How many points is each AP Calculus AB free-response question worth?

Each of the six questions is worth 9 points, and the section is half of your total score.

Do I have to simplify my free-response answers?

Not unless the question tells you to. An unsimplified correct answer is fine, and decimal answers need to be correct to three places unless told otherwise.

What counts as a justification?

A reason stated in terms of a named function or a named theorem, with its conditions checked. A vague statement, such as saying the graph turns around, doesn't count.

Where can I read real scoring guidelines?

The College Board publishes past free-response questions and scoring guidelines on AP Central.

How to earn points on AP Calculus AB free response

Updated November 9, 2026

Free response is half of your AP Calculus AB score. It's scored against a rubric by people reading your handwritten work, so a correct number with no support can lose points.

This guide covers the habits that matter on nearly every question, then works through the justifications that come up most: extrema, the big theorems, tables, and differential equations.

The setup of the section

Six questions, 9 points each. Part A has 2 questions in 30 minutes with a calculator. Part B has 4 questions in 60 minutes with no calculator.
That's 15 minutes per question on average. Later parts of a question often use earlier ones, but many can be done independently, so keep going if you get stuck.
Prompts are on screen, and your answers are handwritten.

Habits that apply to every question

Write the setup on calculator questions. The College Board requires the equation, derivative, or integral you used, followed by the value. See how to use Desmos on the exam.
Use standard math notation. Calculator commands don't count as work.
Say what you mean by "it." Write " $g'$ changes sign," and name the function every time you give a reason.
Give three decimal places unless told otherwise, and don't round until the end.
You don't need to simplify, unless the question asks you to. Extra algebra only adds chances for a slip.
Include units when the question asks for them. The integral of a rate has the units of the amount. A derivative has units of the output per unit of the input.
Use your earlier answer when a later part depends on it, even if you aren't sure it's right, and show how you used it.

Absolute extrema on a closed interval

The Candidates Test works whenever the function is continuous on a closed interval. List the critical points (where the derivative is zero or undefined) and both endpoints, calculate the function at each one, and compare. A sign chart by itself doesn't finish an absolute extremum question, since a relative maximum isn't always the highest point on the interval.

Candidates test

Let

g(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 4x + 2

on the closed interval

[0, 5]

. Find the absolute maximum and minimum values of

g

. Justify your answer.

$g'(x) = x^2 - 5x + 4 = (x - 1)(x - 4)$ , so the critical points are $x = 1$ and $x = 4$ .

The candidates are $g(0) = 2$ , $g(1) = \frac{23}{6}$ , $g(4) = -\frac{2}{3}$ , and $g(5) = \frac{7}{6}$ .

Write: "The absolute maximum value is $\frac{23}{6}$ and the absolute minimum value is $-\frac{2}{3}$ , by comparing $g$ at the critical points and endpoints."

Answer: Maximum $\frac{23}{6}$ , minimum $-\frac{2}{3}$

Check the conditions before you use a theorem

A theorem only applies when its conditions hold. Say that they hold, by name, as part of the answer. If they don't hold, the theorem tells you nothing and the conclusion may be false.

Intermediate Value Theorem: the function must be continuous on $[a, b]$ , and $k$ must be between $f(a)$ and $f(b)$ . Being differentiable on $[a, b]$ implies continuous.
Mean Value Theorem: the function must be continuous on the closed interval and differentiable on the open interval.
Extreme Value Theorem: the function must be continuous on a closed interval.

When the Intermediate Value Theorem doesn't apply

Let

h(x) = x + 1

for

x < 2

, and

h(x) = x + 3

for

x \ge 2

. Then

h(1) = 2

and

h(3) = 6

. Must there be a value

c

(1, 3)

with

h(c) = 4

? Justify your answer.

$h$ jumps at $x = 2$ , since the left-hand limit is $3$ and $h(2) = 5$ . So $h$ isn't continuous on $[1, 3]$ , and the Intermediate Value Theorem doesn't apply.

Check directly. Setting $x + 1 = 4$ gives $x = 3$ , which isn't less than $2$ . Setting $x + 3 = 4$ gives $x = 1$ , which isn't at least $2$ . No such $c$ exists.

Answer: No. Neither piece equals $4$ on $(1, 3)$ ; the Intermediate Value Theorem doesn't apply because $h$ is discontinuous at $x = 2$ .

Working from a table

Table questions test whether you can use a few values well. Read the spacing carefully, because the times are often uneven.

Picking the right interval

The depth of a pond is a differentiable function

h

, in feet, of time

t

, in hours. It has

h(0) = 14

h(3) = 20

h(5) = 16

, and

h(9) = 16

. (a) Estimate

h'(4)

. Include units. (b) Must there be a time

t

(0, 9)

when

h'(t) = 0

? Justify your answer.

(a) Use the closest times around $t = 4$ : $h'(4) \approx \frac{h(5) - h(3)}{5 - 3} = \frac{16 - 20}{2} = -2$ feet per hour.

(b) Compare the average rates. On $[5, 9]$ the average rate is $\frac{16 - 16}{9 - 5} = 0$ .

Write: "Yes. $h$ is differentiable, so it's continuous on $[5, 9]$ and differentiable on $(5, 9)$ . Since $\frac{h(9) - h(5)}{9 - 5} = 0$ , the Mean Value Theorem says there is a time $t$ in $(5, 9)$ with $h'(t) = 0$ ."

Answer: (a) $-2$ feet per hour. (b) Yes, by the Mean Value Theorem on $[5, 9]$ .

A Riemann sum, and whether it's too high or too low

An increasing rate

B(t)

, in gallons per minute, has

B(0) = 5

B(2) = 9

B(3) = 12

, and

B(7) = 20

. Use a left Riemann sum with the three intervals given to approximate

\int_0^7 B(t)\,dt

. Is the approximation too high or too low? Explain.

The widths are $2$ , $1$ , and $4$ . The sum is $2(5) + 1(9) + 4(12) = 67$ . The units are gallons, because a rate in gallons per minute is multiplied by minutes.

Write: "It's too low, because $B$ is increasing, so each left endpoint value is less than the rest of the values on its interval." This argument needs $B$ to be increasing. Without that, a table alone doesn't tell you.

Answer: $67$ gallons, too low

Tangent line approximations

A tangent line estimate is an overestimate if the function is concave down on the interval between the point of tangency and the point you're estimating, and an underestimate if it's concave up there. Give that reason in words.

Over or under

The function

f

has

f(2) = 5

and

f'(2) = -1

, and

f''(x) < 0

for all

x

. Use the line tangent to the graph of

f

x = 2

to approximate

f(2.2)

. Is this an overestimate or an underestimate? Justify your answer.

The tangent line is $y = 5 - (x - 2)$ . At $x = 2.2$ , that gives $4.8$ .

Write: "Overestimate, because $f'' < 0$ , so the graph of $f$ is concave down and lies below its tangent line."

Answer: $f(2.2) \approx 4.8$ , an overestimate

Separable differential equations

Separate the variables before you integrate, and include the constant of integration right away. Published scoring guidelines have often limited the credit for a solution that skips either step.

Choosing the correct root

Find the particular solution

y = f(x)

\frac{dy}{dx} = \frac{5x^4}{2y}

with the initial condition

f(0) = -3

1Separate: $2y\,dy = 5x^4\,dx$ .
2Integrate, with the constant: $y^2 = x^5 + C$ .
3Use the initial condition: $(-3)^2 = 0 + C$ , so $C = 9$ .
4Solve for $y$ : $y = \pm\sqrt{x^5 + 9}$ . Since $f(0) = -3$ is negative, take the negative root.

The solution is valid where $x^5 + 9 > 0$ , which keeps $y$ from being zero and the derivative defined.

Answer: $y = -\sqrt{x^5 + 9}$

Explaining what a number means

Questions that ask you to interpret an answer want the quantity, the units, and the time or interval in one sentence.

A definite integral of a rate: if $C$ is a flow rate in gallons per minute, write " $\int_2^6 C(t)\,dt = 45$ means that 45 gallons flowed in from $t = 2$ to $t = 6$ minutes."
A derivative: if $P$ is a population and $t$ is in hours, write " $P'(3) = -4$ means that at $t = 3$ hours, the population is decreasing at 4 people per hour."
An average value: give the value and name the interval it's averaged over.

More practice with these sentences is in the fundamental theorem and accumulation guide.

Keep going

Practice on exam-style questions

Intersect Prep has 400 multiple-choice questions across all 8 units, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $44 unlocks the course for 6 months.

Try a free lesson See the course

Questions

How many points is each AP Calculus AB free-response question worth?: Each of the six questions is worth 9 points, and the section is half of your total score.
Do I have to simplify my free-response answers?: Not unless the question tells you to. An unsimplified correct answer is fine, and decimal answers need to be correct to three places unless told otherwise.
What counts as a justification?: A reason stated in terms of a named function or a named theorem, with its conditions checked. A vague statement, such as saying the graph turns around, doesn't count.
Where can I read real scoring guidelines?: The College Board publishes past free-response questions and scoring guidelines on AP Central.