Download Review for Final Exam

Final Exam
Review Slides
Example: Absolute Value
Simplify the following expressions using your
knowledge of absolute values:
a. 21  7  14
b.   
c.  3  3
The Order of Mathematical Operations
Order of Operations
1. If the expression is a fraction, simplify the numerator and
denominator individually, according to the guidelines in the
following steps.
2. Parentheses, braces and brackets are all used as grouping
symbols. Simplify expressions within each set of grouping
symbols, if any are present, working from the innermost
outward.
3. Simplify all powers (exponents) and roots.
4. Perform all multiplications and divisions in the expression in
the order they occur, working from left to right.
5. Perform all additions and subtractions in the expression in the
order they occur, working from left to right.
Properties of Exponents and Their Use
Properties of Exponents
Throughout this table, a and b may be taken to represent constants, variables
or more complicated algebraic expressions. The letters n and m represent
integers.
Property
1. Product Rule
a n  a m  a n m
2. Quotient Rule
3. Zero Exponent Rule
an
n m

a
am
a 0  1.
4. Negative Exponent Rule a
n
1
 n
a
Properties of Exponents and Their Use
Power Rules
n m
5. a
 a nm
 
6.  ab 
n
n
 a nb n
n
a
a


7.
   n
b
b
Example: Scientific Notation
The distance from Earth to the sun is approximately
93,000,000 miles. Scientific notation takes advantage of
the observation that multiplication of a number by 10
moves the decimal point one place to the right, and we
can repeat this process as many times as necessary.
7
Thus, in scientific notation, 93,000,000

9.3

10
.
.
.
Example 2: Scientific Notation
a. The mass of an electron, in kilograms, is approximately
0.000000000000000000000000000000911. Scientific
notation takes advantage of the observation that
multiplication of a number by 101 moves the decimal
point one place to the left, and we can repeat this
process as many times as necessary. Thus, in scientific
notation,
0.000000000000000000000000000000911=9.11 1031.
Simplifying Radical Expressions
In the following properties, a and b may be taken to represent
constant variable, or more complicated algebraic expressions.
The letters n and m represent natural numbers.
Property
1. Product Rule
2. Quotient Rule
n
ab  n a  n b
n
a

b
n
a
n
b
Simplifying Radical Expressions
Rationalizing Denominators
Denominator consists of two terms, one or both of which are
square roots.
Once again, remember that we cannot multiply the denominator
by A – B unless we multiply the numerator by this same factor.
Thus, multiply the fraction by
A– B
.
A– B
The factor A – B is called the conjugate radical expression of A + B.
Common Factoring Methods
• Method 1: Greatest common factor.
• Method 2: Factoring by grouping.
• Method 3: Factoring special binomials.
• Method 4: Factoring trinomials.
– Case 1: Leading coefficient is 1.
– Case 2: Leading coefficient is not 1.
Example: The Algebra of Complex Numbers
Simplify the following complex number expressions.
 2  i  1  2i 
 4  3i
The product of two complex numbers leads
to four products via the distributive
property.
Roots and Complex Numbers
We earlier said that if a and b are real numbers, then
a
a
a b  ab and

. If a and b are not real
b
b
numbers, then these properties do not necessarily hold. For
instance:
9 4
 9  4 
  3i  2i 

 36
6
 6i 2
 6
In order to apply either of these two properties, first simplify
any square roots of negative numbers by rewriting them as
pure imaginary numbers.
Types of Equations
There are three types of equations:
1. A conditional equation has a countable
number of solutions. For example, x + 7 = 12
has one solution, 5. The solution set is {5}.
2. An identity is true for all real numbers and has
an infinite number of solutions. For example,
x( x  1)  x 2  x is true for all real number
values of x . The solution set is R.
3. A contradiction is never true and has
no solution. For example, x  6  x is false
for any value of x. The solution set is Ø.
Solving Linear Equations
To solve a linear equation (in x):
1. Simplify each side of the equation separately by removing any
grouping symbols and combining like terms.
2. Add or subtract the same expression(s) on both sides of the
equation in order to get the variable term(s) on one side and the
constant term(s) on the other side of the equation and simplify.
3. Multiply or divide by the same nonzero quantity on both sides of
the equation in order to get the numerical coefficient of the
variable term to be one.
4. Check your answer by substitution in the original equation.
Example: Solving Absolute Value Equations
3x  2  5
Solve:
Step 1: Rewrite the
absolute value
equation without
absolute values.
Step 2: Solve the
two equations
3x  2  5
or
3x – 2 = -5
3x  7
or
3x = -3
7
3
or
x  1
x
Example: Absolute Value Equations
Solve:
4x  3  2  0
|4x + 3| = -2
False, absolute value is never negative. No solution;
the solution set is Ø.
Solve:
|6x – 2| = 0
6x – 2 = 0
6x = 2
x=⅓
If |ax + b| = 0, then ax + b = 0.
Example: Linear Inequalities
Solve the following linear inequality.
10  4( x  2)  (2  x)
Step 1: Distribute.
Step 2: Combine like terms.
Step 3: Divide by 5 . Note
the reversal of the
inequality sign.
10  4 x  8  2  x
4 x  18  2  x
5 x  20
x4
Solution is  4, 
Example: Solving Compound Inequalities
Solve the compound inequality.
12  5  2  x   15
12  10  5 x  15
Note: each inequality
is reversed since we
are dividing by a
negative number!
2  5 x  25
2
 x  5
5
2
5  x 
5
 2
Solution is  5, 
 5
Example: Solving Absolute Value Inequalities
Solve the following absolute value inequality.
5y  3  2  9
Step 1: Subtract 2.
Step 2: Rewrite the
inequality without
absolute values.
Step 3: Solve as compound
inequality.
5y  3  7
7  5 y  3  7
4  5 y  10
4
 y2
5
 4 
Solution is  ,2
5 
Example: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring.
3x 2
x  
5 5
2
Step 1: Multiply both
sides by 5 .
Step 2: Subtract 2 from
both sides so 0 is
on one side.
Step 3: Factor and solve the
two linear equations.
5 x 2  3x  2
5 x 2  3x  2  0
(5 x  2)( x  1)  0
5x  2  0
2
x
5
or
x 1  0
or
x 1
Example: The Quadratic Formula
Solve using the quadratic formula.
4 x 2  7 x  15
c
a2 b
4x  7 x  15  0
b  b 2  4ac
x
2a
x
  7  
7  289
x
8
7  17
x
8
5
x  3,
4
 7   4  4  15
2  4
2
Operations with Rational Expressions
• To add or subtract two rational expressions, a common
denominator must first be found.
• To multiply two rational expressions, the two numerators
are multiplied and the two denominators are multiplied.
• To divide one rational expression by another, the first is
multiplied by the reciprocal of the second.
• No matter which operation is being considered, it is
generally best to factor all the numerators and
denominators before combining rational expressions.
Example: Add/Subtract Rational Expressions
Subtract the rational expression.
5x  2
2x

x2  5x  6 x2  4

5x  2
2x

 x  2  x  3  x  2  x  2 
 x  2
 x  3
5x  2
2x




 x  2   x  2  x  3  x  3  x  2  x  2 
5x2  8x  4
2 x2  6 x


 x  2  x  2  x  3  x  2  x  2  x  3
3x 2  14 x  4

,
 x  2  x  2  x  3
x  2, 2, 3
Step 1: Factor both
denominators.
Step 2: Multiply to
obtain the least
common
denominator
(LCD) and
simplify.
Example: Divide Rational Expressions
Divide the rational expression.
x 2  x  20 x 2  10 x  25

2x
6 x3

 x  4  x  5
2x
3



6 x3
 x  5
2
6 x  x  4  x  5 
3
2 x  x  5
3x 2  x  4 
x 5
2
, x  0,5
2
Example: Complex Rational Expressions
Simplify the complex rational expression.
Step 1: Multiply the numerator
and the denominator by
the LCD  x  z  x .
Step 2: Cancel the common
factor of z.
1
1

xz x
z
1
1

 x  z  x 
x

z
x


z
 x  z  x 
1
x  x  z

z  x  z  x 
z

z  x  z  x 
1

, x  0, z  0, x  z  0
xx  z
Example: Work-Rate Problem
One hose can fill a swimming pool in 10 hours. The owner buys a
second hose that can fill the pool in half the time of the first one.
If both hoses are used together, how long does it take to fill the
pool?
1
The work rate of the first hose is
10
1
The work rate of the second hose is
5
1 1 1
 
Step 1: Set up the problem.
10 5 x
Step 2: Multiply both sides by the
x  2 x  10
LCD 10x , and solve.
3 x  10
1
x  3 hours
3
Example: Solving Radical Equations
Solve the radical equation.

1  x 1  x
1 x  x 1
1 x

2
  x  1
2
1  x  x2  2x  1
0  x 2  3x
0  x  x  3
x  0, 3
x0
Note that 1  ( 3)  ( 3)  1, so -3 is
an extraneous solution.
Recognizing Linear Equations in Two Variables
Linear Equations in Two Variables
A linear equation in two variables, say the variables x
and y, is an equation that can be written in the form
ax  by  c
where a , b, and c are constants and a and b are not
both zero. This form of such an equation is called the
standard form.
Example: Linear Equations
Determine if the equation is a linear equation.
2  x  3 y   5x  x  2 y  1
2 x  6 y  5x  x  2 y  1
3x  6 y  x  2 y  1
3x  6 y  x  2 y  1
4 x  4 y  1
The equation is linear.
Example: Intercepts
Find the x- and y -intercepts of the equation and graph.
3 x  4 y  12
3 0   4 y  12
y  3
y-intercept:  0, 3
3x  4  0   12
x4
x-intercept:  4,0 
Example: Finding Slope Using Two Points
Determine the slopes of the line passing through the following points.
y2  y1
m
x2  x1
8,1 and  2,3
3 1
m
2  8
2
m
10
1
m
5
Slope-Intercept Form of a Line
If the equation of a non-vertical line in x and y is
solved for y, the result is an equation of the form
y  mx  b.
The constant m is the slope of the line, and the line
crosses the y-axis at b ; that is, the y -intercept of
the line is  0,b . If the variable x does not appear in
the equation, the slope is 0 and the equation is
simply of the form y  b .
Point-Slope Form of a Line
Given an ordered pair  x1 , y1  and a real number m,
an equation for the line passing through the
point  x1 , y1  with slope m is
y  y1  m  x  x1  .
Note that m, x1 , and y1 are all constants, and that x
and y are variables. Note also that since the line,
by definition, has slope m, vertical lines cannot be
described in this form.
The Slopes of Parallel and Perpendicular lines
Important
Parallel lines have the same slope. For example:
1
1
m1  and m2  .
2
2
Perpendicular lines have slopes that are negative
reciprocals of each other. For example:
2
3
m1  and m2   .
3
2
Example: Slopes of Parallel Lines
Find the equation, in slope-intercept form, for the line
which is parallel to the line 8 x  2 y  10 and which
passes through the point  1,5 .
8 x  2 y  10
Step 1: Write equation in
2 y  10  8 x
slope-intercept form.
y  4 x  5
Use slope m  4 to write a new equation that passes
through the point  1,5 .
Step 2: Use point-slope form.
y  5  4  x  1
Step 3: Solve for y to obtain
y  5  4 x  4
slope-intercept form.
y  4 x  1
Functions and the Vertical Line Test
Functions
A function is a relation in which every element of the
domain is paired with exactly one element of the range.
Equivalently, a function is a relation in which no two
distinct ordered pairs have the same first coordinate.
Implied Domain of a Function
The domain of the function is implied by the
formula used in defining the function. It is
assumed that the domain of the function
consists of all real numbers at which the
function can be evaluated to obtain a real
number: any values for the argument that result
in division by zero or an even root of a negative
number must be excluded from the domain.
Quadratic Functions and Their Graphs
Vertex Form of a Quadratic Function
2
The graph of the function g  x   a  x  h   k where a,
h and k are real numbers and a  0 is a parabola whose
vertex is (h,k). The parabola is narrower than f  x   x 2
if a  1 and is broader than f  x   x 2 if 0  a  1 . The
parabola opens upward if a is positive and downward if
a is negative.
Commonly Occurring Functions
Piecewise-Defined Function
A piecewise-defined function is a function defined in
terms of two or more formulas, each valid for its own
unique portion of the real number line. In evaluating a
piecewise-defined function f at a certain value for x, it
is important to correctly identify which formula is valid
for that particular value.
Example: Commonly Occurring Functions
Sketch the graph of the function
2 x  2 if x  1
f  x   2
.
 x if x  1
The function f is a linear function on the interval
 , 1 and a quadratic function on the interval
 1,  . To graph f, we graph each portion
separately, making sure that each formula is applied
only on the appropriate interval. The complete
graph appears to the right, with the points f(–4) = 6
and f(2) = 4 noted in particular. Also note the use of
a closed circle at (–1,0) to emphasize that this point
is part of the graph, and the use of an open circle at
(–1,1) to indicate that this point is not part of the
graph.
Example: Variation Problems
For the following phrases, write the general formula that applies.
a. “y varies inversely as the n ͭ ͪ power of x”
y
k
xn
b. “y is directly proportional to the n ͭ ͪ power of x”
y  kx n
c. “y is inversely proportional to the n ͭ ͪ power of x”
k
y n
x
d. “y varies directly as the n ͭ ͪ power of x”
y  kx n
Shifting, Stretching and Reflecting Graphs
Horizontal Shifting
Let f(x) be a function whose graph is known, and let h
be a fixed real number. If we replace x in the definition
of f by x – h, we obtain a new function g  x   f  x  h .
The graph of g is the same shape as the graph of f, but
shifted to the right by h units if h > 0 and shifted to the
left by h units if h < 0.
Shifting, Stretching and Reflecting Graphs
Vertical Shifting
Let f(x) be a function whose graph is known and let k
be a fixed real number. The graph of the function
g  x   f  x   k is the same shape as the graph of f, but
shifted upward if k > 0 and downward if k < 0.
Shifting, Stretching and Reflecting Graphs
Reflecting With Respect to the Axes
Let f(x) be a function whose graph is known.
1. The graph of the function g(x)= –f(x) is the
reflection of the graph f with respect to the x-axis.
2. The graph of the function g(x) = f(–x) is the
reflection of the graph of f with respect to the
y-axis.
Shifting, Stretching and Reflecting Graphs
Stretching and Compressing
Let f(x) be a function whose graph is known, and let a
be a positive real number.
1. The graph of the function g(x) = a f(x) is stretched
vertically compared to the graph of f if a > 1.
2. The graph of the function g(x) = a f(x) is compressed
vertically compared to the graph of f if 0 < a < 1.
Combining Functions Arithmetically
Addition, Subtraction, Multiplication and Division of Functions
1.  f  g  x   f  x   g  x 
2.  f  g  x   f  x   g  x 
3.  f  g  x   f  x   g  x 
f  x
 f 
x

, provided that g  x   0
4.    
g  x
g
The domain of each of these new functions consists of the common
elements (or the intersection of elements) of the domains of f and g
individually.
Composing Functions
Composing Functions
Let f and g be two functions. The composition of f and
g, denoted f g , is the function defined by
 f g  x   f  g  x  .
The domain of f g consists of all x in the domain of g
for which g(x) is in turn in the domain of f. The function
f g is read “f composed with g,” or “f of g.”
Example: Composing Functions
Given f(x) = x2 + 2 and g(x) = x + 5 , find:
Again, we know by definition
 f g  x   f  g  x  
that  f g  x   f  g  x  .
 f  x  5
= (x + 5)2 + 2
= x2 +10x + 25 + 2
= x2 +10x + 27
Finding Inverse Function Formulas
To Find a Formula for f 1
Let f be a one-to-one function, and assume that f is
defined by a formula. To find a formula for f 1, perform
the following steps:
1. Replace f  x  in the definition of f with the variable y.
The result is an equation in x and y that is solved for y.
2. Interchange x and y in the equation.
3. Solve the new equation for y.
4. Replace the y in the resulting equation with f 1  x  .
Long and Synthetic Division
Compare the division of 2 x3  8 x 2  9 x  7 by x  2 below,
using long division on the left and synthetic division on the right.
2 x 2  4 x  1
x  2 2 x 3  8 x 2  9 x  7
  2x 3  4 x 2 
4 x2  9 x  7
  4x 2  8 x 
x  7
   x  2
5
2 2 8  9 7
4 8 2
2 4  1 5
Note: the numbers in blue are the
coefficients of the dividend and the
numbers in pink are the coefficients
of the quotient and remainder.
Synthetic Division
Step 1: Write k and the coefficients of the dividend.
Copy the leading coefficient of the dividend in the first
slot below the horizontal line.
Step 2: Multiply this number by k and write the result
directly below the second coefficient of the dividend.
Step 3: Add the two numbers in that column and write
the result in the second slot below the horizontal line.
Step 4: Repeat the process until the last column is
completed and the last number written down is the
remainder.
Example: Solving Systems by Elimination
Solve the system by the method of elimination.
5 x  2 y  6

2 x  3 y  10
To eliminate a variable, you’ll need to multiply each equation by a unique
constant. Let’s eliminate y. To do so, notice that we’ll have to multiply the first
equation by 3 and the second equation by 2.
3 5 x  2 y  6 
2 2 x  3 y  10
5 2  2 y  6
y  2
15 x  6 y  18
 4 x  6 y  20
19 x  38
x2
Thus, the solution is  2, 2  .