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1.1– RECURSIVE RELATIONS
Sequence – a set of countable terms that have a definite relationship
(finite or infinite)
Recursive Relation
- a relation where one or more initial values are known and a process
is repeated to calculate the value at each subsequent stage using the
values at one or more previous stages.
- The initial value must be known and specified separately.
Note T-shaped figure on page 2
Sequence : {1,4,7,…}
This is an arithmetic sequence.
-one in which there is a common difference.
It can be described with 2 different formulas:
1. Explicit (Non-recursive)
tn = t1 + (n – 1)d
In the above example
2. Recursive
tn = 1 + (n – 1)3
=
t1 = a
tn = tn-1 + d, if n >1, n  N
In the above example
t1 = 1
tn = tn-1 + 3, if n >1, n  N
(Each successive term is the previous term plus 3.)
The values of an arithmetic sequence form a discrete graph since each term
number is a natural number.
Series – the sum of the terms in a sequence.
The sum of the first n terms of the sequence Sn is the nth partial sum.
Consider {5,10,15,20,25,30,35,…}
The series is 5+10+15+20+25+30+35+…
2
The explicit rule is tn = 5 + (n-1)5 =
The sum of the first 7 terms is 140.
The 7th partial sum of the series can be written in sigma notation.
b
t
na

n
indicates a sum, where a is the beginning term
number and b is the ending term number and tn is the rule.
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For our given series, the sigma notation would be
5 j
j 1
Formulas for arithmetic series:
Sn =
n
(t1 + tn)
2
and
Sn =
n
(a + [a + (n – 1)d])
2
t1 = a and tn = a + (n – 1)d
1.2 – MORE RECURSIVE SEQUENCES
Geometric Sequence
-one in which there is a common ratio between the terms.
Ex) {3,9,27,81,…}
It can also be described with 2 different formulas:
1. Explicit (Non-recursive)
tn = arn-1 , if n  N
2. Recursive
t1 = a
tn = rtn-1 , if n >1, n  N
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In order to find the value of an unknown exponent, logarithms can be used.
Ex) 3x = 9
x=
Ex) How long would it take $500 to double at an interest rate of 5% per
year?
This is a geometric sequence therefore:
=
=
Geometric Mean
- A term that is missing between 2 terms of a geometric sequence.
- Calculated by multiplying the 2 outside terms and taking the
square root.
Ex) {8,20,____,125}
Formula for a geometric series
Sn =
a  ar n
, n N
1 r
Fibonacci Sequence:
{1,1,2,3,5,8,…}
Recursive formula-
t1 = 1,
t2 = 1
tn = tn-1 + tn-2 , if n >2, n  N
Many illustrations
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1.4 – LIMITS
The limit of a sequence
-a finite number L that the value tn gets closer to (or approaches) as n
approaches infinity.
Ex) Matt has a 70% on a test. If he makes 100% on every test the
rest of the term, his average will approach 100% but never reach it.
The limit is 100.
Note Inv. 4, p.28
-The sector that remains approaches 1 but will never reach it.
Sequences that go to infinity can be described as:
1. Converging – a sequence in which the terms approach a limit.
1 2 3 4
2 3 4 5
{ , , , ,…} converges to 1
Notation- lim

n
n
=1
n 1
Geometric series converge when r < 1, where r is the common ratio.
The formula for Sn as n approaches infinity if r < 1, can be expressed
as
Sn =
a
.
1 r
Ex) Find the sum of {10,5,2.5,…}.
Sn =
10
= 20
1  0 .5
2. Diverging - a sequence in which the terms do not approach a limit.
{1,1,2,3,5,…} diverges to infinity not a finite #.
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PROBLEMS:
1. A ball is dropped from a cliff, 20m above the ground. It rebounds 70%
of the distance it has fallen. In theory, how far will the ball travel vertically
before coming to rest?
2. A bird on the planet Mutate has a seizure and begins to fall from the sky.
It falls 2.5m in the first second, 7.5m in the second second, 12.5m in the
third second and so on.
a. How far does it fall in the first 8 seconds?
b. If it is initially 2000m high and comes alert again and begins to fly at
1500m, how many seconds was it falling?
3. Determine the sum of:
a.
1 1 1
1
+ + +…+
3 9 27
2187
b. 90+15+
10
c.
8
x4
15
+…
6
9
d.
 3x
2
x 1
4. Write the recursive and explicit(if possible) rule for:
a. {40,30,22.5,16.875, …}
b. {1,2,4,7,13,…}
5. Find t50 and s50 for 4a).
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2.1-COMBINING FUNCTIONS
Function
- a relation such that for every dependent value(x) there is one and
only one independent value(y). i.e. every member of the domain
has one and only one member of the range.
Types:
1.
2.
3.
4.
Linear
Quadratic
Cubic
Quartic
ex)
ex)
ex)
ex)
Finding y-intercepts: Substitute “0” into “x”.
Finding x-intercepts: (zeroes,roots)
1. LinearSubstitute “0” into “y”
2. QuadraticSubstitute “0” into “y” and either factor, complete
the square, graph or use the quadratic formula
Domain - all the possible “x” values of a function
Range – all the possible “y” values of a function
When combining functions, use same rules for adding, subtracting,
multiplying and dividing polynomials.
Note Inv. 1, page 56 and Focus A, page 57,58.
Note: f(x) + g(x) = (f + g)(x) and f(x) – g(x) = (f – g)(x)
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Example 1
Consider f(x) = x2 + 8 and g(x) = 6x.
If h(x) = f(x) – g(x), then h(x) = x2 – 6x + 8
Determine the following:
1. y-intercept
2. x-intercept a) factor
b) quadratic formula
3. vertex( complete the square )
4. sketch
5. domain
{x
6. range
{y
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Example 2
Consider f(x) = 6x2 – x – 12
1. y-intercept
2. x-intercept
factor
quadratic formula
3. vertex(complete the square)
4. sketch
5. domain
6. range
Graphs of Other Polynomial Functions:
Polynomial
- an algebraic expression consisting of 2 or more terms.
Degree
- the greatest power in which a variable is raised. Ex) the degree of
x4 + 3x2 + 1 is 4.
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1. Cubic Function (y = ax3 + bx2 + cx + d)
- If a > 0, then the function goes from the 3rd to the 1st quadrants
- If a < 0, then the function goes from the 2nd to the 4th quadrants
- There are 3 roots to each cubic function
a. 3 different real roots
b. 3 equal real roots
c. 1 real root and 2 complex roots
2. Quartic Function (y = ax4 + bx3 + cx2 + dx + e)
- If a > 0, then the function starts in the 2nd quadrant and ends in the
1st.
- If a < 0, then the function starts in the 3rd quadrant and ends in the
4th.
- There are 4 roots to each quartic function
a. 2 unequal real and 2 equal real
b. 2 unequal real and 2 complex
c. 4 different real
d. 2 pairs of equal
e. 4 equal real
f. 2 equal real and 2 complex
g. 4 complex
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Answer the following:
Given y = x2 - 3x + 4,
Determine the vertex, x and y-intercepts, range,
domain and the nature of the roots. Also, sketch.
Quadratic functions can have roots that are:
1. 2 complex
2. 2 real, unequal
3. 2 real, equal
Consider the following:
f(x) = (x-2)(x+3)2
1. y-intercept
2. x-intercepts
3. degree
4. sketch
Assign pages 65 and 66, #’s 25-29, then page 66, # 30. Check graphs with
calculator.
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2.2 – POLYNOMIAL EQUATIONS AND INEQUALITIES
Solving Quadratic Inequalities:
1. Using the Graph:
a. If the inequality is y>0, the solution will be the x values for
which the curve is above the x axis.
b. If the inequality is y<0, the solution will be the values below
the x axis.
The x-intercepts need to be found.
The inequality x2 – x – 6 > 0 is true when x < -2 and x > 3. See p.67.
The interval notation is x Ε (-  ,-2)  (3,  )
2. Using a Number Line:
Solve 8 + 2x – x2 < 0
a. Solve the equation by finding the intercepts.
b. Mark the roots on a number line.
c. Use test points in each interval to determine if positive or negative.
d. State the intervals that are correct.
Solve x2 – 2x – 3  0
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Solving Polynomial Equations and Inequalities
Note graph on page 70.The roots of y = x3 – 5x2 – 2x + 24 are the xintercepts which are
. The factors would be
Using the same graph, solve x3 – 5x2 – 2x + 24 > 0.
Using the graph on page 71, solve x3 + x2 – 11x – 12  0.
Solving Polynomial Equations by the Factor Theorem
Consider f(x) = x3 – 7x2 – 4x + 28 = 0
1. Find a root of the equation by finding a factor of 28 that will make the
equation = 0.
2. Then decide to use long division or synthetic division.
Long Division:
Synthetic Division:
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Example 2 : x3 – 6x2 – x + 30 = 0
Long Division:
Synthetic Division:
If the leading coefficient is a number other than 1, then the Rational
Root Theorem must be applied.
“ If x = m/n is a rational root of a polynomial equation where a is the
leading coefficient and ao is the constant term, then m/n is a factor of
a/ao.”
Consider 2x3 – 9x2 + 10x – 3 = 0.
The roots will be factors of
3
. The possible rational roots are 1,
2
-1, 3, -3, 1/2, -1/2, 3/2, -3/2. Determine a root that will make the
equation equal 0 and divide.
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Specialized Factoring Techniques:
1. Common Factors:
x3 – x2 – 12x
2. The Sum and Difference of Cubes:
x3 – 64
8 + 125y3
3. Quartic Expressions Factored as Trinomials:
x4 – 5x2 + 4
4x4 – 37x2 + 9
4. Grouping to Find a Common Factor:
x3 – 2x2 – 16x + 32
5. Grouping to Get a Difference of Squares:
x4 + 5x2 + 9
x4 – 6x2 + 1
Solving Inequalities:
1.
2.
3.
4.
Determine the roots.
Draw a number line with the roots put on it.
Use test points to determine the sign of the values in that interval.
Solve the inequality using interval notation.
Example: x3 – 6x2 – x + 30  0
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Finding the Equation of a Polynomial Function Using Given Points
Consider the table on page 82. What is the cubic function that represents
those points?
1. Using y = ax3 + bx2 + cx + d, substitute the points into x and y to
make 4 equations.
2. Use elimination to solve the system of equations for a,b,c and d.
3. State the function.
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2.3 – COMPOSITION OF FUNCTIONS
A composite function, h(x), is created when 2 functions g(x) and f(x) are
combined so that the domain of the second function is restricted by the range
of the first function.
h(x) = f(g(x)) or h(x) = (f ◦ g)(x)
i.e. (f composed with g)
If g(x) = x3 + x2 and f(x) = 2x – 5, then
h(x) = f(g(x))
= f(x3 + x2)
= 2(x3 + x2) – 5
The range of g(x) becomes the domain of f(x). For example in the above
function, if the domain for g(x) is 2, the range value becomes 12. This
becomes the domain for f(x), producing a range value of 19. We can say
that f(g(2)) = 19 or (f ◦ g)(2) = 19 or h(2) = 19.
Consider page 86, # 7aThe value of g(2) = 3. That value becomes f(3) = 7
In 8a, q(-2) = 5 and p(5) = 0.
Assign pages 86,87
TEST!!!!!!
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2.4 – THE LIMIT OF A RATE OF CHANGE
Remember the following concepts learned in “Functions and Relations”:
1. The average velocity on a graph is related to the slope of a secant.
2. The instantaneous velocity is related to the slope of a tangent at a
point.
A tangent can be considered to be equal to a secant with points very close
together.
Consider the example in Inv. 7, page 89.
Consider the slopes of the secants from 1 to 2 seconds as the secants get
closer and closer to being a tangent at 1.
Then consider the slopes of the secants from 0 to 1 seconds as the secants
get closer and closer to being a tangent at 1. See tables on page 90.
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Example:
The displacement of a particle is given by the equation S = 2t2 – 8t, where S
is the displacement in cm and t is the time in s.
a. Find the average velocity between t = 3 and t = 4.
b. Find the instantaneous velocity at t = 3.
Limit Notation:
The limit of the average velocities as time approaches 1 second can be
expressed as:
lim (average velocity)
t1
= instantaneous velocity (at 1 s)
= 0 m/s
This limit can also be expressed as approaching from the left(-) or right(+).
lim (average velocity) = 0 and
t  1
Consider the function y =
lim (average velocity) = 0
t  1
x2  4
on page 92. The limit as x approaches 2 is
x 2
undefined from the equation. The graph shows however that, as x
approaches 2 from both sides, the limit exists as 4.
If the one-sided limits of a function, as x approaches a certain value, are not
the same finite value, the limit does not exist. Consider f(x) on page 92.
Note supplementary pages on limits (pages 23-36).
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– THE SLOPE OF THE TANGENT AND THE
DERIVATIVE
The derivative of a function is the general formula for the slope of the
tangent used to calculate the instantaneous rate of change of a given
function. It is also referred to as the slope function.
On the above graph, consider 2 points - (t , f(t)) and (t + h, f(t + h)). The
distance between the secant points is “h”. To find the slope of a secant
through these 2 points, the equation
y2  y1
is used. This becomes
x2  x1
f ( t  h)  f ( t )
. The “t” in the denominator cancels out for a value of only
( t  h)  t
“h”. The function of the graph is f(t) = -4.9t2 + 9.8t.
Simplify:
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Therefore, lim (-9.8t – 4.9h + 9.8) =
h0
This is the derived function or the slope function. Substituting values into
“t” will determine the value of the slope at the indicated location.
Example:
The slope function at t = 1.2 is –9.8(1.2) + 9.8 = -1.96. This is the
instantaneous velocity at this time.
Example:
Find the slope of the tangent to f(x) = 4x2 - 3x at x = 2.
The notations used to represent the derivative of the function are y’, f’(x) and
dy
.
dx
The general formula for a derivative using first principles is:
f’(x) = lim
h0
f ( x  h)  f ( x )
h
Hand out supplementary page 47.
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Differentiation – process for calculating derivatives using systematic rules.
Function
Derivative
1.
-4.9t2 + 9.8t
1.
2.
4x2 – 3x
2.
3.
3x2
3.
This shortcut rule is known as the Power Rule.
Consider page 100, #19 to #22
Graphing Slope Functions
Graph the function y = x2 – 2x + 8 and its slope function.
Observations:
1. The ________________ of the slope function indicate the x
coordinate of the local _______________ or _______________.
2. The _______________ of the slope function is _____ ______ than the
original function.
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3. The original function is i____________________ when the slope
function is a_______________ the x axis.
A critical point is a point on the graph of a function at which the derivative
is equal to zero or undefined. They are found at the x-intercepts of the slope
function.
An absolute maximum or minimum occurs only for even degree functions.
Answer the following questions for f(x) = x3 – 5x2 + 7x.
1.
2.
3.
4.
Find the x- and y-intercepts.
Find the critical points.
Find the intervals of increase and decrease.
Graph.
Solution:
1. y-intercept:
x-intercept:
0 = x(x2 –5x + 7)
2. critical points:
(find the x-intercepts of the slope function and substitute them into
the original function to find the y values.)
f’(x) =
3. intervals:
(put the x-intercepts of the slope function on a number line and
determine the intervals that are positive and negative. The positive
intervals indicate the original function is increasing and vice
versa)
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4.
More Differentiation Rules
Note supplementary pages 51 to 62.
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3.1 – SIMPLE RATIONAL FUNCTIONS
Rational function
- Defined as f(x) =
p( x )
, where p(x) and q(x) are polynomials and
q( x)
q(x)  0.
- Forms a conic called a hyperbola.
- Has a vertical asymptote and either a horizontal or oblique
asymptote.
An asymptote is a line that a graph approaches values but never crosses.
The function y =
x
-5
-2
1
can be graphed as follows:
x
-1
-
1
4
y
Equation of horizontal asymptote Equation of vertical asymptote –
DomainRange-
0
1
4
1
2
5
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The transformation of y =
1
1
can be expressed as y – k =
, where y = k
x
x h
is the horizontal asymptote and x = h is the vertical asymptote.
These relationships are described as inversely proportional where as the
value of one variable increases, the other decreases.
Construct the graphs of the following:
1. y =
1
x2
2. y =
1
( x  1) 2
3. y =
1
x 1
2
4. y =
1
x 1
2
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Rational Functions Where the Numerator is Not Constant
Graphs that cannot be drawn without lifting your pencil are not continuous.
There can be 3 types of discontinuity:
1. Infinite – vertical asymptotes
Ex) y =
1
x2
2. Jump – value of function abruptly changes
Ex) f(x) =
 x2 , x  0

 x  1, x  0
3. Points or holes – graph has a “hole” caused by a missing domain
element
Ex) y =
x2  4
(page 92)
x 2
Conditions for Continuity:
A function f is said to be continuous at “a” if:
1. lim f(x) exists
x a
2. f(a) is defined
3. lim
f(x) = f(a)
x a
Examples on page 136:
1. f(x) violates the 1st condition.
2. g(x) violates the 2nd condition.
3. h(x) violates the 1st condition.
Vertical Asymptotes:
1. Zeroes of the denominator tell the location of the vertical asymptotes.
2. If the factor that gives the zero does not occur in the numerator, the
function has a vertical asymptote.
3. If the factor also occurs in the numerator, the function has a
discontinuity for that value of x.
Horizontal Asymptotes:
If the degree of the denominator is equal to or greater than the degree of
the numerator, then there are horizontal asymptotes. The value is
determined by dividing each term by the highest power. Also note that as x
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approaches   , the value of
c
approaches 0, where c is a constant and n is
xn
an exponent  1.
Note examples on page 139.
Oblique Asymptotes:
If there is no horizontal asymptote i.e. the degree of the numerator is
greater than the degree of the denominator, there will be an oblique
asymptote. This is found by dividing the numerator by the denominator.
The part of the answer that is not the remainder is the equation of the oblique
asymptote.
Examples: Determine the asymptotes and any points of discontinuity for:
1. f(x) =
2x  1
2 x  3x  2
2
3x 2  x  1
2. f(x) =
x
2
x  7 x  12
3. f(x) =
x5
Solutions:
1. vertical asymptote is x = -2
point of discontinuity is x = ½
horizontal asymptote is y = 0
2. vertical asymptote is x = 0
oblique asymptote is y = 3x – 1
3. vertical asymptote is x = -5
oblique asymptote is y = x + 2
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3.3 – SOLVING RATIONAL EQUATIONS
Simplifying Rational Expressions:
A. Addition and Subtraction
1. Factor any polynomials
2. Determine the common denominator
3. Multiply the numerator by whatever was needed to get the
denominator
4. Combine expressions and simplify
5. State the restrictions by determining what will make the
denominators undefined in the original question
B. Multiplying and Dividing
1. Factor any polynomials
2. Cancel out any common factors in the numerators and
denominators and simplify
3. State the restrictions
4. For division, remember to multiply by the reciprocal
5. Also for division, the restrictions will be in the original question
and for the inverted numerator
Note: page 145, #5 regarding simplifying incorrectly.
Examples:
1
2x  1
1.
- 2
x 3 x  x 6
x 3
x 2  3x
2. 3
 2
2
x  3x  2
x  2 x  3x
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Solving Rational Equations:
Multiply each part of the equation by the common denominator to clear out
the denominators. Simplify and solve.
Examples
x 3
3
x2
- 2
=
x1 x 1
x1
2y  3
8
2y  3
2 =
2y  3 9  4y
2y  3
30
Graphing Rational Functions:
1. Find the value of the horizontal asymptote by dividing each term by
the greatest power of x and finding the limit as x goes to   .
2. Factor
3. From the denominator, find the vertical asymptotes or points of
discontinuity.
4. From the numerator, find the x-intercepts.
5. Find the y-intercept.
6. Sketch these on the graph.
7. Use test points to sketch the curves.
8. In some cases the graph will cross over the horizontal asymptote and
approach back towards it. This can be determined by substituting the
value of the asymptote into f(x) and solving the equation.
Examples
f(x) =
2 x 2  5x  3
x 2  2x  8
31
f(x) =
f(x) =
x 2  4 x  10
x 2
32
Rational Inequalities:
Consider
2 x 2  5x  3
>0
x 2  2x  8
1. Graphically, find out where the graph is positive.
2. Number line
a. Find the x-intercepts and the vertical asymptotes.
b. Plot the points on a number line.
c. Use test points in the intervals.
d. State in interval notation.
Solution:
The factors of the numerator (x-intercepts) are
The factors of the denominator (vertical asymptotes) are
Number line
________________________________________
Interval Notation x  (
TEST!!!!!!!!!!!!!!
) (
) (
)
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3.4 – IRRATIONAL FUNCTIONS
On the same coordinate grid, graph the following functions:
y = x2
y=x
y= x
Note questions 4 – 7 on page 158.
Find the inverse of y = x2 – 5
Since the inverse is a reflection in the line y = x, interchange x and y and
solve for y.
This inverse is not a function. If you restrict the domain, to x  0 or x 
0, then it becomes a function.
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Find the inverse of y = x2 + 6x.
Interchange the x and y and complete the square.
Solve for y.
Discuss page 160, #9 and #10
Sketch:
1. y-3 = x  4
2. 2(y+1) = x  3
Square-Root Functions and Discontinuities
-Square-root functions are discontinuous where f(x)<0.
Consider f(x)= x 2  2 x  8
Where is the graph discontinuous?
35
Solving Irrational Equations:
2x  7 = x – 4
1. Square both sides.
2. Solve for “x”.
3. Check for extraneous roots.
3x  5 = 3 +
x 2
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3.5 – ABSOLUTE-VALUE FUNCTIONS
The function f(x) = x is an absolute value function which can be also called
a piecewise function written as follows:
  x, x  0
f(x) = 
  x, x  0
Absolute value – the distance a number is from 0.
Graph:
f(x) = -2 x  3 + 5
f(x) =  x 2  4
37
Graph:
f(x) = -2 x  3 + 4
and
f ( x)
Absolute Value Inequalities
Example 1 –
x 1  5
a. Number Line - Consider x  5
-How far is x from 0?
-What is the translation in the original statement?
b. Algebra - In order for x –1 to be less than 5 away from 0,
x–15
Example 2 -
AND
2x  3  9
a. Number Line – Consider x  9
x – 1  -5
38
- How far is x from 0?
- What is the translation and stretch factor in the original problem?
b.algebra – In order for 2x + 3 to be greater than 9 away from 0,
2x + 3  9
2x + 3  -9
OR
5  x  2  10
Example 3 -
5  x + 2  10
5  -(x + 2)  10
OR
PIECEWISE FUNCTIONS
Graph g(x) = {
 x
2
 4, x  0
x  4, x  0






















39
3.7- MODELLING WITH EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
Review:
ab=c (exponential form)
logac=b
(logarithmic form)
Properties:
1. logax + logay = loga(xy)
x
y
2. logax – logay = loga( )
3. nlogax = logaxn
Finding the value of an exponent:
2x = 10
Example:
The mummified body of a woman was found to contain 75% of the original
C-14. What was the age of the mummy given the half-life of C-14 is 5700
years?
40
COMPOUND INTEREST
I = Io (1 +
r n
)
n
As the number of compounding periods get larger and larger, what happens
to the value of (1 +
1 n
) ? Note p. 183D
n
This value is approximately 2.71828 or e.
Therefore e = lim
(1 +
n
1 n
)
n
and er =
For continuous compounding interest, I = Ioert
INVESTIGATION 9 – PROPERTIES OF y = ex
Use TI-83.
y-intercept =
domain =
range =
It is increasing, positve and continuous.
Find the derivative at:
a. x=0
b. x=1
c. x=2
d. x=3
How do these values compare to the corresponding y values?
We can say the following-
If y = ex , then y’ = ex .
41
FOCUS M : y = ln x
The inverse of 23 = 8 is log28 = 3. Finding the log means to find the value
of the exponent.
The inverse of y = ex is logey = x or ln y. The x and y values will
interchange for the inverses and form a mirror image in the y=x line.
Example: e2 = 7.3891 is loge7.3891 = 2 or ln 7.3891 = 2
Examples:
1. 42x + 5 = 12
2. 5e-x = 15
4. e2x – 2ex – 8 = 0
5. 3 ln 2x – 4 = 11
Homework: Page 187 – 15,16
Page 190 – 29-32
Review for test: Pages 210-212, 12 –24, omit 23.
42
5.1 – COMPLEX NUMBERS
Complex Number
- a combination of a real number and an imaginary number.
Imaginary Number
- any number of the form bi, where b is a non-zero real number and i
is the imaginary unit, i =  1 . Also, i2 = -1
Consider the roots of y = x2 + 4x + 5. The roots are:
Notice Focus Questions, p. 275
When using the 4 operations in simplifying complex numbers, treat as a
variable. Also in dividing, you may need to rationalize.
Remember i =  1 , i2 = -1, i4 = 1
Assign p. 276, #16-21
Graphing Complex Numbers:
A complex number, a + bi, can be expressed as an ordered pair (a,b). It can
be graphed on a complex plane called an Argand Diagram. This has a
horizontal real axis and a vertical imaginary axis. Note page 277 of text.
A complex number can be represented by a point or a position vector.
The distance between the origin and a complex number z is the absolute
value or modulus of z or z . That distance can be found by the Pythagorean
Thereom.
Example: The modulus of 4 + 5i is 41 .
43
Notice 25 and 27 on pp. 278,279.
The Complex Conjugate of a + bi is a – bi. For any complex number z the
complex conjugate is z .
Assign
p. 278-279 , # 23, 25abcd, 27abc, 29
p. 281-282, # 40-42
p. 284-285, # 49, 52, 53
p. 288, # 72
5.2 – POLAR COORDINATES
Points can be plotted not only on a rectangular coordinate grid but also on
polar paper made up of concentric circles. The center is called the pole and
the polar axis is a horizontal ray pointing right. A point is identified as
P(r,) where r is the distance from P to the pole and  is the measure of the
counterclockwise rotation from the terminal arm.
Plot the points:
a. (2,240) b. (3,-300) c. (5,








5
)
6
d. (-2,3)
44
Converting from Rectangular to Polar Form and Vice-Versa:

Consider P(3,4)


1. r2 = x2 + y2


r=

















2. tan  =




3. (r,) =




Consider (5,150)
1. x = r cos 








2. y = r sin 
3. (x,y) =
45
Converting Complex Numbers:
1. From rectangular(a + bi) to polar:
Find r (using r2 = x2 + y2) and  using tan  =
a= r cos , b = r sin 
y
x
= r cos  + r i sin 
= r(cos  + i sin 
= r cis 
Ex) Convert 4 + 3i to polar form:
2. From polar(r cis ) to rectangular(a + bi):
- Evaluate by using a = r cos  and b = r sin
Ex) Convert 6(cos 60 + i sin 60) to rectangular form:
Ex) Convert 3 cis (
7
) to rectangular form:
4
46
Converting Equations in Rectangular Form into Polar Form:
- Use the following statements:
x2 + y2 = r2
x = r cos 
y = r sin 
Ex) x2 + y2 = 5 x 2  y 2 - 5y
Operations with Complex Numbers in Polar Form:
1. Convert to polar form.
2. Use DeMoivre’s Theorem [(r cis )n = rn cis n] to evaluate powers.
3. To multiply, multiply the radii or moduli and add the angles.
4. To divide, divide the moduli and subtract the angles.
Ex) (1 - i 3 )7
Ex) (2 cis

5
)(7 cis
)
6
4
47
Roots of Complex Numbers:
1
n
1
z = z n = r n cis (

2k
+
), k = 0,1,2,…
n
n
Examples:
Determine all the roots of the equations:
1. z4 = 1
2. z3 = i