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


Unit 9 – Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Unit 9 includes all the topics listed in the title. These are BC only topics (CED – 2019 p. 163 – 176). These topics account for about 11 – 12% of questions on the BC exam.

Comments on Prerequisites: In BC Calculus the work with parametric, vector, and polar equations is somewhat limited. I always hoped that students had studied these topics in detail in their precalculus classes and had more precalculus knowledge and experience with them than is required for the BC exam. This will help them in calculus, so see that they are included in your precalculus classes.

Topics 9.1 – 9.3 Parametric Equations

Topic 9.1: Defining and Differentiation Parametric Equations. Finding dy/dx in terms of dy/dt and dx/dt

Topic 9.2: Second Derivatives of Parametric Equations. Finding the second derivative. See Implicit Differentiation of Parametric Equations this discusses the second derivative.

Topic 9.3: Finding Arc Lengths of Curves Given by Parametric Equations. 

Topics 9.4 – 9.6 Vector-Valued Functions and Motion in the plane

Topic 9.4 : Defining and Differentiating Vector-Valued Functions. Finding the second derivative. See this A Vector’s Derivatives which includes a note on second derivatives. 

Topic 9.5: Integrating Vector-Valued Functions

Topic 9.6: Solving Motion Problems Using Parametric and Vector-Valued Functions. Position, Velocity, acceleration, speed, total distance traveled, and displacement extended to motion in the plane. 

Topics 9.7 – 9.9 Polar Equation and Area in Polar Form.

Topic 9.7: Defining Polar Coordinate and Differentiation in Polar Form. The derivatives and their meaning.

Topic 9.8: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Topic 9.9: Finding the Area of the Region Bounded by Two Polar Curves. Students should know how to find the intersections of polar curves to use for the limits of integration. 


Timing

The suggested time for Unit 9 is about 10 – 11 BC classes of 40 – 50-minutes, this includes time for testing etc.


Previous posts on these topics:

Parametric and Vector Equations

Implicit Differentiation of Parametric Equations

A Vector’s Derivatives

Adapting 2012 BC 2 (A parametric equation question)

Polar Curves

Polar Equations for AP Calculus

Extreme Polar Conditions

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?”

Do you have an answer?

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

Unit 5 – Analytical Applications of Differentiation

Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.

You may want to consider teaching Unit 4 after Unit 5. Notes on Unit 4 are here.

Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation Writing on the AP Calculus Exams and its handout

Topics 5.1

Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.

Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval then it is the absolute (global) maximum or minimum for the function on that interval.

Topic 5.8 Sketching Graphs of Functions and Their Derivatives First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topics 5.10 – 5.11

Optimization is an important application of derivatives. Optimization problems as presented in most textbooks, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not been asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems 

Topic 5.11 Solving Optimization Problems

Topics 5.12

Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.


Timing

Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.

The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context

This is a re-post and update of the third in a series of posts from last year. It contains links to posts on this blog about the differentiation of composite, implicit, and inverse functions for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic. 


Previous posts on these topics include:

Then There Is This – Existence Theorems

What’s a Mean Old Average Anyway

Did He, or Didn’t He?   History: how to find extreme values without calculus

Mean Value Theorem

Foreshadowing the MVT

Fermat’s Penultimate Theorem

Rolle’s theorem

The Mean Value Theorem I

The Mean Value Theorem II

Graphing

Concepts Related to Graphs

The Shapes of a Graph

Joining the Pieces of a Graph

Extreme Values

Extremes without Calculus

Concavity

Reading the Derivative’s Graph

        Other Asymptotes

Real “Real-life” Graph Reading

Far Out! An exploration

Open or closed Should intervals of increasing, decreasing, or concavity be open or closed?

Others

Lin McMullin’s Theorem and More Gold  The Golden Ratio in polynomials

Soda Cans Optimization video

Optimization – Reflections   

Curves with Extrema?

Good Question 10 – The Cone Problem

Implicit Differentiation of Parametric Equations    BC Topic


Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

Limits and Continuity – Unit 1  (8-11-2020)

Definition of t he Derivative – Unit 2  (8-25-2020)

Differentiation: Composite, Implicit, and Inverse Function – Unit 3  (9-8-2020)

Contextual Applications of the Derivative – Unit 4   (9-22-2002)   Consider teaching Unit 5 before Unit 4

Analytical Applications of Differentiation – Unit 5  (9-29-2020) Consider teaching Unit 5 before Unit 4 THIS POST

LAST YEAR’S POSTS – These will be updated in coming weeks

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions 

2019 CED Unit 10 Infinite Sequences and Series


Adapting 2021 AB 4 / BC 4

Four of nine. Continuing the series started in the last three posts, this post looks at the AP Calculus 2021 exam question AB 4 / BC 4. The series considers each question with the aim of showing ways to use the question with your class as is, or by adapting and expanding it.  Like most of the AP Exam questions there is a lot more you can ask from the stem and a lot of other calculus you can discuss.

2021 AB 4 / BC 4

This is a Graph Analysis Problem (type 3) and contains topics from Units 2, 4, and 6 of the current Course and Exam Description. The things that are asked in these questions should be easy for the students, however each year the scores are low. This may be because some textbooks simply do not give students problems like this. Therefore, supplementing with graph analysis questions from past exams is necessary.

There are many additional questions that can be asked based on this stem and the stems of similar problems. Usually, the graph of the derivative is given, and students are asked questions about the graph of the function. See Reading the Derivative’s Graph.

Some years this question is given a context, such as the graph is the velocity of a moving particle. Occasionally there is no graph and an expression for the derivative or function is given.

Here is the 2021 AB 4 / BC 4 stem:

The first thing students should do when they see G\left( x \right)=\int_{0}^{x}{{f\left( t \right)}}dt is to write prominently on their answer page {G}'\left( x \right)=f\left( x \right) and \displaystyle {G}''\left( x \right)={f}'\left( t \right). While they may understand and use this, they must say it.

Part (a): Students were asked for the open intervals where the graph is concave up and to give a reason for their answer. (Asking for an open interval is to remove any concern about the endpoints being included or excluded, a place where textbooks differ. See Going Up.)

Discussion and ideas for adapting this question:

  • Using this or similar graphs go through each of these with your class until the answers and reasons become automatic. There are quite a few other things that may be asked here based on the derivative.
    • Where is the function increasing?
    • Decreasing?
    • Concave down, concave up?
    • Where are the local extreme values?
    • What are the local extreme values?
    • Where are the absolute extreme values?
    • What are the absolute extreme values?
  • There are also integration questions that may be asked, such as finding the value of the functions at various points, such as G(1) = 2 found by using the areas of the regions. Also, questions about the local extreme values and the absolute extreme value including their values. These questions are answered by finding the areas of the regions enclosed by the derivative’s graph and the x-axis. Parts (b) and (c) do some of this.
  • Choose different graphs, including one that has the derivative’s extreme value on the x­-axis. Ask what happens there.

Part (b): A new function is defined as the product of G(x) and f(x) and its derivative is to be found at a certain value of x. To use the product rule students must calculate the value of G(x) by using the area between f(x) and the x-­axis and the value of {f}'\left( x \right) by reading the slope of f(x) from the graph.

Discussion and ideas for adapting this question:

  • This is really practice using the product rule. Adapt the problem by making up functions using the quotient rule, the chain rule etc. Any combination of \displaystyle G,{G}',{G}'',f,{f}',\text{ or }{f}'' may be used. Before assigning your own problem, check that all the values can be found from the given graph.
  • Different values of x may be used.

Part (c): Students are asked to find a limit. The approach is to use L’Hospital’s Rule.

Discussion and ideas for adapting this question:

  • To use L’Hospital’s Rule, students must first show clearly on their paper that the limit of the numerator and denominator are both zero or +/- infinity. Saying the limit is equal to 0/0 is considered bad mathematics and will not earn this point. Each limit should be shown separately on the paper, before applying L’Hospital’s Rule.
  • Variations include a limit where L’Hospital’s Rule does not apply. The limit is found by substituting the values from the graph.
  • Another variation is to use a different expression where L’Hospital’s Rule applies, but still needs values read from the graph.

Part (d): The question asked to find the average rate of change (slope between the endpoints) on an interval and then determine if the Mean Value Theorem guarantees a place where \displaystyle {G}' equals this value. Students also must justify their answer.

Discussion and ideas for adapting this question:

  • To justify their answer students must check that the hypotheses of the MVT are met and say so in their answer.
  • Adapt by using a different interval where the MVT applies.
  • Adapt by using an interval where the MVT does not apply and (1) the conclusion is still true, or (b) where the conclusion is false.

Next week 2021 AB 5.

I would be happy to hear your ideas for other ways to use this questions. Please use the reply box below to share your ideas.


Other Asymptotes

A few days ago, a reader asked if you could find the location of horizontal asymptotes from the derivative of a function. The answer, alas, is no. You can determine that a function has a horizontal asymptote from its derivative, but not where it is. This is because a function and its many possible vertical translations have the same derivative but different horizontal asymptotes. To locate it precisely, we need more information, an initial condition (a point on the curve).

I have written a post about the relationship between vertical asymptotes and the derivative of the function here. Today’s post will discuss horizontal asymptotes and their derivatives.

Definition: a straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the curve at the point approaches the slope of the line. (Merriam-Webster dictionary). Thus, asymptotes can be vertical, horizontal, or slanted.

Horizontal Asymptotes

Horizontal asymptotes are a form of end behavior: they appear as \displaystyle x\to \infty or \displaystyle x\to -\infty . Specifically, if a function has a horizontal asymptote(s), then \displaystyle \underset{{x\to \pm \infty }}{\mathop{{\lim }}}\,{f}'\left( x \right)=0,  At an asymptote, the curve must approach a horizontal line, therefore its slope must be approaching zero as \displaystyle x\to \infty or \displaystyle x\to -\infty .  The converse is not true: If the derivative approaches zero, the function may not have a horizontal asymptote. A counterexample is \displaystyle f\left( x \right)=\sqrt[3]{x}.

In the first four examples that follow, the derivative approaches 0 as \displaystyle x\to \infty and/or \displaystyle x\to -\infty

Example 1: \displaystyle {f}'\left( x \right)=\frac{1}{{1+{{x}^{2}}}}  (Figure 1 in blue).This is the derivative of \displaystyle f\left( x \right)={{\tan }^{{-1}}}(x),\ -\tfrac{\pi }{2}<f\left( x \right)<\tfrac{\pi }{2}. (Figure 1 in red). We see that the derivative approaches zero in both directions, this tells us that there may be horizontal asymptote(s). We must look at the function to find where they are: \displaystyle y=\tfrac{\pi }{2} and \displaystyle y=-\tfrac{\pi }{2}

Figure 1

Example 2: 2008 AB 19 \displaystyle p\left( x \right)=\frac{{5+{{2}^{x}}}}{{1={{2}^{x}}}}. (Figure 2 in Red)

So many students missed this (pre-calculus ?) question on the exam that the question was not counted! As \displaystyle x\to \infty the 2x terms approach infinity at the same rate so the limit is the ratio of their coefficients: the asymptote is y = –1. And as \displaystyle x\to -\infty the 2x terms approach zero so the fractions approaches an asymptote at y = 5. (There is a vertical asymptote at x = 0, not shown.)

The derivative is \displaystyle {p}'\left( x \right)=\frac{{6\ln (2)\cdot {{2}^{x}}}}{{{{{\left( {1-{{2}^{x}}} \right)}}^{2}}}} (Figure 2 blue). The derivative approaching zero in both directions, but the location of the asymptotes must be determined from the equation of the function.

Figure 2

Example 3: \displaystyle g\left( x \right)=\frac{{-3}}{{{{{\left( {x+1} \right)}}^{2}}}} (Figure 3 in blue). This is the derivative of \displaystyle g(x)=\frac{3}{{x+1}}+2 (Figure 2 in red). Notice that replacing the “+2” with a different constant will translate the curve and its asymptote up or down. Here the function has the same asymptote in both directions. (There is a vertical asymptote at x = –1, not shown)

Figure 3

 

Example 4: Consider the function \displaystyle s\left( x \right)={{1.1}^{=x}}\sin \left( x \right)See figure 4.

As \displaystyle x\to \infty this function’s values get closer to zero and the graph appears to have the x-axis as a horizontal asymptote. The interesting thing is that the function’s value at every integral  multiple of π is zero and the function changes sign there. In other words it crosses and re-crosses its asymptote and moves away from and back towards the asymptote. Each time it crosses the asymptote it moves less far from the asymptote, but continues to move towards and away from it. It’s derivative (not shown) exhibits the same phenomenon. 

This feature is difficult to see on a graph since the values become exceedingly small in absolute value very quickly.

Figure 4

Slant Asymptotes

Other lines lines may also be asymptotes, called slant asymptotes. If as \displaystyle x\to \infty or \displaystyle x\to -\infty , the function approaches a non-horizontal line, which appears as a slant asymptote. In this case the derivative approaches a constant equal to the slope of the line.

Example 5: \displaystyle {h}'\left( x \right)=1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}} (Figure 5 in blue) is the derivative of \displaystyle h\left( x \right)=\frac{{{{x}^{2}}+x-5}}{{(x-2)}}=x+3+\frac{1}{{\left( {x-2} \right)}}.

The derivative shows a horizontal asymptote at y = 1, so the function will have a slant asymptote with a slope of 1. This is obvious (I hope) from the second form of the function where the final term approaches zero as \displaystyle x\to \infty and \displaystyle x\to -\infty . For positive numbers, the last term is positive and the function is therefore above line; for negative numbers the last term is negative and the function is below the line. Vertical translations will move the function and its slant asymptote, but not change the asymptote’s slope. (There is a vertical asymptote at x = 2, not shown)

Figure 5

 

Other “Asymptotes”

Example 6: Playing with the idea from examples 5, if the function were \displaystyle k\left( x \right)={{x}^{2}}+x+\frac{1}{{\left( {x-2} \right)}}. (Red in figure 6). The curve will approach the parabola \displaystyle y={{x}^{2}}+x (Figure 6 black dotted) as an “asymptote.” Its derivative \displaystyle {k}'\left( x \right)=2x+1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}} (Figure 6 blue) approaches the line (its own slant asymptote) \displaystyle y=2x+1 (not shown). There is a vertical asymptote at x = 2 (not shown). 

Figure  6

Another interesting feature of this kind of function is this:  \displaystyle k\left( x \right)={{x}^{2}}+x+a+\frac{1}{{x-2}} is the same function translated up or down by an amount a. The resulting functions have the same derivative and the same parabola for their asymptote! Try investigating this situation using Desmos or GeoGebra or some other graphing utility with a slider. 

This type of “asymptote” might make a good topic for an exploration, project, or investigation by a student.


 

 

 

 

 


Revised March 23, 2021.