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 [email protected] .
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 4/17 Test on separable Derivative of an differential inverse equations Worksheets attached 4/20 4/21 4/22 4/23 4/24 Quiz on derivative Watch ap online Review area Test on area Review project of an inverse video of area between curves between curves the calculus between curves packet from conundrum and do problems before break in drop box from Study for test Worksheet video attached
4/27 4/28 4/29 4/30 2019 AP free 2018 free 2017 free 2016 free response response response response questions 1-6 questions 1-6 questions 1-6 questions 1-6
See above link See above link See above link 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
-1 1 Dividing gives the derivative (f (x))’ = f’(푓−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
1 1 1 (h-1)‘(2) = = = ℎ′(ℎ−1(2)) ℎ′(1) 8
AP Calculus Name ______CHAPTER 7 WORKSHEET INVERSE FUNCTIONS 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. x 4 1. f x 2x 2 2. gx x a3 b 3. hx 2 x x3 4
4. Think About It…Find the derivative of y tan x . Notice that the subject derivative has the same sign for all values of x, so is a monotonic function. However, 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. f x x 32 6. gx x 5
In 7-9, find the derivative of the inverse function at the corresponding value.
3 d 1 7. Given f (x) x 2x 1, find f (Note: you may need to use guess and check to solve an dx x 2 equation involved in this problem.)
d 8. Given g(x) 2x5 x3 1, find g 1 dx x1
π π d 1 9. Given h(x) sin x on the interval , find h 2 2 dx x 1 2
SEE OTHER SIDE 10. Selected values of a strictly monotonic function 𝑔(푥) and its derivative 𝑔′(푥) are shown on the table below. 푥 −3 −1 1 4 g(x) 5 1 0 −3 1 1 g’(x) −4 −2 5 6 a) Find g 1 1 b) Find g 1 3
11. Selected values of a strictly monotonic function ℎ(푥) and its derivative ℎ′(푥) are shown on the table below. 푥 −1 0 2 4 h(x) −5 −1 4 7 1 1 h’(x) 3 5 2 6 Let f x be a function such that f x h1x. a) Find f '1 b) Find f '4
True or False? In 12-15, determine whether the statement is true or false. Justify your answer. 12. If f (x) is an even function, then f 1 (x) exists.
13. If the inverse of f exists, then the y-intercept of f is an x-intercept of f -1.
14. If f (x) xn where n is odd, then exists.
15. There exists no function f such that f = f -1.
AP Calculus CHAPTER 7 WORKSHEET ANSWER KEY INVERSE FUNCTIONS Derivatives of Inverse Functions
1. f 'x x3 4x xx2 1 Performing a sign analysis, f 'x 0 if x < 0, but f 'x 0 if x > 0 (except at x = 1), so this is not a strictly monotonic function and it does not have an inverse function.
2 2. g'x 3x a Performing a sign analysis, g'x 0 for all values of x, except at x = −a. So this is a strictly monotonic function and it has an inverse function.
2 3. h'x 1 3x Performing a sign analysis, h'x 0 for all values of x. So this is a strictly monotonic function and it has an inverse function.
4. y' sec2 x . We have y' sec2 x 0, for all values of x included in the domain of sec x. Therefore y tan x is always increasing. But the graph of has vertical tangent lines and it does not pass the horizontal line test: is not a one-to-one function.
5. f x x 32 6. gx x 5
1 f x x 3 1 g x x 5 1 Domain of f x is [0, ) Domain of g 1 x is d 1 1 7. Given f (x) x3 2x 1, f 1 dx x 2 f '1 5
d 1 8. Given g(x) 2x5 x3 1, g 1 undefined dx x1 g'0
π π d 1 1 2 2 3 9. Given h(x) sin x on the interval , , h 2 2 dx 3 3 x 1 2 h' 6
10. 푥 −3 −1 1 4 g(x) 5 1 0 −3 1 1 g’(x) −4 −2 5 6 1 a) g 1 1 5 g'1 1 1 b) g 1 3 g'4 2
11. 푥 −1 0 2 4 h(x) −5 −1 4 7 1 1 h’(x) 3 5 2 6 1 a) f '1 2 h'0 1 b) f '4 6 h'2
12. If f (x) is an even function, then f 1 (x) 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 f (x) xn where n is odd, then exists. TRUE. If where n is odd, its derivative is f '(x) nx n1 where n – 1 is even. So f '(x) 0 for all values of x except x = 0. Therefore is strictly monotonic.
15. There exists no function f such that f = f -1. FALSE. There are many such functions! Some examples: 푦 = 푥, 푦 = −푥, 푦 = −푥 + 푎,…
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:
2 s(t) = .5 (-32)t + 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 #
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.
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
t
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:
AP
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.
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 #
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 2 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.
Eliminate another suspect. You must find an exact answer. The denominator of your answer is the suspect to cross off….
Suspect #
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.”
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.
I am the Folium of Descartes x3 - y 3 + 6xy = 0
Find my slope at 1 1
“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!
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 I
I I
I 1 inch
Eliminate another suspect: To the nearest dozen, how many cookies are in the box?
Suspect #
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 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 1 when illuminated would be the height of the cone. Volume of a cone = 휋푟2ℎ 3
Use your answer to eliminate another suspect.
Suspect #
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? _