Summer Assignment AP Calculus BC June 2016

Dear Student: Your AP Calculus Exam will be on Tuesday, May 9, 2017. That may seem like a long time, but it leaves us less than 150 school days of class next year to prepare. With snow days, half days, exams, and so on, we can only expect 135-140 actual days of class before the exam. This summer assignment is designed to provide a review of algebraic and trigonometric concepts that we covered during Pre-Calc. By covering this material before school begins in the fall, we will be able to continue and expand our study of differential calculus on the first day of school. You will have a test on the summer assignment material within the first week and a half of school. You may work with another student, or students, and you may email me for help. However, any evidence of copying will result in failure of the assignment (a test grade) and may constitute grounds for removal from the class. The important thing is that YOU understand the work. My email address is: cwhite@springfieldschools.com Don’t be shy!!!! Seek help or email me if you need help or would like further explanations. Responses will not be automatic; so do NOT wait until the last minute to email for help.

Due Date for Assignments—In my mailbox in the main office: Thursday 8/4/16 at 12:00 noon The main office is open Monday-Thursday from 8am-12 noon and 1pm-4pm, and closed on Friday. If you are handing it in at the last minute, make sure you get here by 4pm on Thursday, otherwise you will not be able to get into the main office and your assignment will be late. The maximum possible grade for a late assignment is a C. If you cannot drop your assignment at school, you may mail it to the school at: Jonathan Dayton High School 139 Mountain Ave. Springfield, NJ 07081 Attn: Mr. White (AP Calc summer assignment) If you want, you may scan your assignment and convert it to a .pdf file if you know how and email it to me at cwhite@springfieldschools.com

If you have any concerns about getting a summer assignment to me by a due date, please see me before the end of the school year so we can make other arrangements. Failure of a fax machine, the Postal Service, UPS, FedEx, etc. are NOT acceptable excuses for a summer assignment being late! ****Remember the goal is for YOU to understand the material thoroughly so that we can focus on the calculus material immediately when the school year begins. Failure to turn in the assignment will result in you having to drop the course. If you are not willing to work on your own, to hand in material on time or you can not demonstrate an understanding of the material on this assignment, you will not succeed in this course.

AP Calculus: Summer Assignment 2016

Problem Set : Complete the following problems from the textbook. Note: For all problems, work in the section called Exercises, not the section called Quick Review. All work must be shown. Any answer given with no work or explanation will be considered incorrect. Pages 7-9: #14, 16, 20, 32, 34, 36, 44 a-c Pages 48-50: #18, 20 . Pages 62-64: #3, 6, 8, 10, 13, 16, 22, 24, 25, 26, 29, 32, 44, 48 For #8-29, simply find the limits. You do not have to support graphically. For #32, briefly explain why each is true or false. Please note that question 32 has parts a-i with c, d, e, f, g, h and i being in the final column on the page. Pages 71-72: #1-4, 10, 18, 22, 35b, 36b, 50 Pages 80-81: #1, 2, 12, 16, 20, 24, 25, 28, 36, 37 Pages 87-89: #2, 3 (see note V farther along in this document),10a-c, 11a-c, 15, 16

(normal means perpendicular). Page 91 #2, 5, 6, 8. If you are using the .pdf version of the text book, this page was accidentally left out of the pdf file. The problems are:

2) limx→−2

x2 +13x2 − 2x + 5

5) limx→0

12+ x

−12

x 6) lim

x→±∞

2x2 +35x2 + 7

8) limx→0

sin2x4x

Pages 120-121: #2, 4, 8, 9, 14, 16, 17, 24 a-b, 28, 32 --

�

d2ydx 2

means 2nd derivative, which

is just the derivative of the derivative. Pages 140-141: #2, 4, 6, 8, 12, 20 If you are having trouble, there are a few pages of notes after this point. Please read and review these notes!

A few notes on select topics: I. Finding the equation of a line tangent to a curve at a given x-value.

You need two things for the equation of a line: a point and a slope. To find the point, you plug the given x-value into the original function. To find the slope, you plug the given x-value into the derivative. Then use point-slope form to write the equation of the line. Example: Find the equation of the line tangent to the graph of f x( ) = x3 − 2x2 + 5x − 3 at

x = 2.

f 2( ) = 23 − 2 2( )2 + 5 2( )− 3 = 7 ′f x( ) = 3x2 − 4x + 5′f 2( ) = 3 2( )2 − 4 2( )+ 5 = 9

The equation of the tangent line is y − 7 = 9 x − 2( )

II. Finding the equation of the normal line to a curve at a given x-value.

Normal just means perpendicular. The process is identical to the one shown above. The only difference is that when you find the slope using the derivative, you take the opposite reciprocal. If you were asked to find the line normal to the same graph at the same point

as in Example 1, the equation of the line would be y − 7 = − 19x − 2( ) .

III. Asymptotes

A. Vertical Asymptotes: Vertical asymptotes occur at any x-value that makes the denominator of a fully reduced rational function equal zero.

Example: Find the vertical asymptote(s) of f x( ) = x2 − 5x + 4x2 −16

.

The function can be reduced by factoring the numerator and the denominator.

f x( ) = x2 − 5x + 4x2 −16

=x − 4( ) x −1( )x + 4( ) x − 4( ) =

x −1x + 4

. Since the only number that results in a zero

denominator in the reduced expression is -4, there is one vertical asymptote at x = -4. B. Horizontal Asymptotes. Horizontal asymptotes are found by finding the limit of the function as x→ ±∞ . If you’re working with a rational function with polynomials in the numerator, remember the basic rules: - Numerator power is larger: no horizontal asymptote - Denominator power is larger: horizontal asymptote is y = 0 - Powers are the same: use the lead coefficients

IV. Removable discontinuities.

Only point discontinuities are removable. This can be done by changing one y-value in a function and usually writing the function as a piecewise function.

Example: Remove the discontinuity from f x( ) = x2 + 4x + 3x +1

First, factor the denominator. f x( ) = x +1( ) x + 3( )x +1( ) . There are discontinuities where the

denominator equals zero—that is at x = -3 and x = -1. However, when the function is reduced, it becomes f x( ) = x + 3 . If you plug in -1 to the reduced function, the result is 2. This is the y-value when x = -1 that will remove the point discontinuity at x = -1.

The new (extended) function is f x( ) =x2 + 4x + 3

x +1; x ≠ −1

2; x = −1

⎧⎨⎪

⎩⎪

V. Rates of Change

A. Average rate of change. If you are asked to find the average rate of change of a

function over an interval, use the formula f b( )− f a( )

b − a.

Example: Find the average rate of change of the function f x( ) = x3 − 2x on the interval [2, 5].

f 5( )− f 2( )5− 2

= 115− 43

= 1113

= 37

B. Instantaneous rate of change. If you are asked to find the instantaneous rate of change, simply plug the x-value into the derivative.

Example: Find the instantaneous rate of change of the function f x( ) = x3 − 2x at x = 2.

′f x( ) = 3x2 − 2′f 2( ) = 3 2( )2 − 2 =10

VI. Differentiability

A function is differentiable at a certain x-value if it is possible to find the derivative at that value. Functions are not differentiable at the following places:

- any discontinuity - sharp corners - cusps - vertical tangents - infinite oscillations

VII. Derivatives ddx

means find the derivative of⎛⎝⎜

⎞⎠⎟

A. Power Rule: ddxxn = nxn−1

B. Product Rule: ddx

f x( )g x( ) = f x( ) ′g x( )+ g x( ) ′f x( )

C. Quotient Rule: ddx

f x( )g x( ) =

g x( ) ′f x( )− f x( ) ′g x( )g x( )( )2

D. Derivatives of special functions

f x( ) ′f x( ) sin x cos x cos x −sin x tan x sec2 x csc x −csc xcot x sec x sec x tan x cot x −csc2 x ex ex

VIII. Interpreting functions and their derivatives

A. Interpreting a parent function f x( ) a. If f x( ) is increasing, then ′f x( ) is positive (above zero) b. If f x( ) is decreasing, then ′f x( ) is negative (below zero) c. If f x( ) has a minimum or a maximum, then ′f x( )= 0 or is undefined d. If f x( ) is concave up, then ′′f x( ) is positive (above zero) e. If f x( ) is concave down, then ′′f x( ) is negative (below zero) f. If f x( ) has a point of inflection, then ′′f x( ) = 0 or is undefined

B. Interpreting a first derivative ′f x( ) a. If ′f x( ) is positive, then f x( ) is increasing b. If ′f x( ) is negative, then f x( ) is decreasing c. If ′f x( )= 0 or is undefined and changes sign, f x( ) has a min or a max d. If ′f x( ) is increasing, then ′′f x( ) is positive and f x( ) is concave up e. If ′f x( ) is decreasing, then ′′f x( ) is negative and f x( ) is concave down

f. If ′f x( ) has a min or a max, then ′′f x( ) = 0 and f x( ) has a point of inflection

C. Interpreting a second derivative ′′f x( ) a. If ′′f x( ) is positive, then f x( ) is concave up b. If ′′f x( ) is negative, then f x( ) is concave down c. If ′′f x( ) = 0 or is undefined and changes sign, f x( ) has a point of inflection

Here is an example problem that covers finding all the features of a function.

For the function f x( ) = 13x3 + x2 −8x + 5 , find the following:

a) x-coordinates of all minima and maxima. Justify your answer. b) All intervals where the function is increasing/decreasing. c) x-coordinates of all points of inflection. d) All intervals where the function is concave up/concave down.

a) To find the locations of minima and maxima follow these steps:

1. Find the first derivative of the function: !f x( ) = x2 + 2x −8 2. Set the derivative equal to zero and solve:

x2 + 2x −8 = 0x + 4( ) x − 2( ) = 0x = −4 and x = 2

3. Plot these points on a number line. Then choose one number from each section of the number line and plug into the derivative. Follow the interpretations of the first derivative as shown in section VIII-B above to determine features of f x( ) .

Plugging in -5 to the derivative gives 7, plugging in 0 gives -8 and plugging in 3 gives 7. The numeric values are unimportant—just the positive/negative. Anywhere the first derivative is positive, the original function is increasing, and anywhere the first derivative is negative, the original function is decreasing.

4. Interpret your results! Since !f −4( ) = 0 and f x( ) switches from increasing to decreasing at x = -4, there must be a maximum at x = -4. Since !f 2( ) = 0 and f x( ) switches from decreasing to increasing at x = 2 , there

must be a minimum at x = 2. b) As can be seen from the answer to part a, f x( ) is increasing on the intervals

−∞,−4( ) and 2,∞( ) and is decreasing on the interval −4,2( ) . c) Locating points of inflection works similar to minima/maxima, except you use the second derivative.

1. Find the second derivative of the function: !!f x( ) = 2x + 2 . 2. Set this equal to zero and solve:

2x + 2 = 0x = −1

3. Plot this points on a number line. Then choose one number from each section of the number line and plug into the second derivative. Follow the interpretations of the second derivative as shown in section VIII-C above to determine features of f x( ) .

Plugging in -2 to the 2nd derivative gives -2, and plugging in 0 gives 2. The numeric values are unimportant—just the positive/negative. Anywhere the second derivative is positive, the original function is concave up, and anywhere the second derivative is negative, the original function is concave down.

4. Interpret your results! Since !!f −1( ) = 0 and f x( ) switches from concave down to concave up at x = −1

, there must be a point of inflection at x = −1 .

d) As can be seen from the results to part c, f x( ) is concave up on the interval −1,∞( ) and f x( ) is concave down on the interval −∞,−1( ) .