Download AP Calculus AB - Review for AP Calculus AB Exam (2009).

AP Calculus AB
AP Exam Review
Limits
Properties of Limits
1. Scalar Multiple:
lim  k f  x    k lim  f  x  
xc
xc
2. Sum/Difference:
lim  f  x   g  x   
xc
lim  f  x    lim  g  x  
xc
xc
lim  f  x  g  x    lim  f  x  
xc
3. Product:
xc
 f  x 
lim 

xc
 g  x 
4. Quotient:

lim  f  x  
xc
lim  g  x  
xc 
lim  g  x  
xc
if lim  g  x    0
xc
Definition of Limit: as x gets closer and closer to a specific value,
y gets closer and closer to what value?
- recall, the answer to a limit is the y-value (but the y-value does
not have to exist to
have a limit value)
One-Sided Limits
lim
f  x
means x approaches a from the right
lim
f  x
means x approaches a from the left
x  a
x  a
Two-Sided Limit : lim f  x  means x approaches a from both sides
xa
Only exists if
lim
x  a
f  x 
lim
x  a
f  x   L , where L is a Real number
Limits at Infinity
If lim f  x   L or
x
Recall: To evaluate lim
x  
Also:
If
lim
x  a
lim
x  
f  x   L , then y  L
f  x
g  x
, use
is a horizontal asymptote
highest degree term
( x    if an x remains)
highest degree term
small
BIG
 0 and
 
BIG
small
f  x     or lim  f  x     , then x  a is a vertical asymptote
xa
AP Calculus AB
AP Exam Review
Continuity
i) lim  f  x  
xa
lim
x  a
f  x
ii) f  a  exists
iii) lim f  x  
xa
(Roads meet)
(Bridge is built)
f a
(Roads meet at the bridge)
Types of Discontinuity
Jump
Removable
Infinite

3x 2  x
f
x




Example Problem: If 
2x
 f  0  k

, x  0


 and


if f is continuous at x  0 , then k 
(A) 
3
2
(B)  1
(C) 0
(D) 1
(E)
3
2
Two Major Theorems on Continuity
( I ) Intermediate Value Theorem :
If f is continuous on [ a , b ] and k is any number between
f  a  and f  b  , then there is at least one number c
between a and b such that f  c   k .
( II ) Differentiability Implies Continuity
- If the derivative exists at a point, the function is continuous at that point
AP Calculus AB
AP Exam Review
Differential Calculus
Definition of Derivative
(i) f '  a   lim
f a  h  f a
h0
(ii) f '  a  
lim
h
f  x  f a
x  a
xa
(iii) f '  a   lim
f a  h  f a  h
h0
Symmetric Difference Quotient
2h
(good to use given a table of values)
Terminology for Derivative: Slope of the curve
Instantaneous Rate of Change
Slope of the tangent
Rules
Power Rule:
d n
x   n x n 1
dx
d
 f g  f g '  g f '
dx
Product Rule:
(1st times derivative of 2nd
plus
2nd times derivative of 1st)
Quotient Rule:
d f 
g f '  f g'
  
dx  g 
g2
(bottom times derivative of top minus top times
derivative of bottom all over bottom squared)
Chain Rule:
d
f  g  x    f '  g  x   g '  x  (derivative of outside at inside times derivative of
dx
inside)
Trig Derivatives
d
du
sin u  cos u
dx
dx
d
du
tan u  sec 2 u
dx
dx
d
du
sec u  sec u tan u
dx
dx
d
du
cos u   sin u
dx
dx
d
du
cot u   csc 2 u
dx
dx
d
du
csc u   csc u cot u
dx
dx
AP Calculus AB
AP Exam Review
Inverse Trig Derivatives
d
sin  1 u  

dx
d
cos 1 u  

dx
1
du
1  u 2 dx
1
du
1  u 2 dx
d
1 du
tan  1 u  

dx
1  u 2 dx
d
 1 du
cot  1 u  

dx
1  u 2 dx
d
sec1 u  

dx
u
d
 1
du
cs c1 u  

2
dx
u u  1 dx
1
du
u  1 dx
2
Exponential / Log Derivatives
d u
du
e   eu

dx
dx
d
du
a u   a u ln a

dx
dx
d
1 du
 ln u  
dx
u dx
Implicit Differentiation - use when you cannot solve for y easily or solving for y makes
taking the derivative more difficult
- Remember: anytime you take a derivative of a function with y,
dy
you must put
after that term
dx
dy
- Then solve the resulting equation for
dx
** Don’t forget about PRODUCT RULE on these problems!
Example Problem: Find
Answer:
dy
dx
given
y 3  xy  2 y  x 2   2 .
dy
2x  y

2
dx
3y  x  2
Recall also: We use Implicit Differentiation in Related Rates problems
(differentiate variables w/ respect to t)
AP Calculus AB
AP Exam Review
Derivatives of Inverse Functions
If y  f  x  and x  f  y  are inverse functions, then the derivative of the inverse
function is the reciprocal of the derivative of the original function:
Alternative Notation:
d
f
dx
1
 evaluated at x  f  a 
=
dx
1

dy
dy
dx
1
f 'a
** Recall: The specific value where you want to evaluate the derivative of the inverse
function is the y-value of the original function, so you have to find the x-value on the
original function which will give you that specific y-value and use this value to evaluate
the derivative of the original function.
Example Problem: (May use calculator)
Let f  x   x 3  7 x 2  25x  39 and let g be the inverse of f (x). What is the
value of g '  0 ?
(A) 
1
25
Answer: C
(B)
1
25
(C)
1
10
(D) 10
(E) 25
AP Calculus AB
AP Exam Review
Logarithmic Differentiation
Use when function is very complicated OR if you have a function
function
Steps: 1. Take ln of both sides
2. Differentiate
3. Solve for
dy
dx
4. Substitute for y
 x  1
y  2

 x  1
2
Example 1:
1
3
1
 x2  1  3
ln y  ln  2

 x  1
1  x2  1 
1
ln  x 2  1  ln  x 2  1
ln y  ln  2
 

3  x  1
3

1 dy
1
1
1
 2

 2x   2
 2 x 
y dx
3   x  1
 x  1 

dy
1  2x
2x 
 2
 y


2
dx
3   x  1
x

1




dy
1  2x



dx
3   x 2  1

Example 2: y   tan x 
1
2x   x2  1  3

 x2  1   x 2  1 
x
ln y  ln  tan x 
x
ln y  x ln  tan x 
1 dy
 1

 x
sec 2 x    ln  tan x 1

y dx
 tan x

dy   1


x
sec 2 x   ln  tan x 1  y

dx   tan x


dy   1


x
sec 2 x    ln  tan x 1 

dx   tan x


 tan x 
x
(can’t use power rule)
AP Calculus AB
AP Exam Review
Equation of Tangent Line
- Must have point =  a , f  a   , and slope = f '  a 
-
y  f  a   f '  a  x  a 
Equation of Normal Line
- Perpendicular to tangent at point of tangency,  a , f  a  
-
y  f a 
1
 x  a  (same equation with slope negative reciprocal)
f 'a
Linearization
- using the y-value of the tangent line to approximate the function value
Equation of tangent line
y  f  a   f '  a  x  a 
becomes
L  x   f '  a  x  a   f  a 
** Remember x = “ugly” number (one with the decimal) and a = “pretty number” (close to x)
Mean Value Theorem for Derivatives
If f is continuous on [a , b] and differentiable on (a , b), then there exists
a number c between a and b such that
f 'c 
-
Slope of tangent line
Instantaneous rate of change
f b   f  a 
b  a
=
=
slope of secant line
average rate of change
Extreme Value Theorem
If f is continuous on [a , b], then f has both an absolute maximum and an
absolute minimum.
- Max or min must occur at critical points f '  x   0 or f '  x   DNE
or at the endpoints.
AP Calculus AB
AP Exam Review
Curve Sketching Rules
f '  c   0 (horizontal tangent)
Critical points occur when:
f '  c   DNE (corner, cusp, vertical tangent)
(denominator = 0)
Situation
Indicates
f ' x  0
f  x  is increasing
f ' x  0
f  x  is decreasing
f ''  x   0
f  x  is concave up
or
f '  x  is increasing
f ''  x   0
f  x  is concave down
or
f '  x  is decreasing
1st Derivative Test – used to justify max and mins
   0 and f '  c   0
f '  c   0 and f '  c   0
Max at x  c if f ' c

Min at x  c if



(positive to negative)
(negative to positive)
** Must have a sign change to be a max or min, but it is still a critical point
2nd Derivative Test – used to justify max and mins
Max at x  c if
f '  c   0 and f ''  c   0 (horizontal tangent, concave down)
Max at x  c if
f '  c   0 and f ''  c   0 (horizontal tangent, concave up)
Points of Inflection
- where the graph changes concavity
Example Problems: Use the graph of
[sign change in
f ''  x  ]
y  f '  x  , sketched below on [0 , 7]
1. From the graph it follows that ______________.
y  f ' x
A) f is discontinuous at x  4
B) f is decreasing for 4  x  7
C) f is constant for 0  x  4
D) f has a local maximum at x  0
E) f has a local minimum at x  7
2. Which best describes f at x  5 ?
A) f has a root
B) f is a maximum
C) f is a minimum
D) f has a point of inflection
3.
A)
B)
C)
D)
The function is concave downward for which interval?
(0 , 4)
(4 , 5)
(5 , 7)
(4 , 7)
AP Calculus AB
AP Exam Review
Optimization - Maximize or minimize a quantity
- Write an equation that needs to be maximized or minimized (usually has 2 independent
variables)
- Write 2nd equation that relates the 2 independent variables, then substitute into 1st
equation
- Differentiate, find critical points, use sign chart to justify max/min, and check
endpoints
Related Rates – Implicit differentiation where you differentiate with respect to t
-
Write equation that relates variables
Differentiate with respect to t
Substitute known values (usually use 1st equation to find a missing value)
Then solve for correct rate.
Position, Velocity, Acceleration
x t 
position of particle at time t
v t   x ' t 
instantaneous velocity at time t
a  t   v ' t   x '' t 
instantaneous acceleration at time t
Speed = v  t 
Speeding Up -
velocity and acceleration have same sign
Slowing Down - velocity and acceleration have opposite signs
AP Calculus AB
AP Exam Review
Integral Calculus

Definite Integrals:
b
a
f  x  dx
- represents the NET AREA under the curve
(above x-axis is a positive value, below is
a negative value)
** Answer is always a numerical value
Indefinite Integrals:
 f  x  dx
** DON”T FORGET:
+C
- tells you to find ALL possible antiderivatives
Techniques for Integration
Power Rule in Reverse:

u n du 
un 1
 C (Add one to the exponent,
n 1
divide by new exponent)
Substitution Method: - “Chain Rule” for integrals
- composite functions (function inside a function)
 f  g  x   g '  x  dx
 F  g  x   C
Steps:
Let u  g  x 
So we have
du  g '  x  dx
Take derivative of u:
Substitute du
Take antiderivative
Replace u with g  x 
 f  u  g '  x  dx
so
 f  u  du
 f  u  du  F  u   C
F  g  x   C
Separable Differential Equations
dy
dx
Get y with dy and dx with x:
Integrate and solve for y:

1
dy 
g  y
1
 g  y  dy 
f  x g  y
f  x  dx
 f  x  dx
y  F  x  C
AP Calculus AB
AP Exam Review
To Approximate Definite Integrals
LRAM, RRAM, MRAM - rectangular approximation methods
- approximate area under curve or value of a definite integral
(usually use when given values in a table)
Trapezoidal Rule : Recall: A trap 
1
 b1  b2  h ** must use this formula if values of  x
2
are not uniform
** If  x has same values throughout, you can use formula
1
f  x 0   2 f  x1   2 f  x 3   ...  2 f  x n  1   f  x n    x
2
A 
FUNDAMENTAL THEOREM OF CALCULUS
Part I: Derivatives and Integrals are inverses
d
dx

x
a
f  t  dt  f  x 
** Remember, integral must start at a constant, and
only have a variable as the upper limit
** Also, the independent variable for f becomes
upper limit variable
Part II (Evaluation):

b
a
f  x  dx  F  b   F  a 
(antiderivative at upper limit minus
antiderivative at lower limit)
CHAIN RULE Version
If y 

g x
a
f  t  dt , find
dy
.
dx
Steps: Evaluate integral using Part I
Let F  t  be any antiderivative of f  t  , so

Now find
dy
using chain rule,
dx
but since F '  f ,
g x
a
f  t  dt  F  g  x    F  a  , so
y  F  g  x    F  a 
dy
 F '  g  x   g '  x   0 ,
dx
dy
 f  g  x   g '  x 
dx
AP Calculus AB
AP Exam Review
1
b  a
avg f 
Average Value of a Function

b
a
f  x  dx
MEAN VALUE THEOREM FOR INTEGRALS
-
If f is continuous on [a , b], then at some point c in [a , b],
1
b  a
f c 

b
a
f  x  dx
** States that there is an actual y-value on f equals the average y-value of the function
Position, Velocity, Acceleration

Displacement: Net distance =
Total Distance Traveled:

b
a
b
a
v  t  dt
v  t  dt (on a calculator)
x b  x  a  
Final Position = Initial Position + Displacement or
Area Between Curves
 y
b
 y bottom  dx
a
top
b
 r 2 dx
 x

d
or
right
c
b
a
v  t  dt
 x left  dy
Volumes of Revolution
- Disk Method: (around x-axis)
- Washer Method:

b
a

a
  R 2  r 2  dx
or
or
(around y-axis)

d
c
b

a
d
c
d
c
  R 2  r 2  dy
Volumes of Known Cross-Sections


A  x  dx (cross-sections perpendicular to x-axis)
A  y  dy (cross-sections perpendicular to y-axis)
 r 2 dy