Free-response practiceAP Calculus AB

AP Calculus AB FRQ Practice: Four Original Teaching Tasks

Write a complete solution before revealing the worked reasoning. These four independently authored teaching tasks connect sampled data, implicit differentiation, differential equations and an accumulated function. Use the checklists to inspect your methods, conditions and communication, not to calculate an official AP score.

Version 2026-10-28Published 2026-10-28Free · No sign-up

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Read this scope first

Selected AB-level practice for the 2026–27 course. This four-task resource is not a complete six-question exam, an official College Board paper, an official rubric or a validated scoring simulation. Classroom time estimates are editorial. The linked official released questions and scoring guidelines are separate study references.

Author calculations and source-version checks are recorded separately. Independent agent review is distinct from qualified AP educator review. Self-assessment items are local learning criteria, not official marks.

Independent content review: peer checkedQualified educator review: pending

Independent checks and qualified educator review are separate. This original study resource does not claim official endorsement, a validated score prediction or a pass guarantee.

Write, check, and revise

Attempt one task on paper before opening its solutions. Label every subpart, show the setup, and give the reason for each conclusion. A numerical answer alone may hide a missing hypothesis or a sign error.

Use the question-only PDF for a clean attempt, and the separate solution PDF for worked reasoning and self-assessment. These downloads contain the same original tasks as this page.

Keep approximate results distinct from exact results.
Name the theorem and verify its conditions when a justification is requested.
State units in contextual rate and accumulation answers.
Treat each self-assessment statement as a learning check, not as an official point or AP score.

Four goals and their prerequisites

The tasks emphasize different mathematical decisions. Table values support estimates and a theorem argument; an implicit equation supports tangent and related-rate reasoning; an initial value selects a differential-equation solution; graph vertices support FTC and sign analysis.

Task	Main goal	Suggested time
1	Sampled rates, net accumulation and MVT	18 minutes
2	Implicit derivatives, tangent behavior and related rates	20 minutes
3	Separation, initial values and a justified estimate	18 minutes
4	Read a derivative graph through an accumulation function	18 minutes

How this practice relates to the current exam

The current official exam page lists six free-response questions in 90 minutes:two calculator-required questions in 30 minutes and four calculator-not-permitted questions in 60 minutes. The 2026–27 clarifications do not add new AB course content; the announced May 2027 question-count changes concern multiple choice.

This resource deliberately contains only four teaching tasks. Its practice calculator instructions and suggested times do not recreate a complete official section. The linked released 2026 paper and its scoring guidelines retain their own version-specific instructions.

Continue with official released material

After revising these tasks, open the official released-question index and choose a released paper. Use that paper's own calculator directions and published scoring guidelines. Compare whether your written work establishes the requested setup, reason and interpretation rather than copying an answer.

The numerical data, functions, graph vertices, prompts and local checklists below were authored for this resource. No College Board question or official point table is reproduced, and no qualified educator approval is claimed.

Write your own solution before reading the worked reasoning. Explain assumptions, notation and units. The checklist supports self-assessment; it does not assign official marks.

ORIGINAL MULTI-PART TEACHING TASK

Task 1 — A reservoir observed at uneven times

Sampled rates, trapezoidal accumulation, net change and the Mean Value Theorem18 estimated minutes

Calculator use: Practice instruction:calculator optional for arithmetic; show the quotient, trapezoidal setup and integral model. These are not official section-placement instructions.

A reservoir contains 60 liters at time $t=0$ . Water enters at a nonnegative rate $R(t)$ , in liters per minute, and leaves at the constant rate 4 liters per minute for $0\le t\le8$ . The reservoir has enough capacity for this model. Assume $R$ is continuous on $[0,8]$ and differentiable on $(0,8)$ . Only the table values of $R$ are known; do not assume linear interpolation. Let $V(t)$ be the amount of water, in liters.

Original sampled-rate table. Values between sample times are not specified.
t (minutes)	R(t) (liters/minute)
0	3
2	7
5	5
8	1

a Estimate $R^{\prime}(7/2)$ using the average rate of change on $[2,5]$ . Include units and interpret the sign.

Worked reasoning

Use the two specified samples: $R^{\prime}(7/2)\approx [R(5)-R(2)]/(5-2)$ .
The quotient is $(5-7)/3=-2/3$ liters per minute per minute.
The estimate suggests that the inflow rate is decreasing near $t=7/2$ . It is not the rate of change of the reservoir volume, and it is not an exact derivative determined by the table.

Answer: $R^{\prime}(7/2)\approx-2/3\ \text{L/min}^2$ ; the inflow rate is estimated to be decreasing.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Use a difference quotient with both the rate difference and the time difference.
Report units of liters per minute squared, not liters or liters per minute.
Identify the derivative as an estimate and interpret its negative sign for the inflow rate.

b Use trapezoids on $[0,2]$ , $[2,5]$ and $[5,8]$ to estimate $\int_0^8R(t)\,dt$ . Explain what this integral measures.

Worked reasoning

The interval widths are 2,3 and 3 minutes; the partition is not uniform.
The trapezoidal estimate is $2(3+7)/2+3(7+5)/2+3(5+1)/2$ .
The three contributions are 10,18 and 9 liters, so the estimate is 37 liters.
The integral measures the total water entering during the eight-minute interval. Since the true rates between samples are unspecified, the trapezoidal result is an estimate, not a guaranteed exact total.

Answer: $\int_0^8R(t)\,dt\approx37$ liters of incoming water.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Use all three actual interval widths with the average of each pair of endpoint rates.
Obtain 37 from the displayed trapezoidal setup.
Interpret incoming volume over the stated interval and preserve the approximation sign.

c Write an exact integral expression for $V(8)$ , then estimate $V(8)$ using part (b).

Worked reasoning

The net rate of volume change is $V^{\prime}(t)=R(t)-4$ .
Initial amount plus accumulated net change gives $V(8)=60+\int_0^8[R(t)-4]\,dt$ .
The constant outflow removes $4\cdot8=32$ liters. Use the estimate 37 liters for inflow: $V(8)\approx60+37-32=65$ liters.

Answer: Exact expression: $60+\int_0^8[R(t)-4]\,dt$ . Estimated amount:65 liters.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Include the initial 60 liters.
Subtract the outflow over the full eight minutes rather than subtracting 4 only once.
Distinguish the exact integral model from its 65-liter numerical estimate.

d Prove that some $c\in(2,5)$ satisfies $R^{\prime}(c)=-2/3$ . Does this prove that $R^{\prime}(7/2)=-2/3$ exactly? Explain.

Worked reasoning

The assumed continuity on $[0,8]$ implies continuity on $[2,5]$ , and differentiability on $(0,8)$ implies differentiability on $(2,5)$ .
The Mean Value Theorem therefore provides some $c\in(2,5)$ with $R^{\prime}(c)=[R(5)-R(2)]/(5-2)=-2/3$ .
The theorem guarantees an unspecified interior point. It does not identify $c$ as $7/2$ , so it does not convert part (a)'s estimate at that particular point into an exact derivative.

Answer: Such a $c$ exists by the Mean Value Theorem, but its location need not be $7/2$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Verify continuity on the closed interval and differentiability on its interior.
Use the endpoint quotient to establish the guaranteed derivative value.
Separate existence of some interior point from identifying the midpoint.

Lesson: integral data Lesson: integration Lesson: theorem decisions

ORIGINAL MULTI-PART TEACHING TASK

Task 2 — A positive branch of an implicit constraint

Implicit derivatives, tangent approximation, concavity and related rates20 estimated minutes

Calculator use: Practice instruction:no calculator is needed. Keep exact fractions; show implicit differentiation before substituting the point.

Consider the curve $2x^2+xy+3y^2=13$ and its positive, twice-differentiable branch through $(2,1)$ . This branch remains positive for $2\le x\le21/10$ . Parts(a)–(c) use dimensionless coordinates; part (d) introduces a separate length-and-time model with the same algebraic constraint.

a Find $dy/dx$ in terms of $x$ and $y$ . Then find the tangent-line equation at $(2,1)$ .

Worked reasoning

Differentiate every term with respect to $x$ : $4x+y+xy^{\prime}+6yy^{\prime}=0$ . The $xy$ term requires the product rule.
Collect the derivative terms: $(x+6y)y^{\prime}=-(4x+y)$ , so $y^{\prime}=-(4x+y)/(x+6y)$ .
At $(2,1)$ , the denominator is 8 and the numerator is 9, so the slope is $-9/8$ .
The tangent line is $L(x)=1-(9/8)(x-2)$ .

Answer: $y^{\prime}=-(4x+y)/(x+6y)$ ; $y-1=-(9/8)(x-2)$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Differentiate xy with the product rule and y² with the chain rule.
Collect y′ terms before dividing, and note that x+6 y is nonzero at the point.
Use both the slope −9/8 and the point (2,1) in the tangent equation.

b Find $d^2y/dx^2$ at $(2,1)$ . Justify the sign of the result.

Worked reasoning

Differentiate $4x+y+(x+6y)y^{\prime}=0$ once more: $4+2y^{\prime}+6(y^{\prime})^2+(x+6y)y^{\prime\prime}=0$ .
Therefore $y^{\prime\prime}=-[4+2y^{\prime}+6(y^{\prime})^2]/(x+6y)$ .
Substitute $y^{\prime}=-9/8$ , $x=2$ , $y=1$ . The numerator is $4-18/8+6(81/64)=299/32$ ; divide by 8 with the leading minus sign.
Thus $y^{\prime\prime}(2)=-299/256<0$ , which means this branch is concave down at the point.

Answer: $y^{\prime\prime}(2)=-299/256$ , so the branch is concave down there.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Account for both y′ contributions and the 6(y′)²term.
Use the previously found slope and compute the exact fraction −299/256.
Link the negative second derivative to concavity rather than to whether y itself is positive.

c Use the tangent line to estimate $y$ at $x=21/10$ . Is the tangent estimate above or below the actual branch value? Justify the comparison throughout $[2,21/10]$ , not just by giving the sign at one point.

Worked reasoning

Insert $x-2=1/10$ into the tangent line: $L(21/10)=1-9/80=71/80$ .
On the stated positive branch, $x+6y>0$ . For any real value $p=y^{\prime}$ , $4+2p+6p^2=6(p+1/6)^2+23/6>0$ .
The second-derivative expression from part (b) is therefore negative throughout the stated interval. The branch is concave down there, so its tangent line lies above the branch.
The estimate $71/80$ is an overestimate. A negative second derivative only at the initial point would not, by itself, justify the comparison over the entire requested interval.

Answer: $y(21/10)\approx71/80=0.8875$ , an overestimate.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Evaluate the tangent line at the requested input rather than treating the slope as the estimate.
Use positivity of the denominator and strict positivity of the quadratic numerator on the whole interval.
Conclude that concave-down behavior puts this tangent estimate above the branch.

d In a separate model, $x$ and $y$ are positive lengths in centimeters that depend on time $s$ , in seconds, and satisfy the same constraint(with constant 13 square centimeters). At the instant $x=2$ , $y=1$ and $dx/ds=4$ centimeters per second, find $dy/ds$ . Interpret the sign.

Worked reasoning

Differentiate the constraint with respect to time: $(4x+y)\,dx/ds+(x+6y)\,dy/ds=0$ .
At the stated instant, $9\cdot4+8\,dy/ds=0$ .
Hence $dy/ds=-36/8=-9/2$ centimeters per second. The negative sign means the length y is decreasing at that instant.

Answer: $dy/ds=-9/2\ \text{cm/s}$ ; the length y is decreasing.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Differentiate both changing lengths with respect to the same time variable.
Use 9 and 8 for the two coefficient values at the instant.
Give the negative rate −9/2 with units and its contextual interpretation.

Lesson: implicit Lesson: combined rules Lesson: local analysis Lesson: related constraints

ORIGINAL MULTI-PART TEACHING TASK

Task 3 — Select and test a differential-equation solution

Separation of variables, initial conditions, concavity and a unique threshold18 estimated minutes

Calculator use: Practice instruction:exact work requires no calculator. A calculator may be used only to report the optional three-decimal value in part (d).

A positive function $y$ satisfies $dy/dx=(2x+1)/(2y)$ for $0\le x\le2$ , with $y(0)=2$ . The variables in this task are dimensionless. Write supporting differentiation or integration steps, not just a candidate formula.

a Find $y^{\prime}(0)$ , write the tangent line at $x=0$ , and use it to estimate $y(1/5)$ .

Worked reasoning

The differential equation gives $y^{\prime}(0)=1/(2\cdot2)=1/4$ .
The line through $(0,2)$ with this slope is $L(x)=2+x/4$ .
Therefore $y(1/5)\approx L(1/5)=2+1/20=41/20$ .

Answer: $y^{\prime}(0)=1/4$ ; $L(x)=2+x/4$ ; $y(1/5)\approx41/20$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Substitute both x=0 and y=2 into the differential equation.
Use the initial value as the tangent-line intercept.
Label 41/20 as an approximation rather than an exact solution value.

b Use separation of variables to find the particular solution. Explain how the sign of the square root is determined, and verify the differential equation and initial condition.

Worked reasoning

Because the stated solution is positive, its denominator is nonzero. Separate and integrate: $2y\,dy=(2x+1)\,dx$ , so $y^2=x^2+x+C$ .
The initial condition gives $4=C$ . Thus $y=\sqrt{x^2+x+4}$ ; the negative square-root branch would have value −2 at x=0 and violates both the initial condition and positivity.
Differentiating the positive branch yields $y^{\prime}=(2x+1)/(2\sqrt{x^2+x+4})=(2x+1)/(2y)$ . Also $y(0)=\sqrt4=2$ .

Answer: $y(x)=\sqrt{x^2+x+4}$ on the stated interval.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Show separation and integration with a constant of integration.
Determine C=4 and explicitly choose the positive branch.
Verify both the derivative equation and the initial value; one check alone is incomplete.

c Use the particular solution to prove that the tangent estimate in part (a) is an underestimate.

Worked reasoning

Differentiate the first-derivative expression, using $y$ as a function of $x$ : $y^{\prime\prime}=[4y^2-(2x+1)^2]/(4y^3)$ .
From part (b), $y^2=x^2+x+4$ . Hence the numerator is $4(x^2+x+4)-(2x+1)^2=15$ .
Therefore $y^{\prime\prime}=15/(4y^3)>0$ on $[0,2]$ . The graph is concave up, so the tangent line at 0 lies below it for $0<x\le2$ , including $x=1/5$ .

Answer: $41/20<y(1/5)$ ; the tangent estimate is an underestimate.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Differentiate with the chain/quotient rules rather than treating y as constant.
Show the numerator simplifies to 15 and use y>0.
Use concavity over the relevant interval to justify the tangent comparison.

d Find the first $x\in[0,2]$ for which $y(x)=3$ . Give an exact answer; an optional decimal should be rounded to three places.

Worked reasoning

Set $\sqrt{x^2+x+4}=3$ . Squaring gives $x^2+x-5=0$ .
The two algebraic roots are $(-1\pm\sqrt{21})/2$ . Only $(-1+\sqrt{21})/2$ lies in $[0,2]$ .
On this interval, $2x+1>0$ and $y>0$ , so $y^{\prime}>0$ . The solution is strictly increasing, which makes this crossing unique and therefore the first one.
The optional decimal is $1.791$ .

Answer: $x=(-1+\sqrt{21})/2\approx1.791$ , the unique crossing in the stated interval.

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Solve the threshold equation rather than using the tangent approximation as the actual solution.
Discard the negative root using the stated interval.
Justify that the positive crossing is unique by the sign of y′.

Lesson: differential equations Lesson: ode reasoning Lesson: local analysis

ORIGINAL MULTI-PART TEACHING TASK

Task 4 — Recover behavior from a graph of a rate

FTC, signed geometric area, local extrema and absolute extrema18 estimated minutes

Calculator use: Practice instruction:no calculator is needed. Use signed areas of triangles, trapezoids and rectangles and show the endpoint comparisons.

A continuous function $h$ on $[-2,6]$ has a graph consisting exactly of straight-line segments joining consecutive vertices in the table. The table is an original exact description of the graph, not a collection of unspecified samples. Define $G(x)=-2+\int_0^x h(t)\,dt$ . All coordinates and function values in this task are dimensionless.

Original polygonal graph represented by its vertices. Join consecutive rows by straight segments; between x=0 and x=2 the graph is horizontal.
x-coordinate	h(x) at a graph vertex
−2	1
0	−1
2	−1
4	3
6	1

a Find $G^{\prime}(3)$ and $G^{\prime\prime}(3)$ . Explain how each value is read from the given graph.

Worked reasoning

Since h is continuous, the Fundamental Theorem of Calculus gives $G^{\prime}(x)=h(x)$ .
Between 2 and 4, the line through $(2,-1)$ and $(4,3)$ has slope $[3-(-1)]/(4-2)=2$ , so $h(3)=1$ .
Therefore $G^{\prime}(3)=1$ . At this non-corner point, $G^{\prime\prime}(3)=h^{\prime}(3)=2$ . The value of h and the slope of h answer different derivative questions.

Answer: $G^{\prime}(3)=1$ and $G^{\prime\prime}(3)=2$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Use G′=h from continuity and FTC.
Read the line value 1 at x=3 rather than using its slope for G′.
Use the segment slope 2 for G″and note that x=3 is not a corner.

b Compute $G(4)$ using signed areas.

Worked reasoning

On $[0,2]$ , h=−1, so $\int_0^2h(t)\,dt=-2$ .
On $[2,4]$ , the exact signed trapezoidal area is $2[-1+3]/2=2$ . Equivalently, the small negative triangle has area −1/4 and the positive triangle has area 9/4.
Thus $\int_0^4h(t)\,dt=0$ , and $G(4)=-2+0=-2$ .

Answer: $G(4)=-2$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Use signed area, not total unsigned area.
Include the constant −2 in G.
Explain that the integral is exact here because the complete polygonal graph is specified.

c Find every interior local maximum and local minimum of G. Give the x-values and justify each classification.

Worked reasoning

The segment from $(-2,1)$ to $(0,-1)$ crosses the x-axis at −1. The segment from $(2,-1)$ to $(4,3)$ crosses at $5/2$ .
Because $G^{\prime}=h$ , G is increasing on $(-2,-1)$ , decreasing on $(-1,5/2)$ , and increasing on $(5/2,6)$ .
At −1, the derivative changes from positive to negative, giving a local maximum. At 5/2, it changes from negative to positive, giving a local minimum. The other graph vertices are not zeros of h and do not create additional local extrema of G.

Answer: Interior local maximum at $x=-1$ ; interior local minimum at $x=5/2$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Find zeros of h rather than zeros or corners of G without evidence.
State the derivative sign change at each critical point.
Check all interior possibilities rather than identifying only one extremum.

d Find the absolute maximum and absolute minimum values of G on $[-2,6]$ , and where they occur. Justify why the candidate list is complete.

Worked reasoning

G is continuous on the closed interval because h is continuous. The Extreme Value Theorem guarantees absolute extrema. The endpoint/critical-point candidates are $-2,-1,5/2,6$ , since $G^{\prime}=h$ exists throughout the interior and is zero only at −1 and 5/2.
The signed areas on $[-2,0]$ cancel, so $G(-2)=-2$ . On $[-1,0]$ , the integral of h is −1/2; reversing the integral gives $G(-1)=-2+1/2=-3/2$ .
From 0 to 2, the integral is −2; from 2 to 5/2, it is a negative triangle of area −1/4. Hence $G(5/2)=-2-2-1/4=-17/4$ .
Part(b) gives $G(4)=-2$ . The exact trapezoidal area on $[4,6]$ is $2(3+1)/2=4$ , so $G(6)=2$ .
Compare all four candidate values:the absolute minimum is −17/4 at x=5/2, and the absolute maximum is 2 at x=6.

Answer: Absolute minimum $-17/4$ at $x=5/2$ ; absolute maximum 2 at $x=6$ .

Self-assessment checklist

Use these learning checks after writing your response. They are not official scoring criteria.

Use continuity for existence, but also identify the complete endpoint/critical-point candidate list.
Handle the reversed integral for negative inputs with the correct sign.
Compute and compare all four candidate values; local extrema alone do not establish absolute extrema.

Lesson: ftc Lesson: accumulation analysis Lesson: sign analysis Lesson: integral data

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Sources and review notes

The references establish the stated framework and source basis. Original examples and study tasks are independently authored; reference links do not imply endorsement or permission to reproduce official test items.

AP Calculus AB course and current content framework
Version / date: 2026–27 course page, reviewed 2026-10-28
Locator: Course Content and Mathematical Practices; eight-unit AB framework.
AP Calculus AB exam overview
Version / date: Current 2026–27 page / May 2027 exam, reviewed 2026-10-28
Locator: Clarifications and Exam Updates for 2026–27; SectionII:Free Response.
AP Calculus AB and BC Course and Exam Description
Version / date: CED linked by the current 2026–27 course page; retained 2026-10-28
Locator: Topics 3.2,3.6,4.4–4.6,5.1–5.6,6.2,6.4–6.5, and 7.6–7.7. Relevant PDF pages 76,80,90–92,99–104,119,121–122,142–143.
AP Calculus AB and BC CED clarifications and corrections
Version / date: Effective Fall 2026
Locator: PDF pages 1–2:EVT clarification, FUN-7. B.2, and May 2027 MCQ changes.
Official AP Calculus AB released FRQs and scoring resources
Version / date: Current public index, reviewed 2026-10-28
Locator: 2026 released FRQ and Scoring Guidelines links.
2026 AP Calculus AB released free-response questions
Version / date: 2026 released examination
Locator: PDF pages 2 and 6:calculator/noncalculator directions and written-justification instructions. Questions are linked for separate practice, not reproduced here.
2026 AP Calculus AB released scoring guidelines
Version / date: 2026 released scoring guidelines
Locator: PDF page 2:general scoring notes and setup/units distinction; question-specific notes illustrate that criteria are tied to actual official prompts. No official point scheme is assigned to these new tasks.

Lesson: integral data Lesson: integration Lesson: theorem decisions Lesson: implicit Lesson: combined rules Lesson: related constraints Lesson: local analysis Lesson: differential equations Lesson: ode reasoning Lesson: ftc Lesson: accumulation analysis Lesson: sign analysis

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Write, check, and revise

Four goals and their prerequisites

How this practice relates to the current exam

Continue with official released material

a Estimate R′(7/2)R^{\prime}(7/2)R′(7/2) using the average rate of change on [2,5][2,5][2,5]. Include units and interpret the sign.

Worked reasoning

Self-assessment checklist

b Use trapezoids on [0,2][0,2][0,2],[2,5][2,5][2,5] and [5,8][5,8][5,8] to estimate ∫08R(t) dt\int_0^8R(t)\,dt∫08​R(t)dt. Explain what this integral measures.

Worked reasoning

Self-assessment checklist

c Write an exact integral expression for V(8)V(8)V(8), then estimate V(8)V(8)V(8) using part (b).

Worked reasoning

Self-assessment checklist

d Prove that some c∈(2,5)c\in(2,5)c∈(2,5) satisfies R′(c)=−2/3R^{\prime}(c)=-2/3R′(c)=−2/3. Does this prove that R′(7/2)=−2/3R^{\prime}(7/2)=-2/3R′(7/2)=−2/3 exactly? Explain.

Worked reasoning

Self-assessment checklist

a Find dy/dxdy/dxdy/dx in terms of xxx and yyy. Then find the tangent-line equation at (2,1)(2,1)(2,1).

Worked reasoning

Self-assessment checklist

b Find d2y/dx2d^2y/dx^2d2y/dx2 at (2,1)(2,1)(2,1). Justify the sign of the result.

Worked reasoning

Self-assessment checklist

c Use the tangent line to estimate yyy at x=21/10x=21/10x=21/10. Is the tangent estimate above or below the actual branch value? Justify the comparison throughout [2,21/10][2,21/10][2,21/10], not just by giving the sign at one point.

Worked reasoning

Self-assessment checklist

Worked reasoning

Self-assessment checklist

a Find y′(0)y^{\prime}(0)y′(0), write the tangent line at x=0x=0x=0, and use it to estimate y(1/5)y(1/5)y(1/5).

Worked reasoning

Self-assessment checklist

b Use separation of variables to find the particular solution. Explain how the sign of the square root is determined, and verify the differential equation and initial condition.

Worked reasoning

Self-assessment checklist

c Use the particular solution to prove that the tangent estimate in part (a) is an underestimate.

Worked reasoning

Self-assessment checklist

d Find the first x∈[0,2]x\in[0,2]x∈[0,2] for which y(x)=3y(x)=3y(x)=3. Give an exact answer; an optional decimal should be rounded to three places.

Worked reasoning

Self-assessment checklist

a Find G′(3)G^{\prime}(3)G′(3) and G′′(3)G^{\prime\prime}(3)G′′(3). Explain how each value is read from the given graph.

Worked reasoning

Self-assessment checklist

b Compute G(4)G(4)G(4) using signed areas.

Worked reasoning

Self-assessment checklist

c Find every interior local maximum and local minimum of G. Give the x-values and justify each classification.

Worked reasoning

Self-assessment checklist

d Find the absolute maximum and absolute minimum values of G on [−2,6][-2,6][−2,6], and where they occur. Justify why the candidate list is complete.

Worked reasoning

Self-assessment checklist

Download the resource PDFs

Sources and review notes

a Estimate $R^{\prime}(7/2)$ using the average rate of change on $[2,5]$ . Include units and interpret the sign.

b Use trapezoids on $[0,2]$ , $[2,5]$ and $[5,8]$ to estimate $\int_0^8R(t)\,dt$ . Explain what this integral measures.

c Write an exact integral expression for $V(8)$ , then estimate $V(8)$ using part (b).

d Prove that some $c\in(2,5)$ satisfies $R^{\prime}(c)=-2/3$ . Does this prove that $R^{\prime}(7/2)=-2/3$ exactly? Explain.

a Find $dy/dx$ in terms of $x$ and $y$ . Then find the tangent-line equation at $(2,1)$ .

b Find $d^2y/dx^2$ at $(2,1)$ . Justify the sign of the result.

c Use the tangent line to estimate $y$ at $x=21/10$ . Is the tangent estimate above or below the actual branch value? Justify the comparison throughout $[2,21/10]$ , not just by giving the sign at one point.

a Find $y^{\prime}(0)$ , write the tangent line at $x=0$ , and use it to estimate $y(1/5)$ .

d Find the first $x\in[0,2]$ for which $y(x)=3$ . Give an exact answer; an optional decimal should be rounded to three places.

a Find $G^{\prime}(3)$ and $G^{\prime\prime}(3)$ . Explain how each value is read from the given graph.

b Compute $G(4)$ using signed areas.

d Find the absolute maximum and absolute minimum values of G on $[-2,6]$ , and where they occur. Justify why the candidate list is complete.