AP Calculus BC Prerequisite Knowledge -...

AP Calculus BC Prerequisite Knowledge

Please review these ideas over the summer as they come up during our class

and we will not be reviewing them during class. Also, I feel free to quiz you at any

time on these facts and procedures.

When you join me in the fall, I will using the TI

Nspire CX CAS calculator as our demonstration

calculator. While you are welcome to use any

graphing calculator, I won’t be lecturing about the

other calculators in class. You may purchase the TI

Nspire CX CAS on Amazon.com for approximately

$140.

If can wait until the fall, you can order one from

me for $130. There is an order form on the back of this page.

As you look through these pages, feel free to ask me, other math teachers and

your fellow future calculus students for clarifications.

Email: [email protected].

Cell: 313-799-2252

Website: www.toddfadoir.com/apcalc

We’ll be using Calculus: Graphical, Numerical, Algebraic. Third Edition by

Finney, Demana, Waits, & Kennedy.

Have a great summer!

Sincerely, Mr. Fadoir

Prerequisite information for AP Calculus BC

toddfadoir.com/apcalc 2

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Algebra

Find the equation of a line given a slope (m) and

a point ( )0 0,x y

I know your tendency to want to create the equation in the

form y mx b= + , I would like you to start using the form

( )0 0y y m x x= + − .

What is the equation of a line with slope

5m = that goes through the point ( )3,4

( )4 5 3y x= + − . Done. Leave it in this form.

What is the equation of a line that goes through

the points ( )0 0,x y and ( )1 1,x y . 1 0

1 0

y ym

x x

−=

− , then use m in the equation

( )0 0y y m x x= + − .

What is the equation of a line that goes through

the points ( )3,4 and ( )5,8 .

8 4 42

5 3 2m

−= = =

− , then use 2 in the equation

( )8 2 5y x= + − . Also ( )4 2 3y x= + − works.

When graphing polynomials, factor to find the zeroes. If the polynomials have factors with odd powers, the polynomial changes sign at the corresponding roots. If the polynomials have factors with even powers, the polynomials doesn’t change sign at the corresponding root.

Ex. 3 22y x x= − + Factor to find the zeroes

( )2 2y x x= − − There are at 0 and 2x x= = .

Also, the factor that provide the zeroes at 0x = is to an

even power, so sign change at 0x = . The factor that

provides the zero at 2x = is to an odd power, so there is a

sign change at 2x =

Lastly, ( )3 2lim 2x

x x→−∞

− + = ∞ , so the function starts

positive, touches axis at 0x = , then goes through the axis

at 2x = . Let’s confirm by graphing. See later in the document for review of limits.



Limits

When we ask ( )limx a

f x→

, we are asking,

what value does ( )f x take on when we

infinitesimally close to a, as compared to

the question, what is the value of ( )f a .

The value ( )1 1 2f = , even though the ( )1

1lim 1xf x

→=

Limits can be “one-sided”.

( )limx a

f x−→

asks the question, “what value

does f give as we get infinitesimally close to a from the left hand (or negative side).

( )limx a

f x+→

asks the question, “what value

does f give as we get infinitesimally close to a from the right hand (or positive side). When the left-hand limit and the right-hand do not agree, we say the overall limit does not exist.

( )11

lim 2x

f x−→

= , ( )11

lim 1x

f x+→

= , ( )11

limxf x

→ does not exist



Limits

We say the function f is continuous at a if three conditions are met.

1. ( )f a exists.

2. ( )limx a

f x→

exists.

3. ( ) ( )limx af x f a

→=

1f is not continuous at 1x = because ( )1limx af x

→does not exist.

For continuous functions, the value of the function is the value of the limit.

2

2lim 4xx

→=

For some discontinuous functions, the limit’s value can be found through algebraic manipulation.

Example 1:

21 1

1 1lim lim

1x x

x x

x→ →

− −=

− ( )1x − ( ) ( )1

1 1lim

1 21 x xx →= =

++

Example 2:

( )

0 0

0 0

5 5 5 5 5 5lim lim

5 5

5 5lim lim

5 5

x x

x x

x x x

x x x

x x

x x

→ →

→ →

+ − + − + += ⋅

+ +

+ −= =

+ + x ( )1

2 55 5x=

+ +

When the limit can be found through algebraic manipulation, the graph of these functions appear with removable discontinuities.

21

1 1lim

1 2x

x

x→

−=

−



Limits

Some limits can’t be found using the algebraic manipulation technique.

21

1lim

1x

x

x→−

−−

does not exist.

The Intermediate Value Theorem states that if a function is continuous on the

interval [ ],a b and you choose any value

between ( )f a and ( )f b (let’s call it L ),

then there exists a c in the interval ( ),a b

such that ( )f c L= .

Example: Show that ( ) 3f x x= has at least one zero. Condition

check: First we acknowledge that f is continuous on [ ]1,1− and

that ( )1 1f − = − and ( )1 1f = . 0 is a value between 1− and 1.

Consequence: Therefore f must take on the value 0 somewhere

on the interval ( )1,1− .

Some limits have unbounded behavior in a particular direction. While we say their limits do not exist, we also say their limit approach infinity. When a function has a limit that approaches infinity as x approaches a finite value, we say the function has a vertical asymptote.

21

1lim

1x

x

x+→−

−−

does not exist and 2

1

1lim

1x

x

x+→−

−= ∞

−

21

1lim

1x

x

x−→−

−−

does not exist and 2

1

1lim

1x

x

x−→−

−= −∞

−

1f has a vertical asymptote at 1x = −



Limits

We can ask “what happens to a function f when x gets arbitrarily large”, and we use

the notation ( )limx

f x→∞

. If the function gets

closer to a particular value, we say the function has a horizontal asymptote.

( )1lim 0x

f x→∞

= . 0y = is a horizontal asymptote..

Many times, we take the limit of a rational functions; rational functions are functions

in the form ( )( )P x

Q x where ( )P x and ( )Q x

are polynomials.

When the order of ( )P x is larger than

( )Q x , ( )( )

limx

P x

Q x→∞= ∞ .

When the order of ( )P x is smaller than

( )Q x , ( )( )

lim 0x

P x

Q x→∞= .

When the order of ( )P x is equal than

( )Q x , ( )( )

limx

P x

Q x→∞ is found by looking at

the ratio of the coefficients of the highest terms.

Example 1: 3

2

1lim

2 3x

x

x→∞

−= ∞

−

Example 2:

2

3

3 1lim 0

1x

x

x→∞

−=

−

Example 3:

3

3

3 1 3lim 3

1 1x

x

x→∞

−= = −

− −



Functions

Some function follow the rule that

( ) ( )f a f a= − . We call these functions

“even” functions. They have symmetry about the y-axis.

( )1 1f and ( )1 1f − both equal 3− .

( )1 2f and ( )1 2f − both equal 0.

Some function follow the rule that

( ) ( )f a f a= − − . We call these functions

“odd” functions. They have symmetry about the origin. Symmetry about the origin means that the midpoint between two corresponding points is the origin.

( )1 1 3f = − and ( ) ( )1 11 3 1f f− = = − − .

( )1 3 15f = and ( ) ( )1 13 15 3f f− = − = − .

If ( )( )1f f x x− = and ( )( )1f f x x− = , we

say the f and 1f − are inverse functions.

We use inverse functions to “undo” the other

function when solving equations. xe and

( )ln x are inverse functions. 2x and x are

inverse functions. sin x and 1sin x−

(sometimes called arcsin x ) are inverse functions.

Example 1:

( ) ( )( )

10

ln ln 10

ln 10

x

x

e

e

x

=

=

=

Example 2:

ln 10

10

ln 10

x

x

e e

x e

=

=

=

We used the fact that ln and xe are inverse functions to isolate x.



Functions

For every point on function f, the is a

corresponding point on 1f − , the inverse

function of f .

Graphically, the function f is the reflection

of the inverse function 1f − over the line

y x= .

( )1

xf x e= , ( ) ( )2 lnf x x= . Because the point ( )1,e is on the

graph of 1f , then the point ( ),1e is on the graph of 2f .

Because the point ( )1,0 is on the graph of 2f , the point

( )0,1 .

Rules for logs

( ) ( ) ( )

( ) ( )

( ) ( )

ln ln ln

ln ln ln

ln ln

lnlog

ln

ln 1

ln1 0

b

b

a b a b

aa b

b

a b a

aa

b

e

+ = ⋅

− =

= ⋅

=

=

=

Example 1: ( ) ( ) ( ) ( )ln 3 ln 3 ln ln 3e e= ⋅ =

Example 2: ( ) ( ) 1ln 1 ln 2 ln

2

xx x

x

− − − + = +

Example 3:

( ) ( ) ( ) ( )

( )

( )

( )( )

1 1

2 2

1

2

1

2

1 1ln 1 ln 2 ln 1 ln 2

2 2

1 1ln ln

22

x x x x

x x

xx

+ − + = + + +

− − = = + +



Trig (Note: we work exclusively in radians in AP Calculus)

Graph of ( )siny x=

Notes:

• The function has zeroes at ,x n nπ= is an

integer.

• The function has a maximum value of 1.

• The function has a minimum value of -1.

• The function is odd.

Graph of ( )cosy x=

Notes:

• The function has zeroes at ,2

x n nπ

π= +

is an integer.

• The function has a maximum value of 1.

• The function has a minimum value of -1.

• The function is even.

Graph of ( )tany x=

Notes:

• ( ) ( )( )

sintan

cos

xx

x=

• The function has zeroes wherever

( )sin 0x = .

• The function is undefined when

( )cos 0x = .

• The function is increasing.





Graph of ( )secy x=

Notes:

• ( )( )1

seccos

xx

=

• The function has local maximum of -1.

• The function has local minimum of 1.

• The function is undefined wherever

( )cos 0x = .

• The function is even, just like ( )cos x .

Graph of ( )cscy x=

Notes:

• ( )( )1

cscsin

xx

=

• The function has local maximum of -1.

• The function has local minimum of 1.

• The function is undefined wherever

( )cos 0x = .

• The function is odd, just like ( )sin x .

Graph of ( )coty x=

Notes:

• ( ) ( )( )

coscot

sin

xx

x=

• The function is undefined when

( )sin 0x = .

• The function is always decreasing.

• The function is odd, just like ( )sin x .




Graph of ( )1tany x−=

Notes:

• In the normal tangent function, you give an angle, and you get back a ratio. For the inverse tangent function, you give a ratio and get back an angle.

• ( )1tan 14

π− = because tan 14

π =

.

• 2

yπ

= is a horizontal asymptote because

( )1lim tan2x

xπ−

→∞= .

• 2

yπ

= − is a horizontal asymptote because

( )1lim tan2x

xπ−

→−∞= − .


You should be able to state the value of any of the big

six trig functions for any multiple of 6

π from 2π− to

2π and any multiple of 4

π from 2π− to 2π . These are

the values you memorized on the unit circle.

( )cos 0 1= ; ( )sin 0 0= ; ( )tan 0 0= ;

1sin

6 2

π =

; 2

sin4 2

π =

;3

sin3 2

π =

;

3cos

6 2

π =

; 2

cos4 2

π =

;1

cos3 2

π =

;

1tan

6 3

π =

; tan 14

π =

; tan 33

π =

;

cos 02

π =

; sin 12

π =

; tan2

π

is und.;

1sin

6 2

π − = −

; 2

sin4 2

π − = −

;

3sin

3 2

π − = −

; and more…

Trig Pythagorean Identities 2 2sin cos 1x x+ = 2 2tan 1 secx x+ =

2 21 cot cscx x+ =

AP Calculus BC Prerequisite Knowledge -...

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