AP Calculus AB Unit 5 FRQ Practice: Analyzing Functions

Q: What is on AP Calculus AB Unit 5?

Unit 5, Analytical Applications of Differentiation, has 12 topics in the College Board's course description: the Mean Value Theorem; the Extreme Value Theorem, critical points, and global versus local extrema; intervals where a function is increasing or decreasing; relative extrema by the first derivative test; absolute extrema by the candidates test; concavity and points of inflection; the second derivative test; sketching the graphs of a function and its derivatives; connecting f, f′ and f″; setting up and solving optimization problems; and the behavior of implicit relations, such as horizontal and vertical tangents and the second derivative. The course description weights the unit at 15–20% of the multiple-choice section.

Q: How is an AP Calculus AB free-response question scored?

Each question is worth 9 points, spread over its three or four parts. Points are earned for specific pieces of work: a setup, an antiderivative, a value with its units, a conclusion with its reason. The scoring guide lists every point and what earns it, then a worked solution for each part. On the exam the six free-response questions are half of the score.

Q: Can students use a calculator?

Only where the exam allows one. On this set, all four are in Part B, without a calculator, as almost every Unit 5 free-response part on the exam has been.

Q: How were the questions chosen?

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 5 settings the exam leans on most get the most questions here, each in its own context with its own parts.

Q: Are these real AP exam questions?

No. Every question here is original, written by Sigma Prep in the style of the exam: the settings, functions, numbers and wording are our own, and no official question or scoring guideline is copied or reworded. Every value and statement in the scoring guide was checked by an independent computer-algebra check before the set was published.

Q: Is there a scoring guide?

Yes. Each version has its own: open it under the questions, or download the PDF, which prints the questions with room to write and then the scoring guide on its own pages. There is also a PDF without the scoring guide to hand out.

Four exam-style free-response questions on extrema, concavity, the Mean Value Theorem and the graphs of f, f′ and f″, all without a calculator. Each is scored out of 9, with a scoring guide and a printable PDF with room to write, free to use in class. Good practice before the Unit 5 Progress Check in AP Classroom. Multiple choice for this unit: Unit 5 worksheet.

Sigma Prep · sigmaprep.io/worksheets/ap-calculus-ab

Name

AP Calculus AB Unit 5 Free Response: Analyzing Functions

Date Version A

Directions: Show your work in the space under each part: the setup, the values you use and your reasoning. A justification needs a mathematical reason, with the conditions of any theorem or test you use checked. An answer with no supporting work may earn no credit. Give decimal answers correct to three decimal places unless a part says otherwise. Each question is worth 9 points. Unless a question says otherwise, the domain of a function f is every real number x for which f(x) is a real number.

Part B: No calculator is allowed.

Questions 1–4

1.
A differentiable function $f$ , defined for $- 2 \leq x \leq 4$ , satisfies $f (1) = 3$ . Its derivative $f^{'}$ has the graph shown, made up of straight segments and one semicircle.
1. (a)
  Is there a relative extremum of $f$ at $x = - 1$ ? If so, is it a relative maximum or a relative minimum? Give a reason for your answer.
2. (b)
  Find all open intervals contained in $- 2 < x < 4$ on which the graph of $f$ is concave down. Give a reason for your answer.
1. (c)
  Find $\lim_{x \to 1} \frac{2 f (x) + 2 x - 8}{x^{2} - 4 x + 3}$ , or show that the limit does not exist. Justify your answer.
2. (d)
  What is the absolute maximum value of $f$ for $- 2 \leq x \leq 4$ ? Justify your answer.
2.
For $y > 0$ , the equation $y^{2} - 2 y = 8 - 5 \sin x$ defines $y$ as a differentiable function of $x$ , $y = k (x)$ , whose graph is the curve given by the equation.
1. (a)
  Show that $\frac{d y}{d x} = - \frac{5 \cos x}{2 y - 2}$ .
2. (b)
  The point $(0, 4)$ lies on the curve. Find an equation for the tangent line there.
1. (c)
  For $0 \leq x \leq π$ , the curve has exactly one horizontal tangent line. Find the point $(x, y)$ at which it occurs.
2. (d)
  At the $x$ -coordinate of the point from part (c), does $k$ have a relative maximum, a relative minimum, or neither? Justify your answer.
3.
A function $P$ of time $t$ , in weeks, models the number of subscribers to a new podcast, in thousands. It satisfies $P (0) = 2$ and $\frac{d P}{d t} = \frac{P (10 - P)}{4}$ . Under this model $P (t)$ stays strictly between $0$ and $10$ for every $t \geq 0$ .
1. (a)
  What is $\frac{d P}{d t}$ when $t = 0$ ? Using correct units, interpret this value in context.
2. (b)
  Express $\frac{d^{2} P}{d t^{2}}$ in terms of $P$ , and decide whether $\frac{d P}{d t}$ is increasing or decreasing at $t = 0$ . Explain.
1. (c)
  Is the number of subscribers increasing faster when $P = 4$ or when $P = 8$ ? Give a reason for your answer.
2. (d)
  For $0 \leq t \leq 0.5$ , $P (t) < 5$ . Approximate $P (0.5)$ using the tangent line to the graph of $P$ at $t = 0$ , and determine whether this approximation is an overestimate or an underestimate. Give a reason for your answer.
4.
A differentiable function $D$ models the depth of water in a tide pool, in centimeters, at time $t$ hours, for $0 \leq t \leq 8$ . The table shows $D (t)$ at five times.
$\begin{array}{cccccc} t (hours) & 0 & 2 & 4 & 6 & 8 \\ D (t) (centimeters) & 5 & 10 & 7 & 14 & 36 \end{array}$
1. (a)
  Approximate $D^{'} (5)$ with the table values. Then interpret your estimate in context, including units.
2. (b)
  Is there necessarily a number $c$ between $0$ and $4$ with $D^{'} (c) = \frac{1}{2}$ ? Explain why or why not.
1. (c)
  Is there necessarily a time $t$ strictly between $4$ and $6$ when $D (t) = 11$ ? Explain why or why not.
2. (d)
  A second model for the depth of the water in the tide pool is $M (t) = \frac{1}{4} t^{3} - \frac{9}{4} t^{2} + 6 t + 4$ , $0 \leq t \leq 8$ . According to this model, what is the absolute minimum value of the depth of the water in the tide pool on that interval? Justify your answer.
  End of Version A

Free AP Calculus AB worksheets at sigmaprep.io/worksheets/ap-calculus-abOriginal questions written by Sigma Prep. AP is a registered trademark of the College Board, which is not affiliated with this sheet.

Scoring guide · Version A

Question 1Extrema, concavity and a limit from the graph of f′ · no calculator · 9 points

(a) 1 point Relative min/max classification · Topic 5.4

1 point: answer (relative minimum) with a reason: $f^{'}$ changes from negative to positive
Also acceptable: The second derivative test: $f^{'} = 0$ at $x = - 1$ , and $f^{''} > 0$ there because the graph of $f^{'}$ is increasing at that point, so $f$ has a relative minimum there.

Solution

From the graph, $f^{'} (- 1) = 0$ . By the first derivative test, what happens at $x = - 1$ depends on whether $f^{'}$ changes sign there.
Just to the left of $x = - 1$ , $f^{'}$ is negative; just to the right, it is positive.
$f$ has a relative minimum at $x = - 1$ because $f^{'}$ changes from negative to positive at $x = - 1$ .

(b) 2 points Concavity or inflection points · Topic 5.6

1 point: the interval $(0, 3)$
Also acceptable: The same answer with $x = 2$ left out (where $f^{''}$ does not exist): $(0, 2) \cup (2, 3)$ .
1 point: reason: $f^{'}$ is decreasing there

Solution

The graph of $f$ is concave down where $f^{''}$ is negative, that is, where $f^{'}$ is decreasing.
From the graph, $f^{'}$ is decreasing on $(0, 3)$ .

(c) 3 points L'Hospital's rule · Topic 4.7

1 point: the limits of the numerator and the denominator, each $0$ (using $f (1) = 3$ and the continuity of $f$ )
1 point: uses L'Hospital's Rule:
$\lim_{x \to 1} \frac{2 f^{'} (x) + 2}{2 x - 4}$
1 point: answer $- 2$ , with $f^{'} (1) = 1$ from the graph

Solution

Since $f$ is differentiable, it is continuous, so the numerator tends to $2 (3) + 2 (1) - 8 = 0$ ; the denominator is $0$ at $x = 1$ too.
The form is $\frac{0}{0}$ , so L'Hospital's Rule applies: differentiate the numerator and the denominator separately.
From the graph, $f^{'} (1) = 1$ , so the limit is $\lim_{x \to 1} \frac{2 f^{'} (x) + 2}{2 x - 4} =$ $\frac{2 (1) + 2}{- 2} =$ $- 2$ .

(d) 3 points Absolute extremum (candidates test) · Topic 5.5

1 point: considers $f^{'} (x) = 0$ : $x = - 1$ and $x = 2$
Also acceptable: the same list with an endpoint where $f^{'} (x) = 0$ named as well; the endpoints are candidates either way
1 point: justification: the values $f (- 2) =$ $\frac{3}{2},$ $f (- 1) =$ $\frac{1}{2},$ $f (2) =$ $\frac{7}{2},$ $f (4) =$ $\frac{7}{2} -$ $\frac{π}{2}$ at the critical points and both endpoints
Also acceptable: A sign argument: $f^{'} \leq 0$ on $(- 2, - 1)$ , $f^{'} \geq 0$ on $(- 1, 2)$ , and $f^{'} \leq 0$ on $(2, 4)$ , zero at most at isolated points, so the absolute maximum is at $x = - 2$ or $x = 2$ , and of $f (- 2) = \frac{3}{2}$ and $f (2) = \frac{7}{2}$ the greatest is $f (2) = \frac{7}{2}$ .
1 point: answer $\frac{7}{2}$

Solution

$f^{'} (x) = 0$ at $x = - 1$ and $x = 2$ ; with the endpoints $x = - 2$ and $x = 4$ , these are the candidates.
With $f (x) = 3 + \int_{1}^{x} f^{'} (t) d t$ and the signed areas from the graph of $f^{'}$ :
$f (- 2) =$ $\frac{3}{2},$ $f (- 1) =$ $\frac{1}{2},$ $f (2) =$ $\frac{7}{2},$ $f (4) =$ $\frac{7}{2} -$ $\frac{π}{2}$ .
The absolute maximum value of $f$ on $[- 2, 4]$ is $f (2) = \frac{7}{2}$ , the largest of these values.

Question 2An implicit curve: tangent line, horizontal tangent and the second derivative test · no calculator · 9 points

(a) 2 points Implicit differentiation · Topic 3.2

1 point: implicit differentiation
1 point: verifies $\frac{d y}{d x} = - \frac{5 \cos x}{2 y - 2}$

Solution

Differentiate both sides with respect to $x$ (the chain rule on each term in $y$ ):
$2 y \frac{d y}{d x} - 2 \frac{d y}{d x} = - 5 \cos x$ .
So $(2 y - 2) \frac{d y}{d x} = - 5 \cos x$ , and $\frac{d y}{d x} = - \frac{5 \cos x}{2 y - 2}$ .

(b) 1 point Tangent line or tangent-line approximation · Topic 2.2

1 point: answer:
$y = 4 - \frac{5}{6} x$

Solution

Substitute $x = 0$ and $y = 4$ into part (a):
$\frac{d y}{d x} = \frac{- 5 \cos 0}{2 (4) - 2} = - \frac{5}{6}$ .
The tangent line through $(0, 4)$ with slope $- \frac{5}{6}$ :
$y = 4 - \frac{5}{6} x$ .

(c) 3 points Horizontal, vertical or given-slope tangent · Topic 5.12

1 point: sets $\frac{d y}{d x} = 0$
1 point: $x = \frac{π}{2}$
1 point: $y = 3$

Solution

Set $\frac{d y}{d x} = 0$ : the numerator is $0$ , so $\cos x = 0$ ; $\cos x = 0$ for $0 \leq x \leq π$ only at $x = \frac{π}{2}$ .
Substitute $x = \frac{π}{2}$ into the equation of the curve: $y^{2} - 2 y = 3,$ so $(y - 3) (y + 1) = 0$ . Since $y > 0$ , $y = 3$ .
At $(\frac{π}{2}, 3)$ the denominator is not $0$ , so the point is $(\frac{π}{2}, 3)$ .

(d) 3 points Relative min/max classification · Topic 5.7

1 point: considers $\frac{d^{2} y}{d x^{2}}$ (quotient rule)
Also acceptable: A first derivative argument in place of $\frac{d^{2} y}{d x^{2}}$ : near $(\frac{π}{2}, 3)$ the denominator of $\frac{d y}{d x}$ is positive and $\cos x$ changes sign at $x = \frac{π}{2}$ , so $\frac{d y}{d x}$ changes from negative to positive there.
1 point: $\frac{d^{2} y}{d x^{2}} = \frac{5}{4}$ at $(\frac{π}{2}, 3)$
Also acceptable: with the first derivative argument:
$\frac{d y}{d x} = 0$ at $(\frac{π}{2}, 3)$ and the sign of $\frac{d y}{d x}$ on each side of $x = \frac{π}{2}$ stated correctly
1 point: relative minimum, because $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} > 0$ there, or because $\frac{d y}{d x}$ changes from negative to positive

Solution

Differentiate $\frac{d y}{d x} = - \frac{5 \cos x}{2 y - 2}$ with the quotient rule:
$\frac{d^{2} y}{d x^{2}} = \frac{(5 \sin x) (2 y - 2) - (- 5 \cos x) (2 \frac{d y}{d x})}{{(2 y - 2)}^{2}}$ .
At $(\frac{π}{2}, 3)$ , $\frac{d y}{d x} = 0$ , so the second term of the numerator is $0$ :
$\frac{d^{2} y}{d x^{2}} = \frac{5}{2 (3) - 2} = \frac{5}{4}$ .
$\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} > 0$ : by the second derivative test, $k$ has a relative minimum at $x = \frac{π}{2}$ .

Question 3A differential equation analyzed without solving it · no calculator · 9 points

(a) 2 points Derivative value with units interpretation · Topic 4.3

1 point: $\frac{d P}{d t} = 4$ at $t = 0$
1 point: interpretation with units: increasing at $4$ thousand subscribers per week at $t = 0$

Solution

At $t = 0$ , $\frac{d P}{d t} = \frac{2 (10 - 2)}{4} = 4$ .
At time $t = 0$ weeks, the number of subscribers is increasing at a rate of $4$ thousand subscribers per week.

(b) 2 points Second derivative at a point · Topic 5.12

1 point: $\frac{d^{2} P}{d t^{2}} =$ $\frac{1}{4} (10 - 2 P) \frac{d P}{d t} =$ $\frac{P (10 - P) (10 - 2 P)}{16}$
1 point: increasing, because $\frac{d^{2} P}{d t^{2}} = 6$ at $t = 0$

Solution

$\frac{d P}{d t} = \frac{1}{4} ⁣ (10 P - P^{2})$ , so by the chain rule $\frac{d^{2} P}{d t^{2}} =$ $\frac{1}{4} (10 - 2 P) \frac{d P}{d t} =$ $\frac{P (10 - P) (10 - 2 P)}{16}$ .
At $t = 0$ , $P = 2$ :
$\frac{d^{2} P}{d t^{2}} = \frac{2 (8) (6)}{16} = 6$ .
Since $\frac{d^{2} P}{d t^{2}} > 0$ at $t = 0$ , the rate of change $\frac{d P}{d t}$ is increasing at time $t = 0$ .

(c) 2 points Slopes read from a differential equation · Topic 4.3

1 point: $\frac{d P}{d t} = 6$ at $P = 4$ and $\frac{d P}{d t} = 4$ at $P = 8$
1 point: faster when $P = 4$ , with the comparison

Solution

When $P = 4$ , $\frac{d P}{d t} = \frac{4 (10 - 4)}{4} = 6$ ; when $P = 8$ , $\frac{d P}{d t} = \frac{8 (10 - 8)}{4} = 4$ .
Since $6 > 4$ , the number of subscribers is increasing faster when $P = 4$ .

(d) 3 points Over- or underestimate from concavity · Topic 4.6

1 point: approximation: $P (0.5) \approx 4$
1 point: $\frac{d^{2} P}{d t^{2}} > 0$ for $0 \leq t \leq 0.5$ , with the reason
1 point: underestimate, because the graph of $P$ is concave up on the interval

Solution

By part (a), the tangent line at $t = 0$ is $y = 2 + 4 t$ , so $P (0.5) \approx 2 + 4 (0.5) = 4$ .
For $0 \leq t \leq 0.5$ , $0 < P (t) < 5$ , so $P > 0$ , $10 - P > 0$ and $10 - 2 P > 0$ . By part (b), $\frac{d^{2} P}{d t^{2}} = \frac{P (10 - P) (10 - 2 P)}{16} > 0$ for $0 \leq t \leq 0.5$ .
So the graph of $P$ is concave up on $0 \leq t \leq 0.5$ , the tangent line at $t = 0$ lies below the graph there, and the approximation is an underestimate for $P (0.5)$ .

Question 4A table of values: the Mean Value Theorem and the candidates test · no calculator · 9 points

(a) 2 points Table estimate of a derivative · Topic 2.3

1 point: estimate:
$D^{'} (5) \approx \frac{7}{2}$
1 point: interpretation with units: at $t = 5$ , the depth of the water in the tide pool is increasing at about $\frac{7}{2}$ centimeters per hour

Solution

$t = 5$ is between the table values $t = 4$ and $t = 6$ , so use the average rate of change over that interval.
$D^{'} (5) \approx \frac{14 - 7}{2} = \frac{7}{2}$ centimeters per hour.
So at time $t = 5$ hours, the depth of the water in the tide pool is increasing at about $\frac{7}{2}$ centimeters per hour.

(b) 2 points MVT justification · Topic 5.1

1 point: $\frac{D (4) - D (0)}{4} = \frac{1}{2}$
1 point: yes, by the Mean Value Theorem, with the conditions

Solution

$D$ is differentiable, so it is also continuous; the Mean Value Theorem applies on $[0, 4]$ .
The average rate of change is $\frac{7 - 5}{4} = \frac{1}{2}$ .
So there must be a $c$ in $(0, 4)$ with $D^{'} (c) = \frac{1}{2}$ .

(c) 2 points IVT justification · Topic 1.16

1 point: $D$ is differentiable, so it is continuous on $[4, 6]$
1 point: yes, because $D (4) = 7 < 11 < 14 = D (6)$ , by the Intermediate Value Theorem

Solution

A differentiable function is continuous, so the Intermediate Value Theorem applies to $D$ on $[4, 6]$ .
From the table, $D (4) = 7$ and $D (6) = 14$ , and $11$ is between them.
So $D (t) = 11$ for some $t$ in $(4, 6)$ .

Model answer. $D$ is differentiable, so $D$ is continuous on $[4, 6]$ . Since $D (4) = 7 < 11 < 14 = D (6)$ , by the Intermediate Value Theorem there must be a value $t$ , $4 < t < 6$ , at which $D (t) = 11$ .

(d) 3 points Absolute extremum (candidates test) · Topic 5.5

1 point: considers $M^{'} (t) = 0$ : $t = 2$ and $t = 4$
1 point: justification: the values of $M$ at $t = 0$ , $t = 2$ , $t = 4$ , and $t = 8$ compared
Also acceptable: A sign argument: $M^{'} \geq 0$ on $(0, 2)$ , $M^{'} \leq 0$ on $(2, 4)$ , and $M^{'} \geq 0$ on $(4, 8)$ , zero at most at isolated points, so the absolute minimum is at $t = 0$ or $t = 4$ , and of $M (0) = 4$ and $M (4) = 8$ the least is $M (0) = 4$ .
1 point: answer: $4$ (centimeters; units are not required)

Solution

$M^{'} (t) =$ $\frac{3}{4} t^{2} -$ $\frac{9}{2} t +$ $6 =$ $\frac{3}{4} (t - 2) (t - 4)$ , so $M^{'} (t) = 0$ at $t = 2$ and $t = 4$ .
The candidates for the absolute minimum are $t = 0$ , $t = 2$ , $t = 4$ , and $t = 8$ : the endpoints and the times where $M^{'} (t) = 0$ .
$M (0) = 4$ , $M (2) = 9$ , $M (4) = 8$ , and $M (8) = 36$ .
The smallest is $M (0) = 4$ , so the absolute minimum value of $M (t)$ is $4$ centimeters, at $t = 0$ .

What does the AP Calculus AB free response ask on Unit 5?

A lot: Unit 5 is about 15% of the free-response points since 2012.

The graph of f′: about 42% of Unit 5 free-response points; 1 on this set.
An implicitly defined curve: about 22% of Unit 5 free-response points; 1 on this set.
A differential equation, without solving it: about 13% of Unit 5 free-response points; 1 on this set.
The Mean Value Theorem from a table: about 11% of Unit 5 free-response points; 1 on this set.
A table of f, g and their derivatives: about 11% of Unit 5 free-response points; not on this set.

This set has 4 questions in that mix, all without a calculator (about 91% of Unit 5 free-response points are in the no-calculator section), each scored out of 9 with a scoring guide.

Counted from the official AP Calculus AB free-response questions, 2012–2026. No official question is reproduced here.

AP Calculus AB Unit 5 free-response settings by share of exam points

Free-response settings on this set and their share of official AP Calculus AB free-response points, 2012–2026
Setting	Questions on this set	Share of Unit 5 free-response points	Share of all free-response points	Last on the exam
A differential equation analyzed without solving itNo calculator	1 of 4	13%	–	2024
Extrema, concavity and a limit from the graph of f′No calculator	1 of 4	42%	8%	–
An implicit curve: tangent line, horizontal tangent and the second derivative testNo calculator	1 of 4	12%	7%	–
A table of values: the Mean Value Theorem and the candidates testNo calculator	1 of 4	11%	–	2023

A part counts toward a setting when its primary skill is in Unit 5, in the same section, and the setting asks that skill (in the part’s own setting where there is one), on the main exams 2012–2026, by its points; a part that fits two settings is shared between them.

Questions about this free-response set

Unit 5, Analytical Applications of Differentiation, has 12 topics in the College Board's course description: the Mean Value Theorem; the Extreme Value Theorem, critical points, and global versus local extrema; intervals where a function is increasing or decreasing; relative extrema by the first derivative test; absolute extrema by the candidates test; concavity and points of inflection; the second derivative test; sketching the graphs of a function and its derivatives; connecting f, f′ and f″; setting up and solving optimization problems; and the behavior of implicit relations, such as horizontal and vertical tangents and the second derivative. The course description weights the unit at 15–20% of the multiple-choice section.

Each question is worth 9 points, spread over its three or four parts. Points are earned for specific pieces of work: a setup, an antiderivative, a value with its units, a conclusion with its reason. The scoring guide lists every point and what earns it, then a worked solution for each part. On the exam the six free-response questions are half of the score.

Only where the exam allows one. On this set, all four are in Part B, without a calculator, as almost every Unit 5 free-response part on the exam has been.

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 5 settings the exam leans on most get the most questions here, each in its own context with its own parts.

No. Every question here is original, written by Sigma Prep in the style of the exam: the settings, functions, numbers and wording are our own, and no official question or scoring guideline is copied or reworded. Every value and statement in the scoring guide was checked by an independent computer-algebra check before the set was published.

Yes. Each version has its own: open it under the questions, or download the PDF, which prints the questions with room to write and then the scoring guide on its own pages. There is also a PDF without the scoring guide to hand out.

AP is a registered trademark of the College Board, which was not involved in making these questions and does not endorse them. All questions and scoring guides are original to Sigma Prep. Frozen 2026-10-24.

AP Calculus AB Unit 5 FRQ Practice: Analyzing Functions

Part B: No calculator is allowed.

Scoring guide · Version A

Question 1Extrema, concavity and a limit from the graph of f′ · no calculator · 9 points

Question 2An implicit curve: tangent line, horizontal tangent and the second derivative test · no calculator · 9 points

Question 3A differential equation analyzed without solving it · no calculator · 9 points

Question 4A table of values: the Mean Value Theorem and the candidates test · no calculator · 9 points

What does the AP Calculus AB free response ask on Unit 5?

AP Calculus AB Unit 5 free-response settings by share of exam points

More AP Calculus AB free response

Questions about this free-response set

What is on AP Calculus AB Unit 5?+

How is an AP Calculus AB free-response question scored?+

Can students use a calculator?+

How were the questions chosen?+

Are these real AP exam questions?+

Is there a scoring guide?+