### Chapter 1

```Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall
Slide 1- 1
Chapter 1
Prerequisites for Calculus
1.1
Lines
Quick Review
1.
Find the value of y that corresponds to x = 3 in y = - 2 + 4 (x - 3).
2.
Find the value of x that corresponds to y = 3 in y = 3 - 2 (x + 1).
In Exercises 3 and 4, find the value of m that corresponds
to the values of x and y.
3.
x = 5,
m=
y- 3
x- 4
4.
x = - 1, y = - 3, m =
2- y
3- x
y = 2,
Slide 1- 4
Quick Review
In exercises 5 and 6, determine whether the ordered pair is a
solution to the equation.
5. 3 x  4 y  5
6. y  2 x  5
 1
a)  2, 
b)  3,  1 a)  1, 7 
b)  2,1
 4
In exercises 7 and 8, find the distance between the points.
7. 1,0  and  0,1
1

8.  2,1 and 1,  
3

Slide 1- 5
Quick Review
In Exercises 9 and 10, solve for y in terms of x.
9.
4x- 3 y = 7
10. - 2 x + 5 y = - 3
Slide 1- 6
Quick Review Solutions
1.
Find the value of y that corresponds to x = 3 in y = - 2 + 4 (x - 3).
y= - 2
2.
Find the value of x that corresponds to y = 3 in y = 3 - 2 (x + 1).
x= - 1
In Exercises 3 and 4, find the value of m that corresponds
to the values of x and y.
3.
x = 5,
m=
y- 3
x- 4
m= - 1
4.
x = - 1, y = - 3, m =
2- y
3- x
m=
y = 2,
5
4
Slide 1- 7
Quick Review Solutions
In exercises 5 and 6, determine whether the ordered pair is a
solution to the equation.
5. 3 x  4 y  5
6. y  2 x  5
 1
a)  2, 
b)  3,  1 a)  1, 7 
b)  2,1
 4
a) yes
b) no
a) yes
b) no
In exercises 7 and 8, find the distance between the points.
7. 1,0  and  0,1
1

8.  2,1 and 1,  
3

distance = 2
distance = 5/3
Slide 1- 8
Quick Review Solutions
In Exercises 9 and 10, solve for y in terms of x.
9. 4 x - 3 y = 7
10. - 2 x + 5 y = - 3
4
7
2
3
y = xy = x3
3
5
5
Slide 1- 9





Increments
Slope of a Line
Parallel and Perpendicular Lines
Equations of Lines
Applications
…and why.
Linear equations are used extensively in
Slide 1- 10
Increments
If a particle moves from the point (x1 , y1 ) to the point (x2 , y2 ), the
increments in its coordinates are
x  x2  x1 and
y  y2  y1
The symbols x and D y are read delta x and delta y.
The letter D is a Greek capital d for difference.
Neither x nor D y denotes multiplication;
D x is not delta times x nor is D y delta times y.
Slide 1- 11
Example Increments
The coordinate increments from (8, 3) to (-6, 1) are:
x   6  8   14,
y  1  3  2
Slide 1- 12
Slope of a Line
Let P1 ( x1 , y1 ) and P2 ( x2 , y2 ) be points on a nonvertical line, L. The slope
of L is
m=
rise y2 - y1
=
run x2 - x1
A line that goes uphill as x increases has a positive slope. A line that goes
downhill as x increases has a negative slope.
Slide 1- 13
Slope of a Line
A horizontal line has slope zero since all of its points have the same
y -coordinate, making D y = 0.
Dy
is undefined.
Dx
We express this by saying that vertical lines have no slope.
For vertical lines, D x = 0 and the ratio
Slide 1- 14
Parallel and Perpendicular Lines
Parallel lines form equal angles with the x-axis. Hence, nonvertical
parallel lines have the same slope.
m1 = m2
If two nonvertical lines L1 and L2 are perpendicular, their slopes
m1 and m2 satisfy m 1 m2 = - 1, so each slope is the negative reciprocal
of the other: m1 = -
1
1
, m2 = m2
m1
Slide 1- 15
Equations of Lines
The vertical line through the point (a , b) has equation x= a
since every x-coordinate on the line has the same value a.
Similarly, the horizontal line through (a , b) has equation y = b.
Slide 1- 16
Example Equations of Lines
Write the equations of the vertical and horizontal lines through
the point (- 3,8).
x= - 3
and
y= 8
Slide 1- 17
Point Slope Equation
The equation y= m(x - x1 ) + y1 is the point - slope equation of the line
through the point (x1 , y1 ) with slope m.
Slide 1- 18
Example: Point Slope Equation
Write the point-slope equation for the line through (7, -2) and (-5, 8).
8 - ( - 2)
10
10
5
=
= = - 5 - (7) - 12
12
6
We can use this slope with either of the two given points in the point-slope
equation. For ( x1 , y1 ) = (7, - 2) we obtain
The line's slope is m =
5
(x 6
5
y=- x +
6
5
y=- x +
6
y= -
7) + - 2
35
- 2
6
23
6
Slide 1- 19
Equations of Lines
The y -coordinate of the point where a non-vertical line
intersects the y -axis is the y-intercept of the line.
Similarly, the x-coordinate of the point where a non-horizontal
line intersects the x-axis is the x-intercept of the line.
A line with slope m and y -intercept b passes through (0, b)so
y = m ( x - 0) + b, or y = m x + b
Slide 1- 20
Slope-Intercept Equation
The equation y=m x + b is the slope - intercept equation of the line
with slope m and y-intercept b.
Slide 1- 21
General Linear Equation
The equation Ax + By = C (A and B not both 0)
is a general linear equation in x and y.
Although the general linear form helps in the quick identification of lines, the
slope-intercept form is the one to enter into a calculator for graphing.
Slide 1- 22
Example Analyzing and Graphing a
General Linear Equation
Find the slope and y-intercept of the line 2x - 3y = 15. Graph the line.
Solve the equation for y to put the equation in slope-intercept form:
- 3y = - 2x + 15
- 2
15
y=
x +
- 3
- 3
2
y= x - 5
3
[-10, 10] by [-10, 10]
Slide 1- 23
Example Determining a Function
The following table gives values for the linear function f ( x) = mx + b.
Determine m and b.
x
f(x)
-1
-1
1
5
3
11
The graph of f is a line. We know from the table that the following points are on
the line: (- 1, - 1), (1,5), (3,11)
Using the last two points the slope m is: m =
11 - 5
6
= =3
3- 1
2
So f(x) = 3x + b. Because f (1) = 5, we have
f (1) = 3(1) + b
5=3 + b
b=2
Thus, m = 3, b = 2 and f ( x) = 3x + 2
Slide 1- 24
Example Reimbursed Expenses
A company reimburses its sales representatives \$150 per day for lodging and
meals plus \$0.34 per mile driven.
Write a linear equation giving the daily cost C to the company in terms of x,
the number of miles driven.
How much does it cost the company if a sales representative drives 137 miles on
a given day?
Because we know that the relationship is linear, we know that it conforms to the
equation C ( x) = .34 x + 150.
If a sales representative drives 137 miles, then x = 137. Thus,
C (137) = .34(137) + 150
C (137) = 46.58 + 150
C (137) = 196.58
It will cost the company \$196.58 for a sales representative to drive 137 miles a day.
Slide 1- 25
1.2
Functions and Graphs
Quick Review
In Exercises 1 - 6, solve for x.
1.
3x - 1 £ 5 x + 3
2.
x (x - 2) > 0
3.
x- 3 £ 4
4.
x- 2 ³ 5
5. x 2 < 16
6. 9 - x 2 ³ 0
In Exercises 7 and 8, describe how the graph of f can be
transformed to the graph of g.
2
7.
f (x)= x 2 ,
g (x )= (x + 2) - 3
8.
f (x)= x ,
g (x )= x - 5 + 2
Slide 1- 27
Quick Review
In Exercises 9 - 12, find all real solutions to the equations.
9.
f (x )= x 2 - 5
(a ) f (x)= 4
1
10. f (x)=
x
(a ) f (x)= - 5
(b) f (x)= - 6
(b) f (x)= 0
Slide 1- 28
Quick Review
11. f (x )=
x+ 7
(a ) f (x)= 4
12. f (x )=
3
(b) f (x)= 1
x- 1
(a ) f (x)= - 2
(b) f (x)= 3
Slide 1- 29
Quick Review Solutions
In Exercises 1 - 6, solve for x.
1.
3x - 1 £ 5 x + 3
[- 2, ¥ )
2.
x (x - 2) > 0
(- ¥ , 0)È (2, ¥ )
3.
x- 3 £ 4
[- 1, 7]
4.
x- 2 ³ 5
(- ¥ , - 3] È [7, ¥ )
5.
x 2 < 16
(- 4, 4)
6.
9- x2 ³ 0
[- 3,3]
In Exercises 7 and 8, describe how the graph of f can be
transformed to the graph of g .
2
7.
f (x)= x 2 ,
g (x )= (x + 2) - 3
2 units left and 3 units downward
8.
f (x)= x ,
g (x )= x - 5 + 2
5 units right and 2 units upward
Slide 1- 30
Quick Review Solutions
In Exercises 9 - 12, find all real solutions to the equations.
9.
f (x )= x 2 - 5
(a ) f (x)= 4
(a ) - 3,3
(b) f (x)= - 6
(b) no real solution
1
x
(a ) f (x)= - 5
(b) f (x)= 0
10. f (x )=
(a )
- 1
5
(b) no real solution
Slide 1- 31
Quick Review Solutions
11. f (x )=
x+ 7
(a ) f (x)= 4
(a ) 9
12. f (x )=
3
(b) f (x)= 1
(b) - 6
x- 1
(a ) f (x)= - 2
(a ) - 7
(b) f (x)= 3
(b) 28
Slide 1- 32







Functions
Domains and Ranges
Viewing and Interpreting Graphs
Even Functions and Odd functions - Symmetry
Functions Defined in Pieces
Absolute Value Function
Composite Functions
…and why
Functions and graphs form the basis for
understanding mathematics applications.
Slide 1- 33
Functions
A rule that assigns to each element in one set a unique
element in another set is called a function. A function is like
a machine that assigns a unique output to every allowable
input. The inputs make up the domain of the function; the
outputs make up the range.
Slide 1- 34
Function
A function from a set D to a set R is a rule that
assigns a unique element in R to each element in
D.
In this definition, D is the domain of the function and R is a set
containing the range.
Slide 1- 35
Function
The symbolic way to say " y is a function of x " is y = f (x)
which is read as y equals f of x.
The notation f (x) gives a way to denote specific values of a function.
The value of f at a can be written as f (a), read as " f of a."
Slide 1- 36
Example Functions
Evaluate the function f ( x) = 2 x + 3 when x = 6.
f (6) = 2(6) + 3
f (6) = 12 + 3
f (6)= 15
Slide 1- 37
Domains and Ranges
When we define a function y = f (x ) with a formula and the domain is
not stated explicitly or restricted by context, the domain is assumed to be
the largest set of x-values for which the formula gives real y -values the so-called natural domain. If we want to restrict the domain, we must say so.
The domain of C (r )= 2p r is restricted by context because the radius, r ,
must always be positive.
Slide 1- 38
Domains and Ranges
The domain of y = 5 x is assumed to be the entire set of real numbers.
If we want to restrict the domain of y = 5 x to be only positive values,
we must write y = 5 x, x > 0.
Slide 1- 39
Domains and Ranges



The domains and ranges of many real-valued functions of a
real variable are intervals or combinations of intervals. The
intervals may be open, closed or half-open, finite or infinite.
The endpoints of an interval make up the interval’s boundary
and are called boundary points.
The remaining points make up the interval’s interior and are
called interior points.
Slide 1- 40
Domains and Ranges


Closed intervals contain their boundary points.
Open intervals contain no boundary points
Slide 1- 41
Domains and Ranges
Slide 1- 42
Graph
The points (x, y )in the plane whose coordinates are the
input-output pairs of a function y = f (x )make up the
function's graph.
Slide 1- 43
Example Finding Domains and Ranges
Identify the domain and range and use a grapher
to graph the function y = x 2 .
Domain: The function gives a real value of y for every value of x
so the domain is (- ¥ , ¥ ).
Range: Every value of the domain, x, gives a real, positive value of y
so the range is [0, ¥ ).
y = x2
[-10, 10] by [-5, 15]
Slide 1- 44
Viewing and Interpreting Graphs
Graphing with a graphing calculator requires that you
develop graph viewing skills.

Recognize that the graph is reasonable.

See all the important characteristics of the graph.

Interpret those characteristics.

Recognize grapher failure.
Slide 1- 45
Viewing and Interpreting Graphs
Being able to recognize that a graph is reasonable comes
with experience. You need to know the basic functions,
their graphs, and how changes in their equations affect
the graphs.
Grapher failure occurs when the graph produced by a
grapher is less than precise – or even incorrect – usually
due to the limitations of the screen resolution of the
grapher.
Slide 1- 46
Example Viewing and Interpreting
Graphs
Identify the domain and range and use a grapher to
graph the function y =
x2 - 4
Domain: The function gives a real value of y for each value of x ³ 2
so the domain is (- ¥ , - 2]È [2, ¥ ).
Range: Every value of the domain, x,
gives a real, positive value of y
so the range is [ 0, ¥ ).
y=
x2 - 4
[-10, 10] by [-10, 10]
Slide 1- 47
Even Functions and
Odd Functions-Symmetry

The graphs of even and odd functions have important
symmetry properties.
A function y = f ( x)is a
even function of x if f (- x) = f (x )
odd function of x if f (- x)= - f (x )
for every x in the function's domain.
Slide 1- 48
Even Functions and
Odd Functions-Symmetry

The graph of an even function is symmetric about the
y-axis. A point (x,y) lies on the graph if and only if the point
(-x,y) lies on the graph.

The graph of an odd function is symmetric about the origin.
A point (x,y) lies on the graph if and only if the point (-x,-y)
lies on the graph.
Slide 1- 49
Example Even Functions and Odd
Functions-Symmetry
Determine whether y = x3 - x is even, odd or neither.
y = x3 - x is odd because
3
f (- x)= (- x) - (- x) = - x3 + x = - (x3 - x)= - f (x)
y = x3 - x
Slide 1- 50
Example Even Functions and Odd
Functions-Symmetry
Determine whether y = 2 x + 5 is even, odd or neither.
y = 2 x + 5 is neither because
f (- x)= 2 (- x)+ 5 = - 2 x + 5 ¹ f ( x) ¹ - f (x )
y = 2x + 5
Slide 1- 51
Functions Defined in Pieces

While some functions are defined by single formulas, others
are defined by applying different formulas to different parts of
their domain.

These are called piecewise functions.
Slide 1- 52
Example Graphing a Piecewise Defined
Function
Use a grapher to graph the following piecewise function :
2 x  1 x  0
f ( x)   2
x  3 x  0
y = x2 + 3; x > 0
y = 2 x - 1; x £ 0
[-10, 10] by [-10, 10]
Slide 1- 53
Absolute Value Functions
The absolute value function y = x is defined piecewise by the formula
ìïï - x, x < 0
x= í
ïïî x,
x³ 0
The function is even, and its graph is symmetric about the y-axis
Slide 1- 54
Composite Functions
Suppose that some of the outputs of a function g can be used as inputs of
a function f . We can then link g and f to form a new function whose inputs
x are inputs of g and whose outputs are the numbers f (g (x )).
We say that the function f (g (x )) read ( f of g of x )is the composite
of g and f . The usual standard notation for the composite is f o g ,
which is read " f of g."
Slide 1- 55
Example Composite Functions
Given f ( x) = 2 x - 3 and g (x)= 5x, find f o g.
( f o g ) (x ) = f (g (x))
= f (5 x)
= 2 (5 x)- 3
= 10 x - 3
Slide 1- 56
1.3
Exponential Functions
Quick Review
to 3 decimal places.
2
3
1.
5
3.
3- 1.5
2.
3
2
to 4 decimal places.
4.
x3 = 17
6.
x10 = 1.4567
5.
x5 = 24
Slide 1- 58
Quick Review
In Exercises 7 and 8, find the value of investing P dollars for
n years with the interest rate r compounded annually.
7. P = \$500,
r = 4.75%,
n = 5 years
8. P = 1000,
r = 6.3%,
n = 3 years
In Exercises 9 and 10, simplify the exponential expression.
9.
2 2
(x y )
4 3 3
(x y )
- 3
2
- 1
æa 3b- 2 ö÷ æa 4 c- 2 ö÷
çç
10. çç 4 ÷
÷
÷
çè c ø÷ èç b3 ø÷
÷
Slide 1- 59
Quick Review Solutions
to 3 decimal places.
2
3
1.
5
3.
3- 1.5
2.924
2.
3
2
4.729
0.192
to 4 decimal places.
4.
x 3 = 17
6.
x10 = 1.4567 ± 1.0383
2.5713
5.
x 5 = 24
1.8882
Slide 1- 60
Quick Review Solutions
In Exercises 7 and 8, find the value of investing P dollars for
n years with the interest rate r compounded annually.
7. P = \$500,
r = 4.75%,
n = 5 years
8. P = 1000,
r = 6.3%,
n = 3 years
\$630.58
\$1201.16
In Exercises 9 and 10, simplify the exponential expression.
9.
2 2
(x y )
4 3 3
x
( y)
- 3
1
x18 y 5
2
- 1
æa 3b- 2 ö
æa 4 c- 2 ö÷
÷
çç
10. çç 4 ÷
÷
÷
çè c ø
÷ èç b3 ø÷
÷
a2
bc 6
Slide 1- 61




Exponential Growth
Exponential Decay
Applications
The Number e
…and why
Exponential functions model many growth
patterns.
Slide 1- 62
Exponential Function
Let a be a positive real number other than 1. The function
f ( x) = a x
is the exponential function with base a.
The domain of f ( x) = a x is (- ¥ , ¥ ) and the range is (0, ¥ ).
Compound interest investment and population growth are examples
of exponential growth.
Slide 1- 63
Exponential Growth
If a 1 the graph of f looks like the graph
of y = 2 x in Figure 1.22a
Slide 1- 64
Exponential Growth
If 0  a 1 the graph of f looks like the graph
of y = 2-
x
in Figure 1.22b.
Slide 1- 65
Rules for Exponents
If a > 0 and b > 0, the following hold for all real numbers x and y.
x
1. a x ×a y = a x + y
4. a x ×b x = (ab)
ax
2. y = a xa
x
æa ö
a
5. çç ÷
÷
÷ = bx
çè b ø
3. (a
x
y
x
y
y x
) = (a ) = a xy
Slide 1- 66
Half-life
Exponential functions can also model phenomena that produce
decrease over time, such as happens with radioactive decay.
The half-life of a radioactive substance is the amount of time it
takes for half of the substance to change from its original
Slide 1- 67
Exponential Growth and Exponential Decay
The function y = k ×a x , k > 0, is a model for exponential growth
if a > 1, and a model for exponential decay if 0 < a < 1.
Slide 1- 68
Example Exponential Functions
Use a grapher to find the zero's of f (x)= 4x - 3.
f (x)= 4x - 3
[-5, 5], [-10,10]
Slide 1- 69
The Number e
Many natural, physical and economic phenomena are best modeled
by an exponential function whose base is the famous number e, which is
2.718281828 to nine decimal places.
æ
We can define e to be the number that the function f (x)= çç1 +
çè
x
1 ö÷
÷
ø
x÷
approaches as x approaches infinity.
Slide 1- 70
The Number e
The exponential functions y = e x and y = e- x are frequently used as models
of exponential growth or decay.
Interest compounded continuously uses the model y = P ×e r t , where P is the
initial investment, r is the interest rate as a decimal and t is the time in years.
Slide 1- 71
Example The Number e
The approximate number of fruit flies in an experimental population after
t hours is given by Q (t )= 20 e 0.03 t ,
t ³ 0.
a. Find the initial number of fruit flies in the population.
b. How large is the population of fruit flies after 72 hours?
c. Use a grapher to graph the function Q.
a. To find the initial population, evaluate Q (t ) at t = 0.
Q (0)= 20e
0.03(0)
= 20e0 = 20 (1)= 20 flies.
b. After 72 hours, the population size is
Q (72)= 20e
0.03(72)
= 20e 2.16 » 173 flies.
c.
Q(t )= 20e0.03t , t ³ 0
[0,100] by [0,120] in 10’s
Slide 1- 72
Quick Quiz Sections 1.1 – 1.3
You may use a graphing calculator to solve the following problems.
1.
Which of the following gives an equation for the line through (3, - 1)
and parallel to the line: y = - 2 x + 1?
1
7
2
2
1
5
B
y
=
x
( )
2
2
(C) y = - 2 x + 5
(A) y = x +
(D) y = - 2 x - 7
(E) y = - 2 x + 1
Slide 1- 73
Quick Quiz Sections 1.1 – 1.3
You may use a graphing calculator to solve the following problems.
1.
Which of the following gives an equation for the line through (3, - 1)
and parallel to the line: y = - 2 x + 1?
7
1
2
2
5
1
(B) y = x 2
2
(C) y = - 2 x + 5
(A) y = x +
(D) y = - 2 x - 7
(E) y = - 2 x + 1
Slide 1- 74
Quick Quiz Sections 1.1 – 1.3
2.
If f (x )= x 2 + 1 and g (x )= 2 x - 1, which of the
following gives ( f o g )(2)?
(A) 2
(B) 5
(C) 9
(D) 10
(E) 15
Slide 1- 75
Quick Quiz Sections 1.1 – 1.3
2.
If f (x )= x 2 + 1 and g (x )= 2 x - 1, which of the
following gives ( f o g )(2)?
(A) 2
(B) 5
(C) 9
(D) 10
(E) 15
Slide 1- 76
Quick Quiz Sections 1.1 – 1.3
3.
The half-life of a certain radioactive substance is 8 hours. There
are 5 grams present initially. Which of the following gives the
best approximation when there will be 1 gram remaining?
(A) 2
(B) 10
(C) 15
(D) 16
(E) 19
Slide 1- 77
Quick Quiz Sections 1.1 – 1.3
3.
The half-life of a certain radioactive substance is 8 hours. There
are 5 grams present initially. Which of the following gives the
best approximation when there will be 1 gram remaining?
(A) 2
(B) 10
(C) 15
(D) 16
(E) 19
Slide 1- 78
1.4
Parametric Equations




Relations
Circles
Ellipses
Lines and Other Curves
…and why
Parametric equations can be used to obtain
graphs of relations and functions.
Slide 1- 80
Quick Review
In Exercises 1 - 3, write an equation for the line.
1.
the line through the points (1, 8) and (4, 3)
2.
the horizontal line through the point (3, - 4)
3.
the vertical line through the point (2, - 3)
In Exercises 4 - 6, find the x- and y -intercepts of the graph of the relation.
4.
6.
x2
y
+ =1
9 16
2 y2 = x + 1
5.
x2 y 2
=1
16 9
Slide 1- 81
Quick Review
In Exercises 7 and 8, determine whether the given points lie on
the graph of the relation.
7.
2 x2 y + y 2 = 3
(a ) (1, 1)
8.
(b) (- 1, - 1)
æ1
ö
÷
ç
c
,
2
( ) çç
÷
÷
è2
ø
(b) (1, - 3)
(c) (- 1, 3)
9 x 2 - 18 x + 4 y 2 = 27
(a ) (1, 3)
Slide 1- 82
Quick Review
9.
Solve for t.
(a ) 2 x + 3t = - 5
(b) 3 y - 2t = - 1
10. For what values of a is each equation true?
(a )
a2 = a
(b) a 2 = ± a
(c)
4a 2 = 2 a
Slide 1- 83
Quick Review Solutions
In Exercises 1 - 3, write an equation for the line.
1.
the line through the points (1, 8) and (4, 3) y = -
2.
the horizontal line through the point (3, - 4)
3.
the vertical line through the point (2, - 3)
5
29
x+
3
3
y= - 4
x= 2
In Exercises 4 - 6, find the x- and y -intercepts of the graph of the relation.
4.
5.
6.
x2
+
9
x2
16
y2
=1
16
y2
=1
9
2 y2 = x + 1
x = - 3, 3;
y = - 4, 4
x = - 4, 4;
no y -intercepts
x = - 1;
y= -
1
2
,
1
2
Slide 1- 84
Quick Review Solutions
In Exercises 7 and 8, determine whether the given points lie on
the graph of the relation.
7.
2x2 y + y 2 = 3
(a ) (1, 1) Yes
8.
(b) (- 1, - 1)
No
æ1
ö
ç
(c) çç , - 2÷÷÷ Yes
è2
ø
9 x 2 - 18 x + 4 y 2 = 27
(a ) (1, 3) Yes
(b) (1, - 3) Yes
(c) (- 1, 3) No
Slide 1- 85
Quick Review Solutions
9.
Solve for t.
(a ) 2 x + 3t = - 5
t=
- 2x - 5
3
(b) 3 y - 2t = - 1
t=
3y + 1
2
10. For what values of a is each equation true?
(a )
a2 = a
(b) a 2 = ± a
(c)
4a 2 = 2 a
a³ 0
All Reals
All Reals
Slide 1- 86
Relations

A relation is a set of ordered pairs (x, y) of real numbers.

The graph of a relation is the set of points in a plane that
correspond to the ordered pairs of the relation.

If x and y are functions of a third variable t, called a
parameter, then we can use the parametric mode of a grapher
to obtain a graph of the relation.
Slide 1- 87
Parametric Curve, Parametric Equations
If x and y are given as functions
x = f (t ),
y = g (t )
over an interval of t -values, then the set of points (x, y )= ( f (t ), g (t ))
defined by these equations is a parametric curve. The equations are
parametric equations of the curve.
Slide 1- 88
Relations
The variable t is a parameter for the curve and its domain I is the
parameter interval. If I is a closed interval, a £ t £ b, the point
( f (a), g (a)) is the initial point of the curve and the point ( f (b), g (b))
is the terminal point of the curve. When we give parametric equations
and a parameter interval for a curve, we say that we have parametrized
the curve. A grapher can draw a parametrized curve only over a closed
interval, so the portion it draws has endpoints even when the curve being
graphed does not.
Slide 1- 89
Example Relations
Describe the graph of the relation determined by x = t , y = 1- t 2 .
Set x1 = t , y1 = 1- t 2 , and use the parametric mode
of the grapher to draw the graph.
Slide 1- 90
Circles

In applications, t often denotes time, an angle or the distance a
particle has traveled along its path from a starting point.

Parametric graphing can be used to simulate the motion of a
particle.
Slide 1- 91
Example Circles
Describe the graph of the relation determined by
x = 3cos t , y = 3sin t , 0 £ t £ 2p .
Find the initial points, if any, and indicate the direction in which the curve
is traced.
Find a Cartesian equation for a curve that contains the parametrized curve.
x 2 + y 2 = 9 cos 2 t + 9sin 2 t = 9 (cos 2 t + sin 2 t )= 9 (1)= 9
Thus, x 2 + y 2 = 9
This represents the equation of a circle with radius 3 and center at the origin.
As t increases from 0 to 2p the curve is traced in a counter-clockwise direction
beginning at the point (3,0).
x2 + y 2 = 9
Slide 1- 92
Ellipses
Parametrizations of ellipses are similar to parametrizations of circles.
Recall that the standard form of an ellipse centered at (0, 0) is
x2 y 2
+ 2 = 1.
2
a
b
x2 y 2
For x = a cos t and y = a sin t , we have 2 + 2 = 1
a
b
which is the equation of an ellipse with center at the origin.
Slide 1- 93
Lines and Other Curves

Lines, line segments and many other curves can be defined
parametrically.
Slide 1- 94
Example Lines and Other Curves
If the parametrization of a curve is x = t , y = t + 2, 0 £ t £ 2 graph the curve.
Find the initial and terminal points and indicate the direction in which the curve
is traced. Find a Cartesian equation for a curve that contains the parametrized
curve. What portion of the Cartesian equation is traced by the parametrized curve?
x = t , y = t + 2, 0 £ t £ 2
Initial point (0, 2) Terminal point (2, 4)
Curve is traced from left to right
If x = t and y = t + 2, then by substitution y = x + 2.
The graph of the Cartesian equation is a line through (0, 2) with m = 1.
The segment of that line from (0, 2) to (2, 4)is traced by the parametrized curve.
Slide 1- 95
1.5
Functions and Logarithms
Quick Review
In Exercises 1 - 4, let f (x)=
3
x - 1, g (x )= x 2 + 1, and
evaluate the expression.
( f o g )(1)
3. ( f o g )(x)
1.
2. (g o f )(- 7)
4. (g o f )(x )
In Exercises 5 and 6, choose parametric equations and a parameter
interval to represent the function on the interval specified.
1
5. y =
,
x³ 2
6. y = x, x < - 3
x- 1
Slide 1- 97
Quick Review
In Exercises 7 - 10, find the points of intersection of the two curves.
7. y = 2 x - 3, y = 5
8. y = - 3 x + 5, y = - 3
(a ) y = 2 x , y = 3
10. (a ) y = e- x , y = 4
9.
(b) y = 2 x , y = - 1
(b) y = e- x , y = - 1
Slide 1- 98
Quick Review Solutions
In Exercises 1- 4, let f (x)=
3
x - 1, g (x )= x 2 + 1, and
evaluate the expression.
1.
3.
( f o g )(1)
2. (g o f )(- 7) 5
1
( f o g )(x)
x
2
3
4. (g o f )(x )
2
3
(x - 1) + 1
In Exercises 5 and 6, choose parametric equations and a parameter
interval to represent the function on the interval specified.
5.
1
y=
,
x- 1
x = t, y =
x³ 2
1
, t³ 2
t- 1
6.
y = x, x < - 3
x = t, y = t, t < - 3
Slide 1- 99
Quick Review Solutions
In Exercises 7 - 10, find the points of intersection of the two curves.
(4,5)
7.
y = 2 x - 3, y = 5
8.
æ8
y = - 3 x + 5, y = - 3 çç , çè3
(a ) y = 2 x , y = 3
10. (a ) y = e- x , y = 4
9.
ö
3÷
÷
÷
ø
(1.58,3)
(-1.39, 4)
(b) y = 2 x , y = - 1 None
(b) y = e- x , y = - 1 None
Slide 1- 100






One-to-One Functions
Inverses
Finding Inverses
Logarithmic Functions
Properties of Logarithms
Applications
…and why
Logarithmic functions are used in many applications
including finding time in investment problems.
Slide 1- 101
One-to-One Functions




A function is a rule that assigns a single value in its range to
each point in its domain.
Some functions assign the same output to more than one input.
Other functions never output a given value more than once.
If each output value of a function is associated with exactly
one input value, the function is one-to-one.
Slide 1- 102
One-to-One Functions
A function f (x) is one - to - one on a domain D if f (a)¹ f (b)
whenever a ¹ b.
Slide 1- 103
One-to-One Functions
The horizontal line test states that the graph of a one-to-one function
y = f (x ) can intersect any horizontal line at most once.
If it intersects such a line more than once it assumes the same y -value
more than once and is not a one-to-one function.
Slide 1- 104
Inverses



Since each output of a one-to-one function comes from just
one input, a one-to-one function can be reversed to send
outputs back to the inputs from which they came.
The function defined by reversing a one-to-one function f is
the inverse of f.
Composing a function with its inverse in either order sends
each output back to the input from which it came.
Slide 1- 105
Inverses
The symbol for the inverse of f is f - 1 , read "f inverse."
The -1 in f
- 1
is not an exponent; f
- 1
1
(x) does not mean
f (x)
If ( f o g )(x )= (g o f )(x ), then f and g are inverses of one another;
otherwise they are not.
Slide 1- 106
Identity Function
The result of composing a function and its inverse in either order
is the identity function.
Slide 1- 107
Example Inverses
Determine via composition if f (x)=
x and g (x)= x 2 , x ³ 0
are inverses.
( f o g )(x)= f (x 2 )= x 2 = x = x
(g o f )(x)= g ( x ) =
2
( x) = x
Slide 1- 108
Writing f -1as a Function of x.
Solve the equation y = f (x ) for x in terms of y.
Interchange x and y. The resulting formula
will be y = f - 1 (x ).
Slide 1- 109
Finding Inverses
Slide 1- 110
Example Finding Inverses
Given that y = 4 x - 12 is one-to-one, find its inverse.
Graph the function and its inverse.
Solve the equation for x in terms of y.
1
x= y + 3
4
Interchange x and y.
1
y= x+ 3
4
f (x)= 4 x - 12
f - 1 (x ) =
1
x+ 3
4
Notice the symmetry about the line y = x
[-10,10] by [-15, 8]
Slide 1- 111
Base a Logarithmic Function
The base a logarithm function y = log a x is the inverse of
the base a exponential function y = a x (a  0, a ¹ 1).
The domain of log a x is (0, ¥ ), the range of a x .
The range of log a x is (- ¥ , ¥ ), the domain of a x .
Slide 1- 112
Logarithmic Functions

Logarithms with base e and base 10 are so important in applications that
calculators have special keys for them.

They also have their own special notations and names.
y = loge x = ln x is called the natural logarithm function.
y = log x = log x is often called the common logarithm function.
10
Slide 1- 113
Inverse Properties for ax and loga x
Base a : a loga x = x, a  1,
x> 0
Base e : eln x = x, ln e x = x, x  0
Slide 1- 114
Properties of Logarithms
For any real numbers x  0 and y  0,
Product Rule : log a xy = log a x + log a y
Quotient Rule : log a
x
= log a x - log a y
y
y
Power Rule : log a x = y log a x
Slide 1- 115
Example Properties of Logarithms
Solve the following for x.
2 x = 12
2 x = 12
ln 2 x = ln12 Take logarithms of both sides
x ln 2 = ln12 PowerRule
ln12 2.302585
x=
=
» 3.32193
ln 2 .693147
Slide 1- 116
Example Properties of Logarithms
Solve the following for x.
e x + 5 = 60
e x + 5 = 60
e x = 55
Subtract 5
ln e x = ln 55 Take logarithm of both sides
x = ln 55 » 4.007333
Slide 1- 117
Change of Base Formula
log a x =
ln x
ln a
This formula allows us to evaluate log a x for any
base a  0, a ¹ 1, and to obtain its graph using the
natural logarithm function on our grapher.
Slide 1- 118
Example Population Growth
The population P of a city is given by P = 105,300e0.015t
where t = 0 represents 1990. According to this model,
when will the population reach 150,000?
P = 105,300e0.015t ,
P = 150, 000
150, 000 = 105,300e0.015t
150,000
= e0.015t
105,300
æ150, 000 ÷
ö
0.015t
ç
ln ç
=
ln
e
÷
çè105,300 ÷
ø
Solve for t
Take logarithm of both sides
0.353822 = 0.015 t
Inverse property
0.353822
t=
» 23.588133 years t = 0 is 1990, so
0.015
the population will reach 150,000 in the year 2013.
Slide 1- 119
1.6
Trigonometric Functions
Quick Review
In Exercises 1 - 4, convert from radians to degrees or
1.
3.
p
3
- 40°
2.
- 2.5
4.
45°
In Exercises 5 - 7, solve the equation graphically in the
given interval.
5.
sin x = 0.6,
0 £ x < 2p
6.
cos x = - 0.4,
0 £ x < 2p
7.
tan x = 1,
-
p
3p
£ x<
2
2
Slide 1- 121
Quick Review
8.
Show that f (x)= 2 x 2 - 3 is an even function. Explain why
its graph is symmetric about the y -axis.
9.
Show that f (x)= x 3 - 3 x is an odd function. Explain why
its graph is symmetric about the origin.
10. Give one way to restrict the domain of the function f (x )= x 4 - 2
to make the resulting function one-to-one.
Slide 1- 122
Quick Review Solutions
In Exercises 1 - 4, convert from radians to degrees or
1.
p
3
60°
2.
- 2.5
- 143.24°
2p
p
4. 45°
9
4
In Exercises 5 - 7, solve the equation graphically in the
given interval.
3.
- 40°
5.
sin x = 0.6,
0 £ x < 2p
x » 0.6435, 2.4981
6.
cos x = - 0.4,
0 £ x < 2p
x » 1.9823, 4.3009
7.
tan x = 1,
-
-
p
3p
£ x<
2
2
x » 0.7854, 3.9270
Slide 1- 123
Quick Review Solutions
8.
Show that f (x)= 2 x 2 - 3 is an even function. explain why
its graph is symmetric about the y -axis.
2
f (- x)= 2(- x ) - 3= 2 x 2 - 3= f (x )
The graph is symmetric about the y -axis because if a point
(a, b) is on the graph, then so is the point (- a, b).
Slide 1- 124
Quick Review Solutions
9.
Show that f (x)= x 3 - 3 x is an odd function. Explain why
its graph is symmetric about the origin.
3
f (- x )= (- x ) - 3(- x )= - x 3 + 3 x = - f (x )
The graph is symmetric about the origin because if a point
(a, b)is on the graph, then so is the point (- a, - b).
10. Give one way to restrict the domain of the function f (x )= x 4 - 2
to make the resulting function one-to-one.
x³ 0
Slide 1- 125






Graphs of Trigonometric Functions
Periodicity
Even and Odd Trigonometric Functions
Transformations of Trigonometric Graphs
Inverse Trigonometric Functions
…and why
Trigonometric functions can be used to model periodic
behavior and applications such as musical notes.
Slide 1- 126

The radian measure of the angle ACB at the center of the unit
circle equals the length of the arc that ACB cuts from the unit
circle.
Slide 1- 127

An angle of measure θ is placed in standard position at the
center of circle of radius r,
Slide 1- 128
Trigonometric Functions of θ
The six basic trigonometric functions of q are
defined as follows:
y
r
x
cosine: cos q =
r
y
tangent: tan q =
x
sine: sin q =
r
y
r
secant: sec q =
x
x
cotangent: cot q =
y
cosecant: csc q =
Slide 1- 129
Graphs of Trigonometric Functions

When we graph trigonometric functions in the coordinate plane, we usually
Slide 1- 130
Angle Convention
From now on in this book, it is assumed that all angles are measured in radians
unless degrees or some other unit is stated explicitly. When we talk about the angle
p
p
p
we mean radians ( which is 60°), not degrees.
3
3
3
Slide 1- 131
Periodic Function, Period
A function f (x ) is periodic if there is a positive number p such
that f (x + p )= f (x ) for every value of x. The smallest value
of p is the period of f .
The functions cos x, sin x, sec x and csc x are periodic with
period 2p . The functions tan x and cot x are periodic with
period p .
Slide 1- 132
Even and Odd Trigonometric Functions

The graphs of cos x and sec x are even functions because their
graphs are symmetric about the y-axis.

The graphs of sin x, csc x, tan x and cot x are odd functions.
y = cos x
y = sin x
Slide 1- 133
Example Even and Odd Trigonometric
Functions
Show that csc x is an odd function.
1
1
csc(- x)=
=
 - csc x
sin (- x) - sin x
Slide 1- 134
Transformations of Trigonometric Graphs

The rules for shifting, stretching, shrinking and reflecting the
graph of a function apply to the trigonometric functions.
Vertical stretch or shrink
Vertical shift
y = a f (b (x + c ))+ d
Horizontal stretch or shrink
Horizontal shift
Slide 1- 135
Example Transformations of
Trigonometric Graphs
Determine the period, domain, range and draw the graph of
y = - 2sin (4 x + p )
æ æ p ö÷
ö
ç
÷
We can rewrite the function as y = - 2sin ç4 ççx + ÷
÷÷
÷
çè çè
4ø
ø
2p
. In our example b = 4,
b
2p p
so the period is
= . The domain is (- ¥ , ¥ ).
4 2
The graph is a basic sin x curve with an amplitude of 2.
Thus, the range is [ - 2, 2].
The period of y = a sin bx is
The graph of the function is shown together
with the graph
of the sin x function.
[-5, 5] by [-4,4]
Slide 1- 136
Inverse Trigonometric Functions


None of the six basic trigonometric functions graphed in
Figure 1.42 is one-to-one. These functions do not have
inverses. However, in each case, the domain can be restricted
to produce a new function that does have an inverse.
The domains and ranges of the inverse trigonometric functions
become part of their definitions.
Slide 1- 137
Inverse Trigonometric Functions
Function
Domain
y = cos- 1 x
- 1£ x£ 1
y = sin- 1 x
- 1£ x£ 1
y = tan - 1 x
- ¥ < x< ¥
y = sec- 1 x
x³ 1
y = csc- 1 x
x³ 1
y = cot - 1 x
- ¥ < x< ¥
Range
0£ y£ p
p
p
- £ y£
2
2
p
p
< y<
2
2
0 £ y £ p, y ¹
-
p
2
p
p
£ y£ ,y¹ 0
2
2
0< y< p
Slide 1- 138
Inverse Trigonometric Functions

The graphs of the six inverse trigonometric functions are shown here.
Slide 1- 139
Example Inverse Trigonometric
Functions
æ 1÷
ö
ç
Find the measure of sin ç- ÷
çè 2 ÷
ø
- 1
æ 1÷
ö
ç
Put the calculator in degree mode and enter sin ç- ÷
.
çè 2 ÷
ø
The calculator returns - 30°.
ö
- 1æ 1÷
ç
Put the calculator in radian mode and enter sin ç- ÷
.
çè 2 ÷
ø
- 1
The calculator returns - .52359877556 radians.
p
This is the same as radians.
6
Slide 1- 140
Quick Quiz Sections 1.4 – 1.6
You should solve the following problems without using a graphing caluclator.
1. Which of the following is the domain of f ( x)   log  x  3 ?
2
(A)  ,  
(B)  ,3
(C)  3,  
(D) [3, )
(E) (  ,3]
Slide 1- 141
Quick Quiz Sections 1.4 – 1.6
You should solve the following problems without using a graphing caluclator.
1. Which of the following is the domain of f ( x)   log  x  3 ?
2
(A)  ,  
(B)  ,3
(C)  3,  
(D) [3, )
(E) (  ,3]
Slide 1- 142
Quick Quiz Sections 1.4 – 1.6
2. Which of the following is the range of f ( x)  5cos  x     3?
(A)  ,  
(B)  2,4
(C)  8, 2
(D)  2,8
 2 8
(E)   , 
 5 5
Slide 1- 143
Quick Quiz Sections 1.4 – 1.6
2. Which of the following is the range of f ( x)  5cos  x     3?
(A)  ,  
(B)  2,4
(C)  8, 2
(D)  2,8
 2 8
(E)   , 
 5 5
Slide 1- 144
Quick Quiz Sections 1.4 – 1.6
3. Which of the following gives the solution of tan x  -1 in   x 
(A) 
(B)
(C)


3
?
2
4
4

3
3
(D)
4
5
(E)
4
Slide 1- 145
Quick Quiz Sections 1.4 – 1.6
3. Which of the following gives the solution of tan x  -1 in   x 
(A) 
(B)
(C)


3
?
2
4
4

3
3
(D)
4
5
(E)
4
Slide 1- 146
Chapter Test
In Exercises 1 and 2, write an equation for the specified line.
1.
through (4, - 12) and parallel to 4 x + 3 y = 12
2.
the line y = f (x ) where f has the following values:
-2 2
x
f(x) 4 2
4
1
1
5
3. Determine whether the graph of the function y = x is
symmetric about the y -axis, the origin or neither.
4.
x4 + 1
Determine whether the function y = 3
is even, odd or neither.
x - 2x
Slide 1- 147
Chapter Test
In Exercises 5 and 6, find the (a ) domain and (b) range, and
(c)graph the function.
5. y = 2sin (3x + p )- 1
6. y = ln (x - 3)+ 1
Slide 1- 148
Chapter Test
7.
Write a piecewise formula for the function.
1
1
2
Slide 1- 149
Chapter Test
8.
x = 5cos t ,
y = 2sin t ,
for a curve.
0 £ t £ 2p
is a parametrization
(a ) Graph the curve.
Identify the initial and terminal points, if any.
Indicate the direction in which the curve is traced.
(b) Find a Cartesian equation for a curve that contains
the parametrized curve.
What portion of the graph of the Cartesian equation
is traced by the parametrized curve?
Slide 1- 150
Chapter Test
9.
Give one parametrization for the line segment with
endpoints (- 2, 5) and (4,3).
10. Given f (x)= 2 - 3x,
(a ) find f - 1 and show that ( f ○ f - 1 )(x)= ( f - 1 ○ f )(x )
(b) graph f and f - 1 in the same viewing window
Slide 1- 151
Chapter Test Solutions
In Exercises 1 and 2, write an equation for the specified line.
4
20
1. through (4, - 12) and parallel to 4 x + 3 y = 12
y = - x3
3
2. the line y = f (x ) where f has the following values:
x
-2 2
f(x) 4
3.
2
4
1
x+ 3
2
1
1
5
Determine whether the graph of the function y = x is
symmetric about the y -axis, the origin or neither.
4.
y= -
Origin
x4 + 1
Determine whether the function y = 3
is even, odd
x - 2x
or neither.
Odd
Slide 1- 152
Chapter Test Solutions
In Exercises 5 and 6, find the (a ) domain and (b) range, and
(c)graph the function.
5. y = 2sin (3x + p )- 1
(a ) All Reals (b) [- 3, 1]
[-π, π] by [-5, 5]
6.
y = ln (x - 3)+ 1
(a )(3, ¥ )
(b) All Reals
[- 2, 10] by [- 2, 5]
Slide 1- 153
Chapter Test Solutions
7.
Write a piecewise formula for the function.
1
1
2
ìïï 1- x, 0 £ x < 1
f (x)= í
ïïî 2 - x, 1£ x £ 2
Slide 1- 154
Chapter Test Solutions
8.
x = 5cos t ,
y = 2sin t ,
for a curve.
0 £ t £ 2p
is a parametrization
(a ) Graph the curve.
Identify the initial and terminal points, if any.
Indicate the direction in which the curve is traced.
Initial Point (5, 0)
Terminal Point (5, 0)
Slide 1- 155
Chapter Test Solutions
8.
(b) Find a Cartesian equation for a curve that contains
2
2
æx ÷
ö æy ö÷
ç
the parametrized curve. ç ÷
+ çç ÷
=1
çè 5 ÷
ø èç 2 ø÷
What portion of the graph of the Cartesian equation
is traced by the parametrized curve? All
Slide 1- 156
Chapter Test Solutions
9.
Give one parametrization for the line segment with
endpoints (- 2, 5) and (4,3). (A possible answer)
x = - 2 + 6t ,
y = 5 - 2t ,
0 £ t£ 1
10. Given f (x )= 2 - 3 x,
(a ) find f - 1 and show that ( f ○ f - 1 )(x)= ( f - 1 ○ f )(x )
2- x
3
æ2 - x ö
æ2 - x ö
- 1
÷
÷
ç
f ( f (x ))= f ç
= 2 - 3çç
÷= 2 - (2 - x)= x
÷
÷
çè 3 ÷
ç
ø
è 3 ø
(a ) f - 1 (x)=
f
- 1
( f (x)) =
f
- 1
2 - (2 - 3 x) 3 x
=
=x
(2 - 3x)=
3
3