQU2Topic 2.2Medium No calc
The limit lim(h→0) ((2+h)³ − 8)/h is the definition of the derivative of a function at a point. What is its value?
A8
B6
C12
D4
Explanation
The expression matches f′(a) = lim(h→0) (f(a+h) − f(a))/h with f(x) = x³ and a = 2. Thus the limit equals f′(2). Since f′(x) = 3x², f′(2) = 3(2²) = 12. Answering 8 mistakes the limit for the function value f(2) = 2³. Answering 6 uses an incorrect power-rule result 3x (dropping the square) evaluated at 2. Answering 4 comes from using x² = 2² without the coefficient 3.
Question 4 of 10
MCQU6Topic 6.7Medium No calc
Evaluate ∫₀π/4 sec²(x) dx.
A2
B0
C1
D√2/2
Explanation
An antiderivative of sec²(x) is tan(x). By the Evaluation part of the FTC, ∫₀^(π/4) sec²(x) dx = tan(π/4) − tan(0) = 1 − 0 = 1. The value √2/2 comes from using sin(π/4) instead of tan(π/4), confusing the antiderivative. Getting 0 mistakenly treats the two endpoint values as equal, and 2 would double-count the lower limit.
Question 5 of 10
MCQU1Topic 1.15Easy No calc
Evaluate lim(x→∞) (3x² − 5x + 1)/(2x² + 7).
A3/7
B0
CThe limit is +∞
D3/2
Explanation
The numerator and denominator are both degree 2, so the end-behavior limit equals the ratio of the leading coefficients, 3/2. Dividing every term by x² sends −5x, 1, and 7 relative to x² to 0, leaving 3/2. Answering 0 treats the polynomial as if the denominator dominated (which happens only when the denominator's degree is larger). Pairing 3 with the constant 7 to get 3/7 confuses the constant term with the leading coefficient.
Question 6 of 10
MCQU3Topic 3.1Medium No calc
If f(x) = (2x³ − 5)⁴, then f′(x) =
A24x²(2x³ − 5)⁴
B24x²(2x³ − 5)³
C24x³(2x³ − 5)³
D4(2x³ − 5)³
Explanation
By the chain rule, d/dx[u⁴] = 4u³·u′ where u = 2x³ − 5, so u′ = 6x². Thus f′(x) = 4(2x³ − 5)³·6x² = 24x²(2x³ − 5)³. Writing 4(2x³ − 5)³ forgets to multiply by the inner derivative. Leaving the power at 4 fails to reduce the exponent when applying the power rule.
Question 7 of 10
MCQU5Topic 5.8Medium No calc Diagram
The graph of f′ consists of two line segments: one from (0, −2) up to (2, 2) crossing the x-axis at x = 1, and one from (2, 2) down to (4, −2) crossing the x-axis at x = 3. On the interval [0, 4], at which x does f have a local maximum?
Ax = 2
Bx = 3
Cx = 1
Dx = 0
Explanation
f has a local maximum where f′ changes from positive to negative. From the described segments, f′ > 0 on (1, 3) and f′ < 0 on (3, 4), so f′ switches + to − at x = 3 → a local maximum. At x = 1, f′ changes − to +, giving a local minimum instead. x = 2 is where f′ reaches its own peak value, not where f is maximized.
Question 8 of 10
MCQU8Topic 8.7Easy No calc Diagram
The base of a solid is the region bounded by y = √x, the x-axis, and the line x = 4. Cross sections perpendicular to the x-axis are squares. What is the volume of the solid?
A16
B16/3
C8
D8π
Explanation
For solids with known cross sections, V = ∫ A(x) dx where A(x) is the cross-sectional area. Each square has side equal to the height of the base, s = √x, so A(x) = s² = (√x)² = x, and V = ∫₀⁴ x dx = [x²/2]₀⁴ = 8. The value 8π comes from treating each slice as a disc, A = π(√x)², inserting a factor of π that belongs to circular cross sections, not squares. The value 16/3 = ∫₀⁴ √x dx forgets to square the side length before integrating. The value 16 evaluates x²/2 at x = 4 without the 1/2, i.e. mis-antidifferentiates x as x² rather than x²/2.
Question 9 of 10
MCQU4Topic 4.7Medium No calc
Evaluate lim(x→3) (x² − x − 6)/(x² − 9).
A0
B1
C5/6
DThe limit does not exist.
Explanation
Direct substitution gives 0/0, an indeterminate form, so L'Hôpital's rule (or factoring) applies. Factoring: (x² − x − 6)/(x² − 9) = (x − 3)(x + 2)/[(x − 3)(x + 3)] = (x + 2)/(x + 3) for x ≠ 3, which at x = 3 gives 5/6. Concluding the limit does not exist misreads the indeterminate form 0/0 as an automatic failure; the common factor (x − 3) cancels, so the limit exists.
Question 10 of 10
MCQU4Topic 4.2Medium No calc Word
A particle moves along a horizontal line so that its position at time t seconds (t ≥ 0) is s(t) = t³ − 6t² + 9t meters. Which statement correctly describes the motion at t = 4?
AThe particle is slowing down, because its acceleration is zero at this instant.
BThe particle is speeding up, because its velocity and acceleration are both positive.
CThe particle is at rest, because its velocity is equal to zero at this instant.
DThe particle is slowing down, because its velocity is positive while its acceleration is negative.
Explanation
Velocity is v(t) = s′(t) = 3t² − 12t + 9 and acceleration is a(t) = v′(t) = 6t − 12. At t = 4, v(4) = 48 − 48 + 9 = 9 > 0 and a(4) = 24 − 12 = 12 > 0. A particle is speeding up exactly when velocity and acceleration have the same sign, so it is speeding up here. Claiming the particle is at rest confuses this instant with the times where v = 0 (t = 1 and t = 3); at t = 4 the velocity is not zero.
Free Response 1 · Section II
FRQRate and AccumulationU6 Calc
A reservoir supplies water to a small town. For 0 ≤ t ≤ 8 (where t is measured in hours after 6:00 A.M.), water flows INTO the reservoir at a rate of E(t) = 90 + 45·sin(t/3) gallons per hour, and water is released OUT of the reservoir at a rate of L(t) = 30 + 2t² gallons per hour. At time t = 0 the reservoir contains 200 gallons of water. (a) How many gallons of water flow into the reservoir during the 8-hour period 0 ≤ t ≤ 8? Show the setup for your calculation. (b) Find the average rate, in gallons per hour, at which water flows into the reservoir over the interval 0 ≤ t ≤ 8. (c) At time t = 5, is the amount of water in the reservoir increasing or decreasing? Give a reason for your answer that uses a rate computed at t = 5. (d) For 0 ≤ t ≤ 8, find the time t at which the amount of water in the reservoir is greatest. Justify your answer.
Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.
Model answer
Let W(t) be the number of gallons in the reservoir at time t, so W(0) = 200 and W′(t) = E(t) − L(t).
(a) Total water that flows in = ∫₀⁸ E(t) dt = ∫₀⁸ (90 + 45·sin(t/3)) dt. Using a calculator, ∫₀⁸ (90 + 45·sin(t/3)) dt ≈ 975.059 gallons. (By hand the antiderivative is 90t − 135·cos(t/3), giving 720 − 135(cos(8/3) − 1) ≈ 975.059.)
(b) Average inflow rate = (1/(8 − 0))·∫₀⁸ E(t) dt = 975.059/8 ≈ 121.882 gallons per hour.
(c) The amount of water is changing at the net rate W′(t) = E(t) − L(t). E(5) = 90 + 45·sin(5/3) ≈ 134.793 gal/hr and L(5) = 30 + 2(5²) = 80 gal/hr. W′(5) = 134.793 − 80 ≈ 54.793 > 0. Since W′(5) > 0, the amount of water in the reservoir is INCREASING at t = 5.
(d) The amount W(t) has critical points where W′(t) = E(t) − L(t) = 0, i.e. where 90 + 45·sin(t/3) = 30 + 2t². Solving on 0 ≤ t ≤ 8 (calculator) gives one solution, t ≈ 6.856. Check the sign of W′: at t = 0, W′(0) = 90 − 30 = 60 > 0 (water increasing), and for t just past 6.856, W′ < 0 (for example W′(8) = E(8) − L(8) ≈ 110.577 − 158 ≈ −47.423 < 0). So W′ changes from positive to negative at t ≈ 6.856, meaning W has a local maximum there. Because it is the only critical point and W increases on [0, 6.856) and decreases on (6.856, 8], the greatest amount of water occurs at t ≈ 6.856 hours (about 12:51 P.M.). The maximum amount is W(6.856) = 200 + ∫₀^6.856 (E(t) − L(t)) dt ≈ 619.978 gallons.
Scoring · 6 points
(a) Correct definite integral ∫₀⁸ E(t) dt written as the setup for total inflow — 1 point.
(a) Correct numerical value ≈ 975.059 gallons with units — 1 point.
(b) Average inflow computed as (1/8)·∫₀⁸ E dt ≈ 121.882 gal/hr with units — 1 point.
(c) Uses W′(5) = E(5) − L(5) ≈ 54.793 and concludes increasing because the value is positive — 1 point.
(d) Sets E(t) = L(t) and finds the critical time t ≈ 6.856 — 1 point.
(d) Justifies a maximum by showing W′ changes from positive to negative (or compares endpoint/critical values) — 1 point.