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29 questions · suggested time 62 minutes. Record one letter per question.
What is lim as x approaches 3 of (x²−9)/(x−3)?
Answer C. For x≠3, the quotient equals x+3 after factoring x²−9=(x−3)(x+3). A limit uses nearby inputs, so its value is 3+3=6 even though the original quotient is undefined at 3.
Review focus: Evaluate a limit by algebraic simplification. Unit 1: Limits and continuity.
The graph consists of the two indicated closed line segments. What is lim as x approaches 0 of f(x)?
Graph description: Continuous line segments join (−2,3) to (0,−1) to (2,3). f (solid): straight segments through (-2, 3), (0, -1), (2, 3). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer A. From both sides, the segment heights approach −1 at x=0. A corner can prevent differentiability without preventing a limit or continuity. Therefore the two-sided limit is −1.
Review focus: Read a two-sided limit from a supplied graph. Unit 1: Limits and continuity.
A function f is continuous on [−1,2], with f(−1)=4 and f(2)=−2. Which conclusion is guaranteed?
Answer D. The value 1 lies strictly between −2 and 4, so continuity and the intermediate value theorem give an interior input with value 1. Continuity does not guarantee monotonicity, differentiability, or uniqueness of a zero.
Review focus: Apply the intermediate value theorem with its hypotheses. Unit 1: Limits and continuity.
What is lim as h approaches 0 of [√(9+h)−3]/h?
Answer B. Rationalizing gives 1/[√(9+h)+3] for h≠0. Its limit is 1/(3+3)=1/6. Equivalently, this is the derivative of √x at x=9.
Review focus: Interpret and evaluate a derivative-definition limit. Unit 2: Derivative definitions and rules.
The table gives exact values for differentiable functions f and g. If h=f/g near x=2, what is h′(2)?
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 2 | 3 | −1 | 4 | 2 |
Answer C. The quotient rule gives h′(2)=[(−1)·4−3·2]/4²=−10/16=−5/8. Dividing the derivatives alone, omitting the squared denominator, or reversing the subtraction gives a different result.
Review focus: Apply the quotient rule to supplied derivative data. Unit 2: Derivative definitions and rules.
For f(x)=3sin x−2cos x, with x in radians, what is f′(π/2)?
Answer A. f′(x)=3cos x+2sin x. At π/2, cosine is 0 and sine is 1, so f′(π/2)=2. The derivative of −2cos x is positive 2sin x.
Review focus: Differentiate trigonometric functions in radians. Unit 2: Derivative definitions and rules.
Differentiable functions f and g have the exact values shown. What is the derivative of f(g(x)) at x=1?
| x | f′(x) | g(x) | g′(x) |
|---|---|---|---|
| 1 | 4 | 3 | −3 |
| 3 | −2 | 1 | 5 |
Answer D. The inner output is g(1)=3. Thus the outer derivative must be read at 3: f′(g(1))g′(1)=f′(3)(−3)=(−2)(−3)=6. The product f′(1)g′(1) uses the wrong outer input.
Review focus: Track the intermediate input in the chain rule. Unit 3: Composite, implicit and inverse derivatives.
On the curve x²+xy+y²=13, what is dy/dx at the point (1,3)?
Answer B. The point lies on the curve because 1+3+9=13. Differentiating gives 2x+y+(x+2y)y′=0. At (1,3), y′=−(2+3)/(1+6)=−5/7; the nonzero denominator supports the local y-branch.
Review focus: Differentiate an implicit relation at a valid point. Unit 3: Composite, implicit and inverse derivatives.
A differentiable function f has a differentiable inverse near output 5. Given f(2)=5, f′(2)=3, f(5)=9 and f′(5)=7, what is (f⁻¹)′(5)?
Answer A. Since f(2)=5, the inverse maps 5 to 2. Its derivative is 1/f′(2)=1/3. The derivative f′(5) concerns a different input and is not used here.
Review focus: Use the input-output correspondence for an inverse derivative. Unit 3: Composite, implicit and inverse derivatives.
A particle has position s(t)=t³−6t²+9t for 0≤t≤4. On which intervals is its speed increasing?
Answer C. Velocity is 3(t−1)(t−3), and acceleration is 6(t−2). Speed increases where their signs agree. Both are negative on (1,2), and both are positive on (3,4). On the other listed subintervals their signs are opposite.
Review focus: Relate velocity and acceleration signs to changing speed. Unit 4: Contextual differentiation.
A 10-metre ladder rests against a vertical wall. Its bottom moves horizontally away from the wall at 2 m/s. When the bottom is 6 metres from the wall, at what rate is the top’s height changing?
Answer A. Let x be the bottom distance and y the height, so x²+y²=100. At x=6, y=8. Differentiation gives xx′+yy′=0, hence y′=−6·2/8=−3/2 m/s. The negative sign denotes downward motion.
Review focus: Set up and interpret a signed related rate. Unit 4: Contextual differentiation.
A differentiable function satisfies f(4)=7 and f′(4)=−2. What estimate for f(4.1) is given by the tangent line at x=4?
Answer D. Linearization gives f(4.1)≈f(4)+f′(4)(4.1−4)=7−2(0.1)=6.8. The quantity −0.2 is the estimated change, not the estimated function value.
Review focus: Use a local linear model without confusing value and rate. Unit 4: Contextual differentiation.
The supplied graph is f′, not f. Assume f is differentiable throughout an open interval containing [−2,3]. At which input does f have an interior local minimum?
Graph description: The derivative is linear through (−2,2),(0,0),(1,−1),(3,1). Its zeros are 0 and 2. f′ (solid): straight segments through (-2, 2), (0, 0), (1, -1), (3, 1). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer B. At x=2 the derivative changes from negative to positive, so f changes from decreasing to increasing and has a local minimum. At 0 the sign change gives a local maximum. The lowest plotted derivative value at 1 is not the lowest value of f.
Review focus: Identify a local minimum from a supplied derivative graph. Unit 5: Analytical differentiation.
The supplied continuous piecewise-linear graph is f′. At which interior inputs does f change concavity?
Graph description: The derivative is linear through (0,1),(2,3),(4,−1),(6,1). f′ (solid): straight segments through (0, 1), (2, 3), (4, -1), (6, 1). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer D. f′ increases on (0,2), decreases on (2,4), and increases on (4,6). Therefore f changes from concave up to down at 2 and from down to up at 4. Zeros of f′ at 3.5 and 5 concern possible extrema of f, not concavity changes.
Review focus: Infer concavity changes from the trend of a derivative graph. Unit 5: Analytical differentiation.
A function f is continuous on [0,5] and differentiable on (0,5). Its only critical inputs in (0,5) are 1 and 3. Using the table, what is its absolute maximum value on [0,5]?
| x | 0 | 1 | 3 | 5 |
|---|---|---|---|---|
| f(x) | 2 | 5 | −1 | 4 |
Answer A. Continuity on the closed interval ensures an absolute maximum. With all interior critical inputs supplied, compare their values and the endpoint values: 2,5,−1,4. The largest value is 5, attained at input 1.
Review focus: Compare all candidates for an absolute extremum. Unit 5: Analytical differentiation.
A rectangle has perimeter 40 metres. What is its greatest possible area?
Answer C. If one side is x, the other is 20−x. Area A=x(20−x)=100−(x−10)², which is at most 100 for 0<x<20 and attains that value at x=10. Equivalently, A′=20−2x vanishes at the maximum.
Review focus: Optimize a quantity subject to a geometric constraint. Unit 5: Analytical differentiation.
The function f is continuous on [0,4] and differentiable on (0,4), with f(0)=1 and f(4)=9. Which conclusion follows from the mean value theorem?
Answer B. The average rate is (9−1)/(4−0)=2. The theorem guarantees at least one interior derivative equal to 2, without identifying c or making the derivative constant. The function f(x)=1+2x also shows that f(c)=2c need not hold anywhere.
Review focus: Apply the mean value theorem with precise quantifiers. Unit 5: Analytical differentiation.
A synthetic flow rate is sampled at the times in the table. What is the composite trapezoid estimate of the amount flowing from t=0 to t=6?
| t | 0 | 1 | 3 | 6 |
|---|---|---|---|---|
| Rate | 2 | 4 | 3 | 1 |
Answer D. Use widths 1,2,3: the trapezoid estimate is 1(2+4)/2+2(4+3)/2+3(3+1)/2=3+7+6=16 L. Unequal time spacing requires separate interval weights.
Review focus: Construct a trapezoid sum with unequal interval widths. Unit 6: Integration and accumulation.
Let H(x)=∫ from 1 to x² of √(1+t³) dt, for x near 1. What is H′(1)?
Answer A. The integrand is continuous near the relevant endpoint. FTC and the chain rule give H′(x)=√(1+x⁶)·2x, so H′(1)=2√2. The integral itself is zero at 1, but its derivative need not be zero.
Review focus: Combine FTC with a changing upper bound. Unit 6: Integration and accumulation.
Suppose f is continuous, ∫₀²f(x)dx=3 and ∫₂⁵f(x)dx=−4. What is ∫₅⁰2f(x)dx?
Answer C. Additivity gives ∫₀⁵f=3−4=−1. Reversing bounds changes this to 1, and multiplying the integrand by 2 gives 2.
Review focus: Use additivity, scaling and reversed integral bounds. Unit 6: Integration and accumulation.
What is ∫₀¹6x(x²+1)²dx?
Answer B. Set u=x²+1, so du=2x dx and the bounds become 1 and 2. The integral is 3∫₁²u²du=[u³]₁²=8−1=7.
Review focus: Evaluate a definite integral by substitution. Unit 6: Integration and accumulation.
What is ∫ from −2 to 0 of 1/(x²+4x+8) dx?
Answer D. The denominator is (x+2)²+4. A primitive is (1/2)arctan((x+2)/2). At the upper and lower endpoints the normalized inputs are 1 and 0, so the value is (1/2)(π/4)=π/8.
Review focus: Complete a square and retain the inverse-trigonometric scale. Unit 6: Integration and accumulation.
Let A(x)=1+∫₀ˣg(t)dt, where g is the supplied piecewise-linear graph. What is A(6)?
Graph description: The integrand is linear through (0,0),(2,4),(4,0),(6,−2). g (solid): straight segments through (0, 0), (2, 4), (4, 0), (6, -2). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer A. The positive triangle on [0,4] has area (1/2)·4·4=8. The negative triangle on [4,6] contributes −(1/2)·2·2=−2. Adding the initial value gives A(6)=1+8−2=7.
Review focus: Compute accumulation from signed geometric areas. Unit 6: Integration and accumulation.
A solution of y′=2xy satisfies y(0)=3. What is y(1)?
Answer C. The positive solution separates as dy/y=2x dx, giving ln y=x²+C and y=Ce^(x²). The initial condition gives C=3, hence y(1)=3e. Differentiating 3e^(x²) verifies the equation.
Review focus: Solve a separable initial-value equation. Unit 7: Differential equations.
The table lists slopes at four points of a slope field. Which differential equation agrees with all four values?
| x | y | dy/dx |
|---|---|---|
| 0 | 1 | 2 |
| 1 | 1 | 3 |
| 0 | 2 | 4 |
| 1 | 2 | 5 |
Answer B. Substituting the four points into x+2y gives 2,3,4,5 respectively. Each other proposed rule disagrees at least at one supplied point. These checks distinguish these four candidates, not every conceivable formula.
Review focus: Identify a differential equation from supplied slope data. Unit 7: Differential equations.
Which function satisfies T′=−(T−10)/5 and T(0)=30?
Answer D. For T=10+20e^(−t/5), T(0)=30 and T′=−4e^(−t/5)=−(T−10)/5. The constant 10 is the equilibrium; exponential decay applies to the difference from that equilibrium, not directly to T.
Review focus: Verify an exponential model with an equilibrium shift. Unit 7: Differential equations.
What is the average value of f(x)=3x² on [1,3]?
Answer B. The integral is [x³]₁³=26. Divide by interval length 2 to get average value 13. Averaging the endpoint values gives 15, which is not generally the integral average.
Review focus: Distinguish integral average from endpoint average. Unit 8: Applications of integration.
What is the area of the region enclosed by y=x² and y=x+2?
Answer A. Intersections satisfy x²=x+2, so x=−1 and 2. The line is above the parabola between them. Area is ∫₋₁²(x+2−x²)dx=[x²/2+2x−x³/3]₋₁²=10/3−(−7/6)=9/2.
Review focus: Determine intersections and integrate the correct region width. Unit 8: Applications of integration.
A solid has base bounded by y=√x, y=0, x=0 and x=2. Its cross sections perpendicular to the x-axis are rectangles whose heights are four times their bases in the plane region. What is the volume?
Answer C. At x the base length is √x and the rectangular height is 4√x, so section area is 4x. Volume is ∫₀²4x dx=8. Integrating the base length alone would give plane area, not solid volume.
Review focus: Build a cross-section area before integrating volume. Unit 8: Applications of integration.
13 questions · suggested time 38 minutes. Record one letter per question.
For x≠1.2, let F(x)=(eˣ−exp(1.2))/(x−1.2). At x=1.2, define F(1.2)=k²+2, where k>0. Which value of k makes F continuous at 1.2? Round to three decimal places.
Answer B. The quotient is the derivative definition for eˣ at 1.2, so its limit is exp(1.2)≈3.320116923. Continuity requires k²+2=exp(1.2), hence k=√(exp(1.2)−2)≈1.148963412. The positive restriction selects 1.149. Keep exp(1.2) unrounded until after subtracting 2 and taking the square root; substituting x=1.2 directly into the quotient gives 0/0, not the limit.
Review focus: Connect a limit with continuity and determine a parameter. Unit 1: Limits and continuity.
Let f(x)=e⁻ˣ−x². The function is continuous on [0,1], with f(0)>0 and f(1)<0. Which value is the zero of f in this interval, to three decimal places?
Answer C. The intermediate value theorem guarantees a zero. On (0,1), f′(x)=−e⁻ˣ−2x<0, so there is only one. Solve e⁻ˣ−x²=0 numerically with the bracket [0,1], obtaining x≈0.7034674225. A refinement bracket [0.70346,0.70348] has opposite signs and fixes the requested rounding to 0.703. The answer is a numerical zero of the continuous function, not an inference from a coarse graph alone.
Review focus: Use continuity and a numerical root with a justified interval. Unit 1: Limits and continuity.
For f(x)=√(1+x²) ln(2+x), what is f′(1.7), rounded to three decimal places? Here ln denotes the natural logarithm.
Answer D. Differentiate both factors: f′(x)=[x/√(1+x²)]ln(2+x)+√(1+x²)/(2+x). Evaluate at 1.7, where the logarithm and square root are defined. The two contributions are approximately 1.127696822 and 0.533056295, summing to 1.660753117, so the answer is 1.661. Alternatively a numerical derivative near 1.7 can check this value, but omitting either product-rule term gives a different result.
Review focus: Calculate and interpret a derivative from a supplied formula. Unit 2: Derivative definitions and rules.
The graph of g(x)=e^(−x²)+cos x has a tangent line at x=0.8. What is the x-intercept of that tangent line, to three decimal places? Use radian mode.
Answer A. The tangent is y=g(0.8)+g′(0.8)(x−0.8), with g′(x)=−2x e^(−x²)−sin x. Its x-intercept is 0.8−g(0.8)/g′(0.8). Numerical evaluation gives approximately 1.584100153, which rounds to 1.584. This solves the tangent-line equation, not g(x)=0; the tangent and the original curve need not share their x-intercepts. Retain the full numerical slope and height until the final division.
Review focus: Connect a numerical derivative to a tangent-line equation. Unit 2: Derivative definitions and rules.
Let f(x)=x+ln x for x>0, and let h be its inverse. What is h′(2.5), rounded to three decimal places?
Answer B. First find the correct original input u=h(2.5), rather than evaluating f′ at 2.5. Solving u+ln u=2.5 gives u≈1.8726470404. Since f′(x)=1+1/x>0 on x>0, the inverse exists and h′(2.5)=1/f′(u)=u/(u+1)≈0.6518890118. Thus the answer is 0.652. The numerical inverse input 1.873 is not the derivative.
Review focus: Combine numerical inversion with the inverse derivative rule. Unit 3: Composite, implicit and inverse derivatives.
A particle moves along a line with velocity v(t)=sin(t²)−0.2t meters per second. At t=1.2 seconds, which statement correctly gives its acceleration (rounded to three decimal places) and whether its speed is increasing or decreasing? Use radians.
Answer C. Acceleration is a(t)=2t cos(t²)−0.2, so a(1.2)≈0.113016901 m/s². Velocity is v(1.2)≈0.751458348 m/s, which is positive. At nonzero velocity, speed increases when velocity and acceleration have the same sign; here both are positive. Hence acceleration is 0.113 m/s² and speed increases. The velocity value itself is not acceleration, and dropping the chain factor 2t changes the acceleration.
Review focus: Interpret velocity and acceleration together in context. Unit 4: Contextual differentiation.
A spherical balloon has radius 3.2 centimeters at t=4 seconds. Its volume is increasing at the rate dV/dt=9e^(−t/5) cubic centimeters per second. At that instant, what is dr/dt in centimeters per second, rounded to four decimal places?
Answer A. Differentiate V=(4/3)πr³ with respect to time: dV/dt=4πr² dr/dt. Substitute the supplied instantaneous radius and t=4, giving dr/dt=9e^(−4/5)/(4π·3.2²)≈0.03142657868 cm/s. Thus the answer is 0.0314. No integration of the volume rate is needed because the radius at the relevant instant is already supplied. The numerator alone is a volume rate, with different units.
Review focus: Use a related-rate equation with contextual units. Unit 4: Contextual differentiation.
For 0≤x≤2, a rectangle has vertices (0,0), (x,0), (x,4−x²+sin x), and (0,4−x²+sin x). What is the largest possible area of the rectangle, rounded to three decimal places? Use radians.
Answer D. The area is A(x)=x(4−x²+sin x), not merely the height. Solve A′(x)=4−3x²+sin x+x cos x=0 on [0,2], giving x≈1.3279079761. Then A≈4.2590095597. Endpoint areas are A(0)=0 and A(2)=2sin2≈1.818594854. To ensure all critical candidates are covered, note A″(x)=−6x+2cos x−x sin x and A‴(x)=−6−3sin x−x cos x<0 on [0,2] (the last positive contribution is at most 2). Thus A″ decreases from positive to negative; A′ first rises, then falls from positive to negative and has just one zero. The maximum area is 4.259, not the maximizing x-coordinate.
Review focus: Find and justify a global maximum using numerical critical points. Unit 5: Analytical differentiation.
The Mean Value Theorem applies to f(x)=ln(1+x²) on [1,4]. Which value c in (1,4) satisfies the theorem conclusion, rounded to three decimal places?
Answer B. The average slope is [ln17−ln2]/3=ln(17/2)/3≈0.713355388. Solve f′(c)=2c/(1+c²)=ln(17/2)/3 for c in (1,4), obtaining c≈2.3842288023. The other positive algebraic solution is below 1 and is not an eligible Mean Value Theorem point for this interval. Moreover f′ decreases on (1,4), so the eligible solution is unique. The answer is 2.384; the average slope itself is not c.
Review focus: Locate a Mean Value Theorem point numerically. Unit 5: Analytical differentiation.
A quantity has initial value Q(0)=5 and rate Q′(t)=e^(−t²/3) cos t for 0≤t≤2. What is Q(2), rounded to three decimal places? Use radians.
Answer C. The accumulation theorem gives Q(2)=5+∫₀²e^(−t²/3)cos t dt. Numerical integration gives the signed integral approximately 0.8384159880, so Q(2)≈5.838415988 and rounds to 5.838. The integrand becomes negative after π/2; integrating its absolute value would answer a different question. A calculator may evaluate the integral directly; refine the integration tolerance and retain additional digits before adding the initial value.
Review focus: Evaluate a non-elementary signed accumulation numerically. Unit 6: Integration and accumulation.
A pump rate R(t), in liters per minute, was measured at the times in the table. Using the trapezoidal rule on exactly the three intervals indicated by these times, what is the estimated volume pumped from t=0 to t=6.5 minutes?
| t (minutes) | R(t) (liters/minute) |
|---|---|
| 0 | 2.1 |
| 1.5 | 3.8 |
| 4 | 3.2 |
| 6.5 | 1.4 |
Answer A. Use each actual width: (1.5)(2.1+3.8)/2+(2.5)(3.8+3.2)/2+(2.5)(3.2+1.4)/2=4.425+8.750+5.750=18.925 liters. A common-width formula is invalid because the intervals have unequal lengths. This is the requested trapezoidal estimate; the finite table alone does not establish the exact volume or the direction of its error.
Review focus: Build a numerical integral from unequal-width supplied data. Unit 6: Integration and accumulation.
In the first quadrant, a region is bounded by y=e^(−x²), y=x/2, and the y-axis. What is its area, rounded to three decimal places?
Answer D. The positive intersection solves e^(−b²)=b/2, giving b≈0.8960502208. The exponential is above the line between 0 and b. Thus area=∫₀ᵇ(e^(−x²)−x/2)dx≈0.5037516725, rounding to 0.504. The intersection is unique for x≥0 because the difference has derivative −2x e^(−x²)−1/2<0. Use the stored intersection as the bound rather than rounding b to three decimals before integrating. Integrating only the exponential or reporting b itself does not give the enclosed area.
Review focus: Find an intersection and use it as a numerical area boundary. Unit 8: Applications of integration.
A solid has base in the xy-plane between y=ln(1+x) and y=x/3, for 0≤x≤2. Cross sections perpendicular to the x-axis are squares whose sides span the vertical base width. What is the volume, rounded to three decimal places?
Answer B. The side length is d(x)=ln(1+x)−x/3, which is nonnegative on [0,2]: d(0)=0 and d′(x)=1/(1+x)−1/3≥0 there. Square area is d(x)², so V=∫₀²[ln(1+x)−x/3]²dx≈0.2268571581 cubic units, rounding to 0.227. Numerical integration with a tighter tolerance confirms the rounding. Integrating d alone would give a two-dimensional base area, and using πd² or half d² would describe different sections.
Review focus: Construct and numerically evaluate a cross-sectional volume. Unit 8: Applications of integration.
2 questions · suggested time 30 minutes. Each question has 9 practice-rubric points.
CALCULATOR PERMITTED. Questions 1–2 share a 30-minute section. A botanical garden models attendance continuously for 0≤t≤6 hours after opening. Visitors enter at E(t)=18+10exp(−t²/8) people/hour and leave at L(t)=8+3t people/hour. Initially 120 people are present. The rates are synthetic averages, and capacity and access restrictions do not alter them on this interval. Let P(t) be the modeled attendance. Do not round a continuous model prediction to a whole-person count before completing the calculations.
Find the modeled total number of people entering during the six hours. Show the definite integral.
Find the time and attendance at the absolute maximum of P on [0,6]. Justify that it is the absolute maximum.
Find the average rate of change of attendance over the six hours. Explain why averaging the incoming rate alone would answer a different question.
Find P″(2), with units, and interpret its sign as a statement about the rate of change of attendance.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Total entry=∫₀⁶[18+10exp(−t²/8)]dt≈132.998608895 people. This is incoming attendance, not net change or final attendance.
P′=N=10+10exp(−t²/8)−3t. N′=−(5t/2)exp(−t²/8)−3<0 on [0,6]; N(0)>0 and N(6)<0. Thus N has exactly one zero r≈3.853983891771 hours, and P increases before r and decreases afterward. P(r)=120+∫₀ʳN(t)dt≈159.973240080 people. The sign change on the whole interval establishes the absolute maximum, including comparison with its endpoint behavior.
Average change=[P(6)−P(0)]/6=(1/6)∫₀⁶(E−L)dt. Total departures are 102, so net change is 30.998608895 and the average is 5.166434816 people/hour. Averaging E counts incoming flow without subtracting outgoing flow.
P″(2)=E′(2)−L′(2)=−5exp(−1/2)−3≈−6.032653299 people/hour². At that time the rate at which attendance is changing is decreasing. This does not by itself mean attendance is decreasing; in fact N(2)>0.
Review focus: Numerical accumulation, rates and global optimization. Unit 8: Applications of integration.
CALCULATOR PERMITTED. Questions 1–2 share a 30-minute section. A runner moves on a straight track for 0≤t≤3 seconds with velocity v(t)=3+sin(t²) meters/second. Position at t=0 is 2 meters relative to a fixed marker. Trigonometric arguments use radians. The formula is a stipulated short-time model, not measured data.
Find the total distance traveled on [0,3], showing why your integral represents distance.
Find the time at which the runner first reaches position 8 meters. State an equation, identify a bracket and justify uniqueness.
Is the runner’s speed increasing or decreasing at t=1? Give a calculus justification.
Find the average velocity over [0,3]. Explain why there must be a time c in (0,3) when instantaneous velocity equals that average; do not numerically solve for c.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Since 2≤v(t)≤4, the runner never reverses direction. Distance=∫₀³[3+sin(t²)]dt≈9.773562527 meters. Here displacement and distance coincide because velocity is positive throughout.
Solve 2+∫₀ᵀ[3+sin(t²)]dt=8, or ∫₀ᵀv=6. The integrals to 1 and 2 are approximately 3.310268302 and 6.804776489, giving a bracket (1,2). Position is continuous and strictly increasing because v≥2. Consequently the first and only time is T≈1.704410619444 seconds.
Acceleration is a(t)=2t cos(t²); a(1)=2cos(1)≈1.080604612>0. Also v(1)=3+sin(1)>0, so the derivative of speed equals acceleration there and speed is increasing.
Average velocity=(1/3)∫₀³v≈3.257854176 meters/second. The position s(t)=2+∫₀ᵗv is continuous on [0,3] and differentiable on (0,3), because v is continuous. The mean value theorem gives s′(c)=[s(3)−s(0)]/3, hence v(c) equals the computed average.
Review focus: Numerical motion, an integral equation and theorem justification. Unit 8: Applications of integration.
4 questions · suggested time 60 minutes. Each question has 9 practice-rubric points.
NO CALCULATOR. Questions 3–6 share a 60-minute section. The supplied graph defines a continuous function g by straight segments joining (0,2), (2,0), (4,−2), and (6,2). These are exact segments, not interpolations of unknown measurements. Define H(x)=1+∫₀ˣg(t)dt for 0≤x≤6.
Graph description: Exact continuous piecewise-linear integrand g, descending from (0,2) through (2,0) to (4,−2), then ascending to (6,2). The final segment crosses the horizontal axis at x=5. g (solid): straight segments through (0, 2), (2, 0), (4, -2), (6, 2). Axes use the labeled scales; no values outside the specified domain are supplied.
Find H(4) and H(5), using signed geometric areas.
Find the absolute maximum and minimum of H on [0,6], giving values and every location. Justify your answer.
Give the open intervals of concavity of H and identify its inflection inputs. Does H″(4) exist?
For 0<x<√6, let K(x)=H(x²). Find K′(√3), showing the graph lookup.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Areas from 0 to 2 and from 2 to 4 are +2 and −2, so H(4)=1. From 4 to 5 the negative triangle has area −1, hence H(5)=0. The initial value must be included.
H′=g is positive on (0,2), negative on (2,5), and positive on (5,6). Candidate values are H(0)=1,H(2)=3,H(5)=0,H(6)=1. Thus the absolute maximum is 3 at x=2 and absolute minimum is 0 at x=5.
g decreases on (0,4), so H is concave down there; g increases on (4,6), so H is concave up there. Thus x=4 is an inflection input. The left and right slopes of g are −1 and 2, so H″(4)=g′(4) does not exist, although H′(4)=g(4)=−2 exists.
K′(x)=2xH′(x²)=2xg(x²). At x=√3, g(3)=−1, so K′(√3)=−2√3.
Review focus: Supplied integrand graph, signed area and FTC. Unit 6: Integration and accumulation.
NO CALCULATOR. A differentiable branch of the curve x²+xy+y²=19 passes through (2,3). Coordinates are in meters when describing the motion in part (c).
Find dy/dx and the tangent line at (2,3).
Find d²y/dx² at (2,3) and classify local concavity of this branch near the point.
A particle moves along the curve through (2,3), with dx/dt=3 meters/second at that instant. Find the rate of change of its distance from the origin.
Find all points on the curve with a horizontal tangent. Check that the derivative denominator is nonzero there.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Differentiation gives 2x+y+xy′+2yy′=0, so y′=−(2x+y)/(x+2y). At (2,3), y′=−7/8 and the tangent line is y−3=−(7/8)(x−2).
Differentiate 2x+y+(x+2y)y′=0 to obtain 2+2y′+2(y′)²+(x+2y)y″=0. At the point the numerator is 2−7/4+49/32=57/32, and x+2y=8. Thus y″=−57/256. The branch has a continuous second derivative near the point since x+2y stays nonzero there; it is locally concave down.
dy/dt=(dy/dx)(dx/dt)=−21/8. If ρ=√(x²+y²), then dρ/dt=(x dx/dt+y dy/dt)/ρ. At the point this is (6−63/8)/√13=−15/(8√13) meters/second. Its distance is decreasing.
Horizontal requires 2x+y=0 with x+2y≠0. Substitute y=−2x to get 3x²=19. The points are (√(19/3),−2√(19/3)) and (−√(19/3),2√(19/3)). At each, x+2y=−3x≠0.
Review focus: Implicit derivatives, tangent geometry and related rates. Unit 3: Composite, implicit and inverse derivatives.
NO CALCULATOR. Let y be the solution of y′=t(1+y) with y(0)=1. Consider 0≤t≤1.
Find y′(0) and y″(0). Explain why a horizontal initial tangent does not make the solution constant.
Solve the initial-value problem by separation of variables. Show how the initial condition determines the constant.
Use the tangent line to the solution at t=1 to approximate y(9/10). Give an exact expression for the approximation.
Is the tangent-line approximation in part (c) below or above the exact y(9/10)? Justify using concavity on the interval between 9/10 and 1.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
y′(0)=0. Along a solution, y″=1+y+ty′, so y″(0)=2. The positive second derivative means the solution initially bends upward; y=1 constant would not satisfy the equation at positive t.
Near the initial point 1+y>0, so ∫dy/(1+y)=∫t dt gives ln|1+y|=t²/2+C. The initial value gives C=ln2 and the positive branch yields 1+y=2exp(t²/2). Therefore y=2exp(t²/2)−1, which verifies the equation and initial condition on [0,1]. The omitted constant solution y=−1 does not satisfy this initial condition.
From part (b), y(1)=2√e−1. The differential equation gives y′(1)=1+y(1)=2√e. The tangent line is L(t)=2√e−1+2√e(t−1). Thus L(9/10)=(9/5)√e−1.
The exact solution has 1+y=2exp(t²/2)>0. Substituting y′=t(1+y) into the second derivative gives y″=(1+y)(1+t²)>0 throughout [9/10,1]. A strictly convex differentiable graph lies above its tangent line away from the tangency point, so the approximation is below the exact value.
Review focus: Separation, local behavior and tangent-line approximation. Unit 7: Differential equations.
NO CALCULATOR. The region R is bounded by y=2x−x² and y=0 for 0≤x≤2. All lengths are in coordinate units. Let F(x)=∫₀ˣ(2t−t²)dt.
Find the area of R, justifying the order of its boundaries.
A solid has base R. Its cross sections perpendicular to the x-axis are rectangles whose base is the vertical width of R and whose other side has length 1+x. Find its volume.
Use the mean value theorem to justify that there is c in (0,2) with F′(c)=2/3, and find all such c.
Instead rotate R about the horizontal line y=−1. Set up and evaluate the volume using washers. Identify both radii.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
2x−x²=x(2−x)≥0 on [0,2]. Area=∫₀²(2x−x²)dx=[x²−x³/3]₀²=4/3 square units.
The area of a section is (2x−x²)(1+x)=2x+x²−x³. Volume=∫₀²(2x+x²−x³)dx=4+8/3−4=8/3 cubic units.
F is continuous on [0,2] and differentiable on (0,2), since its integrand is a polynomial. Its average rate is [F(2)−F(0)]/2=(4/3)/2=2/3. The mean value theorem supplies c. Solving 2c−c²=2/3 gives c=1±1/√3, both in (0,2).
Outer radius is 1+2x−x² and inner radius is 1. Volume=π∫₀²[(1+2x−x²)²−1]dx. Writing w=2x−x² gives integrand 2w+w²; the two integrals are 8/3 and 16/15. Thus volume=56π/15 cubic units.
Review focus: Area, cross sections, washers and the mean value theorem. Unit 8: Applications of integration.
Each free-response rubric entry is worth one point. Accept equivalent exact expressions. For calculator results, accept values within 0.001 of the stated reference unless an exact value is requested; retain unrounded intermediate values. This tolerance is a convention for this practice form, not a universal official scoring rule.
Award one point for each correct multiple-choice answer and zero for an incorrect or omitted answer. For free response, use the stated criteria rather than rewarding a matching final number alone. Accept equivalent valid mathematics. Unless a criterion explicitly depends on a correct earlier value, award later method points for mathematically consistent carry-forward work; do not award a correctness point for a wrong value. Record the specific error rather than subtracting the same point twice.
Keep the two raw totals separate: multiple choice out of 42 and free response out of 54. An optional practice-only weighted percentage is 50×(MC/42)+50×(FR/54). This is not an AP 1–5 conversion or a passing threshold. An independently reviewed explanation is stronger evidence than self-marking after reading a solution.
For each missed or uncertain item, record its ID, your original reasoning, the error type, a corrected argument and a later unfamiliar retry. Distinguish a method-selection gap from algebra, graph reading, missing justification, calculator operation or premature rounding. Repeating this exposed form does not create a fresh independent test.
Each multiple-choice item has one primary unit label. Questions can use several ideas; this table is not proof that every learning objective is tested.
| Primary unit | Questions |
|---|---|
| 1. Limits and continuity | 5 |
| 2. Derivative definitions and rules | 5 |
| 3. Composite, implicit and inverse derivatives | 4 |
| 4. Contextual differentiation | 5 |
| 5. Analytical differentiation | 7 |
| 6. Integration and accumulation | 8 |
| 7. Differential equations | 3 |
| 8. Applications of integration | 5 |
Return to graphs, tables and accumulation, derivative reasoning, integration, calculus synthesis or differential equations. The full mathematics curriculum and additional examination forms remain in development.
Also available: the original Calculus BC practice form.
Format checked 25 October 2026 against the official AB exam overview; unit organization checked against the official AB course overview. For the actual digital environment and device rules, use the official Bluebook guidance and calculator policy.
