AP Calculus AB Study Guide & Notes

This article consists of the scoring guidelines for the free-response section of the Calculus AB Mock Exam #1. You may view those questions here.

General Scoring Notes for AP Calculus Free Response Questions:

The model solution is presented using standard mathematical notation.
Answers (numeric or algebraic) need not be simplified. Answers given as a decimal approximation should be correct to three places after the decimal point. Within each individual free-response question, at most one point is not earned for inappropriate rounding.

Question 1 (9 points, graphing calculator required)

A tank contains 500 gallons of water at time $t = 0$ hours. Water flows into the tank at a rate modeled by $I (t)$ gallons per hour, and water flows out of the tank at a rate modeled by $O (t)$ gallons per hour. Values of $I (t)$ and $O (t)$ are given in the table below for selected values of $t$ .

$t (hours) I (t) (gallons per hour) O (t) (gallons per hour) 05045255525626085848106555$

Note: Your calculator should be in radian mode.

(a) Use the data in the table to approximate $\frac{d I}{d t}$ at $t = 6$ hours. Show the computations that lead to your answer. Using correct units, interpret the meaning of $\frac{d I}{d t}$ at $t = 6$ in the context of this problem.

Model Solution:

$\frac{d I}{d t}_{t = 6} \approx \frac{I ( 8 ) - I ( 5 )}{8 - 5} = \frac{58 - 62}{3} = \frac{- 4}{3} \approx - 1.333$

At time $t = 6$ hours, the rate at which water flows into the tank is decreasing at approximately $1.333$ gallons per hour per hour.

Points:

1 point for an approximation with supporting work

1 point for a correct interpretation with appropriate units

Total for part (a): 2 points

Scoring Notes:

The approximation may use any correct difference quotient involving the data in the table. For example, $\frac{I ( 8 ) - I ( 2 )}{8 - 2}$ earns the first point.
To earn the first point, the response must show at least a difference and a quotient.
The numerical value need not be simplified to earn the first point, but if simplified, it must be correct to three decimal places.
To earn the second point, the response must include correct units (gallons per hour per hour, or equivalent such as gal/hr²) and interpret the meaning in context.
A response that states the rate is "changing" without specifying "decreasing" may still earn the second point if the sign of the derivative is clearly indicated.

(b) Using a trapezoidal sum with the four subintervals indicated by the data in the table, approximate $\int_{0}^{10} [I (t) - O (t)] d t$ . Using correct units, interpret the meaning of $\int_{0}^{10} [I (t) - O (t)] d t$ in the context of this problem.

Model Solution:

The subintervals are $[0, 2]$ , $[2, 5]$ , $[5, 8]$ , $[8, 10]$ with widths 2, 3, 3, and 2 respectively.

$Trapezoidal sum = \frac{1}{2} \cdot 2 \cdot [(I (0) - O (0)) + (I (2) - O (2))] + \frac{1}{2} \cdot 3 \cdot [(I (2) - O (2)) + (I (5) - O (5))] + \frac{1}{2} \cdot 3 \cdot [(I (5) - O (5)) + (I (8) - O (8))] + \frac{1}{2} \cdot 2 \cdot [(I (8) - O (8)) + (I (10) - O (10))]$

$Trapezoidal sum = 1 \cdot [5 + 3] + 1.5 \cdot [3 + 2] + 1.5 \cdot [2 + 10] + 1 \cdot [10 + 10] = 8 + 7.5 + 18 + 10 = 43.5$

The value $\int_{0}^{10} [I (t) - O (t)], d t$ represents the net change in the amount of water in the tank over the 10-hour period, measured in gallons.

Points:

1 point for a trapezoidal Riemann sum

1 point for the correct answer

1 point for a correct interpretation using appropriate units

Total for part (b): 3 points

Scoring Notes:

The first point is earned for a correct trapezoidal sum setup. This may be presented in various equivalent forms.
If the setup has at most one error in the coefficients or function values, the response earns the first point but not the second.
The answer must be $43.5$ to earn the second point (assuming the first point was earned).
The third point requires mentioning "net change" or "change" in the amount of water, with units of gallons.
A response that states "total amount of water" instead of "net change" does not earn the third point.

(c) For $0 \leq t \leq 10$ , the amount of water in the tank at time $t$ is given by $W (t) = 500 + \int_{0}^{t} [I (x) - O (x)] d x$ gallons. Is the amount of water in the tank increasing or decreasing at time $t = 7$ hours? Give a reason for your answer.

Model Solution:

$W^{'} (t) = I (t) - O (t)$

At $t = 7$ , we need to determine the sign of $I (7) - O (7)$ .

From the table, $I (5) = 62$ , $I (8) = 58$ , so $I (7)$ is between these values. Similarly, $O (5) = 60$ , $O (8) = 48$ , so $O (7)$ is between these values.

Estimating: $I (7) \approx 60$ and $O (7) \approx 54$ , so $I (7) - O (7) > 0$ .

Therefore, the amount of water in the tank is increasing at $t = 7$ hours because $W^{'} (7) = I (7) - O (7) > 0$ .

Points:

1 point for considering

W^{'} (7)

I (7) - O (7)

1 point for the correct answer with appropriate justifications.

Total for part (c): 2 points

Scoring Notes:

The first point is earned for considering $W^{'} (7)$ , $I (7) - O (7)$ , or equivalent.
The response need not evaluate $I (7)$ or $O (7)$ explicitly to earn the first point.
A response that states "increasing because $I (t) > O (t)$ near $t = 7$ " earns both points.
A response that provides numerical estimates for $I (7)$ and $O (7)$ and compares them earns both points if the conclusion is correct.
The second point requires earning the first point and stating a correct conclusion with justification.

(d) At time $t = 10$ hours, the outflow rate changes. For $t > 10$ , the outflow rate is modeled by $O (t) = 55 + 3 (t - 10)$ gallons per hour. Using this model and the value $I (10) = 65$ from the table, find the rate of change of the amount of water in the tank at time $t = 10$ hours. Indicate units of measure.

Model Solution:

$\frac{d W}{d t}_{t = 10} = I (10) - O (10) = 65 - [55 + 3 (10 - 10)] = 65 - 55 = 10$

The rate of change of the amount of water in the tank at $t = 10$ hours is $10$ gallons per hour.

Points:

1 point for the correct answer with supporting work

1 point for the appropriate units.

Total for part (d): 2 points

Scoring Notes:

The first point is earned for the answer of 10 with supporting work showing $I (10) - O (10)$ .
The second point requires correct units (gallons per hour or equivalent).
A response that uses $O (10) = 55 + 3 (10 - 10) = 55$ to compute $I (10) - O (10) = 10$ earns both points.
A response that incorrectly uses $O (10)$ from the table (55) but still computes $65 - 55 = 10$ earns both points since this is coincidentally correct.

Question 2 (9 points, graphing calculator required)

Let $R$ be the region in the first quadrant bounded by the graph of $y = \frac{4}{1 + x ^{2}}$ , the $x$ -axis, and the vertical line $x = 3$ .

Note: Your calculator should be in radian mode.

(a) Find the area of region $R$ .

Model Solution:

$Area = \int_{0}^{3} \frac{4}{1 + x ^{2}} d x = 4 arctan (x)_{0}^{3} = 4 arctan (3) - 4 arctan (0) = 4 arctan (3) \approx 5.032$

Points:

1 point for the integral

1 point for the answer

Total for part (a): 2 points

Scoring Notes:

The first point is earned for the integral $\int_{0}^{3} \frac{4}{1 + x ^{2}} d x$ or equivalent.
The response may use a calculator to evaluate the integral directly.
The second point requires the answer $5.031$ or $5.032$ .
A response that presents only the antiderivative $4 arctan (x)$ without limits or evaluation does not earn the second point.

(b) Find the volume of the solid generated when region $R$ is revolved about the $x$ -axis.

Model Solution:

$V = π \int_{0}^{3} (\frac{4}{1 + x ^{2}})^{2} d x = π \int_{0}^{3} \frac{16}{( 1 + x ^{2} ) ^{2}} d x$

Using the antiderivative: $\int \frac{1}{( 1 + x ^{2} ) ^{2}} d x = \frac{1}{2} [\frac{x}{1 + x ^{2}} + arctan (x)] + C$

$V = 16 π \cdot \frac{1}{2} [\frac{x}{1 + x ^{2}} + arctan (x)]_{0}^{3} = 8 π [(\frac{3}{10} + arctan (3)) - 0]$

$= 8 π (0.3 + 1.2490) = 8 π (1.549) \approx 38.918$

Points:

1 point for the integrand

1 point for the limits and the constant

1 point for the answer

Total for part (b): 3 points

Scoring Notes:

The first point is earned for the integrand $(\frac{4}{1 + x ^{2}})^{2}$ or its equivalent in a definite integral.
The second point requires correct limits (0 and 3) and the constant $π$ .
The third point requires one of the following answers: $38.917$ , $38.918$ , or $38.919$ .
A response is eligible for the third point only if the first two points are earned.

(c) The region $R$ is the base of a solid. For this solid, each cross section perpendicular to the $x$ -axis is a rectangle whose height is three times the length of its base in region $R$ . Find the volume of this solid.

Model Solution:

The base of each rectangle has length $\frac{4}{1 + x ^{2}}$ and height $3 \cdot \frac{4}{1 + x ^{2}} = \frac{12}{1 + x ^{2}}$ .

Area of cross section: $A (x) = \frac{4}{1 + x ^{2}} \cdot \frac{12}{1 + x ^{2}} = \frac{48}{( 1 + x ^{2} ) ^{2}}$

$V = \int_{0}^{3} \frac{48}{( 1 + x ^{2} ) ^{2}}, d x = 48 \cdot \frac{1}{2} [\frac{x}{1 + x ^{2}} + arctan (x)]_{0}^{3}$

$V = 24 [(\frac{3}{10} + arctan (3)) - 0] = 24 (1.549) \approx 37.176$

Points:

1 point for the cross-sectional area

1 point for the integral with limits

1 point for the answer

Total for part (c): 3 points

Scoring Notes:

The first point is earned for the area expression $\frac{4}{1 + x ^{2}} \cdot \frac{12}{1 + x ^{2}}$ or its equivalent.
A response that presents the integrand $\frac{48}{( 1 + x ^{2} ) ^{2}}$ earns the first point.
The second point requires the integral with correct limits.
The third point requires one of the following answers: $37.175$ , $37.176$ , or $37.177$ .

(d) Find the average value of the function $f (x) = \frac{4}{1 + x ^{2}}$ on the interval $[0, 3]$ .

Model Solution:

$Average value = \frac{1}{3 - 0} \int_{0}^{3} \frac{4}{1 + x ^{2}} d x = \frac{1}{3} \cdot 4 arctan (3) \approx \frac{1}{3} (5.032) \approx 1.677$

Points:

1 point for the average value formula and the answer

Total for part (c): 1 point

Scoring Notes:

The point is earned for $\frac{1}{3} \int_{0}^{3} \frac{4}{1 + x ^{2}}, d x$ or equivalent, along with the correct answer.
The response may import the value from part (a) if it was earned.
The answer must be one of the following: $1.676$ , $1.677$ , or $1.678$ .
A response that presents $\frac{1}{3} \cdot 5.032$ earns the point.

Question 3 (9 points, calculator not allowed)

Let $R$ be the region in the first quadrant bounded by the graph of $y = \frac{4}{1 + x ^{2}}$ , the $x$ -axis, and the vertical line $x = 3$ .

Note: Your calculator should be in radian mode.

The graph of $f^{'}$ , the derivative of the function $f$ , is shown in the figure below. The function $f$ is twice differentiable for all real numbers.

(a) On what open intervals contained in $(0, 10)$ is the function $f$ increasing? Justify your answer.

Model Solution:

The function $f$ is increasing on the interval $(3, 7)$ because $f^{'} (x) > 0$ on this interval.

Points:

1 point for the answer

1 point for a correct justification

Total for part (a): 2 points

Scoring Notes:

The first point is earned for the answer $(3, 7)$ .
Endpoints may be included without penalty.
The second point requires a justification based on $f^{'} (x) > 0$ on $(3, 7)$ .
A response that states " $f$ is increasing where the graph of $f^{'}$ is above the $x$ -axis" earns the second point.

(b) At what values of $x$ does $f$ have a relative maximum or relative minimum? Use the First Derivative Test to justify your answer.

Model Solution:

$f$ has a relative minimum at $x = 3$ because $f^{'}$ changes from negative to positive at $x = 3$ .

$f$ has a relative maximum at $x = 7$ because $f^{'}$ changes from positive to negative at $x = 7$ .

Points:

1 point for identifying

x = 3

as a minimum and

x = 7

as a maximum

1 point for a correct justification using the First Derivative Test

Total for part (b): 2 points

Scoring Notes:

The first point requires identifying both $x = 3$ as a relative minimum and $x = 7$ as a relative maximum.
A response that identifies only one critical point correctly earns 0 points for the first point.
The second point requires using the First Derivative Test (sign changes in $f^{'}$ ).
A response that uses the Second Derivative Test does not earn the second point.

(c) On what open intervals contained in $(0, 10)$ is the graph of $f$ concave up? Justify your answer.

Model Solution:

The graph of $f$ is concave up on $(1, 4)$ and $(6, 9)$ because $f^{'}$ is increasing on these intervals (equivalently, $f^{''} (x) > 0$ on these intervals).

Points:

1 point for identifying

x = 3

as a minimum and

x = 7

as a maximum

1 point for a correct justification using the First Derivative Test

Total for part (c): 2 points

Scoring Notes:

The first point requires identifying both $x = 3$ as a relative minimum and $x = 7$ as a relative maximum.
A response that identifies only one critical point correctly earns 0 points for the first point.
The second point requires using the First Derivative Test (sign changes in $f^{'}$ ).
A response that uses the Second Derivative Test does not earn the second point.

(d) Let $g (x) = x^{2} f (x)$ . Write an expression for $g^{'} (x)$ in terms of $x$ , $f (x)$ , and $f^{'} (x)$ .

Model Solution:

$g^{'} (x) = 2 x f (x) + x^{2} f^{'} (x)$

Points:

1 point for the product rule

1 point for the correct expression

Total for part (d): 2 points

Scoring Notes:

The first point is earned for applying the product rule.
A response with a correct product rule structure but a differentiation error (such as $g^{'} (x) = 2 x f (x) + x^{2}$ earns the first point but not the second.
The second point requires the completely correct expression.

(e) Given that $\int_{0}^{10} f^{'} (x), d x = 8$ , find $f (10) - f (0)$ .

Model Solution:

By the Fundamental Theorem of Calculus: $\int_{0}^{10} f^{'} (x), d x = f (10) - f (0) = 8$

Points:

1 point for a correct answer with justification

Total for part (e): 1 points

Scoring Notes:

The point is earned for the answer $8$ with a reference to the Fundamental Theorem of Calculus or equivalent reasoning.
A response that simply states $f (10) - f (0) = 8$ without justification earns the point.

Question 4 (9 points, calculator not allowed)

A farmer has $800 feet$ of fencing and wants to enclose a rectangular area and then divide it into three equal pens with fencing parallel to one side of the rectangle.

(a) Write an expression for the total area $A$ of the three pens in terms of $x$ , where $x$ is the length of fencing perpendicular to the dividing fences.

Model Solution:

Let $y$ be the length parallel to the dividing fences.

The fencing constraint is: $4 x + 2 y = 800$ , which gives $y = 400 - 2 x$ .

The total area is: $A (x) = x y = x (400 - 2 x) = 400 x - 2 x^{2}$

Points:

1 point for the constraint equation

1 point for the expression for the area of the three pens

Total for part (a): 2 points

Scoring Notes:

The first point is earned for establishing the relationship $4 x + 2 y = 800$ or equivalent.
A response may use different variable names.
The second point requires $A = x y$ where $y$ is expressed in terms of $x$ , or the equivalent expression $400 x - 2 x^{2}$ .
A response that presents $A (x) = 400 x - 2 x^{2}$ without showing the constraint earns the second point but not the first.

(b) Find the dimensions that maximize the total area of the three pens. Justify that your answer gives the maximum area.

Model Solution:

$A^{'} (x) = 400 - 4 x$

Setting $A^{'} (x) = 0$ : $400 - 4 x = 0 \Rightarrow x = 100$

$A^{''} (x) = - 4 < 0$

Since $A^{''} (100) < 0$ , there is a maximum at $x = 100$ .

When $x = 100$ , $y = 400 - 2 (100) = 200$ .

The dimensions that maximize the area are $x = 100$ feet and $y = 200$ feet.

Points:

1 point for setting

A (x)

equal to zero

1 point for the correct answer

1 point for an appropriate justification

Total for part (b): 3 points

Scoring Notes:

The first point is earned for finding $A^{'} (x)$ and setting it equal to 0.
The second point requires both dimensions: $x = 100$ and $y = 200$ .
The third point requires a justification that this gives a maximum. Acceptable justifications include:
- the Second Derivative Test: $A^{''} (100) < 0$ ,
- the First Derivative Test: $A^{'}$ changes from positive to negative at $x = 100$ ,
- and the Endpoints Test: Comparing $A (100)$ with $A (0)$ and $A (200)$
A response that only states " $x = 100$ is the only critical point" without further justification does not earn the third point.

(c) A ladder $15 feet$ long is leaning against a wall. The base of the ladder is sliding away from the wall at a rate of $2 feet/second$ . How fast is the top of the ladder sliding down the wall when the base of the ladder is $9 feet$ from the wall?

Model Solution:

Let $x$ be the distance from the base of the ladder to the wall, and let $y$ be the height of the top of the ladder on the wall.

By the Pythagorean theorem: $x^{2} + y^{2} = 1 5^{2} = 225$

Differentiating with respect to time $t$ : $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$

When $x = 9$ : $9^{2} + y^{2} = 225 \Rightarrow y^{2} = 144 \Rightarrow y = 12$

Substituting: $2 (9) (2) + 2 (12) \frac{d y}{d t} = 0$

$36 + 24 \frac{d y}{d t} = 0$

$\frac{d y}{d t} = - \frac{36}{24} = - \frac{3}{2}$

The top of the ladder is sliding down at a rate of $\frac{3}{2}$ feet per second (or 1.5 feet per second).

Points:

1 point for an equation relating

x

and

y

1 point for differentiating with respect to time

1 point for finding

y

when

x = 9

1 point for the correct answer

Total for part (): 4 points

Scoring Notes:

The first point is earned for $x^{2} + y^{2} = 225$ or equivalent.
The second point requires differentiating both sides with respect to time.
The third point requires finding $y = 12$ when $x = 9$ .
The fourth point requires the answer $- \frac{3}{2}$ or $- 1.5$ . The negative sign is essential.
Units are not required for the fourth point.

Question 5 (9 points, calculator not allowed)

Consider the differential equation $\frac{d y}{d x} = \frac{x ( y - 1 )}{2}$ .

(a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated.

Model Solution:

[Grid showing points at coordinates: $(- 1, 0)$ , $(0, 0)$ , $(1, 0)$ , $(- 1, 1)$ , $(0, 1)$ , $(1, 1)$ , $(- 1, 2)$ , $(0, 2)$ , $(1, 2)$ ]

At each point $(x, y)$ , the slope is $\frac{x ( y - 1 )}{2}$ :

$(- 1, 0)$ : slope $= \frac{( - 1 ) ( - 1 )}{2} = \frac{1}{2}$
$(0, 0)$ : slope $= 0$
$(1, 0)$ : slope $= \frac{1 ( - 1 )}{2} = - \frac{1}{2}$
$(- 1, 1)$ : slope $= 0$
$(0, 1)$ : slope $= 0$
$(1, 1)$ : slope $= 0$
$(- 1, 2)$ : slope $= \frac{( - 1 ) ( 1 )}{2} = - \frac{1}{2}$
$(0, 2)$ : slope $= 0$
$(1, 2)$ : slope $= \frac{1 ( 1 )}{2} = \frac{1}{2}$

Points:

1 point for the slope field

Total for part (a): 1 point

Scoring Notes:

The point is earned if at least 7 of the 9 slope segments are correct.
Slope segments should be short line segments with appropriate directions.
Horizontal segments at all points on the line $y = 1$ and at points where $x = 0$ .

(b) Find $\frac{d ^{2} y}{d x ^{2}}$ in terms of $x$ and $y$ . Determine whether the solution curve that passes through the point $(0, 2)$ is concave up or concave down at that point. Give a reason for your answer.

Model Solution:

$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} [\frac{x ( y - 1 )}{2}] = \frac{1}{2} [(y - 1) + x \frac{d y}{d x}]$

$= \frac{1}{2} [(y - 1) + x \cdot \frac{x ( y - 1 )}{2}] = \frac{1}{2} (y - 1) [1 + \frac{x ^{2}}{2}]$

$= \frac{( y - 1 ) ( 2 + x ^{2} )}{4}$

At $(0, 2)$ : $\frac{d ^{2} y}{d x ^{2}}_{(0, 2)} = \frac{( 2 - 1 ) ( 2 + 0 )}{4} = \frac{2}{4} = \frac{1}{2} > 0$

The solution curve is concave up at $(0, 2)$ because $\frac{d ^{2} y}{d x ^{2}} > 0$ at that point.

Points:

1 point for

\frac{d ^{2} y}{d x ^{2}}

1 point for the correct answer with appropriate justification

Total for part (b): 2 points

Scoring Notes:

The first point is earned for a correct expression for $\frac{d ^{2} y}{d x ^{2}}$ in terms of $x$ and $y$ .
The expression may be in various equivalent forms.
The second point requires evaluating $\frac{d ^{2} y}{d x ^{2}}$ at $(0, 2)$ , determining its sign, and concluding correctly.
A response that states that the solution curve is concave up because $\frac{d ^{2} y}{d x ^{2}} > 0$ without showing the evaluation earns only the second point if the first point was earned.

(c) Find the particular solution $y = f (x)$ to the differential equation with the initial condition $f (0) = 3$ .

Model Solution:

$\frac{d y}{d x} = \frac{x ( y - 1 )}{2}$

Separating variables: $\frac{d y}{y - 1} = \frac{x}{2} d x$

Integrating both sides: $\int \frac{d y}{y - 1} = \int \frac{x}{2} d x$

$ln ∣ y - 1∣ = \frac{x ^{2}}{4} + C$

$y - 1 = A e^{x^{2} /4}$

$y = 1 + A e^{x^{2} /4}$

Using $f (0) = 3$ : $3 = 1 + A e^{0} = 1 + A$

$A = 2$

Therefore: $y = 1 + 2 e^{x^{2} /4}$

Points:

1 point for separation of variables

1 point for the antiderivative of $\frac{1}{y - 1}$

1 point for the antiderivative of $\frac{x}{2}$

1 point for the constant of integration and using the initial condition

f (0) = 3

1 point for solving for

y

Total for part (c): 5 points

Scoring Notes:

The first point is earned for correctly separating variables.
The second point requires $ln ∣ y - 1∣$ or equivalent.
The third point requires $\frac{x ^{2}}{4}$ or equivalent.
The fourth point requires including a constant and using the initial condition to find its value.
The fifth point requires solving for $y$ in the form $y = 1 + 2 e^{x^{2} /4}$ or equivalent.
A response with no separation of variables earns 0 out of 5 points.
A response with no constant of integration can earn at most 3 out of 5 points.

(d) For the solution found in part (c), determine whether $f$ has a relative minimum, a relative maximum, or neither at $x = 0$ . Justify your answer.

Model Solution:

From part (c): $y = f (x) = 1 + 2 e^{x^{2} /4}$

$f^{'} (x) = 2 e^{x^{2} /4} \cdot \frac{x}{2} = x e^{x^{2} /4}$ $

$f^{'} (0) = 0$

$f^{''} (x) = e^{x^{2} /4} + x \cdot e^{x^{2} /4} \cdot \frac{x}{2} = e^{x^{2} /4} (1 + \frac{x ^{2}}{2})$

$f^{''} (0) = e^{0} (1 + 0) = 1 > 0$

Since $f^{''} (0) > 0$ , $f$ has a relative minimum at $x = 0$ by the Second Derivative Test.

Points:

1 point for a correct answer with appropriate justification

Total for part (d): 1 point

Scoring Notes:

The point is earned for stating "relative minimum" with a correct justification.
Acceptable justifications include the following:
- The Second Derivative Test: $f^{''} (0) > 0$
- The First Derivative Test: $f^{'}$ changes from negative to positive at $x = 0$
A response may import $f$ from part (c) even if it was incorrect, and earn the point for a consistent answer with correct reasoning.

Question 6 (9 points, calculator not allowed)

Let $f$ be a twice-differentiable function. Selected values of $f$ , $f^{'}$ , and $f^{''}$ are given in the table below.

$x 1234 f (x) 4665 f^{'} (x) 20 - 1 - 2 f^{''} (x) - 3 - 2 14$

(a) Approximate $f^{'} (2.5)$ using the data in the table. Show the computations that lead to your answer.

Model Solution:

$f^{'} (2.5) \approx \frac{f ^{'} ( 3 ) - f ^{'} ( 2 )}{3 - 2} = \frac{- 1 - 0}{1} = - 1$

Points:

1 point for an approximation with supporting work

Total for part (a): 1 point

Scoring Notes:

The point is earned for using a difference quotient with values from the table for $f^{'}$ .
Other approximations such as $\frac{f ^{'} ( 4 ) - f ^{'} ( 2 )}{4 - 2} = \frac{- 2 - 0}{2} = - 1$ also earn the point.
The response must show at least a difference and a quotient.

(b) Evaluate $\int_{1}^{4} f^{''} (x) d x$ using the data in the table. Show the work that leads to your answer.

Model Solution:

By the Fundamental Theorem of Calculus: $\int_{1}^{4} f^{''} (x) d x = f^{'} (4) - f^{'} (1) = - 2 - 2 = - 4$

Points:

1 point for using the Fundamental Theorem of Calculus

1 point for the answer

Total for part (b): 2 points

Scoring Notes:

The first point is earned for recognizing that $\int_{1}^{4} f^{''} (x) d x = f^{'} (4) - f^{'} (1)$ .
The second point requires the answer $- 4$ .
A response that states $\int_{1}^{4} f^{''} (x) d x = - 4$ with no supporting work earns only the second point.

(c) Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.2)$ . Is this approximation greater than or less than $f (1.2)$ ? Give a reason for your answer.

Model Solution:

The tangent line at $x = 1$ is

$y = f (1) + f^{'} (1) (x - 1) = 4 + 2 (x - 1)$

$f (1.2) \approx 4 + 2 (1.2 - 1) = 4 + 2 (0.2) = 4.4$

Since $f^{''} (1) = - 3 < 0$ , $f$ is concave down at $x = 1$ , so the tangent line lies above the graph of $f$ .

Therefore, the approximation $4.4$ is greater than $f (1.2)$ .

Points:

1 point for the approximation

1 point for the comparison with appropriate justification

Total for part (c): 2 points

Scoring Notes:

The first point is earned for $f (1.2) \approx 4.4$ .
The second point requires stating "greater than" with a reason based on concavity.
Acceptable reasons include that $f^{''} (1) < 0$ " or that $f$ is concave down.

(d) Let $g$ be the function defined by $g (x) = \int_{1}^{x} f (t), d t$ for $1 \leq x \leq 4$ . On what intervals, if any, is $g$ decreasing? Justify your answer.

Model Solution:

$g^{'} (x) = f (x)$

$g$ is decreasing where $g^{'} (x) = f (x) < 0$ .

From the table, we know $f (1) = 4 > 0$ , $f (2) = 6 > 0$ , $f (3) = 6 > 0$ , $f (4) = 5 > 0$ .

Since all given values of $f$ are positive, and $f$ is continuous, there is no interval in $[1, 4]$ where $f (x) < 0$ .

Therefore, $g$ is not decreasing on any interval in $[1, 4]$ .

Points:

1 point for considering that

g^{'} (x) = f (x)

1 point for a correct answer with appropriate justification

Total for part (d): 2 points

Scoring Notes:

The first point is earned for recognizing $g^{'} (x) = f (x)$ or stating that $g$ decreases where $f < 0$ .
The second point requires concluding that $g$ is not decreasing on any interval, with justification that all table values of $f$ are positive.
A response that incorrectly concludes $g$ is decreasing on some interval does not earn the second point.

(e) The function $h$ is defined by $h (x) = f (2 x)$ . Find $h^{'} (2)$ .

Model Solution:

Using the chain rule:

$h^{'} (x) = f^{'} (2 x) \cdot 2$

$h^{'} (2) = f^{'} (2 \cdot 2) \cdot 2 = f^{'} (4) \cdot 2 = (- 2) (2) = - 4$

Points:

1 point for the chain rule

1 point for the answer

Total for part (e): 2 points

Scoring Notes:

The first point is earned for correctly applying the chain rule: $h^{'} (x) = f^{'} (2 x) \cdot 2$ .
The second point requires the answer $- 4$ .
A response that presents $h^{'} (2) = 2 f^{'} (4)$ earns the first point and may earn the second point for $- 4$ .

FiveHive

FiveHive

9.2 - Mock Exam #1 (Scoring Guidelines)

General Scoring Notes for AP Calculus Free Response Questions:

Question 1 (9 points, graphing calculator required)

Question 2 (9 points, graphing calculator required)

Question 3 (9 points, calculator not allowed)

Question 4 (9 points, calculator not allowed)

Question 5 (9 points, calculator not allowed)

Question 6 (9 points, calculator not allowed)