Download Advanced Placement AB Calculus NAME

Advanced Placement AB Calculus
NAME ____________________
Line Review:
It is very important that you can write equations of lines given:
1)
two points on the line
2)
a point on the line and the line’s slope (m).
The most frequently used form of a line in this class will be point-slope form of a linear equation.
You should know the following:
 The slope of a line going through  x1 , y1  and
Slope (m) =
 x2 , y2 
rise
y
y  y1

 2
.
run
x
x2  x1
is determined by
x1
 x2 
 Remember, vertical lines have an undefined slope while horizontal lines have a slope of zero.
 The point-slope form of a line going through x1, y1  and having slope of m is y  y1  m  x  x1  .
 Parallel lines have the same slope.
 Perpendicular lines have slopes that are negative reciprocals of each other.
 The slope-intercept form of a line with slope m and y-intercept b is y  mx  b .
EX 1] Find the slope of the line that goes through 4,3 and (2,5).
m 
y
5  (3)
8


 4
x
2  4
2
m= 4
EX 2] Using the point-slope equation, write an equation for the line through the point (2,3) with
3
slope
.
2
y  y1  m  x  x1 
Note: No need to write in slope3
y  3 
 x  2
intercept form,
2
unless you would prefer.
EX 3] Write an equation for the line through  2,1 and (3,4).
1st
Find the slope of theline.
m 
y
4  (1)
5


1
x
3  (2)
5
2nd Use one point on theline and the slope.
y  4  1( x  3)
EX 4] Write an equation for the line through  1,2 that is
a) parallel and b) perpendicular
to the line L: y  3x  4 .
mL  3
1
, (1, 2)
3
1
y  2 
 x  (1) 
3
1
y  2 
 x  1
3
a) m  3, (1, 2)
b) m 
y  2  3  x  (1) 
y  2  3( x  1)
EX 5] Find the slope and y-intercept of the line 8 x  5 y  20 .
Graph the line.
5 y   8 x  20
How?
Rewrite 8 x  5 y  20
in y  mx  b form.
So, m 
8
x  4
5
y 
8
5
b  4
y
x
8 x  5 y  20
Area and Volume:
b1
Area Formulas:
w
h
b
A 
s
l
1
bh
2
h
r
s
b2
A   r2
A  s2
A  lw
A 
1
h  b1  b2 
2
Volume Formulas:
r
h
h
r
h
w
r
l
V  lwh
V   r 2h
V 
1 2
r h
3
V 
4 3
r
3
Function Notation:
Throughout this class, we will be using function notation. EX] f ( x)  3x  4
The “x” represents the inputs of the function more formally called
the domain and the independent variable.
The “ f (x) ” represents the outputs of the function more formally called
the range and the dependent variable.
If I want to input a ‘3’ into the function, I write f (3)  3(3)  4 . So, with a domain (input) of 3,
we get a range (output) of 5.
It will be important to be able to identify the domain and range for a given function.
 When asking for the domain of a function, I want to know for which x-values the function is defined.
 It is often easier to find where the function is undefined.
Function
y = x2
Domain (x)
(-,)
Range (y)
[0, )
(-, 0)  (0, )
(-, 0)  (0, )
x
[0, )
[0, )
4  x
(-, 4]
[0, )
y = 1  x2
[-1,1]
[0,1]
y=
y=
y=
1
x
NOTE: ‘ [ ’ means include the endpoint in the domain.
‘ ( ’ means the endpoint is not included in the domain.
 f ( x) 
 .
You must be able to deal with rational functions 
 g ( x) 

A rational function will be undefined when the denominator (g(x)) is equal to zero.

The zeros of a rational function
o occur where the graph crosses the x-axis (x-intercepts) and
o are where the numerator  f (x) is equal to zero and the denominator is not equal to zero.
EX 1] Find the domain and real zeros of the given functions.
x  2
x3
a) f ( x)  2
b) g ( x)  2
x  3x  2
x 1
Domain
x2  1  0
( x  1)( x  1)  0
x  1
 ,  1   1,1  1, 
Zero(s)
x3  0
x  0
Domain
x 2  3x  2  0
( x  2)( x  1)  0
x  1, 2
 ,1  1, 2  2, 
x2  9
d) j ( x) 
2x  1
3x  2
c) h( x)  2
x  9
Domain
x2  9  0

  , 
Zero(s)
x  2  0
x  2
Zero(s)
3x – 2 = 0
3x = 2
x = 2/3
Domain
2x  1  0
2x  1
x  12
  , 12   12 , 
Zero(s)
x2  9  0
(x + 3)(x – 3) = 0
x  3
Note how I set up the appropriate equations (numerator = 0 for zeros; denominator  0 for undefined values).
Then, I factored to solve equations.

Solving equations by factoring is a really good skill to know and you should practice it on this review.

Verify your solutions with your calculator.

A composite function f g ( x) or f  g is read " f of g" .
To simplify a composite function, replace the x in f (x) with g (x ) and simplify.
EX 2] Find the formula for f g (x) if g ( x)  x 2 and f ( x)  x  7 . Then, find f g (2).
 
f 2  
f g ( x)  f x 2  x 2  7
f g 2 
EX 3] If f ( x)  x 2 , find
2
f (4)  4  7   3
f ( x  h)  f ( x )
.
h


x  h   x 2  x 2  2 xh  h2  x 2  2 xh  h 2  h(2 x  h)  2 x  h
f ( x  h)  f ( x )

h
h
h
h
h
2
Exponents and Radicals:
It is very important that you remember the Rules for Exponents.
Rules for Exponents:
If a  0 and b  0 , the following hold true for all real numbers x and y.
ax  ay = ax + y
1)
ax
 ax  y
y
a
2)
 
3) a x
y
 
 ay
x
 a xy
4) (ab)x = ax  bx
a
5)  
b
x

ax
bx
Also, remember how to work with:
Rational Exponents: a
x
y
 a

y
Negative Exponents: a  2 
1
a2
x

and
a
y
x
1
 a2
2
a
Zero Exponents: a0 = 1, a  0
When you are asked to simplify expressions there should be
 no parenthesis,
 no negative exponents, or
 no powers with the same base in the answer.
EX]
9
a 
9
b 
9
ab
Solving Equations with Exponents:
EX] Solve
1
 32 x  2 for x.
3
31  3( 2 x
 2)
Get LIKE bases!
 1  2x  2
Set exponents equal to each other.
 3  2x
Solve for x.
x 
3
2
Logarithmic Functions:
Logarithmic Function – For x  0 , a  0 , and a  1 , y  log a x iff x  a y .
f ( x)  log a x is called the logarithmic function with base a.
NOTE: A logarithm is an EXPONENT! Logarithms and exponents are inverse functions!!!
y
y  x
y  ex
y  ln ( x)
x
b) Write log 32 4 
EX] a) Write log 4 64  3 in exponential form.
2
43  64
EX] a) Write 7 2 
log 7
2
in exponential form.
5
32 5  4
1
in logarithmic form.
49
b) Write e x  17 in exponential form.
1
 2
49
log e 17  x
NOTE: x  ln 17
Common Logarithmic Function – the logarithmic function with base “10”.
Natural Logarithmic Function – the function defined by
Properties of Logarithms
1) log a 1  0
2) log a a  1
3) log a a x  x
4) If log a x  log a y, then x  y.
f ( x)  log e x  ln x , x  0 .
Properties of Natural Logarithms
1) ln1  0
2) ln e  1
3) ln e x  x
4) If ln x  ln y, then x  y.
EX] Evaluate each expression without using a calculator.
a) log 2 32
b) ln e3
Let log 2 32  x .
Let ln e3  x .
(Go exponential!) 2 x  32
(Go exponential!) e x  e3
x  3
(Get like bases.)
2x  25
NOTE: You could have used Property #3 above.
 x  5
Change-of-Base Formula: Let a, b, and x be positive real numbers such that a  1 and b  1 then,
log b x
log a x is given by log a x 
.
log b a
Properties of Logarithms
Let a be a positive number such that a  1, and let n be a real number.
If u and v are positive real numbers, the following properties are true.
1. loga  uv   loga u  loga v
1. ln (uv)  ln u  ln v
u
2. log a    log a u  log a v
v
u
2. ln    ln u  ln v
v
3. log a u n  n log a u
3. ln u n  n ln u
EX] Expand each logarithm.
a) ln
3x  5
7
b) log10 5x 3 y
1
= ln 3x  5 2  ln 7
= log 5  log x3  log y
1
ln  3 x  5   ln 7
2
= log 5  3log x  log y
=
EX] Write each expression as the logarithm of a single quantity.
a) ln x  3ln ( x  1)
b)
= ln x  ln  x  1
3
(Condense each logarithm.)
1
1
ln x 2 
ln y 3
3
2
2
3
= ln x 3  ln y 2
2
= ln
x
x
 1
= ln
3
x3
y
3
2
 ln
3
x2
y3
 ln
3
x2
y y
EX] Find the EXACT value of the logarithm log 5 75  log 5 3 without using a calculator.
75
 log 5 25
3
 5x  25
log 5 75  log 5 3  log 5
5x  52
 log5 75  log5 3  2
Let log 5 25  x .
x  2
Trigonometric Functions:

An angle has three parts: an initial ray, a terminal ray, and a vertex (the point of intersection
of the two rays).

Standard position for an angle occurs if the initial ray of the angle coincides with the positive
x-axis and its vertex is at the origin.

Positive angles are measured counterclockwise, and negative angles are measured clockwise.

Coterminal angles have the same terminal ray.

180 =  radians
EX] -45 is coterminal with 315.
Please, please, please, I beg you to know your unit circle with radian measures. (We will only use radian
measures in calculus.) I’m sure you got familiar with it in your Precalculus class. Remember???
 Know these:
hypotenuse
opposite
sin  =
csc  =
opposite
hypotenuse

cos  =
adjacent
hypotenuse
sec  =
hypotenuse
adjacent
tan  =
opposite
adjacent
cot  =
adjacent
opposite
You should know the graphs of sin  , cos  , and tan  , and be able to picture them quickly
from [-2, 2]. (In case you were wondering, the “[-2, 2]” represents the domain values.
The “[ ]” means including endpoints.)
y
y
y
1

y  sin ( x)
Domain: all reals
Range:  1,1
Period: 2
Function
y  a sin (bx  c)  d or
y  a cos (bx  c)  d
x

1
x

x
3
y  tan ( x)
y  cos ( x)
Domain: all reals
Range:  1,1
Period: 2
Period Amplitude
2
a
b
y  a tan (bx  c)  d or
y  a cot (bx  c)  d

b
Not
applicable
y  a csc (bx  c)  d or
y  a sec (bx  c)  d
2
b
Not
applicable
Domain: all x  2  n
Range:  , 
Period: 
Vertical Shift
Phase Shift
bx  c  0
Asymptotes

bx  c  
2
Asymptotes
bx  c  0, 
d
d
d

Remember to draw triangles to represent the problem.
EX] Determine all six trigonometric functions for the angle whose terminal side occurs at (-3,4).
y
(-3,4)
5 (Use Pythagorean Theorem to find the length of this side.)
4

-3
x

sin  
y
4

r
5
csc 
r
5

y
4
cos 
x
3
 
r
5
sec  
r
5
 
x
3
tan  
y
4
 
x
3
cot  
x
3
 
y
4
As mentioned before, the unit circle WILL be helpful. Know the trig values of the basic angles.
You should be quick with the sin x, cos x, and tan x.
1
1
1
2
30
3
2
60
3
2
1
2

6




3
3 1
, 

2 2

4
1
x
45
2
2
2
2
 2 2


 2 , 2 


1
 ,
2
3
2


Graphs of Common Functions:
Here are the graphs of very common functions.
Know the characteristics of these graphs:
Increasing vs Decreasing;
Domain;
Range; etc…
Be able to sketch these basic functions!
y
y
f ( x)  c
y
f ( x)  x
f ( x)  x
2
2
x
x
x
2
2
y
y
f ( x) 
2
2
x
f ( x)  x
y
2
2
2
2
x
x
x
2
2
y
f ( x)  b
x
(b  1)
2
y
y
f ( x)  b
2
2
f ( x)  x 3
x
x
(b  1)
2
f ( x)  log b x
x
2
x
2
2
y
y
y  cos ( x)
y  sin ( x)
1
y
x
1
x
3
2
1
x
1
y  tan ( x)