# Graph Analysis Questions (Type 3)

### AP  Questions Type 3: Graph Analysis

The long name is “Here’s the graph of the derivative, tell me things about the function.”

Students are given either the equation of the derivative of a function or a graph identified as the derivative of a function with no equation is given. It is not expected that students will write the equation of the function from the graph (although this may be possible); rather, students are expected to determine key features of the function directly from the graph of the derivative. They may be asked for the location of extreme values, intervals where the function is increasing or decreasing, concavity, etc. They may be asked for function values at points. They will be asked to justify their conclusions.

The graph may be given in context and students will be asked about that context. The graph may be identified as the velocity of a moving object and questions will be asked about the motion. See Linear Motion Problems (Type 2)

Less often the function’s graph may be given, and students will be asked about its derivatives.

What students should be able to do:

• Read information about the function from the graph of the derivative. This may be approached by derivative techniques or by antiderivative techniques.
• Find and justify where the function is increasing or decreasing.
• Find and justify extreme values (1st and 2nd derivative tests, Closed interval test a/k/a Candidates’ test).
• Find and justify points of inflection.
• Find slopes (second derivatives, acceleration) from the graph.
• Write an equation of a tangent line.
• Evaluate Riemann sums from geometry of the graph only. This usually involves familiar shapes such as triangles or semicircles.
• FTC: Evaluate integral from the area of regions on the graph.
• FTC: The function, g(x), may be defined by an integral where the given graph is the graph of the integrand, f(t), so students should know that if,

$\displaystyle g\left( x \right)=g\left( a \right)+\int_{a}^{x}{{f\left( t \right)dt}}$, then  $\displaystyle {g}'\left( x \right)=f\left( x \right)$  and  $\displaystyle {g}''\left( x \right)={f}'\left( x \right)$.

In this case, students should write $\displaystyle {g}'\left( x \right)=f\left( x \right)$ on their answer paper, so it is clear to the reader that they understand this.

Not only must students be able to identify these things, but they are usually asked to justify their answer and reasoning. See Writing on the AP Exams for more on justifying and explaining answers.

There are numerous ideas and concepts that can be tested with this type of question. The type appears on the multiple-choice exams as well as the free-response. Between multiple-choice and free-response this topic may account for 15% or more of the points available on recent tests. It is very important that students are familiar with all the ins and outs of this situation.

As with other questions, the topics tested come from the entire year’s work, not just a single unit. In my opinion many textbooks do not do a good job with integrating these topics, so be sure to use as many actual AP Exam questions as possible. Study past exams: look them over and see the different things that can be asked.

The Graph Analysis problem may cover topics primarily from primarily from Unit 4, Unit 5, and Unit 8 of the CED

For previous posts on this subject see October 1517192426 (my most read post), 2012 and January 2528, 2013

Free-response questions:

• Function given as a graph, questions about its integral (so by FTC the graph is the derivative):  2016 AB 3/BC 3, 2018 AB3
• Table and graph of function given, questions about related functions: 2017 AB 6,
• Derivative given as a graph: 2016 AB 3 and 2017 AB 3
• Information given in a table 2014 AB 5
• 2021 AB 4 / BC 4
• 2021 AB 5 (b), (c), (d)
• 2022 AB3/BC3 – graph analysis, max/min

Multiple-choice questions from non-secure exam. Notice the number of questions all from the same year; this is in addition to one free-response question (~25 points on AB and ~23 points on BC out of 108 points total)

• 2012 AB: 2, 5, 15, 17, 21, 22, 24, 26, 76, 78, 80, 82, 83, 84, 85, 87
• 2012 BC 3, 11, 12, 15, 12, 18, 21, 76, 78, 80, 81, 84, 88, 89

A good activity on this topic is here. The first pages are the teacher’s copy and solution. Then there are copies for Groups A, B, and C. Divide your class into 3 or 6 or 9 groups and give one copy to each. After they complete their activity have the students compare their results with the other groups.

Revised March 12, 2021, March 18, 2022

# Linear Motion (Type 2)

### AP Questions Type 2: Linear Motion

We continue the discussion of the various type questions on the AP Calculus Exams with linear motion questions.

“A particle (or car, person, or bicycle) moves on a number line ….”

These questions may give the position equation, the velocity equation (most often), or the acceleration equation of something that is moving on the x– or y-axis as a function of time, along with an initial condition. The questions ask for information about the motion of the particle: its direction, when it changes direction, its maximum position in one direction (farthest left or right), its speed, etc.

The particle may be a “particle,” a person, car, a rocket, etc.  Particles don’t really move in this way, so the equation or graph should be considered a model. The question is a versatile way to test a variety of calculus concepts since the position, velocity, or acceleration may be given as an equation, a graph, or a table; be sure to use examples of all three forms during the review.

Many of the concepts related to motion problems are the same as those related to function and graph analysis (Type 3). Stress the similarities and show students how the same concepts go by different names. For example, finding when a particle is “farthest right” is the same as finding when a function reaches its “absolute maximum value.” See my post for Motion Problems: Same Thing, Different Context for a list of these corresponding terms. There is usually one free-response question and three or more multiple-choice questions on this topic.

The positions(t), is a function of time. The relationships are:

• The velocity is the derivative of the position $\displaystyle {s}'\left( t \right)=v\left( t \right)$.  Velocity has direction (indicated by its sign) and magnitude. Technically, velocity is a vector; the term “vector” will not appear on the AB exam.
• Speed is the absolute value of velocity; it is a number, not a vector. See my post for Speed.
• Acceleration is the derivative of velocity and the second derivative of position, $\displaystyle {{s}'}'\left( t \right)={v}'\left( t \right)=a\left( t \right)$ It, too, has direction and magnitude and is a vector.
• Velocity is the antiderivative of acceleration.
• Position is the antiderivative of velocity.

What students should be able to do:

• Understand and use the relationships above.
• Distinguish between position at some time and the total distance traveled during the time period.
• The total distance traveled is the definite integral of the speed (absolute value of velocity) $\displaystyle \int_{a}^{b}{{\left| {v\left( t \right)} \right|dt}}$.
•  Be sure your students understand the term displacement; it is the net distance traveled or distance between the initial position and the final position. Displacement is the definite integral of the velocity (rate of change): $\displaystyle \int_{a}^{b}{{v\left( t \right)dt}}$
• The final position is the initial position plus the displacement (definite integral of the rate of change from xa to x = t): $\displaystyle s\left( t \right)=s\left( a \right)+\int_{a}^{t}{{v\left( x \right)dx}}$ Notice that this is an accumulation function equation (Type 1).
• Initial value differential equation problems: given the velocity or acceleration with initial condition(s) find the position or velocity. These are easily handled with the accumulation equation in the bullet above but may also be handled as an initial value problem.
• Find the speed at a given time. Speed is the absolute value of velocity.
• Find average speed, velocity, or acceleration
• Determine if the speed is increasing or decreasing.
• When the velocity and acceleration have the same sign, the speed increases. When they have different signs, the speed decreases.
• If the velocity graph is moving away from (towards) the t-axis the speed is increasing (decreasing). See the post on Speed.
• There is also a worksheet on speed here
• The analytic approach to speed: A Note on Speed
• Use a difference quotient to approximate the derivative (velocity or acceleration) from a table. Be sure the work shows a quotient.
• Riemann sum approximations.
• Units of measure.
• Interpret meaning of a derivative or a definite integral in context of the problem

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

This may be an AB or BC question. The BC topic of motion in a plane, (Type 8: parametric equations and vectors) will be discussed in a later post.

The Linear Motion problem may cover topics primarily from primarily from Unit 4, and also from Unit 3, Unit 5, Unit 6, and Unit 8 (for BC) of the CED

Free-response examples:

• Equation stem 2017 AB 5,
• Graph stem: 2009 AB1/BC1,
• Table stem 2019 AB2
• Equation stem 2021 AB 2
• Equation stem 2022 AB6 – velocity, acceleration, position, max/min

Multiple-choice examples from non-secure exams:

• 2012 AB 6, 16, 28, 79, 83, 89
• 2012 BC 2, 89

Revised March 15, and May 11, 2022

# Rate & Accumulation (Type 1)

### The Free-response Questions

There are ten general types of AP Calculus free-response questions. This and the next nine posts will discuss each of them.

NOTE: The numbers I’ve assigned to each type DO NOT correspond to the CED Unit numbers. Many AP Exam questions intentionally have parts from different Units. The CED Unit numbers will be referenced in each post.

## AP Questions Type 1: Rate and Accumulation

These questions are often in context with a lot of words describing a situation in which some quantities are changing. There are usually two rates acting in opposite ways (sometimes called an in-out question). Students are asked about the change that the rates produce over a time interval either separately or together.

The rates are often fairly complicated functions. If the question is on the calculator allowed section, students should store the functions in the equation editor of their calculator and use their calculator to do any graphing, integration, or differentiation that may be necessary.

The main idea is that over the time interval [a, b] the integral of a rate of change is the net amount of change

$\displaystyle \int_{a}^{b}{{{f}'\left( t \right)dt}}=f\left( b \right)-f\left( a \right)$

If the question asks for an amount, look around for a rate to integrate.

The final (accumulated) amount is the initial amount plus the accumulated change:

$\displaystyle f\left( x \right)=f\left( {{{x}_{0}}} \right)+\int_{{{{x}_{0}}}}^{x}{{{f}'\left( t \right)dt}}$

where $\displaystyle {{x}_{0}}$ is the initial time, and $\displaystyle f\left( {{{x}_{0}}} \right)$ is the initial amount. Since this is one of the main interpretations of the definite integral the concept may come up in a variety of situations.

What students should be able to do:

• Be ready to read and apply; often these problems contain a lot of words which need to be carefully read and understood.
• Understand the question. It is often not necessary to do as much computation as it seems at first.
• Recognize that rate = derivative.
• Recognize a rate from the units given without the words “rate” or “derivative.”
• Find the change in an amount by integrating the rate. The integral of a rate of change gives the amount of change (FTC):

$\displaystyle \int_{a}^{b}{{{f}'\left( t \right)dt}}=f\left( b \right)-f\left( a \right)$

• Find the final amount by adding the initial amount to the amount found by integrating the rate. If $\displaystyle {{x}_{0}}$ is the initial time, and $\displaystyle f\left( {{{x}_{0}}} \right)$  is the initial amount, then final accumulated amount is

$\displaystyle f\left( x \right)=f\left( {{{x}_{0}}} \right)+\int_{{{{x}_{0}}}}^{x}{{{f}'\left( t \right)dt}}$,

• Write an integral expression that gives the amount at a general time. BE CAREFUL, the dt must be included in the correct place. Think of the integral sign and the dt as parentheses around the integrand.
• Find the average value of a function
• Use FTC to differentiate a function defined by an integral.
• Explain the meaning of a derivative or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) what the numerical argument means in the context of the question.
• Explain the meaning of a definite integral or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) how the limits of integration apply in the context of the question.
• Store functions in their calculator recall them to do computations on their calculator.
• If the rates are given in a table, be ready to approximate an integral using a Riemann sum or by trapezoids. Also, be ready to approximate a derivative using a quotient from the numbers in the table.
• Do a max/min or increasing/decreasing analysis.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

The Rate – Accumulation question may cover topics primarily from Unit 4, Unit 5, Unit 6 and Unit 8 of the CED.

Typical free-response examples:

• 2013 AB1/BC1
• 2015 AB1/BC1
• 2018 AB1/BC1
• 2019 AB1/BC1
• 2022 AB1/BC1 – includes average value, inc/dec analysis, max/min analysis
• One of my favorites Good Question 6 (2002 AB 4)

Typical multiple-choice examples from non-secure exams:

• 2012 AB 8, 81, 89
• 2012 BC 8 (same as AB 8)

Updated January 31, 2019, March 12, 2021, March 11, 2022

# Extreme Average

A recent post on the AP Calculus bulletin board observed that the maximum value of the average value of a function on an interval occurred at the point where the graph of the average value and the function intersect. I am not sure if this concept is important in and of itself, but it does make an interesting exercise.

For a function f(x), we may treat its average value as a function, A(x), defined for all x ∈ [a, b], interval [a, x] as

$\displaystyle A\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} & {x\ne a} \\ {f\left( a \right)} & {x=a} \end{array}} \right.$

Graphically, the segment drawn at y = A(x) is such that the regions between the line and the function above and below the segment have equal areas. See figure 1 in which the red curve is the function, and the blue curve is the average value function. The two shaded regions have the same area.

Regardless of the starting value, the function and its average value start at the same value. If the function is increasing the average value is less than the function and increasing. When the function starts to decrease, the average value will continue to increase for a while. When the two graphs nest intersect, the process starts over, and the average value will now start to decrease. Therefore, the intersection value is when the average value function change from increasing to decreasing and this is its (local) maximum value. See Figure 2.

This continues until the graphs intersect again after the function starts to increase: a (local) minimum value of the average value function. The process continues with the extreme values of the average value function (blue graph) occurring at its intersections with the function. Figure 3

This can be proved by finding the extreme values of the average value function by considering its derivative. Begin by finding its derivative using the product rule (or quotient rule) and the FTC.

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)+\left( {-\tfrac{1}{{{{{\left( {x-a} \right)}}^{2}}}}} \right)\int_{a}^{x}{{f\left( t \right)dt}}$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)-\tfrac{1}{{x-a}}\left( {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} \right)$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}\left( {f\left( x \right)-A\left( x \right)} \right)$

The critical points of a(x) occur when its derivative is equal to zero (or undefined). This is when $f\left( x \right)=A\left( x \right)$ (or when x = a, the endpoint). This is where the graphs intersect.

## How to use this in your class

This is not a concept that is likely to be tested on the AP Calculus Exams. Nevertheless, it is an easy enough idea to explore when teaching the average value of a function and at the same time reviewing some earlier concepts such as product (or quotient) rule, the FTC (differentiating an integral), and some non-ordinary simplification.

You could have your students use their own favorite function and show that the extreme values of its average value occur where the average value intersects the function. This is good practice in equation solving on a calculator since the points do not occur at “nice” numbers. Here’s an example.

If $\displaystyle f\left( x \right)=\sin \left( x \right)$, then its average value on the interval $\displaystyle [0,\infty )$ is

$\displaystyle A\left( x \right)=\tfrac{1}{x}\int_{0}^{x}{{\sin \left( t \right)dt}}=\frac{{-\cos \left( x \right)+1}}{x}$.

The intersections of f(x) and A(x) can be found by solving

$\displaystyle\sin \left( x \right)=\frac{{-\cos \left( x \right)+1}}{x}$

The extreme values of $\displaystyle \frac{{-\cos \left( x \right)+1}}{x}$ may also be found using a calculator.

The points are the same. the first is approximately (2.331, 0.725) and the second is (6.283, 0) or (2π, 0). This second is reasonable since at 2π the sine function has completed one period and its average value zero. (See figure 3 again.).

Other questions you could ask (for my function anyway) are what is the absolute maximum and how can you be sure? Why are all the minimums zero?

The message on the AP Calculus discussion boards that inspired this post was started by Neema Salimi an AP Calculus teacher from Georgia. He made the original observation. You can read his original post and proof, and comments by others here.

## The Rule of Four

Not much has been heard of the Rule of Four lately. The Rule of Four suggests that mathematical concepts should be looked at graphically, numerically, analytically, and verbally. It has not gone away. The Rule of Four has a new name: multiple representations. (In the latest Course and Exam Description, you will find it in Mathematical Practices (p. 14), specifically practices 2.B, 2.C, 2.D, 2.E, 3.E, 3.F, 4.A, and 4.C)

I have used the Rule of Four in this post. The post started with a verbal discussion of the concept and how the result can be seen graphically. That was followed by analytic proof. At the end is a numerical example.

Other posts on the average value of a function:

Finding the average value of a function on an interval is Topic 8.1 in the Course and Exam Description (p. 149)

Average Value of a Function – or How do you average an infinite number of numbers?

Most Triangles Are Obtuse! An obvious observation, but here’s how to figure the exact proportion of obtuse to acute triangles.

Half-full or Half-empty Visualizing the average value of a function

What’s a Mean Old Average Anyway? Be sure to distinguish between the average rate of change, the average value of a function, and the mean value theorem.

# Open or Closed?

About this time of year, you find someone, hopefully one of your students, asking, “If I’m finding where a function is increasing, is the interval open or closed?”

This is a good time to teach some things about definitions and theorems.

The place to start is to ask what it means for a function to be increasing. Here is the definition:

A function is increasing on an interval if, and only if, for all (any, every) pairs of numbers x1 < x2 in the interval, f(x1) < f(x2).”

(For decreasing on an interval, the second inequality changes to f(x1) > f(x2). All of what follows applies to decreasing with obvious changes in the wording.)

1. Notice that functions increase or decrease on intervals, not at individual points. We will come back to this in a minute.
2. Numerically, this means that for every possible pair of points, the one with the larger x-value always produces a larger function value.
3. Graphically, this means that as you move to the right along the graph, the graph is going up.
4. Analytically, this means that we can prove the inequality in the definition.

For an example of this last point consider the function f(x) = x2. Let x2 = x1 + h where h > 0. Then in order for f(x1) < f(x2) it must be true that

$\displaystyle {{x}_{1}}^{2}<{{\left( {{{x}_{1}}+h} \right)}^{2}}$

$\displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}$

$\displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2}$

$\displaystyle 0<2{{x}_{1}}h+{{h}^{2}}$

This can only be true if $\displaystyle {{x}_{1}}\ge 0$ Thus, x2 is increasing only if$\displaystyle {{x}_{1}}\ge 0$

Now, of course, we rarely, if ever, go to all that trouble. And it is even more trouble for a function that increases on several intervals.  The usual way of finding where a function is increasing is to look at its derivative.

Notice that the expression $\displaystyle 0<{{x}_{1}}^{2}+2{{x}_{1}}h+{{h}^{2}}-{{x}_{1}}^{2}$ looks a lot like the numerator of the original limit definition of the derivative of x2 at x = x1, namely $\displaystyle 0<{{\left( {{{x}_{1}}+h} \right)}^{2}}-{{x}_{1}}^{2}$. If h > 0, where the function is increasing the numerator is positive and the derivative is positive also. Turning this around we have a theorem that says, If $\displaystyle {f}'\left( {{{x}_{1}}} \right)>0$ for all x in an interval, then the function is increasing on the interval. That makes it much easier to find where a function is increasing, we simplify find where its derivative is positive.

There is only a slight problem in that the theorem does not say what happens if the derivative is zero somewhere on the interval. If that is the case, we must go back to the definition of increasing on an interval or use a different method. For example, the function x3 is increasing everywhere, even though its derivative at the origin is zero.

Let’s consider another example. The function sin(x) is increasing on the interval $\displaystyle [-\tfrac{\pi }{2},\tfrac{\pi }{2}]$ (among others) and decreasing on $\displaystyle [\tfrac{\pi }{2},\tfrac{3\pi }{2}]$. It bothers some that $\displaystyle \tfrac{\pi }{2}$ is in both intervals and that the derivative of the function is zero at x = $\displaystyle \tfrac{\pi }{2}$. This is not a problem. Sin($\displaystyle \tfrac{\pi }{2}$) is larger than all the other values is both intervals, so by the definition, and not the theorem, the intervals are correct.

It is generally true that if a function is continuous on the closed interval [a,b] and increasing on the open interval (a,b) then it must be increasing on the closed interval [a,b] as well.

Returning to the first point above: functions increase or decrease on intervals not at points. You do find questions in books and on tests that ask, “Is the function increasing at x = a.” The best answer is to humor them and answer depending on the value of the derivative at that point. Since the derivative is a limit as h approaches zero, the function must be defined on some interval around x = a in which h is approaching zero. So, answer according to the value of the derivative on that interval.

You can find more on this here.

Case Closed.

Slightly revised from a posted published on November 2, 2012.

# Extreme Values

Every function that is continuous on a closed interval must have a maximum and a minimum value on the interval. These values may all be the same (y = 2 on [-3,3]); or the function may reach these values more than once (y = sin(x)).

If the function is defined on a closed interval, then the extreme values are either (1) at an endpoint of the interval or (2) at a critical number. This is known as the Extreme Value Theorem. Thus, one way of finding the extreme values is to simply find the value of the function at the endpoints and the critical points and compare these to find the largest and smallest. This is called the Candidates’ Test or the Closed Interval Test. It is a good one to “play” with: do some sketches of the different situation above; discuss why the interval must be closed.

On an open or closed interval, the shapes can change if the first derivative is zero or undefined at the point where two shapes join. In this case the point is a local extreme value of the function – a local maximum or minimum value. Specifically:

• If the first derivative changes from positive to negative, the shape of the function changes from increasing to decreasing and the point is a local maximum.  If the first derivative changes from negative to positive, the shape of the function changes from decreasing to increasing and the point is a local minimum.

This is a theorem called the First Derivative Test. By finding where the first derivative changes sign and in which direction it changes (positive to negative, or negative to positive) we can locate and identify the local extreme value precisely.

• Another way to determine if a critical number is the location of a local maximum or minimum is a theorem called the Second Derivative Test.

If the first derivative is zero (and specifically not if it is undefined) and the second derivative is positive, then the graph has a horizontal tangent line and is concave up. Therefore, this is the location of a local minimum of the function.

Likewise, if the first derivative is zero at a point and the second derivative is negative there, the function has a local maximum there.

If both the first and second derivatives are zero at a point, then the second derivative test cannot be used, for example y = x4 at the origin.

The mistake students make with the second derivative test is in not checking that the first derivative is zero. If “justify your answer” is required, students should be sure to show that the first derivative is zero as well as the sign of the second derivative.

In the case where both the first and second derivatives are zero at the same point the function changes direction but not concavity (e.g.  (x) = xat the origin), or changes concavity but not direction (e.g.  (x) = xat the origin).

This is a revised version of a post published on October 22, 2012

A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, students find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

 Feature the function ${y}'$> 0 is increasing ${y}'$ < 0 is decreasing ${y}'$ changes – to + has a local minimum ${y}'$changes + to – has a local maximum ${y}'$ increasing is concave up ${y}'$ decreasing is concave down ${y}'$ extreme value has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked.

Here is the same graph with the second derivative features marked.

The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x-2 because its derivative changes from positive to negative at x = -2.”

 Conclusion Justification y is increasing ${y}'$> 0 y is decreasing ${y}'$< 0 y has a local minimum ${y}'$changes  – to + y has a local maximum ${y}'$changes + to – y is concave up ${y}'$increasing y is concave down ${y}'$decreasing y has a point of inflection ${y}'$extreme values

For notes on asymptotes see Asymptotes and the Derivative and Other Asymptotes.

Originally posted on October 26, 2012, and my single most viewed post over the years.