Hello AP Calculus students,

There are several options for turning in your work: Use a google doc, or share an email, or screen shot

your work on an ipad and share it with me at ted.grice@bvasd.net .

I will send out information on Remind or you can check school wires. Also check your school email from

time to time. I will have video lessons from the AP program on these 2 weeks of lessons online as they

are what is needed to finish up the course before taking the AP test. Book work will continue through

the ipad and as always answers are available online.To have continuity of learning here is a brief

overview of the next 2 weeks

AP Online video is found on area here:

https://www.youtube.com/watch?v=EPKKUzsnPBg&list=PLoGgviqq4844keKrijbR_EPKRNIW6hahV&inde

x=12

The ap free response is found here:

2019: https://apstudents.collegeboard.org/ap/pdf/ap19-frq-calculus-ab.pdf

2018: https://apcentral.collegeboard.org/pdf/ap18-frq-calculus-ab.pdf

2017: https://apcentral.collegeboard.org/pdf/ap-calculus-ab-frq-2017.pdf

2016: https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap16_frq_calculus_ab.pdf

4/16 Test on separable differential equations

4/17 Derivative of an inverse Worksheets attached

4/20 Quiz on derivative of an inverse

4/21 Watch ap online video of area between curves and do problems in drop box from video

4/22 Review area between curves packet from before break Study for test

4/23 Test on area between curves

4/24 Review project the calculus conundrum Worksheet attached

4/27 2019 AP free response questions 1-6 See above link

4/28 2018 free response questions 1-6 See above link

4/29 2017 free response questions 1-6 See above link

4/30 2016 free response questions 1-6 See above link

Stay Safe! Mr. Grice

Finding the derivative of an inverse:

Recall if f(x) and g(x) are inverses then f(g(x)) = g(f(x)) = x

Also g(x) notation wise for an inverse would be f-1(x)

Note: d/dx (x) = 1 and the chain rule for the derivative of f-1(x) yields d/dx f(f-1(x)) = f’(f-1(x)) •(f-1(x)’

Therefore taking the derivative of f(f-1(x)) = x yields f’(f-1(x)) •(f-1(x))’ = 1

Dividing gives the derivative (f-1(x))’ =

1

f’(𝑓−1(x))

Example:

The function h is given by h(x) = x5 + 3x – 2 and h(1) =2.

If h-1 is the inverse of h, what is the value of (h-1)‘(2)?

Solution:

h’(x) = 5x4 + 3

h(1) = 2 h-1(2) = 1

(h-1)‘(2) = 1

ℎ′(ℎ−1(2)) =

1

ℎ′(1) =

1

8

AP Calculus CHAPTER 7 WORKSHEET INVERSE FUNCTIONS

Name ________________________________

Seat # ______ Date ____________________

Derivatives of Inverse Functions

In 1-3, use the derivative to determine whether the function is strictly monotonic on its entire domain

and therefore has an inverse.

1. 24

24

xx

xf 2. baxxg 3

3. 32 xxxh

4. Think About It…Find the derivative of xy tan . Notice that the subject derivative has the same

sign for all values of x, so xy tan is a monotonic function. However, xy tan is not a one-

to-one function. Why?

In 5-6, (a) “delete” part of the graph of the function shown so that the part that remains is one-to-one.

Then, (b) find the inverse of the remaining part and (c) state its domain. (Note: there is more than one

correct answer for these questions!)

5. 23 xxf 6. 5 xxg

In 7-9, find the derivative of the inverse function at the corresponding value.

7. Given 12)( 3 xxxf , find 2

1

x

fdx

d (Note: you may need to use guess and check to solve an

equation involved in this problem.)

8. Given 12)( 35 xxxg , find 1

1

x

gdx

d

9. Given xxh sin)( on the interval

2

π,

2

π find

21

1

x

hdx

d

SEE OTHER SIDE

10. Selected values of a strictly monotonic function 𝑔(𝑥) and its derivative 𝑔′(𝑥) are shown on the

table below.

𝑥 −3 −1 1 4

g(x) 5 1 0 −3

g’(x) −4 5

1

6

1 −2

a) Find 11 g

b) Find 31 g

11. Selected values of a strictly monotonic function ℎ(𝑥) and its derivative ℎ′(𝑥) are shown on the

table below.

𝑥 −1 0 2 4

h(x) −5 −1 4 7

h’(x) 3 2

1

6

1 5

Let xf be a function such that xhxf 1 .

a) Find 1' f

b) Find 4'f

True or False? In 12-15, determine whether the statement is true or false. Justify your answer.

12. If )(xf is an even function, then )(1 xf exists.

13. If the inverse of f exists, then the y-intercept of f is an x-intercept of f -1.

14. If nxxf )( where n is odd, then )(1 xf exists.

15. There exists no function f such that f = f -1.

AP Calculus CHAPTER 7 WORKSHEET INVERSE FUNCTIONS

ANSWER KEY

Derivatives of Inverse Functions

1. 14' 23 xxxxxf

Performing a sign analysis, 0' xf if x < 0, but 0' xf if x > 0 (except at x = 1), so this

is not a strictly monotonic function and it does not have an inverse function.

2. 23' axxg Performing a sign analysis, 0' xg for all values of x, except at x = −a. So this is a strictly

monotonic function and it has an inverse function.

3. 231' xxh Performing a sign analysis, 0' xh for all values of x. So this is a strictly monotonic function

and it has an inverse function.

4. xy 2sec' . We have 0sec' 2 xy , for all values of x included in the domain of xsec .

Therefore xy tan is always increasing. But the graph of xy tan has vertical tangent lines

and it does not pass the horizontal line test: xy tan is not a one-to-one function.

5. 23 xxf 6. 5 xxg

31 xxf

Domain of xf 1 is ),0[

51 xxg

Domain of xg 1 is ),0[

7. Given 12)( 3 xxxf , 5

1

1'

1

2

1

ff

dx

d

x

8. Given 12)( 35 xxxg ,

0'

1

1

1

gg

dx

d

x

undefined

9. Given xxh sin)( on the interval

2

π,

2

π,

3

32

3

2

6'

1

21

1

h

hdx

d

x

10.

𝑥 −3 −1 1 4

g(x) 5 1 0 −3

g’(x) −4 5

1

6

1 −2

a)

51'

111

gg

b) 2

1

4'

131

gg

11.

𝑥 −1 0 2 4

h(x) −5 −1 4 7

h’(x) 3 2

1

6

1 5

a)

20'

11'

hf

b)

62'

14'

hf

12. If )(xf is an even function, then )(1 xf exists.

FALSE. An even function has symmetry with respect to the y-axis and, therefore, cannot be one-

to-one.

13. If the inverse of f exists, then the y-intercept of f is an x-intercept of f -1.

TRUE. Switching x and y coordinates will result in switching x and y intercepts.

14. If nxxf )( where n is odd, then )(1 xf exists.

TRUE. If nxxf )( where n is odd, its derivative is 1)(' nnxxf where n – 1 is even. So

0)(' xf for all values of x except x = 0. Therefore nxxf )( is strictly monotonic.

15. There exists no function f such that f = f -1.

FALSE. There are many such functions! Some examples: 𝑦 = 𝑥, 𝑦 = −𝑥, 𝑦 = −𝑥 + 𝑎,…

The Calculus Conundrum So there I was… I was talking to another calculus teacher, Mrs. AP Calca, and she relayed this story to me. Gather around and I will tell you a tale about the horrendous events of this past weekend. You may notice my bedraggled appearance accented by the dark circles under my eyes. I haven’t slept a wink in two days. Someone committed some dastardly deeds which scared me half to death. I am sorry to say that the culprit may be someone you have heard of before. Read the account which follows and see whether you can determine the identity of the perpetrator from the following list of suspects. Cross of suspects as you go. Whomever is not crossed off is the guilty suspect. Or said

another way… whomever you derive to be guilty will become the prime suspect... (get it?)

1. Georg Riemann 2. Michel Rolle 3. I Sac Newton 4. Pierre de Fermat 5. Marquis de L’Hospital 6. Wilhelm Leibniz 7. Rene Descartes 8. Aria Betweencurves 9. Derry Vative 10. Int E Gruel

On Friday afternoon, I left the building quite late after working on a set of calculus tests. As I walked to my car, I passed by the building and narrowly escaped being struck in the head by a falling object. Sadly, that object was the bust of Sir Isaac Newton, a cherished gift from my calculus class. As I picked up Sir Isaac’s remains, I discovered a note glued to one of the shards. It read:

I was afraid not to solve the problem. I pulled a piece of paper from my bag, hurriedly scratched off a solution and located the brick. I hopped into my car and squealed out of the parking lot.

s(t) = .5 (-32)t2 + 25 and s(t) = 0 when it hits the ground.

Help me determine the identity of the perpetrator. Solve the previous problem in this space. To eliminate a suspect, divide your answer to the problem by -10 and cross off the name that corresponds to that number on the list.

Suspect #

t

When I woke up the next morning, I decided the previous day’s episode must have been a silly prank. So I returned to school to plan Monday’s lesson. Of course, since it was Saturday, the building was locked so I had to work in a trailer that was sitting outside. It was an extremely hot day for late April, but the trailer was nice and cool inside. At least, it was cool in the trailer at first. But as I worked, I realized that the air conditioning must have shut down because it just got hotter and hotter by the minute. I tolerated it for about an hour and then, unable to stand it any longer, I packed up my things to leave. I tried to open the door of the trailer but it was jammed. Then a note appeared under the door. Here’s how it read:

Time (min) 0 10 20 30 40 50 60

Temp (F) 68 72.64 74.97 76.31 82.9 85.38 87.5

Eliminate another suspect: It’s the digit that appears in your answer twice.

Suspect #

I solved the problem as quickly as I could and slid it under the door. But I guess it wasn’t fast enough. Here’s the note I got back:

At that point I started to panic. I thought to myself, “Maybe I can escape through the window.” But the window turned out to be jammed too, and I had nothing to break it with but my bare fist. Finally, the trailer door clicked open. I looked out the window and saw a lone figure running off through the trees. I dropped everything, ran out the door, and made a beeline for my car. As I climbed into the driver’s seat, the doors automatically locked! There on the steering wheel was . . . I bet you can guess . . . another note! I turned the key and . . . nothing.

AP

Eliminate two more suspects. Find the answers to 3 decimal places and then do the following:

For question #1, take the first digit of your answer. For question #2, double the digit in the hundredths place.

Suspect # Suspect #

2

I worked furiously because I just wanted to get out of there. When I got my answers, I beeped the horn and a big white flag unfurled from a tree across the street. Here’s what it said:

Within seconds, the motor of my car cranked on. I gunned the engine and sped off. Much to my dismay, the radio came on and I heard these words:

Congratulations! You need a bigger challenge. Since you can’t write and drive at the

same time you have to do this one in your head. Look out the window of your car. I

replaced the orange ping pong ball on your antenna. You needed a new way to find

your car in Walmart’s parking lot. I carved a parabola out of the cover of your

teacher’s edition. Notice how nicely it spins around the antenna in the wind, you old

windbag.

Find the volume of this solid of revolution if the equation of the parabola is

x = y and the antenna is the y-axis. Write your answer on the windshield with

lipstick. It better be up there by the time you cross the Yough or you may just go

for a swim.

This was easy for me because I had worked that very problem just the night before, and I had the answer memorized. I fumbled through my purse for lipstick. Reluctantly, I scribbled the answer on the windshield. I crossed the Yough and then I heard a siren and saw flashing lights in my rear view mirror. The cop pulled me over and said, “Lady, are you drunk?!? Why are you writing numbers on your windshield in lipstick?" I told him I was being tormented by a lunatic calculus student who locked me in a trailer, broke my Isaac Newton, and stole my ping pong ball. He said, “Step out of your vehicle and walk this line.” After a humiliating couple of minutes walking backward and forward on the side of the road, the cop let me go with this observation, “You need help lady.”

Eliminate another suspect. You must find an exact answer. The denominator of your answer is the suspect to cross off….

Suspect #

Finally, I arrived home and pulled into my driveway. Exhausted, I dragged myself up to the front door and found a note hanging on the knob. Here’s what it said:

Eliminate another suspect: Find each answer. Then find the sum of the three answers. Take the first digit of that sum and cross of that suspect.

Suspect #

I went inside and securely bolted the door. I was relieved to find nothing out of place and my precious canary, Archimedes, unharmed. I went into the kitchen for a cup of tea to soothe my nerves. I gulped it down. As I sat down at the kitchen table, I saw, for the first time, two packages. One was beautifully wrapped and tied with a shiny pink ribbon shaped like this:

The other was shabbily constructed and covered with aluminum foil. I was afraid to open either one. Then I saw the writing on the ribbon.

“This is ridiculous,” I thought. I started to reach for the other package when I heard a resounding boom at my kitchen window. A large wad of putty was smooshed against the glass. Hanging from the wad was a sign that looked like this:

DO

NOW!

I am the Folium of Descartes

x3 - y 3 + 6xy = 0

Find my slope at 1 1

Eliminate another suspect: Find the exact value of the problem. Square the answer and cross off the suspect whose number is the same as the denominator.

Suspect #

So I worked the problem. I didn’t know what would happen to me if I didn’t do it. I taped my answer to the window and waited. Nothing happened, so I figured it was safe to open the other box. Under the aluminum foil, I found this note:

I was sick of this game. I knew the police would never believe me, and out of desperation I was gaining courage. I decided to take matters into my own hands. I turned out the lights and went down to the basement. I opened the door a wee bit and peered around the yard. No one was in sight but I knew the punk was nearby. I exited quickly, my flashlight in the waistband of my pants. I could make out the outline of a body up against the concrete wall in the back corner of the yard. I knew all about estimating rates and distances.

I I

I I

I 1 inch

Eliminate another suspect: To the nearest dozen, how many cookies are in the box?

Suspect #

I knew if the beam of a flashlight were aimed at a wall, it would form a cone whose altitude would change at the same rate that I approached the wall. If the volume of the cone of light was decreasing at a rate of 16 cubic feet per second, I figured the radius of the beam would increase at a rate of 7 ft/sec and the height decrease at 6 ft/sec. I asked myself, “How far (to the nearest foot) from the scoundrel would I have to be for all of these facts to come together and illuminate him (or her??) in an 8 foot radius of light?” The distance I would be away from the suspect

when illuminated would be the height of the cone. Volume of a cone = 1

3𝜋𝑟2ℎ

I got within 5 feet and – to my dismay – I had miscalculated. My tormentor sprang up and ran off into the night. And that was the end of my ordeal.

I have related this tale to you, hoping that you might help me discover the identity of my assailant. If you followed my tale and possess the mathematical skills to answer all the problems correctly, you will now know the identity of the culprit and be a true calculus nerd!

Who Dun It? _

Use your answer to eliminate another suspect.

Suspect #