MAC 1140
Distance Learning
Computer Mediated Learning
Pre-calculus Algebra
Course Topics Supplemental
Dr. Lori Holdren
Revised Edition, May 2006
Copyright © 2003
MCC. All rights reserved.
TABLE OF CONTENTS
TOPIC
Page #
Graphs & Transformations of Basic Functions…………………………………………………
3
Graphs of Polynomial Functions…………………………………………………………………
5
Graphing Calculator Usage with Polynomial & Rational Functions………………………….
6
Practice Problems on Polynomial Functions……………………………………………………
8
Solutions to Practice Problems on Polynomial Functions…………………………………….
8
Review for Exam #1……………………………………………………………………………….
9
Solutions to Review for Exam #1…………………………………………………………………
10
Polynomial & Rational Inequalities ………………………………………………………………
12
Polynomial & Rational Inequality Worksheet …………………………………………………..
17
Solutions to Polynomial & Rational Inequality Worksheet …………………………………….
18
Applications of Exponential & Logarithmic Functions………………………………………….
21
Applications of Exponential & Logarithmic Functions Practice Problems……………………
25
Solutions to Applications of Exponential & Logarithmic Functions Practice Problems…….
26
Matrices – Calculator Usage……………………………………………………………………..
28
Review for Exam #2 ………..…………………………………………………………………….
29
Solutions to the Review for Exam #2 ……….………………………………………………….
32
Matrix Applications…………………………………………………………………………………
38
Matrix Application Worksheet…………………………………………………………………….
39
Solutions to Matrix Application Worksheet………………………………………………………
40
Systems of Non-Linear Inequalities………………………………………………………………
41
Practice Problems for Non-Linear Systems……………………………………………………..
41
Solutions to Practice Problems for Non-Linear Systems………………………………………
42
Review for Exam #3 ……………………………………………………………………………….
43
Solutions to Review for Exam #3 …………………………………………………………………
45
Review for the MAC 1140 Final Exam .…………………………………………………………
50
Solutions to Review for the MAC 1140 Final Exam …………………………………………….. 55
2
Graphs and Transformations of Basic Functions
You are responsible for “knowing” the following graphs and their properties. “Knowing” means
that you can make a table of values for each, graph each of them quickly, and can recognize
their basic shapes.
y
1.
Identity Function: f(x) = x
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
2.
Cubic Function: f ( x ) = x 3
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
4.
(– ∞ , ∞ )
[0, ∞ )
(0, 0)
(0, 0)
Even Function
(– ∞ , ∞ )
(– ∞ , ∞ )
(0, 0)
(0, 0)
Odd Function
Square Root Function: f ( x ) = x
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
[0, ∞ )
[0, ∞ )
(0, 0)
(0, 0)
none
y=x
-2
-1
0
2
x
y
Quadratic Function: f ( x ) = x 2
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
3.
(– ∞ , ∞ )
(– ∞ , ∞ )
(0, 0)
(0, 0)
Odd Function
x
-2
-1
0
2
x
-2
-1
0
1
2
y = x2
4
1
0
1
4
x
y
x
-2
-1
0
1
2
y = x3
-8
-1
0
1
8
x
y
x
-1
0
1
4
9
y= x
Not
Real
0
1
2
3
x
3
y
5.
Cubed Root Function: f ( x ) = x
3
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
(– ∞ , ∞ )
(– ∞ , ∞ )
(0, 0)
(0, 0)
Odd Function
x
-8
-1
0
1
8
y= 3 x
-2
-1
0
1
2
x
y
6.
Absolute Value Function: f ( x ) = x
Properties
Domain:
Range:
x-intercept:
y-intercept:
Symmetry:
x
(– ∞ , ∞ )
[0, ∞ )
(0, 0)
(0, 0)
Even Function
Y= x
-2
-1
0
1
2
2
1
0
1
2
x
Summary of Transformation Rules
Given any basic function y = f(x):
Transformation
Behavior of Graph
y = f(x) + d
y = f(x) – d
y = f(x + c)
y = f(x – c)
y = af(x)
graph of f(x) moves up d units
graph of f(x) moves down d units
graph of f(x) moves left c units
graph of f(x) moves right c units
graph of f(x) stretches vertically if a > 1, shrinks vertically if a < 1, flips
y = f(bx)
vertically if a < 0
graph of f(x) stretches horizontally if b < 1, shrinks horizontally if b > 1,
flips horizontally if b < 0
Order of Operations for Transformations:
1. Horizontal translation (right or left)
2. Stretching or Shrinking (horizontal and/or vertical)
3. Reflecting (horizontal and/or vertical)
4. Vertical translation (up or down)
4
Graphs of Polynomial Functions
To graph a polynomial function, you should determine the following:
1) End behavior
2) Any x-intercepts (also called real zeros) and y-intercepts.
3) A table of values including the intercepts and points between all x-intercepts. Use
your calculator to help you with this and the graph.
Example 1
Graph f ( x ) = x 3 + 2x 2 − x − 2
9 End behavior indicates that it should fall to the left and rise to the right
9 X-intercepts are found by solving: x 3 + 2x 2 − x − 2 = 0
x 2 ( x + 2) − ( x + 2) = 0
( x + 2)( x 2 − 1) = 0 ⇒ x = −2, − 1, 1
x-int: (–2, 0), (–1, 0), (1, 0)
9 The y-intercept is found by computing f(0) = –2 so we have (0, –2)
9 The table of values is as follows (use your calculator to get values)
x
y
–2
0
-1.5
4.375
9 The graph would be:
-1
0
0
–2
1
0
2
12
y
x
Example 2
Graph f ( x ) = −2x 4 + 4 x 2
9 End behavior indicates that it should fall to the left and rise to the right
9 X-intercepts are found by solving: − 2x 4 + 4 x 2 = 0
− 2 x 2 ( x 2 − 2) = 0
− 2x 2 = 0 or x 2 − 2 = 0 ⇒ x = 0, ± 2
(
x − int s : (0, 0), ± 2, 0
9 Compute f(0) = 0 and get a y-int of (0, 0)
)
5
9 The table of values is as follows (use your calculator to get values)
x
y
– 2
0
-1
2
0
0
1
2
2
0
9 The graph would be:
y
x
Graphing Calculator Usage with Polynomial & Rational Functions
You are responsible to be able to do the following with your graphing calculator. You should
always use your calculator while you are working through the computer lessons.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Graph any function
Look at the table of values for any function
Adjust your window to be able to view a graph in its entirety
Use your calculator’s graph and table to assist you in hand drawing a graph
Find zeros (x-intercepts) of functions
Find intersections of functions
Find max and min values of a function
Find function values
Look at a graphed function to determine domain & range and where a function is
increasing, decreasing, constant
The following are some examples of calculator-specific problems that may encompass
some of the above:
Example 1
a) Graph f ( x ) = x 3 + 13 x 2 + 10 x − 4 in the standard window
Answer: Your graph should appear as:
b) Based on your knowledge of end behavior, are you looking at the entire graph?
6
Answer: No, the graph should fall to the left and it does not in the standard window.
c) Adjust the window until you can see the graph in its entirety.
Answer: One window that will work is x: [-20, 20], Xscl = 2 and y: [-100, 250], Yscl = 25
NOTE: To return the calculator back to its original window, hit ZOOM 6: ZStandard
Example 2
Use your calculator to find all real solutions to three decimal places: 6 x 3 − 19 x 2 + 16 x − 4 = 0
Solution: The graph should appear as follows.
It is difficult to see how many times the graph crosses the x-axis. So, you will need to enlarge
the area where the graph looks like it may cross. One way to do this is by using the function
ZOOM: Box. For the T-I 83, the sequence would be as follows:
1: Select ZOOM 1: Zbox
2. Move the cursor with the arrow keys to a place to start drawing a box and hit
ENTER. The use the cursor keys to draw a small box surrounding the x-axis
near where you see the potential zeros.
3. Hit ENTER again when the box is complete.
The resulting graph should be similar to the following and show you that the graph crosses 3
times. You will need to run the Zero program found under 2nd TRACE (CALC) menu option 2.
1. Select CALC 2: zero
2. Move the cursor to the left of where the graph crosses the x-axis for the first
point and hit ENTER.
3. Move the cursor to the right of where the graph crosses the x-axis for the first
point and hit ENTER twice. (the second time is for the GUESS)
4. Read off the x-value given by the calculator
Answer: x ≈ 0.5, 0.667, 2.0
7
Practice Problems on Polynomial Functions
1. Graph f ( x ) = x 4 − 6 x 3 + 9 x 2
2. Graph f ( x ) = x 3 − 3 x + 2
3. Graph f ( x ) = − x 3 + 4 x
4. Graph f ( x ) = − x 4 + 4 x 3 − 4 x 2
5. Graph f ( x ) = x 3 + 3 x 2 − x − 3
6. Graph f(x) = − ( x − 9)4 + 8( x − 9) + 12 on your calculator in the standard window.
a) Based on your knowledge of end behavior, are you looking at the entire graph?
b) Adjust the window until you can see the graph in its entirety.
7. Use your calculator to solve to 3 decimal places: − ( x − 9)4 + 8( x − 9) + 12 = 0
8. Use your calculator to find all real zeros to 3 decimal places: g( x ) = − x 3 + 12 x 2 − 12 x + 8
9. Determine the intervals over which f ( x ) = x − 2 + x + 2 is increasing, decreasing or constant
Solutions to Practice Problems on Polynomial Functions
y
x
x
0
1
1.5
1
3
y
y
0
4
5.1
4
0
x
x
0
1
1.5
1
3
y
0
4
5.1
4
0
x
-1
0
1
2
3
y
9
0
-1
0
-9
2.
1.
y
y
x
x
-2
-1
0
1
2
y
0
-3
0
3
0
3.
5.
x
4.
y
x
x
-3
-2
-1
0
1
y
0
3
0
-3
0
6. a) No
b) One window that works is x:
[-10, 15] and y: [-20,20]
7. Run the zero program for each zero: x ≈ 7.778, 11.357
8. First, find a window to display the entire graph, such as x: [-10, 15] and y: [-100, 200] with
Xscl of 1 and Yscl of 10. Then note that there is only one zero: x ≈ 10.973
9. Graph and follow along graph from left to right to determine:
Increasing: (2, ∞)
Decreasing: (– ∞, 2)
Constant: (2, 2)
8
REVIEW FOR EXAM #1
This review is intended to aid you in studying for the exam. This should not be the only thing that
you do to prepare. Be sure to also look over your notes from the computer lessons, book, your
homework, etc.
3x
1. Determine the domain: a) f(x) = 4 − x 2
b) h(x) = 2x + 5 c) p( x ) = 2
x + 5x + 6
f (a + h) − f ( a )
h
2.
Given f(x) = x 2 − 3 , compute f(3), f(– 2) and
3.
Graph a) f(x) = x − 4 + 2
4.
Given f(x) = 2 – x2 and g(x) = – 3x + 5, find: a) f – g
5.
Graph f(x) = x 3 − 16 x 2 − 8 x + 24 on your calculator in the standard viewing window. Use
your knowledge of end behavior to explain why you are not seeing the entire graph.
6.
Find the intercepts & sketch the graph: p( x ) = −2x 4 − x 3 + 3 x 2
7.
Find the intercepts & sketch the graph: f ( x ) = x 3 + x 2 − 4 x − 4
8.
Use your knowledge of end behavior and zeros (x-ints) to determine which equation each
of the graphs represents. a. y = -(x + 1)2(x – 1) b. y = -(x + 1)2(x – 1)2 c. y = (x + 1)(x –
1)2
y
⎧1
c) f ( x ) = ⎨
⎩ x
b) y = − x 3
1
9.
10.
if x ≥ 0
b) g o f
y
x
if x < 0
c) f-1(x), when x ≥ 0
y
x
x
2
3
Determine all asymptotes for the following rational functions:
x −9
x3 + 4
4 − x2
a) f ( x ) = 2
c) h( x ) = 2
b) g( x ) = 2
x − 2x − 8
x + 2x − 3
x +x−6
a) How many vertical asymptotes does 9a) have? Why?
b) Does the graph of 9b) cross its horizontal asymptote?
11.
Graph 9a, 9b, and 9c
9
SOLUTIONS TO REVIEW FOR EXAM #1
1.
a) f ( x ) = 4 − x 2 ⇒ there are no restrictions, so D: (–∞, ∞)
b) h( x ) = 2x + 5 ⇒ 2x + 5 ≥ 0 ⇒ x ≥ −
c) p( x ) =
2.
3x
x + 5x + 6
2
f (3 ) = 3 2 − 3 = 6
5
⎡ 5
⇒ D : ⎢− , ∞ )
2
⎣ 2
⇒ x 2 + 5 x + 6 = 0 ⇒ ( x + 3)( x + 2) = 0 ⇒ D : x ≠ −3,−2
f ( −2) = ( −2)2 − 3 = 1
(
) (
)
f (a + h) − f (a) (a + h)2 − 3 − a2 − 3 a2 + 2ah + h2 − 3 − a2 + 3 2ah + h2
=
=
=
= 2a + h
h
h
h
h
y
y
y
x
x
3.
a)
b)
x
y
4.
2
4
4
2
c)
x
y
6
4
-2
8
( )
b) g o f = g(2 − x ) = −3(2 − x ) + 5 = −6 + 3 x
-1
1
0
0
1
-1
x
-2
0
y=1
0
0
x
0
1
4
0
1
2
y= x
2
-8
a) f – g = 2 − x 2 − (− 3 x + 5 ) = 2 − x 2 + 3 x − 5 = − x 2 + 3 x − 3
2
2
c) x = 2 − y 2 ⇒ y 2 = 2 − x ⇒ y = 2 − x ⇒
5.
6.
x
2
+5 =
9
3
3x 2 − 1
f −1( x ) = 2 − x
The graph should fall to left and rise to the right, which isn’t
seen in the calculator graph.
(
)
3
p( x ) = −2x 4 − x 3 + 3 x 2 = 0 ⇒ − x 2 2x 2 + x − 3 = − x 2 (2x + 3)( x − 1) = 0 ⇒ Zeros : x = 0, − , 1
2
x
y
–2
-12
–1.5
0
–1
2
0
0
0.5
0.5
1
0
2
–28
y
x
10
7.
(
)
f ( x ) = x 3 + x 2 − 4 x − 4 = 0 ⇒ x 2 ( x + 1) − 4( x + 1) = x 2 − 4 ( x + 1) = ( x + 2)( x − 2)( x + 1) = 0
So, x-ints are: (–2, 0), (2, 0) and (–1, 0)
y
x
y
x
-3
-10
-2
0
-1.5
0.875
-1
0
1
-6
2
0
8.
a. 3 because graph is cubic with leading coefficient as negative (rises to left, falls to right)
b. 1 because graph is 4th degree with leading coefficient as negative (falls to left and right)
c. 2 because graph is cubic with leading coefficient as positive (falls to left, rises to right)
9.
a) v.a.:
(2 − x)(2 + x) 2 − x
=
⇒ only get x – 4 = 0 or x = 4 as v.a.
( x + 2)( x − 4) x − 4
h.a.: deg(num)=deg(den) ⇒ take ratio of leading coefficients: y =
−1
⇒ y = −1
1
b) v.a.: x 2 + 2x − 3 = 0 ⇒ ( x − 1)( x + 3) = 0 ⇒ x = 1; x = −3
h.a.: deg(num)< deg(den) ⇒ get x-axis as h.a. y = 0
c) v.a.: x 2 + x − 6 = 0 ⇒ ( x − 2)( x + 3) = 0 ⇒ x = 2; x = −3
h.a.: deg(num)>deg(den) so no h.a.
o.a.: deg(num)=deg(den)+1, so use long division of den by num to get y = x – 1
10. a) only once, since expression simplifies
b) yes, because g( x) = 0 ⇒ x − 9 = 0 or crosses at x = 9.
y
y
x
y
-2 Undef - hole
2
0
3
1
5
-3
6
-2
11. a)
x
y
c)
x
x
-8
-5
-4
-2
0
1
3
4
5
b)
x
x
-5
4
-2
-1
0
2
3
y
-1.2
-2.6
3.7
2.5
3
-1.4
-0.5
y
-10.2
-8.7
-10
1
-0.7
-1.3
5.2
4.8
5.4
11
Polynomial & Rational Inequalities
Polynomial Inequalities
A polynomial inequality will have one of the following forms (or can be put into one of these
forms):
f(x) > 0
f(x) ≥ 0
f(x) < 0
f(x) ≤ 0
Examples: x2 – 16 > 0
x2 + 1 ≤ 4x
6x3 – 7x2 – 5x >0
The solution to a polynomial inequality will be the set of values that satisfy it, which will consist
of one or more intervals. For f(x) > 0 and f(x) ≥ 0 we will want to find the values where f(x) is
positive, or where the graph of f(x) is above the x-axis. For f(x) < 0 and f(x) ≤ 0 we will want to
find the values where f(x) is negative, or where the graph of f(x) is below the x-axis.
To Solve a Polynomial Equation:
1) Get the polynomial on the left side of the inequality in descending order and zero on the
right side of the inequality.
2) Find the critical values. These are the numbers that satisfy f(x) = 0. These are
important as the f(x) will be either positive or negative in the intervals around and
between these values.
3) Set up a number line using the critical values and divide it into intervals based on the
critical values.
4) Determine the sign (+ or –) of the polynomial in each interval. This can be done by
using a test point from inside each interval or by looking at the graph of f(x) on your
calculator to see where it is above (+) or below (–) the x-axis.
5) State answer based on inequality given.
Example 1
Solve: x2 – 16 > 0
Critical values are where: x2 – 16 = 0 or where x2 = 16, so x = –4, 4
–4
4
In looking at the graph from your calculator Ymin= –18, Ymax = 10, Yscl=2, we see that the
graph is positive on (–∞,–4) and (4, ∞) and negative on (–4, 4). Note: since the inequality
symbol is >, we use ( ) for all interval endpoints.
+
–
–4
+
4
12
Since x2 – 16 > 0 we want the intervals where it is positive! So, the solution is:
(–∞,–4) ∪ (4, ∞)
If you do not want to use the graphical method, you must test a point in each of the
intervals. You cannot pick the endpoints. The test point will represent all points for that
interval and alert you to the functions sign. For example, we could do:
Test x = –5 for (–∞,–4): (–5)2 – 16 = 9 > 0*
Test x = 0 for (–4, 4): (0)2 – 16 = – 16 < 0
Test x = 5 for (4, ∞): (5)2 – 16 = 9 > 0*
Again, we find the intervals with positive values on (–∞,–4) ∪ (4, ∞).
Example 2
Solve: x2 + 1 ≤ 4x
First, get zero on the right side: x2 – 4x + 1 ≤ 0
Critical values are where: x2 – 4x + 1 = 0
Need to use the Quadratic Formula since x2 –4x + 1 is not factorable:
x=
− ( −4) ± ( −4) 2 − 4(1)(1) 4 ± 12 4 ± 2 3
=
=
= 2± 3
2(1)
2
2
Critical values are x = 2 − 3, 2 + 3
2− 3
2+ 3
In looking at the graph from your calculator in the standard window, we see that the graph
is negative on 2 − 3, 2 + 3 and positive on − ∞, 2 − 3 and on 2 + 3, ∞
[
]
]
(
+
2− 3
–
[
)
+
2+ 3
Since x2 – 4x + 1 ≤ 0 we want the intervals where it is negative! So, the solution is:
[2 −
3, 2 + 3
]
Note: [ ] are used since the inequality symbol is ≤ !
If you want to do the test point method, you can pick points inside each interval and
substitute them into the inequality to see the signs of each interval.
13
Example 3
Solve: 6x3 – 7x2 – 5x >0
Critical values are where: 6x3 – 7x2 – 5x = 0 This can be easily factored and solved:
1 5
x(2x +1)(3x-5) = 0, so x = 0, − ,
2 3
−
1
2
5
3
0
In looking at the calculator graph from your calculator with Xmin= –5, Xmax = 5, Xscl=1 and
⎛ 1 ⎞
⎛5 ⎞
Ymin= –5, Ymax = 5, Yscl=1, we see that the graph is positive on ⎜ − , 0 ⎟ and on ⎜ , ∞ ⎟
⎝ 2 ⎠
⎝3 ⎠
1⎞
⎛
⎛ 5⎞
–
+
–
+
and negative on ⎜ − ∞, − ⎟ and on ⎜ 0, ⎟ .
2⎠
⎝
⎝ 3⎠
−
1
2
0
5
3
Since 6x3 – 7x2 – 5x >0, we want intervals that are positive! So the solution is:
⎛ 1 ⎞
⎜ − , 0⎟ ∪
⎝ 2 ⎠
⎛5 ⎞
⎜ , ∞⎟
⎝3 ⎠
Again, if you prefer, you can test a point from each interval into the polynomial expression
to determine the sign for each interval.
Rational Inequalities
A rational inequality has one of the following forms (or can be put into one of these forms):
f (x)
>0
g( x )
Examples:
f (x)
<0
g( x )
( x + 3)(2 − x )
( x − 1)
2
f (x)
≥0
g( x )
<0
f (x)
≤0
g( x )
x+2
≥1
x−4
2x + 5 x + 1
≤
x +1 x −1
14
To Solve a Rational Equation:
1) Get the rational inequality into one of the above forms with zero on the right.
2) Find the critical values. For rational inequalities, the critical values occur where the
numerator is equal to zero and where the denominator is equal to zero. We have to
remember, though, that if there is ≤ or ≥ as the symbol that the critical number from the
denominator cannot be used as an interval endpoint as it is not in the domain.
3) Set up a number line using the critical values and divide it into intervals based on the
critical values.
4) Determine the sign (+ or –) of the rational expression in each interval. This can be
done by using a test point from inside each interval or by looking at the graph of f(x) on
your calculator to see where it is above (+) or below (–) the x-axis. We will use the
second method in all examples.
5) State answer based on inequality given.
Example 1
Solve:
( x + 3)(2 − x )
( x − 1) 2
<0
First set the numerator and denominator equal to zero:
From numerator x = –3, 2 and from the denominator x = 1. So critical values are
x = –3, 2, 1
–3
1
2
In looking at the graph from your calculator in the standard window, we see that the graph
is negative on (–∞,–3) and on (2, ∞) and positive on (–3, 1) and on (1, 2) .
–
+
–3
+
1
–
2
Since the inequality is < 0, we want the negative intervals: (–∞,–3) ∪ (2, ∞)
Example 2
Solve:
x+2
≥1
x−4
First, we have to get it into the correct format!
x+2−x+4
6
x+2
x+2
x+2 x−4
≥1 ⇒
−1≥ 0 ⇒
−
≥0⇒
≥0⇒
≥0
x−4
x−4
x−4 x−4
x−4
x−4
15
Next, we find the critical values. Here there is only one: x = 4 from the denominator. Note
that at x = 4 the rational expression is undefined, so x ≠ 4 in any interval.
4
In looking at the graph from your calculator in the standard window, we see that it is
negative on (–∞, 4) and positive on (4, ∞).
–
+
4
Since the inequality is ≥ 0, we want the positive interval:
Example 3
Solve:
2x + 5 x + 1
≤
x +1 x −1
(4, ∞)
First, we have to get it into the correct format!
2x + 5 x + 1 2x + 5 x + 1
(2x + 5)( x − 1) ( x + 1)( x + 1)
≤
⇒
−
≤0⇒
−
≤0⇒
x +1 x −1
x +1 x −1
( x + 1)( x − 1) ( x + 1)( x − 1)
2x 2 + 3 x − 5 − ( x 2 + 2x + 1)
x2 + x − 6
( x + 3)( x − 2)
≤0⇒
≤0⇒
≤0
( x + 1)( x − 1)
( x + 1)( x − 1)
( x + 1)( x − 1)
So, setting the numerator and denominator equal to zero, the critical values are:
x = –3, 2, –1, 1. Note that x ≠ –1, 1 since they are values where the expression is
undefined. Thus, the intervals must not contain these values.
–3
–1
1
2
In looking at the graph on your calculator in the standard window, we see that it is negative
on [–3, –1) and on (1, 2] and positive on (–∞,–3], (–1, 1) and on (2, ∞) .
+
–
–3
+
–1
–
1
+
2
Since we have the inequality ≤ 0, the solution is:
[–3, –1) ∪ (1, 2]
16
Polynomial & Rational Inequality Worksheet
Solve each of the following inequalities. Be sure to show your number line test results with the
appropriate signs.
1.
x2 – 7x ≥ –12
2.
25x – x3 ≤ 0
3.
(x2 + x + 4)(x – 1) > 0
4.
16x2 + 25 < 40x
5.
3−x
≥0
x+4
6.
2x
x 2 − 25
<0
7.
x+6
>4
x −1
8.
3
5
≤
x +1 x − 3
17
Solutions to Polynomial & Rational Inequality Practice Problems
1.
x2 – 7x ≥ –12 ⇒ x2 – 7x + 12 ≥ 0
x2 – 7x + 12 = 0 ⇒ (x – 3)(x – 4) = 0, so critical values are x = 3, 4
+
–
+
3
4
Solution: (–∞, 3] ∪ [4, ∞)
2.
25x – x3 ≤ 0 ⇒ 25x – x3 = 0 ⇒ x(25 – x2) = 0 ⇒ x(5 + x)(5 – x) = 0, so critical values are
x = 0, –5, 5
+
–
–5
+
0
–
5
Solution: [–5, 0] ∪ [5, ∞)
3.
(x2 + x + 4)(x – 1) > 0
(x2 + x + 4)(x – 1) = 0 ⇒ x = 1 and need QF for others
− 1 ± 12 − 4(1)(1) − 1 ± i 3
=
, so no other critical values except x = 1!
x=
2(1)
2
Standard Window
–
+
1
Solution: (1, ∞)
18
4.
16x2 + 25 < 40x
16x2 – 40x + 25 < 0
5
4
16x2 – 40x + 25 = 0 ⇒ (4x – 5)2 = 0, so only one critical value at x =
+
+
5
4
Standard Window
The graph is always positive, and we wanted negative values, so solution is the empty set:
∅
5.
3−x
≥0
x+4
3–x=0
x+4=0
Critical values are x = 3, – 4 (x ≠ – 4)
–
+
–
–4
3
Standard Window
Solution: (–4, 3]
6.
2x
x − 25
2
<0
2x = 0
x2 – 25 = 0
Critical values are x = 0, –5, 5
–
+
–5
–
0
+
5
Standard Window
Solution: (–∞,–5) ∪ (0, 5)
19
7.
x+6
x+6
x + 6 4( x − 1)
x + 6 − 4x + 4
10 − 3 x
> 4⇒
−4>0⇒
−
>0⇒
>0⇒
>0
x −1
x −1
x −1
x −1
x −1
x −1
10
,1
3
So critical values come from 10 – 3x = 0 and x – 1 = 0, so x =
–
+
–
10
3
1
Standard Window
⎛ 10 ⎞
Solution: ⎜1, ⎟
⎝ 3 ⎠
8.
3
5
3
5
3( x − 3)
5( x + 1)
3x − 9 − 5x − 5
≤0 ⇒
≤
⇒
−
≤0 ⇒
−
≤0
x +1 x − 3
x +1 x − 3
( x + 1)( x − 3) ( x + 1)( x − 3)
( x + 1)( x − 3)
− 2x − 14
⇒
≤ 0 So critical values come from –2x – 14 = 0 and (x + 1)(x – 3) = 0
( x + 1)( x − 3)
x = –7, –1, 3
Note that x ≠ –1, 3
+
–
–7
+
–1
–
3
Standard Window
If you cannot tell from the graph of the sign to the left of –7, look at the table of values. You
can see that the graph is positive to the left of –7.
Solution: (–7, –1) ∪ (3, ∞)
20
Applications of Exponential & Logarithmic Functions
You will need to be familiar with the following formula:
nt
r⎞
⎛
Compound Interest: A = P⎜1 + ⎟ , where A is the amount in the account at time t, P is the
⎝ n⎠
principle (amount invested), r is the interest rate as a decimal, ns the number of compoundings
per year and t is time in years.
You will need to memorize the following formulas:
Continuous Compounding Interest: A = Pert , where A is the amount in the account at time t,
P is the principle (amount invested), rs the interest rate as a decimal and t is time in years.
Exponential Growth: N = N0ert and Exponential Decay: N = N0e −rt , where N is the amount
at time t, N0 is the initial amount, r is the growth or decay rate as a decimal and t is time.
EXAMPLES
1. $20,000 is invested in an account that pays 3.75% per year, compounded daily.
a) Find the amount after 10 years
( 365 )(10 )
Solution:
0.0375 ⎞
⎛
≈ $29,099.27
A = 20000 ⎜1 +
⎟
365 ⎠
⎝
(Always round to the nearest cent!)
b) How long will it take the investment to double? Round to one decimal place.
Solution:
365 t
365 t
⎛ 0.0375 ⎞
⎛ 0.0375 ⎞
⎛ 0.0375 ⎞
⇒ 2 = ⎜1 +
⇒ ln 2 = ln ⎜1 +
40000 = 20000 ⎜1 +
⎟
⎟
⎟
365 ⎠
365 ⎠
365 ⎠
⎝
⎝
⎝
ln 2
⎛ 0.0375 ⎞
⇒ t ≈ 18.5 years
ln 2 = 365 t ln ⎜1 +
⎟⇒t=
365 ⎠
⎛ 0.0375 ⎞
⎝
365 ln ⎜1 +
⎟
365 ⎠
⎝
365 t
⇒
2. If $13,500 is invested in an account that pays 6.9% per year, compounded continuously, how
long will it take before the amount reaches $25,000? Round to one decimal place.
Solution:
25000
50
⎛ 50 ⎞
= e0.069 t ⇒
= e0.069 t ⇒ ln⎜ ⎟ = ln e0.069 t ⇒
13500
27
⎝ 27 ⎠
⎛ 50 ⎞
ln⎜ ⎟
27
⎛ 50 ⎞
ln⎜ ⎟ = 0.069 t ⇒ t = ⎝ ⎠ ≈ 8.9 years
0.069
⎝ 27 ⎠
25000 = 13500e0.069 t ⇒
21
3. Find the interest rate on an account, compounded quarterly, which grew from $8000 to
$10,000 in 12 years. Round to the nearest tenth of a percent.
Solution:
r⎞
⎛
10000 = 8000⎜1 + ⎟
⎝ 4⎠
48
4(12 )
10000 ⎛
r⎞
⇒
= ⎜1 + ⎟
8000 ⎝ 4 ⎠
48
5 ⎛
r⎞
⇒ = ⎜1 + ⎟
4 ⎝ 4⎠
⎛ 5
⎞
5
r
5
r
= 1 + ⇒ 48 − 1 = ⇒ 4⎜⎜ 48 − 1⎟⎟ = r ⇒
4
4
4
4
⎝ 4 ⎠
48
⇒
48
5 48 ⎛
r⎞
= ⎜1 + ⎟
4
⎝ 4⎠
48
⇒
r ≈ 1.86%
4. A population of a city grows according to the formula P( t ) = 105,300e0.015 t , where t is in
years.
a) What is the initial population?
Solution: 105,300 people (number in front of e is always the initial amount when t = 0)
b) How many people will there be after 5 years?
Solution: P(5) = 105,300e( 0.015 )(5 ) ≈ 113,501 people
c) Find the number of years that it will take the population to reach 120,000. Round to one
decimal place.
Solution:
120,000
⎛ 120,000 ⎞
= e0.015 t ⇒ ln⎜
⎟ = ln e0.015 t ⇒
105,300
⎝ 105,300 ⎠
⎛ 120,000 ⎞
ln⎜
⎟
105,300 ⎠
⎛ 120,000 ⎞
⎝
ln⎜
= t ⇒ t ≈ 8.7 years
⎟ = 0.015 t ⇒
0.015
⎝ 105,300 ⎠
120,000 = 105,300e0.015 t ⇒
5. A culture starts with 100 bacteria. After 5 hours the count is 300.
a) Find the formula to represent the number of bacteria, N, at time t hours. Round k to 3 d.p.
Solution: 300 = 100ek (5 ) ⇒
300
ln 3
= e5k ⇒ ln 3 = ln e5k ⇒ ln 3 = 5k ⇒ k =
≈ 0.220
100
5
So, N = 100e0.220 t
b) Find the number of bacteria after 2 days.
Solution: N( 48) = 100e0.220( 48 ) ≈ 3,856,113 bacteria (Note: 2 days = 48 hours)
22
c) After how many hours, to two decimal places, will the population double?
Solution:
200 = 100e0.220 t ⇒ 2 = e0.220 t ⇒ ln 2 = ln e0.220 t ⇒ ln 2 = 0.220 t ⇒ t =
ln 2
≈ 3.15 hours
0.220
6. The half-life of Radium-226 is 1620 years.
a) Find the decay formula with k to 3 significant figures.
Solution:
5 = 10e−k (1620 ) ⇒ 0.5 = e −1620k ⇒ ln 0.5 = ln e −1620k ⇒ ln 0.5 = −1620k ⇒ k =
ln 0.5
≈ 0.000427
− 1620
So, N = 10e −0.000427 t
b) If 10 g is present initially, how much will remain after 1000 years
Solution: and N = 10e −0.000427(1000 ) = 6.52 grams
c) How long, to two decimal places, will it take the sample to decay to 3.5 grams?
Solution:
3.5 = 10e −0.000427 t ⇒ 0.35 = e −0.000427 t ⇒ ln 0.35 = ln e −0.000427 t ⇒ ln 0.35 = −0.000427 t ⇒
ln 0.35
t=
≈ 2458.6 years
− 0.000427
7. Use Newton’s Law of Cooling: T( t ) = Ts + D 0 e −kt , where T is the temperature of an object at
time t, Ts is the temperature of the area surrounding the object, and D0 is the initial
temperature difference between the object and the surrounding area, to solve the following:
A cup of coffee has a temperature of 200°F and is placed in a room that has a temperature
of 70°F. After 10 minutes the temperature of the coffee is 150°F.
a) Determine the formula for the cooling coffee, finding k to four significant figures.
Solution:
80
80
= e −10k ⇒ ln
= ln e −10k ⇒
150 = 70 + (200 − 70)e −k (10 ) ⇒ 80 = 130e −10k ⇒
130
130
8
1
8
= −10k ⇒ k = −
≈ 0.04855
So, T( t ) = 70 + 130e −0.04855 t
ln
ln
10 13
13
b) Find, to one decimal place, the coffee’s temperature after 15 minutes.
Solution: T(15) = 70 + 130e−0.04855(15 ) = 132.8o F
23
c) When the coffee will have cooled off to 100°F
Solution:
100 = 70 + 130e −0.4855 t ⇒ 30 = 130e−0.4855 t ⇒
30
3
= e −0.4855 t ⇒ ln
= ln e−0.4855 t ⇒
130
13
3
ln
3
13
ln
= −0.04855 t ⇒ t =
13
− 0.04855
≈
30.2 min utes
8. Sound is measured by a logarithmic scale called decibels. The intensity level β, measured in
⎛ I ⎞
decibels,is defined as: β = 10 log⎜⎜ ⎟⎟ , where I is the intensity of the sound being measures
⎝ I0 ⎠
and
I0 = 10–12 Watts/m2, which is the intensity of a sound barely audible to the human ear.
a) Find the decibel level of a jet engine during takeoff if its intensity is 100 W/m2.
⎛ 100 ⎞
Solution: β = 10 log⎜ −12 ⎟ = 140 db
⎝ 10 ⎠
b) What is the intensity of a whisper that is 30 dB?
Solution:
⎛ I ⎞
30 = 10 log⎜ −12 ⎟
⎝ 10
⎠
⎛ I ⎞
3 = log⎜ −12 ⎟
⎝ 10 ⎠
I
103 = −12
10
(
)
I = 103 10 −12 =
10 −9 W / m2
24
Applications of Exponential and Logarithmic Functions Practice Problems
1. $10,000 is invested in an account that pays 5.5% per year, compounded quarterly.
a) Find the amount after 5 years
b) How long will it take the investment to double? Round to one decimal place.
2. If $7,500 is invested in an account that pays 4% per year, compounded continuously, how long will it
take before the amount reaches $10,000? Round to one decimal place.
3. Find the interest rate on an account, compounded monthly, which grew from $12,500 to $21,000 in 8
years. Round to the nearest tenth of a percent.
4. A population of frogs in a pond grows according to the formula p = 85e 0.18 t , where t is in years.
a) What is the initial population of frogs?
b) How many frogs will there be after 3 years?
c) Find the number of years that it will take the frog population to reach 1000. Round to one decimal
place.
5. A culture starts with 8600 bacteria. After 1 hour the count is 10,000.
a) Find the formula to represent the number of bacteria, N, at time t hours. Round k to 3 d.p.
b) Find the number of bacteria after 1 day.
c) After how many hours will the population triple?
6. The half-life of strontium-90 is 25 years. How long will it take a 40-mg sample to decay to a mass of
5 mg? Round k to 3 decimal place and the final answer to one decimal place.
7. Newton’s Law of Cooling provides a formula useful in homicide investigations to determine the time
of death of a person killed. The formula used is: T( t ) = Ts + D 0 e −0.1947 t , where T is the
temperature of the body at time t hours, Ts is the temperature of the area surrounding the body, and
D0 is the initial temperature difference between the body and the surrounding area. If a person is
discovered dead in a room that has a temperature of 60°F, and the body now has a temperature of
72°F, how long ago did the murder take place? Assume the person had a normal temperature of
98.6°F at the time of death. Round to the nearest hour.
8. Chemist use the formula, pH = − log[H + ] to measure the pH of a substance, where H+ is the
concentration of hydrogen ions measured in moles per liter.
a) The hydrogen ion concentration of a blood sample is 3.16 x 10–8, find its pH to 1 decimal place.
b) The most acidic rain fall occurred in Scotland in 1974. It had a pH of 2.4, what was its
hydrogen ion concentration? Round to three significant figures.
9. In 1935 the American geologist Charles Richter defined the magnitude of an earthquake to be:
⎛I⎞
M = log⎜ ⎟ where I is the intensity of the earthquake (measured by the amplitude of a seismograph
⎝S⎠
reading taken 100 km from the epicenter of the earthquake) and S is the intensity of a “standard”
earthquake (whose amplitude is 1 micron = 10–4cm)
In 1906, the San Franciso earthquake had an estimated magnitude of 8.3 on the Richter Scale. In
1989, the Loma Prieta earthquake had a magnitude of 7.1 on the Richter Scale. How many more
times intense, to one decimal place, was the 1906 earth quake than the 1989 one?
25
Solutions to Exponent & Logarithm Applications Practice Problems
⎛ 0.055 ⎞
1. a) A = 10000⎜1 +
⎟
4 ⎠
⎝
4( 5 )
≈ $13140.67
⎛ 0.055 ⎞
20000 = 10000⎜1 +
⎟
b)
4 ⎠
⎝
4t
⎛ 0.055 ⎞
⇒ 2 = ⎜1 +
⎟
4 ⎠
⎝
ln 2
⎛ 0.055 ⎞
ln 2 = 4t ln⎜1 +
≈
⎟⇒t=
4 ⎠
⎛ 0.055 ⎞
⎝
4 ln⎜1 +
⎟
4 ⎠
⎝
2. 10000 = 7500e0.04 t ⇒
⎛ 0.055 ⎞
⇒ ln 2 = ln⎜1 +
⎟
4 ⎠
⎝
12( 8 )
⇒
21000 ⎛
r ⎞
= ⎜1 + ⎟
12500 ⎝ 12 ⎠
4t
⇒
12.7 years
10000
4
= e0.04 t ⇒ ln = ln e0.04 t
7500
3
r ⎞
⎛
3. 21000 = 12500⎜1 + ⎟
⎝ 12 ⎠
96
4t
4
4
⇒ ln = 0.04t ⇒ t = 3 ≈ 7.2 years
3
0.04
ln
96
⇒ 96
42 96 ⎛
r ⎞
= ⎜1 + ⎟
25
⎝ 12 ⎠
96
⇒
⎛ 42 ⎞
42
r
r
42
= 1+
⇒
= 96
− 1 ⇒ r = 12⎜⎜ 96
− 1⎟⎟ ≈ 6.5%
12
12
25
25
⎝ 25
⎠
4. a) 85 frogs ( Note: this is always p(0) )
b) p = 85e0.18( 3 ) ≈ 146 frogs
c) 1000 = 85e0.18 t ⇒
1000
200
= e0.18 t ⇒ ln
= ln e0.18 t
85
17
5. a) 10000 = 8600ek (1) ⇒
200
200
17 ≈ 13.7 yrs
⇒ ln
= 0.18 t ⇒ t =
17
0.18
ln
10000
10000
50
= ek ⇒ ln
= ln ek ⇒ ln
= k ≈ 0.151
8600
8600
43
So, N = 8600e 0.151t
b) N(24) = 8600e0.151( 24 ) ≈ 322,390 bacteria
c) 25800 = 8600e0.151t ⇒
25800
ln 3
= e0.151t ⇒ ln 3 = ln e0.151t ⇒ ln 3 = 0.151t ⇒ t =
8600
0.151
t ≈ 7.3 hours
26
6. 20 = 40e −k ( 25 ) ⇒ 0.5 = e − 25k ⇒ ln 0.5 = ln e − 25k ⇒ ln 0.5 = −25k ⇒ k =
ln 0.5
≈ −0.028
25
So, N = 40e −0.028 t
5 = 40e−k ( 25 ) ⇒ 0.125 = e−0.028 t ⇒ ln 0.125 = ln e −0.028 t ⇒ ln 0.125 = −0.028t ⇒ t =
So, t ≈ 74.3 years
7. 72 = 60 + (98.6 − 60)e −0.1947 t ⇒ 12 = 38.6e−0.1947 t ⇒
ln
12
= ln e −0.1947 t ln
38.6
ln 0.125
− 0.028
12
= e −0.1947 t ⇒
38.6
12
12
38.6 ≈ 6 hours
= −0.1947 t ⇒ t =
38.6
− 0.1947
ln
8. a) pH = − log 3.16 × 10 −8 ≈ 7.5
b) 2.4 = − log H+ ⇒ −2.4 = log H+ ⇒ H+ = 10 −2.4 ≈ 0.00398 moles/liter
I
⎛I⎞
9. 7.1 = log⎜ ⎟ ⇒ 107.1 = ⇒ I = 107.1S
S
⎝S⎠
I
⎛I⎞
8.3 = log⎜ ⎟ ⇒ 108.3 = ⇒ I = 108.3 S
S
⎝S⎠
108.3 S
107.1S
=
108.3
107.1
≈
15.8 times more intense
27
MATRICES – Calculator Usage
You should be able to compute matrix determinants by hand and with your calculator. Here are
some calculator examples.
4
0 − 9⎤
⎡3
⎢5
6
8 − 7⎥⎥
A= ⎢
⎢− 1 2
0
0⎥
⎢
⎥
⎣ 1 − 3 − 8 14 ⎦
Example See text page 551, problem #12a.
To enter matrix A into the calculator:
1) Hit 2nd
x–1 to get the Matrix Menu
2) Use the left arrow key
>
to move to the EDIT Menu
3) The first matrix [ A ] should be high-lighted dark, so hit enter to select it
4) Next type in the size of [ A ] by hitting
4
ENTER
4
ENTER
5) Enter the row entries by hitting the correct number then ENTER repeatedly until all
rows are entered.
6) Hit 2nd CLEAR to get to the main screen (optional)
7) Repeat step 1)
8) Use left arrow key to move to MATH Menu and select option 1:det
9) Repeat step 1)
10) Select 1: [ A ] from the NAMES Menu
11) Hit
ENTER
and the display should show the determinant to be – 1640.
Practice
You should go to your homework and redo the following using your calculator.
Page 550: 11, 13,19, 23
28
Review for Exam #2
This review is intended to aid you in studying for the exam. This should not be the only thing that
you do to prepare. Be sure to also look over your notes from the computer lessons, book, your
homework, etc.
1.
Given: x 4 − 6 x 2 + 8 = 0 ,
a) Find all possible rational zeros.
b) Determine the possible number of positive and negative zeros.
c) Find all the zeros. (Use your calculator to help locate a zero, then use synthetic division
to help find the remaining zeros, or factor to find the zeros.)
2.
Use the Intermediate Value theorem to show that h( x ) = x 4 − 4 x 2 + 5 x − 2 has a zero on
[–3, –2]. Then use your calculator to approximate the zero to two decimal places.
3.
Find a polynomial f(x) of degree 3, with real coefficients given that 3 – i and 2 are zeros
and f(1) = –45.
4.
Find all the zeros then write as the product of linear factors:
a. x 3 + 27 = 0
b. x 4 + 3 x 3 + 23 x 2 − 37 x − 58 =0
5.
a) Given f(x) = 3 x 3 − 2x + 8 , use the remainder theorem to find f(–2)
b) Use the factor theorem to determine if (x – 3) is a factor of x 4 − 8 x 3 + 16 x 2 − x − 6
6.
Solve: (Be sure to show number line test results!)
a) 20 − x − x 2 ≤ 0
b) x 3 + 3 x 2 > 2x + 6
7.
Solve: (Be sure to show number line test results!)
5
2
2x − 5
a) 2
≥0
b)
<
x −1 x + 3
x −9
8.
Find the domain & range:
a) f ( x ) = 3 x + 2
9.
b) g( x ) = log3 ( x − 2)
Graph f ( x ) = 3 x + 2 and g( x ) = log3 ( x − 2) on the same axis. (Note they are inverse
functions!)
10. Evaluate without using a calculator:
a) log
5
5
b) log6 5 6
c) log2
11. Approximate to 3 decimal places:
a) log 125
b) ln 67
12. a) Write as a single logarithm: log3 x − 4 log3 y +
1
− ln e 4
8
c) log13 199
1
log3 z
2
29
⎛ 64 x 3 ⎞
⎟
12. b) Expand as much as possible: log4 ⎜
⎜ yz2 ⎟
⎝
⎠
13. Solve: (state exact answers, then approximate to 3 d.p. as needed)
a) e2 x −3 − 12 = 50
b) 21− x = 32 x + 5
c) ln(2x – 3) = 1
d) log8 (x + 5 ) − log8 ( x − 2) = 1 e) log2 ( x − 2) + log2 ( x + 1) = 2
14. Solve to 3 decimal places using your calculator: log x = x2 – 2
15. If you invest $20,000 for 8 years compounded quarterly and the amount at the end of the 8
years is $32,524, what was the interest rate? Round to the nearest hundredth of a percent.
16. The population of a certain city is modeled by: P( t ) = 12,534e0.018 t , where t is in years
starting in 1990.
a) What is the initial population? What is the growth rate?
b) What will the population be in 2003?
c) How long will it take the population to reach 20,000? (round to nearest year)
17. How long will it take an investment of $1000 to double in value at an interest rate of 8.5%
compounded continuously? (round to nearest tenth of a year)
18. A culture starts with 8600 bacteria. After 1 hour, there are 10,000 bacteria.
a) Find the exponential growth formula N(t) for this population, round k to 3 sign. fig.
b) Find the number of bacteria present after 2 hours
c) How long will it take for the population to triple? (round to nearest tenth of an hour)
19. If 250 mg of a radioactive element decays to 200 mg in 48 hours, find the half-life of the
element. (round k to 3 significant figures; round answer to nearest hour)
20. The half-life of Carbon-14, used to date artifacts, is 5730 years. The burial cloth of a
mummy contains 59% of the C-14 it originally contained. How old is the mummy?
(round k to 3 significant figures; round answer to nearest tenth of a year)
21. Newton’s Law of Cooling provides a formula useful in homicide investigations to determine
the time of death of a person killed. The formula used is: T( t ) = Ts + D 0 e −0.1947 t , where T
is the temperature of the body at time t hours, Ts is the temperature of the area
surrounding the body, and D0 is the initial temperature difference between the body and the
surrounding area. If a Math Professor is discovered dead in his office that has a
temperature of 72°F, and the body now has a temperature of 88°F, how long ago did the
murder take place? Assume the professor had a normal temperature of 98.6°F at the time
of death. Round to the nearest hour.
22. Chemist use the formula, pH = − log[H + ] to measure the pH of a substance, where H+ is
the concentration of hydrogen ions measured in moles per liter.
a) The hydrogen ion concentration of lemon juice is 5 x 10–3 moles per liter, find its pH to 1
decimal place.
b) Beer has a pH of 4.6 what was its hydrogen ion concentration? Round to three
significant figures.
30
23. Sound is measured by a logarithmic scale called decibels. The intensity level β, measured
⎛ I ⎞
in decibels,is defined as: β = 10 log⎜⎜ ⎟⎟ , where I is the intensity of the sound being
⎝ I0 ⎠
–12
2
measured and I0 = 10 Watts/m , which is the intensity of a sound barely audible to the
human ear.
a) Find the decibel level of traffic at a busy intersection if its intensity is 2.0 x 10– 5 W/m2.
Round to the nearest decibel.
b) What is the intensity of a power mower that is 106 dB? Round to 3 significant figures.
–5
24. Find the determinant of the matrix C = ⎡⎣ –11
7 ⎤
4 ⎦.
⎡ 1
25. Find the cofactor of the element 5 in the matrix A =⎢⎢ –11
⎣ –9
⎡ –1
26. Find the determinant of the matrix B = ⎢⎢ –2
⎣ –3
4
3
5
0
–3
–2
0
5
0
4
7
2
⎤
⎥.
⎥
⎦
⎤
⎥
⎥
⎦
27. Use the properties of determinants to find the determinant of the matrix:
⎡1
⎢2
A =⎢
⎢3
⎢
⎣4
0
0
0
0
3
4
5
6
4⎤
5⎥⎥
6⎥
⎥
7⎦
28. Use determinants to determine whether the following system has exactly one solution, is
inconsistent, or is dependent.
3x – 6y = 7
2x + 5y = 13
29. Use determinants to determine whether the following system has exactly one solution, is
inconsistent, or is dependent.
3x – 6y = 7
–6x + 12y = 14
30.
Use Cramer’s Rule to solve this system of linear equations:
6x + 9y – 4z = –5
2x + 3y – 4z = 1
3x – 3y + 6z = –1
31
SOLUTIONS TO REVIEW #2
1.
a) possible rational zeros: ±1, ±2, ±4, ±8
b) possible # positive real zeros: 2 or 0 (since 2 sign changes in f(x))
possible # negative real zeros: 2 or 0 (since 2 sign changes in f(–x))
c) Calculator graph and table show x = 2, –2 are zeros
-2 1
0 -6 0 8
-2 4 4 -8
1 -2 -2 4 0
2 1 -2 -2 4
2 0 -4
1 0 -2 0
(
)
So, (x + 2)( x − 2) x 2 − 2 = 0
x2 − 2 = 0 ⇒ x = ± 2
Zeros: x = ± 2 , ±2
2.
h( x ) = x 4 − 4 x 2 + 5 x − 2
h( −3) = ( −3) 4 − 4( −3) 2 + 5( −3) − 2 = 28 and
h( −2) = ( −2) 4 − 4( −2) 2 + 5( −2) − 2 = −12 , since there is a sign change if f(x) by IVT there is a
zero on [–3, –2].
3.
x = 3−i
(
f ( x ) = 9(x
x = 3+i
(
)
x = 2 ⇒ f ( x ) = a( x − 3 + i)( x − 3 − i)( x − 2) = a x 2 − 6 x + 10 ( x − 2)
)
(
+ 22x − 20 ) ⇒ f ( x ) = 9 x
)
f ( x ) = a x 3 − 8 x 2 + 22x − 20 ⇒ f (1) = a 13 − 8(1)2 + 22(1) − 20 = −45 ⇒ −5a = −45 ⇒ a = 9
4.
3
− 8x 2
d
3
− 72x 2 + 198 x − 180
i
a) x 3 + 27 = 0 ⇒ ( x + 3) x 2 − 3 x + 9 = 0 ⇒ x + 3 = 0 or x 2 − 3 x + 9 = 0 ⇒ x = −3 or
x=
−( −3) ±
b−3g
2
− 4(1)(9)
2(1)
F
GH
( x + 3) x +
3 + 3i 3
2
⇒x=
−3 ± −27
−3 ± 3i 3
⇒x=
, x = −3
2
2
I F x + 3 − 3i 3 I = 0
JK GH
JK
2
b) Use calculator graph to identify two rational zeros to use in synthetic division: x = –1, 2
-1 1
3 23 -37 -58
-1 -2 -21 58
1 2 21 -58
0
2 1 2 21 -58
2 8 58
1 4 29
0
−4 ± 4 2 − 4(1)(29) −4 ± −100
x + 1 x − 2 x + 4 x + 29 = 0 ⇒ x + 4 x + 29 = 0 ⇒ x =
=
2(1)
2
−4 ± 10i
x=
⇒ x = −2 ± 5i, − 1, 2
x + 1 ( x − 2) x + 2 + 5i x + 2 − 5i =0
2
b gb
gd
2
i
2
b g
5.
a)
0 -2
8
-6 12 -20
3 -6 10 -12
b
gb
g
-2 3
So, f(–2) = – 12
32
3 1 -8 16 -1 -6
3 -15 3 6
1 -5 1
2 0
b)
6.
Since remainder = 0,
(x – 3) is a factor.
a) 20 − x − x 2 ≤ 0 First, find the critical values: 20 − x − x 2 = 0 ⇒ (5 + x )( 4 − x ) = 0
So, x = – 5, 4
+
–
+
–5
4
Want negative intervals, so get:
b) x 3 + 3 x 2 > 2x + 6
First get critical values: x 3 + 3 x 2 − 2x − 6 = 0
x 2 ( x + 3) − 2( x + 3) = 0 ⇒ x 2 − 2 (x + 3 ) = 0 ⇒ x = ± 2, − 3
(
–
+
–
− 2
–3
a)
2x − 5
x −9
2
≥0
)
+
2
Want positive intervals, so get:
7.
[– 5, 4]
(− 3, − 2 )∪ (
2, ∞
)
Need to find crit values: 2x – 5 = 0, x2 – 9 = 0
So get, x =
5
, –3, 3
2
This graph is harder to see, so you may want to look at the
5
signs in the table of values between
and 3.
2
–
+
–3
–
5
2
+
3
5⎤
⎛
Want positive intervals and x ≠ –3, 3: ⎜ − 3, ⎥ ∪ (3, ∞ )
2⎦
⎝
33
b)
5
2
<
x −1 x + 3
First, need to get it in the correct format:
5
2
5
2
5( x + 3)
2( x − 1)
5 x + 15 − 2x + 2
<0
<0⇒
−
<0⇒
−
<0⇒
−
x −1 x + 3
x −1 x + 3
( x − 1)( x + 3) ( x + 3)( x − 1)
( x + 3)( x − 1)
3 x + 17
< 0 so, 3x + 17 = 0, x + 3 = 0, x – 1 = 0
( x + 3)( x − 1)
17
Thus, critical values are: x = −
, –3, 1
3
–
+
–
+
−
17
3
–3
1
17 ⎞
⎛
Want negative intervals, so solution is: ⎜ − ∞, − ⎟ ∪ (− 3, 1)
3 ⎠
⎝
8.
a) f(x) = 3x + 2 shifts y = 3x up 2 which doesn’t effect the domain, but does the range
domain: (–∞, ∞) range: (2, ∞)
b) g(x) = log3(x – 2) shifts y = log3(x) right 2, effecting the domain but not the range
domain: (2, ∞) range: (–∞, ∞)
y
9.
x
-1
0
1
2
a)
y
2.3
3
5
11
b)
x
3
5
11
y
0
1
2
x
10. a) log
b) log6 5
c) log2
1
x = 1 ⇒ x = 2 ⇒ log
2
1
1
6 = x ⇒ 6 x = 5 6 ⇒ x = ⇒ log6 5 6 =
5
5
x
5
5 = x ⇒ 5 = 5 ⇒ 51 2x = 51 ⇒
1
− ln e 4 ⇒ −3 + 4 = 1
8
11. Use calculator:
a) 2.097
sin ce 2 −3 =
b) 4.205
5
5=2
1
and e 4 = e 4
8
c) 2.064 (use change of base formula)
34
12. a) log3 x − 4 log3 y +
⎛ xz1 2 ⎞
1
log3 z = log3 x − log3 y 4 + log3 z1 2 = log3 ⎜ 4 ⎟
⎜ y ⎟
2
⎝
⎠
⎛ 64 x3 ⎞
⎟ = log4 64 + log4 x3 − log4 y − log4 z2 = 3 + 3 log4 x − log4 y − 2 log4 z
b) log4 ⎜
⎜ yz2 ⎟
⎝
⎠
3 + ln 62
≈ 3.564
2
b) 21− x = 32 x + 5 ⇒ ln 21− x = ln 32 x + 5 ⇒ (1 − x ) ln 2 = (2x + 5) ln 3 ⇒ ln 2 − x ln 2 = 2x ln 3 + 5 ln 3
ln 2 − 5 ln 3
⇒ ln 2 − 5 ln 3 = 2x ln 3 + x ln 2 ⇒ ln 2 − 5 ln 3 = x(2 ln 3 + ln 2) ⇒ x =
≈ −1661
.
2 ln 3 + ln 2
e+3
c) ln(2x – 3) = 1 ⇒ e1 = 2x − 3 ⇒ e + 3 = 2x ⇒ x =
≈ 2.859
2
x+5
⎛ x +5⎞
d) log8 (x + 5 ) − log8 ( x − 2) = 1 ⇒ log8 ⎜
= 8 ⇒ 8 x − 16 = x + 5 ⇒ 7 x = 21
⎟ = 1⇒
x−2
⎝ x − 2⎠
x = 3 note: must check answer in original equation
13. a) e2 x −3 − 12 = 50 ⇒ e2 x −3 = 62 ⇒ ln e2 x −3 = ln 62 ⇒ 2x − 3 = ln 62 ⇒ x =
(
)
e) log2 ( x − 2) + log2 ( x + 1) = 2 ⇒ log2 x 2 − x − 2 = 2 ⇒ 22 = x 2 − x − 2 ⇒ x 2 − x − 6 = 0
(x – 3)(x+2) = 0; x – 3 = 0 or x + 2 = 0; x = 3 or x = 2, but x = 2 does not check, so
only get x = 3
14.
Enter the equations and observe the graph. As we know that the y = log x continues
toward negative infinity as x gets smaller, we note that the graphs cross in 2 places. Run
the intersect program from the CALC menu to find both.
x ≈1.47, 0.010
15.
F rI
32524 = 20000G 1 + J
H 4K
32
F rI
⇒ 16252
.
= G1 + J ⇒
H 4K
r
16252
.
− 1 = ⇒ r = 4e 16252
.
− 1j ≈ 0.0612 ⇒
4
4( 8 )
32
32
32
16252
.
=
32
FG1 + r IJ
H 4K
32
⇒ 32 16252
.
= 1+
r
4
6.12%
16. a) P(0) = 12,534 people
b) P(13) = 12534e 0.018(13 ) ≈
15839 people
35
c) 20000 = 12534e 0.018 t ⇒
t=
IJ
K
FG
H
20000
10000
10000
= e 0.018 t ⇒ ln
= ln e 0.018 t ⇒ ln
= 0.018 t
12534
6267
6267
1
10000
ln
≈ 26.0 years
0.018
6267
17. 2000 = 1000e.085 t ⇒ 2 = e.085 t ⇒ ln 2 = ln e.085 t ⇒.085 t = ln 2 ⇒ t =
18. a) 10000 = 8600e k (1) ⇒
ln 2
≈ 8.2 years
.085
10000
10000
= e k ⇒ ln
= ln e k = k ⇒ k ≈ 0.151
8600
8600
N( t) = 8600e 0.151t
b) N( t) = 8600e 0.151( 2) ≈ 11632 bacteria
c) 25800 = 8600e 0.151t ⇒
25800
ln 3
= e 0.151t ⇒ ln 3 = ln e 0.151t ⇒ t =
≈ 7.3 hours
8600
0.151
200
ln 0.8
= e − k ( 48 ) ⇒ ln 0.8 = ln e − k ( 48 ) ⇒ −48k = ln 0.8 ⇒ k =
≈ 0.00465
250
−48
ln 0.5
⇒ 0.5 = e −0.00465 t ⇒ ln 0.5 = −0.00465 t ⇒ t =
≈ 149.1 years
−0.00465
19. 200 = 250e − k ( 48 ) ⇒
125 = 250e −0.00465 t
ln 0.5
≈ 0.000121
−5730
ln 0.59
⇒ ln 0.59 = ln e −0.000121t ⇒ t =
≈
−0.000121
20. 0.5N0 = N0 e −5730k ⇒ 0.5 = e −5730k ⇒ ln 0.5 = ln e −5730k ⇒ k =
0.59N0 = N0 e −0.000121t ⇒ 0.59 = e −0.000121t
4360.6 years
21. 88 = 72 + (98.6 − 72)e −0.1947 t ⇒ 16 = 26.6e −0.1947 t ⇒
16
16
= e −0.1947 t ⇒ ln
= ln e −0.1947 t
26.6
26.6
16
16
26.6 ≈ 2.6 hours
ln
= −0.1947 t ⇒ t =
26.6
−0.1947
22. a) pH = − log 5 × 10 −3 ≈ 2.3
ln
d
i
b) 4.6 = − log H+ ⇒ −4.6 = log H+ ⇒ H+ = 10 −4.6 = 2.51 × 10 −5 moles/liter
FG 2.0 × 10 IJ ≈ 73dB
H 10 K
F I IJ ⇒ 10.6 = logFG I IJ ⇒ 10
106 = 10 logG
H 10 K
H 10 K
23. a) β = 10 log
b)
−5
−12
−12
−12
10.6
=
I
⇒ I = 10 10.610 −12 ≈
10 −12
3.98 × 10 −2 W/m2
36
24.
−5 7
= ( −5)(4) − (7)( −11) = 57
−11 4
25.
LM 1
MM11
N9
OP
1
7 P ⇒ min or of 5 =
9
2PQ
0 4
5
0
4
2
= −34 ⇒ cofactor of 5 = ( +1)( −34) = − 34
1 4 0
26.
2 3 3 = 1C 11 + 4C 12 + 0C 31 = 1( −9) + 4(5) = 11
3 5 2
M11 =
3 3
2 3
= −9 M12 =
= −5 ⇒ C 11 = (1)( −9) = −9
5 2
3 2
C 21 = ( −1)( −5) = 5
27. Det(A) = 0 since there is a column of zeros!
28.
5 1
= 10 − 3 = 7 ≠ 0 ⇒ exactly one solution
3 2
29. det A =
3
−6
−6 12
6
9
= 0; det A x =
−4
7
−6
14 12
= 168 ≠ 0 ⇒ Inconsistent System
−5
9
−4
6 −5 −4
30. det A = 2 3 −4 = −120; det A x = 1 3 −4 = −48; A y = 2
3 −3 6
−1 −3 6
3
6
9
det A z = 2 3
3 −3
1
−4 = 152
−1
6
−5
−48
2
152
19
120
1 = 120 ⇒ x =
= ;y=
=− ;z=
= −1
−120 5
−120
15
−120
−1
FG 2 , − 19 , − 1IJ
H5 5 K
37
Matrix Applications
I.
Solving Independent, Consistent Equations Using A–1
Let A be a coefficient matrix for a system of equations, X be a column matrix containing the
variables from the system of equations and B be a column matrix with the for a n x n
system of equations. If AX = B has a unique solution, then X = A–1B.
Example: Solve the system using inverse matrices:
⎡ 1 3 − 1⎤
A = ⎢⎢ 3 − 1 2⎥⎥ , B =
⎢⎣ 2 − 1 1⎥⎦
⎡− 3⎤
⎢ 1⎥ and X =
⎢ ⎥
⎢⎣ − 1⎥⎦
⎡x⎤
⎢y⎥
⎢ ⎥
⎢⎣ z ⎥⎦
x + 3y – z = –3
3x – y + 2z = 1
2x – y + z = –1
⎡ − 2⎤
So, X = A–1B = ⎢⎢ 1 ⎥⎥ or x = –2, y = 1 and z = 4.
⎢⎣ 4 ⎥⎦
You can do the computation of A–1B with your calculator. Note that this only works for
independent, consistent systems.
II.
Encoding and Decoding Messages
It is possible to use a matrix to encode a message and its inverse to decode a message.
First, you need to decide the length of the matrices you will use. If you want to encode a
message using a 3 x 3 matrix, you need to code 3 letters of the message at a time; if you
want to encode using a 4 x 4 matrix, you need to code 4 letters of the message at a time;
etc. The matrix you will encode by must have an inverse. To encode the letters use:
A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, G = 7, H = 8, I = 9, J = 10, K = 11, L = 12, M = 13,
N = 14, O = 15, P = 16, Q = 17, R = 18, S = 19, T = 20, U = 21, V = 22, W = 23, X = 24,
Y = 25, Z = 26 and a space = 0.
⎡− 1 2 − 1⎤
Example: Encode the message “Meet me at noon” using the 3 x 3 matrix A = ⎢⎢ 0 3 − 1⎥⎥
⎢⎣ 0 4 − 1⎥⎦
You need to code: [ M E E ] [ T _ M ] [ E _ A ] [ T _ N ] [ O O N]
to be: [ 13 5 5 ] [ 20 0 13 ] [ 5 0 1 ] [ 20 0 14 ] [ 15 15 14 ]
Next, multiply each by the above matrix using your calculator. The result is:
[ –13 61 –23 ] [ –20 92 –33 ] [–5 14 –6 ] [–20 96 –34] [–15 131 –44 ]
2
⎡ −1
⎢
If this code was sent to someone who knew A = ⎢ 0 − 1
⎢⎣ 0 − 4
–1
these coded 1 x 3 matrices by A to decode the message.
–1
− 1⎤
1⎥⎥ , they could multiply
3⎥⎦
38
Matrix Application Worksheet
For questions 1 – 3, solve using an inverse matrix and your calculator. Be sure to write what you
are using as A, B and X and what you are computing on your calculator.
1.
6x – 8y + 5z = 14
3x + 3y – 3z = – 9
9x – 8y – 4z = –55
2.
2x – 3y + 4z – w = 21
x + 5y = 2
–2y + 3z + w = 12
–3x +4w = –5
3.
x+ y+ z= 0
x – y + 5z = –14
4x + y + z = –12
4.
5.
⎡− 1
Encode the following message using A = ⎢⎢ 0
⎢⎣ 0
–1
Decode the following message using A
2 − 1⎤
3 − 1⎥⎥ : SEND HELP
4 − 1⎥⎦
2
⎡ −1
⎢
= ⎢ 0 −1
⎢⎣ 0 − 4
− 1⎤
1⎥⎥ :
3⎥⎦
[ –25 179 –61 ] [ 0 48 –13 ] [ –4 44 –13 ] [ –20 40 –20 ]
39
Solutions to Matrix Application Worksheet
1.
5⎤
⎡ 6 −8
⎢
A= ⎢ 3
3 − 3 ⎥⎥ B =
⎢⎣ 9 − 8 − 4 ⎥⎦
2.
⎡ 2 −3
⎢ 1
5
A= ⎢
⎢ 0 −2
⎢
0
⎣− 3
3.
⎡ 1 1
A = ⎢⎢ 1 − 1
⎢⎣ 4
1
4.
[ S E N ] = [ 19 5 14 ], [ D __ H ] = [ 4 0 8 ], [ E L P ] = [ 5 12 16 ]
5.
4
0
3
0
⎡ 14 ⎤
⎢ − 9 ⎥ and X =
⎢
⎥
⎢⎣− 55⎥⎦
− 1⎤
0⎥⎥
B=
1⎥
⎥
4⎦
1⎤
5 ⎥⎥ B =
1 ⎥⎦
⎡ 21 ⎤
⎢2⎥
⎢ ⎥ and X =
⎢ 12 ⎥
⎢ ⎥
⎣− 5⎦
⎡ 0 ⎤
⎢− 14⎥ and X =
⎢
⎥
⎢⎣ − 12⎥⎦
1⎤
5 ⎥⎥
1 ⎥⎦
⎡ 1 1
[ 5 12 16 ] ⎢⎢ 1 − 1
⎢⎣ 4
1
1⎤
5 ⎥⎥ = [ 81 9 81]
1 ⎥⎦
2
⎡ −1
⎢
[ –4 44 –13 ] ⎢ 0 − 1
⎢⎣ 0 − 4
⎡ 1⎤
X = A B = ⎢⎢4⎥⎥
⎢⎣8 ⎥⎦
–1
⎡x⎤
⎢y⎥
⎢ ⎥
⎢z⎥
⎢ ⎥
⎣w ⎦
⎡x⎤
⎢y⎥
⎢ ⎥
⎢⎣ z ⎥⎦
⎡ 1 1
[19 5 14] ⎢⎢ 1 − 1
⎢⎣ 4
1
2
⎡ −1
⎢
[ –25 179 –61 ] ⎢ 0 − 1
⎢⎣ 0 − 4
⎡x⎤
⎢y⎥
⎢ ⎥
⎢⎣ z ⎥⎦
⎡7⎤
⎢− 1⎥
–1
X=A B= ⎢ ⎥
⎢2⎥
⎢ ⎥
⎣4⎦
⎡ − 4⎤
X = A B = ⎢⎢ 5 ⎥⎥
⎢⎣ − 1⎥⎦
= [ 80 28 58 ]
− 1⎤
1⎥⎥ = [25 15 21 ]
3⎥⎦
− 1⎤
1⎥⎥ = [ 4 0 9 ]
3⎥⎦
–1
⎡ 1 1
[ 4 0 8 ] ⎢⎢ 1 − 1
⎢⎣ 4
1
1⎤
5 ⎥⎥ = [ 36 12 12 ]
1 ⎥⎦
2
⎡ −1
⎢
[ 0 48 –13 ] ⎢ 0 − 1
⎢⎣ 0 − 4
− 1⎤
1⎥⎥ = [ 0 4 9 ]
3⎥⎦
2
⎡ −1
⎢
[ –20 40 –20 ] ⎢ 0 − 1
⎢⎣ 0 − 4
− 1⎤
1⎥⎥ = [ 20 0 0 ]
3⎥⎦
25 15 21 0 4 9 4 0 9 20 0 0 = YOU DID IT
40
Systems of Non-Linear Inequalities
Graphing of systems on non-linear inequalities are similar to graphing systems of linear
inequalities. You need to graph each inequality separately, noting in which area each graph
would shade individually, then shade the overlap of all the inequalities.
y
Example 1
x2 + y2 ≤ 9
Graph the solution set: x 2 y 2
+
≥1
4 16
x
Sketch each of the graphs based on its
properties. Pick a point, not on the graph,
to test in the inequality to decide which way
shade. Often, (0, 0) is a good point.
02 02
+
≥ 1 is a false statement, we want to shade
4 16
inside of the circle and outside of the ellipse.
As 02 + 02 ≤ 9 is a true statement and
y
Example 2
Graph the solution set:
x2 y2
−
≤1
4
9
x
x2 y2
+
<1
9 25
Sketch each of the graphs based on its
properties. Note that the second graph is
drawn in using a dashed border as the inequality is just < . Pick a point, not on the graph, to test
in the inequality to decide which way shade. Again,(0, 0) is a good point to use.
02 02
02 02
−
≤ 1 and
+
< 1 are true statements, we want to shade where the center
4
9
9 25
portion of the hyperbola and the inside of the ellipse intersect.
As
Practice Problems – Sketch the solution set to each of the following
1.
x2 y2
+
≤1
16 9
2
2
x
y
+
≤1
9 16
x ≤ y 2 − 4y
2.
y2
x +
≤1
9
2
4 y − 9 x ≤ 36
3x + y < 1
2
3.
2
x 2 + y 2 ≤ 25
4. x 2 y 2
+
≥1
4 16
41
Solutions to Practice Problems of Non-Linear Inequalities
1.
y
x
2.
y
x
3.
y
x
4.
y
x
42
REVIEW FOR EXAM #3
This review is intended to aid you in studying for the exam. This should not be the only thing that
you do to prepare. Be sure to also look over your notes from the computer lessons, book, your
homework, etc.
1.
Solve: a) 5x – 6y = 4
3x + 7y = 8
b) 2x – 6y = 10
x = 3y + 5
c) 3m – 4n = 2
6m – 8n = 6
2.
Solve by Gaussian Elimination
with Back-Substitution:
a)
b) x + 3y + z = 0
x+ y– z=0
x – 2y – 4z = 0
3.
Solve by Gauss-Jordon
Elimination:
a) 2x – 3y
= 12
3y + z = –2
5x
– 3z = 3
4.
⎡ −1
Let A = ⎢⎢ 4
⎢⎣− 2
3⎤
2⎥⎥ , B =
− 1⎥⎦
⎡ 1
⎢− 2
⎣
2⎤
⎡− 3
and C = ⎢
⎥
1⎦
⎣ 1
a) Find 3C + B
5.
⎡ 1
⎢ 3
Given A = ⎢
⎢− 2
⎢
⎣ 1
−1
1
2
0
x + 3y – z = –3
3x – y + 2z = 1
2x – y + z = –1
b)
− 1⎤
4⎥⎦
b) Find AB, by hand
2⎤
1⎥⎥
and B =
1⎥
⎥
1⎦
⎡ 1 −2
⎢− 2
1
⎢
⎢⎣ 2
3
x + 3y – 3z = –5
2x – y + z = –3
–6x + 3y – 3z = 4
0
1
0
3
0
1
c) Find BC – 2B
0⎤
1⎥⎥
1⎥⎦
a) Find the dimension of the matrix AB
b) Determine AB23 and AB31
c) Find BA
6.
⎡ 1
Determine if A and B are inverse matrices: A = ⎢
⎣− 2
7.
⎡ 2 − 4⎤
Find A-1 by hand: A = ⎢
3⎥⎦
⎣ 1
8.
Solve using inverse matrices:
2⎤
1⎥⎦
2⎤
⎡ 1
⎢ 5 − 5⎥
B= ⎢
2
1⎥
⎢−
⎥
5⎦
⎣ 5
3x – 3y + z + w = 27
x – 2y + 2z – 2w = 14
–3x + z = –10
x+y=0
43
9.
Write the equation in standard form of the circle whose center is at (0, 0) and that has a
radius of 14 units.
10. Write the equation in standard form of the circle whose center is at (9, –3) and that has a
radius of 15 units.
11. Write the equation in standard form of the circle whose center is at (–6, 1) and that passes
through the point (–2, 4).
12. Find the equation in standard form of the parabola that opens to the right, has its vertex at
(– 4, 5), and goes through the point (0, –3).
13. Graph the circle whose equation is x2 + y2 – 8x – 4y = 5. State the center and radius.
14. Graph the parabola whose equation is (y + 1)2 = –8(x + 2). State the vertex, focus, length
of latus rectum, directrix and direction which the parabola opens.
15. Find the standard form for the equation of the ellipse whose center is at (0, 0), whose
major axis, of length 20, is along the y-axis, and whose minor axis intersects the ellipse at
the point (5, 0).
16. Find the standard form for the equation of the ellipse whose vertices are at (3, 2) and
(–7, 2) and whose minor axis is of length 6.
17. Graph the ellipse whose equation is 4x2 + y2 = 36. Determine the foci coordinates.
18. Graph the ellipse whose equation is 4x2 + 9y2 – 32x + 54y = –1. Determine the foci
coordinates.
19. Find the standard form for the equation of the hyperbola whose center is at (0, 0), which
has vertices at the points (– 4, 0) and (4, 0), and which passes through the point (8, 6).
2
y2
20. Graph the hyperbola whose equation is x – = 1. Find the foci.
25
21. Graph the hyperbola whose equation is
16
(y – 4)2
25
–
(x – 1)2
16
= 1. Find the foci.
22. Put into standard for & determine the center and vertices: 4x2 – 49y2 – 8x – 196y = 388.
x 2 + y 2 = 25
23. Solve: a)
3x − 4y = 0
24. Solve: a)
y ≥ x2 − 3
x 2 + 4 y 2 < 16
b)
x+y=5
y = 3 − x2
c)
2x 2 + 5 y 2 = 39
3 x 2 − y 2 = −1
x 2 + y 2 ≤ 16
b)
y ≥3−x
44
SOLUTIONS TO REVIEW FOR EXAM #3
1.
35 x − 42y = 28
(5 x − 6 y = 4)7 18 x + 42y = 48
168
28
76
⎛ 76 ⎞
⇒
⇒x=
⇒ 5⎜ ⎟ − 6 y = 4 ⇒ −6 y = −
⇒y=
a)
(3x + 7y = 8)6
53
53
53
⎝ 53 ⎠
53 x = 76
⎛ 76 28 ⎞
⎜ , ⎟
⎝ 53 53 ⎠
b)
2x − 6 y = 10
x = 3y + 5
⇒ 2(3 y + 5) − 6 y = 10 ⇒ 10 = 10 ⇒ dependent system
(3y + 5, y )
m + 8n = −4
(3m − 4n = 2)( −2) − 66m
− 8n = 6 ⇒ inconsistent system ∅
⇒
c)
6m − 8n = 6
0=2
2.
⎡1
⎢
a) ⎢3
⎢2
⎣
⎡1
⎢
⎢0
⎢
⎣0
⎡1 3
− 1 − 3⎤ ⎡ 1 3
− 1 − 3⎤
−1 − 3 ⎤
⎥ ⎢
⎥
⎢
⎥
− 1 2 1 ⎥ = ⎢0 − 10 5 10 ⎥ − 3R1 + R2 = ⎢0 − 10 5 10 ⎥
=
− 1 1 − 1⎥⎦ ⎢⎣0 − 7 3 5 ⎥⎦ − 2R1 + R2 ⎢⎣0
0
− 5 − 20⎥⎦ − 7R2 + 10R3
x + 3 y − z = −3
3 − 1 − 3⎤
x + 3(1) − 4 = −3
⎥
1
1
1
⇒ (–2, 1, 4)
1
10 ⎥ − 0.1R2 ⇒ y + z = 10 ⇒ y + (4 ) = 10 ⇒
x = −2
2
2
2
⎥
0 1 4 ⎦ − 02R3
z=4
y =1
3
⎡1 3
1 0⎤ ⎡ 1 3 1 0⎤
⎥ ⎢
⎥
⎢
b) ⎢1 1 − 1 0⎥ = ⎢0 2 2 0⎥ R1 − R2 =
⎢1 − 2 − 4 0⎥ ⎢0 5 5 0⎥ R1 − R3
⎦ ⎣
⎦
⎣
⎡ 1 3 1 0⎤
x + 3y + z = 0
⎢
⎥
⎢0 1 1 0⎥ 0.5R2 ⇒ y + z = 0
⎢0 0 0 0⎥ 5R2 − 2R3
0=0
⎣
⎦
dependent system! y = −z ⇒ x + 3( −z ) + z = 0 ⇒ x = 2z ⇒ (2z, − z, z )
3.
⎡2 − 3 0 12 ⎤ ⎡2 − 3
⎢
⎥ ⎢
1 − 2⎥ = ⎢0 3
a) ⎢0 3
⎢5 0 − 3 3 ⎥ ⎢0 0
⎣
⎦ ⎣
⎡2 − 3 0 12 ⎤
⎡2
⎢
⎥
⎢
⎢0 3 0 − 6⎥ R2 − R3 = ⎢0
⎢0 0 1 4 ⎥
⎢0
⎣
⎦
⎣
⎡2
0 12 ⎤
⎥
⎢
1 − 2⎥
= ⎢0
11 44 ⎥⎦ 5R1 − 2R3 ⎢⎣0
0 0 6 ⎤ R1 + R2 ⎡1
⎥
⎢
3 0 − 6⎥
= ⎢0
⎢0
0 1 4 ⎥⎦
⎣
− 3 0 12 ⎤
⎥
3 1 − 2⎥
=
0 1 4 ⎥⎦ (1 / 11)R3
0 0 3 ⎤ 0.5R1
⎥
1 0 − 2⎥ (1 3)R2 ⇒ (3, –2, 4)
0 1 4 ⎥⎦
⎡ 1
⎡ 1 3 − 3 − 5⎤
3 − 3 − 5⎤ ⎡ 1 3
−3 −5 ⎤
⎢
⎥ ⎢
⎥
⎢
⎥
7
7 ⎥ − 2R1 + R2 = ⎢0 − 7 7 7 ⎥
⇒
b) ⎢ 2 − 1 1 − 3⎥ = ⎢0 − 7
⎢− 6 3 − 3 4 ⎥ ⎢0 21 − 21 − 26⎥ 6R1 + R3 ⎢0 0
0 − 5⎥⎦ 3R2 + R3
⎣
⎦ ⎣
⎦
⎣
0 = –5 indicates an inconsistent system, so answer is: ∅
45
4.
⎡− 3 − 1⎤ ⎡ 1 2⎤ ⎡− 8 − 1⎤
a) 3 ⎢
=
+
4 ⎥⎦ ⎢⎣− 2 1⎥⎦ ⎢⎣ 1 13 ⎥⎦
⎣ 1
⎡ −1
b) ⎢⎢ 4
⎢⎣− 2
3⎤
⎡ ( −1)(1) + (3)( −2)
⎡ 1 2⎤ ⎢
⎥
4(1) + 2(2)
=
2⎥ ⎢
− 2 1⎥⎦ ⎢
⎣
⎢⎣( −2)(1) + ( −1)( −2)
− 1⎥⎦
1⎤
( −1)(2) + (3)(1)⎤ ⎡− 7
⎥
⎢
4(2) + 2(1)⎥ = ⎢ 0 10⎥⎥
( −2)(2) + ( −1)(1)⎥⎦ ⎢⎣ 0 − 5⎥⎦
⎡− 3 3 ⎤
c) Use your calculator: BC – 2B = ⎢
⎥
⎣ 11 4⎦
5.
a) A: 4 x 3 B: 3 x 5
AB: 4 x 5
b) AB23 = 3(0) + 1(0) = 1(3) = 3
c) B: 3 x 5
6.
7.
A: 4 x 3
AB31 = (–2)(1) + 2(–2) + 1(2) = –4
Cannot multiply since dimensions do not match, AB does not exist
⎡− 3 3 ⎤
⎡ 1 0⎤
AB = ⎢
≠ I2 = ⎢
⎥
⎥ So, they are not inverses!
⎣ 11 4⎦
⎣0 1⎦
⎡2 − 4 1
⎢
⎣1 3 0
3
⎡
⎢ 1 0 10
⎢
1
⎢0 1 −
10
⎣
⎡10 0 3 4⎤ 5R1 − 2R2
0 ⎤ ⎡2 − 4 1 0 ⎤
=⎢
=
⎥=⎢
⎥
⎥
1⎦ ⎣0 − 10 1 − 2⎦ R1 − 2R2 ⎣ 0 20 − 2 4⎦
2⎤
2⎤
⎡ 3
⎥
⎢
⎥
(
)
1
10
R
1
5
⇒ A −1 = ⎢ 10 5 ⎥
⎥
1 (1 20 )R2
1 1
⎥
⎢−
⎥
5⎦
⎣ 10 5 ⎦
8.
LM 3
1
A=M
MM−3
N1
9.
C(0, 0); r = 14 ⇒ x 2 + y 2 = r 2 ⇒
−3 1
−2 2
0 1
1 0
OP
−2 P
0P
P
0Q
1
LM 27 OP
14 P
B=M
MM−10PP
N0Q
LM x OP
y
X=M P
MM z PP
Nw Q
LM 4 OP
−4
B=M P
MM 2 PP
N1Q
⇒ X = A −1
x 2 + y 2 = 196
10. C(9, –3); r = 15 ⇒ ( x − h) 2 + ( y − k ) 2 = r 2 ⇒
( x − 9) 2 + ( y + 3) 2 = 225
11. C(–6, 1); point (–2, 4) ⇒ ( x − h) 2 + ( y − k ) 2 = r 2 ⇒
( −2 + 6) 2 + (4 − 1) 2 = r 2 ⇒ 25 = r 2
(x + 6)2 + (y – 1)2 = 25
12. opens right; V(–4, 5); point (0, –3) ( y − k ) 2 = 4p( x − h) ⇒ ( −3 − 5 ) 2 = 4p(0 + 4)
b
64 = 16p ⇒ p = 4 ⇒ y − 5
g
2
= 16( x + 4)
46
13.
b
x 2 − 8 x + y 2 − 4 y = 5 ⇒ x 2 − 8 x + 16 + y 2 − 4 y + 4 = 5 + 16 + 4 ⇒ x − 4
g + by − 2g
2
2
= 25
y
C(4, 2); r = 5
x
14. Vertex: (–2, –1); 4p = –8 or p = –2; focus: (–4, –1)length of latus rectum = 8;
y
directrix: x = 0; opens left.
x
15. a =
20
x2
y2
= 10 ⇒ a 2 = 100; (5, 0) ⇒ b = 5 ⇒ b 2 = 25 ⇒
+
=1
2
25 100
( x + 2) 2 ( y − 2) 2
6
+
=1
16. C(–2, 2); major axis length = 3 – (–7) = 10 or a = 5; b = = 3 ⇒
2
25
9
17.
x2 y 2
+
= 1 ⇒ C(0, 0); b = 3; a = 6 ⇒ a 2 = b 2 + c 2 ⇒ 36 = 9 + c 2 ⇒ c 2 = 27 ⇒ c = 3 3
y
9 36
Foci: 0, ± 3 3
e
j
x
18. 4 x 2 + 9 y 2 − 32x + 54 y = −1 ⇒ 4( x 2 − 8 x + 16 − 16) + 9( y 2 + 6 y + 9 − 9) = −1 ⇒
b x − 4g + b y + 3g
4b x − 4g − 64 + b y + 3g − 81 = −1 ⇒ b x − 4g + b y + 3g = 144 ⇒
36
16
2
2
2
2
2
2
=1
C(4, –3); a = 6; b = 4; a 2 = b 2 + c 2 ⇒ 36 = 16 + c 2 ⇒ c 2 = 20 ⇒ c = 2 5 ⇒
y
Foci: 4 ± 2 5 ,−3
e
j
x
x2 y 2
x2 y 2
82 62
36
2
19. V(4, 0) means a = 4 2 − 2 = 1 ⇒ 2 − 2 = 1 ⇒ − 2 = −3 ⇒ b = 12 ⇒
−
=1
16 12
a
b
b
b
4
47
20.
x2 y 2
−
= 1 ⇒ a = 5; b = 4; a 2 = b 2 + c 2 ⇒ 25 = 16 + c 2 ⇒ c 2 = 9 ⇒ c = 3 ; Foci: (8, 0), (-8, 0)
25 16
y
Center: (0, 0)
x
by − 4g − bx − 1g
2
21.
25
16
2
e
j
= 1 ⇒ c 2 = 25 + 16 = 41 ⇒ foci: 1, 4 ± 41 ; Center: (1, 4)
y
x
22. 4 x 2 − 8 x − 49 y 2 − 196 y = 388
4( x 2 − 2x + 1 − 1) − 49( y 2 + 4 y + 4 − 4) = 388
4( x − 1) 2 − 4 − 49( y + 2) 2 + 196 = 388
4( x − 1) 2 − 49( y + 2) 2 = 196
( x − 1) 2 ( y + 2) 2
−
= 1⇒
49
4
e
j
c 2 = 49 + 4 = 53 ⇒ foci: 1 ± 51, − 2
Vertices: (–6, – 2) and (8, –2); Center: (1, –2)
2
x 2 + y 2 = 25
4
16 2
⎛4 ⎞
23. a)
⇒ 3x = 4y ⇒ x = y ⇒ ⎜ y ⎟ + y 2 = 25 ⇒
y + y 2 = 25
3
3
9
3x − 4y = 0
⎝
⎠
25 2
4
4
y = 25 ⇒ y 2 = 9 ⇒ y = ±3 ⇒ x = (3 ) = 4, x = (− 3 ) = −4
9
3
3
y = (4, 3), (4, –3)
b)
x+y=5
y =3−x
2
⇒ x + 3 − x 2 = 5 ⇒ 0 = x 2 − x + 2 ⇒ use quadratic formula to see there is no
solutions.
∅
48
2x + 5 y = 39
2
c)
2
3 x − y = −1
2
2
(
3x 2 − ± 7
(
)(
7, 2 ,
6 x 2 + 15 y 2 = 117
⇒ − 6 x 2 + 2y 2 = 2
17 y 2 = 119 ⇒ y 2 = 7 ⇒ y = ± 7
)2 = −1 ⇒ 3x 2 − 7 = −1 ⇒ 3x 2 = 6 ⇒ x 2 = 2 ⇒ x = ±
)(
)(
7, − 2 , − 7, 2 , − 7, − 2
2
)
24. a)
b)
49
Review for the MAC 1140 Final Exam
This review is intended to aid you in studying for the exam. This should not be the only thing that
you do to prepare. Be sure to also look over your class notes, textbook, HW, old exams, etc.
1.
For the function f(x) = x2 + 4x, find f(–2), f(5), and f(2z).
2.
Find the difference quotient
3.
Find the domain: a) f ( x ) =
4.
Given: f(x) = x2 + 8x + 15, find the: a) vertex b) axis of symmetry c) intercepts d) graph
5.
Graph: a) f(x) = | 2x | + 1
6.
Given the functions f(x) = 5x – 1 and g(x) = x3 + 7x, find a) (f + g)(x) b) f o g
7.
Use the following table of ordered pairs on the graph of the function y = f(x) to find the
corresponding ordered pairs on the graph of y = f–1(x).
x
y
–5
8
–1
2
5
3
7
11
8.
Given: f ( x ) = x 3 + 2x 2 − 4 x − 8
a) Determine the end behavior
b) Find all possible rational zeros
c) Use Descartes’ Rule of Signs to determine the possible number of positive and
negative real zeros.
d) Find all of the zeros & write in factored form
9.
Use the Intermediate Value Theorem to show that P(x) = 7x3 + 2x2 – 18x + 15 has at least
one zero between –3 and –2.
f ( x + h) − f ( x )
for this function: f(x) = x2 – 2x
h
3x + 7
2x + 9 x − 5
2
b) y = 9x – 4
c) g( x ) = x 2 − 4
b) g( x ) = − x − 4
c) f-1(x)
10. Find the zeros:
a) 3 x 4 + 14 x 3 + 14 x 2 − 8 x − 8 = 0
b) P( x ) = x 4 + 3 x 3 − 30 x 2 − 6 x + 56
c) R( x ) = 8 x 3 + 18 x 2 + 45 x + 27
11. Find a polynomial f(x) of degree three with real coefficients given that it has 1 – 2i and 4 as
some of its zeros and f(3) = –8.
12. Solve: a) x2 – 2x > 8
b)
5−x
x2 − 9
≤0
50
13 – 15 For each of the following rational expressions:
a) Determine any intercepts
b) Determine any asymptotes
c) Sketch the graph
13.
f (x) =
x −3
25 − x 2
14. g( x ) =
4x2 − 8
15. h( x ) =
x 2 + 2x − 3
x2 − 9
x+2
16. Use your calculator to find the local maximum and minimum of to two decimal places:
f ( x ) = x 3 − 5 x 2 + 2x + 5
17. Find the domain & range: a) f ( x ) = 2 x +1 − 5
18. Graph: a) f ( x ) = 2 x +1 − 5
b) g( x ) = log4 ( x + 1)
b) g( x ) = log4 ( x + 1)
19. Evaluate without using a calculator:
a) logπ 3 π
b) logb 1 − ln(e3 )
⎛ 1 ⎞
c) log3 ⎜ ⎟
⎝ 27 ⎠
20. Approximate to three decimal places:
a) log 243
b) ln 10.92
c) log12 400
21. Write as a single logarithm: 5 log4 a − 3 log4 b − log4 c
⎛ 81x 2 y ⎞
22. Expand as much as possible: log3 ⎜ 3 ⎟
⎜ z ⎟
⎝
⎠
23. Solve: (state exact answers, then approximate to 3 d.p. as needed.)
a) e3 x −1 + 14 = 42
b) 4 x + 3 = 51− x
d) log4 ( x + 3) + log4 ( x − 3) = 2
e) 2 ln( x + 3) − ln( x ) = 2 ln( 4)
c) ln(3 x + 5) = 2
24. Solve to 3 decimal places using the graphing features of your calculator: log( x − 1) = 2x − 4
25. Which is a better 10-year investment? $1000 invested at 12% per year compounded
quarterly or $1000 invested at 11.5% compounded continuously?
26. Determine the interest rate, to the nearest tenth of a percent, of an investment of $12,500,
compounded semi-annually, which doubles in 15 years. Round to the nearest tenth of a
percent.
27. If a 100-mg tablet of asthma medication is taken when none of the drug is initially present in
a person’s system, the amount of it in the bloodstream at time t minutes is predicted to be:
51
A( t ) = 100[1 − 0.9 t ] for 0 ≤ t ≤ 10. Determine the time needed for 50 mg of the medication to
enter the bloodstream.
28. A liquid solution initially contains 8 grams of a radioactive material. After 14 days, 5.5
grams remain. Find an exponential decay equation that represents the amount of the
substance after t days. Round k to 5 decimal places. Use your formula to:
a) Determine the amount remaining, to 1 decimal place, after 30 days
b) Determine the half-life, to 1 decimal place, of the solution.
⎡− 4 7⎤
29. Find the determinant of the matrix B = ⎢
⎥
⎣ − 3 2⎦
⎡ 1
30. Find the determinant of the matrix A = ⎢⎢ 9
⎣ 1
3
8
1
–7
0
3
⎤
⎥
⎥
⎦
⎡ 1
31. Find the cofactor of the element 4 in the matrix A = ⎢⎢ –11
⎣ –9
0
5
0
32. Use Cramer’s Rule to solve this system of linear equations:
4
7
2
⎤
⎥.
⎥
⎦
5x + y = –1
3x + 2y = 6
33.
Use determinants to determine whether the following system has exactly one solution, is
inconsistent, or is dependent.
3x – 6y = 7
–6x + 12y = –14
34.
Use Cramer’s Rule to solve this system of linear equations:
x+y– z=0
2x + y – 2z = 1
3x – 4z = 1
⎡ 1 − 3⎤
⎡ 1
35. Let A = ⎢⎢ 1
2⎥⎥ , B = ⎢
⎣− 2
⎢⎣− 2 − 1⎥⎦
a) Find 2C + 3B
36. Given the matrix A =
1⎤
⎡ 2 − 1⎤
and C = ⎢
⎥
⎥
1⎦
⎣ 1 4⎦
b) Find AC, by hand
⎡ 3
⎢ –2
⎣ –4
3 –1
–2
1
–5 2
multiplicative inverse of the matrix A.
c) Find BC – 3C
⎤
⎥ , show that the matrix B
⎦
=
⎡1
⎢0
⎣2
–1
1
2 –1
3
0
⎤
⎥
⎦
is the
52
⎡ 4 6 ⎤
⎥ (by hand)
37. Find the inverse: A = ⎢
⎣ 3 5 ⎦
38. Solve: a) 5x – 4y = – 22
y = 5x – 2
⎡1
b) B = ⎢ 0
⎣2
–1
b) 5x – 3y = –2
10x – 6y = 3
1
2 –1
3
0
⎤
⎥
⎦
(by calculator)
c) 5x – 3y = 0
10x = 6y
39. Use the Gaussian Method to solve this system of linear equations.
3x + y + 2z = 12
x + 2y – z = –1
2x + 2y + 3z = 10
40. Solve by Gauss-Jordan Elimination:
3x + y + 2z= –4
– 3y – 2z= –5
2y + 5z = –4
41. Solve using an inverse matrix and your calculator:
2x – 5y + 3z = –18
3x + 2y – z = –12
x – 3y – 4z = –4
42. Find the vertex, focus and directrix, then graph:
b) (x − 2)2 = −8( y + 3)
a) y 2 = 16 x
43. Find the center and radius, then graph:
a) x 2 + y 2 = 25
b) x 2 + y 2 − 2x + 8 y − 3 = 0
44. Find the vertices, foci and center, then graph:
b) 4 x 2 + 9 y 2 + 16 x − 18 y + 4 = 0
a) 4 x 2 + 25 y 2 = 100
45. Find the center, vertices and foci, then graph:
a)
y2 x2
−
=1
9 36
b) 9 x 2 − y 2 − 36 x + 6 y − 9 = 0
46. Find the equation of the ellipse with vertices of (7, 3) and (–3,3) and with minor axis of
length 8.
47. Find the equation of the hyperbola with foci (–5, 2) and (1, 2) and with conjugate axis of
length 4.
48. Solve: a) y = x2 + 1
y = –2x + 4
b) 5x2 – y2 = 20
x2 – y2 = 4
53
49. Graph the solution set:
x2 y2
b)
+
≥1
4 16
x2 y2
−
≤1
16 25
a) x + y ≤ 9
2
2
y > x +1
⎛ n ⎞
50. Find the first 4 terms and the 100th term of the sequence: an = ( −1)n +1⎜
⎟
⎝ n + 1⎠
51. Find the first five terms: an = 2an−1 + 1, a1 = 1
5
52. Compute:
∑ 2k −1
k =1
53. Find the nth term and the 10th term of the arithmetic sequence with a =
3 and d =
3.
20
54. Write the first three terms and the last term, then compute the sum:
∑ (1 − 2n)
n =0
55. A given geometric sequence has first term a1 = –3 and fifth term a5 = –7203. Find a3 and an.
56. Find the sum of the first 10 terms of the geometric sequence: 4, –12, 36, –108, . . .
∞
n
⎛ 1⎞
57. Compute: ∑ ⎜ ⎟
n =0 ⎝ 3 ⎠
1
58. Use induction to prove: 1 + 4 + 7 + … + (3n – 2) = 2 n(3n – 1)
59.
⎛ 24 ⎞
Compute ⎜⎜ ⎟⎟
⎝5⎠
60. Compute using the binomial theorem: a) (2a + 5 )6
(
b) x 2 − 3 y
)
5
54
Answer Key to Review for the MAC 1140 Final Exam
1.
2.
3.
f(– 2) = ( −2)2 + 4( −2) = −4 ; f(5) = 52 + 4(5) = 45 ; f(2z) = (2z )2 + 4(2z ) = 4z 2 + 8z
[
][
]
f ( x + h) − f ( x ) ( x + h)2 − 2( x + h) − x 2 − 2x
x 2 + 2xh + h2 − 2x − 2h − x 2 + 2x
=
=
=
h
h
h
2xh + h2 − 2h
= 2x + h − 2
h
a) 2x 2 + 9 x − 5 ≠ 0; (2x − 1)( x + 5) ≠ 0; D : x ≠
1
,−5
2
b) y = 9x – 4; graph is a line; D: ( −∞, ∞ )
c) x 2 − 4 > 0 ⇒ ( x + 2)( x − 2) > 0 ⇒ graph to see where > 0 ⇒ D : ( −∞,−2) ∪ (2, ∞ )
4.
a) vertex: x = −
b
8
=−
= −4 f(– 4) = –1 V(–4, –1)
2a
2(1)
b) axis of sym: x = –1
c) x-int: x 2 + 8 x + 15 = 0 ⇒ ( x + 3)( x + 5) = 0 ⇒ x = −3, − 5 ⇒ ( −3, 0) and ( −5, 0)
y
x
y
d) –6 5
–5 0
x
–4 –1
–3 0
–2 5
y
y
x
–2
0
2
x
y
5
1
5
x
4
5
8
x
5.
a)
6.
a) ( f + g) = (5 x − 1) + x 3 + 7 x = x 3 + 12x − 1 b) f o g = f x 3 + 7 x = 5 x 3 + 7 x − 1 = 5 x 3 + 35 x − 1
7.
(
x
y
8 –5
2 –1
3 5
11 7
)
8.
b)
y
0
–1
–2
(
) (
)
a) falls to left, rises to right
b) ±1, ±2, ±4, ±8
c) possible positive real: 1; since 1 sign change in f(x)
55
possible negative real: 2 or 0; since 2 sign changes in f(–x)
d) f ( x ) = x 2 ( x + 2) − 4( x + 2) = ( x 2 − 4)( x + 2)
= ( x + 2)( x − 2)( x + 2) = 0 ⇒ x = 2,−2 (occurs twice )
factored form : f ( x ) = ( x − 2)( x + 2)2; zeros : 2, − 2
9.
P( −3) = 7( −3)3 + 2( −3)2 − 18( −3) + 15 = −102 P( −2) = 7( −2)3 + 2( −2)2 − 18( −2) + 15 = 3
Since P(–3) is negative and P(–2) is positive, by IVT there must be a zero on (–3, –2)
10. a) Use your calculator’s graph and rational zeros theorem to locate zeros to use in
2
–2
3
14
14
–8
–8
synthetic division: x = –2 and −
–6
–16
4
8
3
3
8
–2
–4
0
(
)
Doing synthetic division by x = –2 we get (x + 2) x 3 + 8 x 2 − 2x − 4 = 0 . Continue with
2
3
8
–2
–4
2
synthetic division by x = − :
−
3
3
–2
–4
4
3
6
–6
0
2
⎛
⎞
So, we get (x + 2)⎜ x + ⎟ 3 x 2 + 6 x − 6 = 0 . Now use the quadratic formula on
3⎠
⎝
(
3x2 + 6x − 6 = 0
x=
)
− 6 ± 62 − 4(3)( −6) − 6 ± 108 − 6 ± 6 3
=
=
= −1 ± 3
2(3)
6
6
Thus, the zeros are: x = −1± 3 , −
2
, and –2
3
b) Similarly to a), use your calculator’s graph to determine zeros x = –7 and 4 to use in
synthetic division. The result should be P(x) = ( x − 4)( x + 7) x 2 − 2
(
)
Now, finish solving x 2 − 2 = 0 ⇒ x 2 = 2 ⇒ x = ± 2 So, zeros are x = –7, 4, and ± 2
c) Similarly to a) & b), use your calculator to find zero x = −
(
)
(
3
to use in synthetic division.
4
3⎞
3⎞
⎛
⎛
The result is Q(x) = ⎜ x + ⎟ 8 x 2 + 12x + 36 = 4⎜ x + ⎟ 2x 2 + 3 x + 9
4⎠
4⎠
⎝
⎝
)
Use the Quadratic Formula on 2x 2 + 3 x + 9 = 0
− 3 ± 32 − 4(2)(9) − 3 ± − 63 − 3 ± 3i 7
=
=
x=
2(2)
4
4
Zeros: x =
3
− 3 ± 3i 7
, −
4
4
11. Having x = 1 – 2i as a zero means that x = 1 + 2i is also a zero. So we need to multiply:
(x – 1 – 2i)(x – 1 + 2i)(x – 4) = x 2 − 2x + 5 (x − 4 ) = x 3 − 6 x 2 + 13 x − 20
(
)
(
)
56
(
)
(
)
f ( x ) = a x 3 − 6 x 2 + 13 x − 20 ⇒ f (3) = a 33 − 6(3)2 + 13(3) − 20 = −8 ⇒ −8a = 8 ⇒ a = −1
So, f ( x ) = − x 3 + 6 x 2 − 13 x + 20
12. a) x2 – 2x > 8 → x2 – 2x – 8 > 0 → (x – 4)(x + 2) > 0 Crit #s: x = 4, –2
Now, graph on calculator to see signs of intervals.
+
–
+
(–∞,–2) ∪ (4, ∞)
–2
4
b)
5−x
≤ 0 Set numerator and denominator = 0 to get crit #s: x = 5, 3, –3
x2 − 9
Now, graph on calculator & use table to see signs of intervals.
+
–
+
–
(–3, 3) ∪ [5, ∞)
–3
3
5
13. a) x-int: set numerator = 0: x – 3 = 0, so x = 3 or (3, 0) is x-int
y-int: compute f(0) =
0−3
3
= − , so
5
25 − 0
2
3 ⎞
⎛
⎜ 0, − ⎟ is y-int
25 ⎠
⎝
b) V.A.: set denominator = 0: 25 − x 2 = 0 ⇒ 25 = x 2 ⇒ ±5 = x
H.A.: since deg(num) < deg(den), we get the x-axis as the H.A.: y = 0
y
13. c)
Use your calculator to get a table of values.
x
x
y
-6
0.8
-5.5
1.6
14. a) x-int: 4 x 2 − 8 = 0 ⇒ 4 x 2 = 8 ⇒ x 2 = 2 ⇒ x = ± 2
y-int: g(0) =
-4
-0.8
(
0
-0.1
or ± 2, 0
4
0.1
5.5
-0.5
6
-0.1
)
8
⎛ 8⎞
or ⎜ 0, ⎟
3
⎝ 3⎠
b) V.A.: x 2 + 2x − 3 = 0 ⇒ ( x + 3)( x − 1) = 0 ⇒
x = −3, x = 1
H.A.: since deg(num)=deg(den), take y = ratio of leading coefficients: y =
4
so y = 4
1
y
c)
x
x
y
-6
-6.5
-4
11.2
-2
-2.6
0
8/3
2
1.6
5
2.8
57
15. a) x-int: x 2 − 9 = 0 ⇒ x 2 = 9 ⇒ x = ±3 ⇒ ( ±3, 0)
b) V.A.: x + 2 = 0 or x = –2
y-int: h(0) = −
9⎞
9 ⎛
⎜ 0, − ⎟
2⎠
2 ⎝
H.A.: none, but has an Oblique Asymptote
O.A.: has since deg(num) = deg(den) + 1; need to divide numerator by denominator,
ignoring the remainder. You should get : y = x – 2
y
c)
x
y
x
-5
-5.3
-3
0
0
-9/2
3
0
16. Enter Y1 = x3 – 5x2 + 2x + 5, then use [2nd][TRACE] the CALC menu and run the minimum
and maximum program. Min: x ≈ 0.21, Max: x ≈ 3.12
17. Consider the properties of exponential and log functions & their graphs
b) D: (− 1, ∞ ) R: (− ∞, ∞ )
a) D: (− ∞, ∞ ) R: (− 5, ∞ )
18. a)
y
x
x
-1
0
1
2
y
b)
y
-4
-3
-1
3
>
x
19. a) logπ 3 π = x ⇒ π x = π1 3 ⇒ logπ 3 π =
1
3
x
0
3
7
y
0
1
1.5
( )
b) logb 1 − ln e3 = 0 − 3 = −2
1
⎛ 1 ⎞
⎛ 1 ⎞
= 3−3 ⇒ log3 ⎜ ⎟ = −3
c) log3 ⎜ ⎟ = x ⇒ 3 x =
27
⎝ 27 ⎠
⎝ 27 ⎠
20. a)2.386
b) 2.391
c) log12 400 =
ln 400
≈ 2.411
ln 12
⎛ a5 ⎞
21. Use log rules; log4 ⎜ 3 ⎟
⎜b c⎟
⎝
⎠
2
⎛ 81x y ⎞
22. log3 ⎜ 3 ⎟ = log3 81 + 2 log3 x + log3 y − 3 log3 z = 4 + 2 log3 x + log3 y − 3 log3 z
⎜ z ⎟
⎝
⎠
23. a) e3 x −1 = 28 ⇒ ln e3 x −1 = ln 28 ⇒ 3 x − 1 = ln 28 ⇒ x =
1 + ln 28
≈ 1.444
3
b) ln 4 x +3 = ln 51− x ⇒ ( x + 3) ln 4 = (1 − x ) ln 5 ⇒ x ln 4 + 3 ln 4 = ln 5 − x ln 5 ⇒
58
x ln 4 + x ln 5 = ln 5 − 3 ln 4 ⇒ x(ln 4 + ln 5) = ln 5 − 3 ln 4 ⇒ x =
c) ln(3 x + 5) = 2 ⇒ e2 = 3 x + 5 ⇒ x =
(
ln 5 − 3 ln 4
≈ −7.637
ln 4 + ln 5
e2 − 5
≈ 0.796
3
)
d) log4 x 2 − 9 = 2 ⇒ 16 = x 2 − 9 ⇒ 25 = x 2 ⇒ x = −5 or 5 ⇒ x = 5 (the other answer
doesn’t check in original equation)
⎛ (x + 3 )2 ⎞
(
x + 3 )2
2
⎜
⎟
e) ln
= ln 4 ⇒
= 16 ⇒ x 2 + 6 x + 9 = 16 x ⇒ x 2 − 10 x + 9 = 0 ⇒
⎜ x +1 ⎟
x
⎝
⎠
(x − 9)(x − 1) = 0 ⇒ x = 9, 1 both answers check
( )
24.
Enter each side of the equation and look at the graph for points of intersection. It is easier
to see that there are two points of intersection by doing a ZOOM 1: Zbox. Then, run the
program CALC (2nd TRACE) 5: Intersect, to find each x-value of intersection.
x = 2, 1.01
4(10 )
⎛ 0.12 ⎞
= $3262.04
25. Inv 1: A = 1000⎜1 +
⎟
4 ⎠
⎝
So, Investment 1 is the better option.
r⎞
⎛
26. 25000 = 12500⎜1 + ⎟
⎝ 2⎠
(
)
2(15 )
r⎞
⎛
⇒ 2 = ⎜1 + ⎟
⎝ 2⎠
Inv 2: A = 1000e0.115(10 ) = $3158.19
30
⇒
30
2=
30 ⎛
⎜1 +
⎝
r⎞
⎟
2⎠
30
⇒ 30 2 = 1 +
r
⇒
2
r = 2 30 2 − 1 ≈ 0.0467 = 4.67%
27.
[
]
A( t ) = 100 1 − 0.9 t = 50 ⇒ 1 − 0.9 t = 0.5 ⇒ 0.5 = 0.9t ⇒ ln 0.5 = ln 0.9t ⇒
ln 0.5
ln 0.5 = t ln 0.9 ⇒ t =
≈ 6.6 years
ln 0.9
5. 5
ln 0.6875
= e −14k ⇒ ln 0.6875 = −14k ⇒ k =
≈ 0.02676
− 14
8
N( t ) = 8e −0.02676 t , so N(30) = 8e−0.02676( 30 ) ≈ 3.6 grams
28. a) 5.5 = 8e −k (14 ) ⇒
59
b) 4 = 8e −0.02676 t ⇒ 0.5 = e −0.02676 t ⇒ ln 0.5 = −0.02676 t ⇒ t =
29.
ln 0.5
≈ 25.9 days
− 0.02676
−4 7
= ( −4)(2) − (7)( −3) = 13
−3 2
30. Expanding on row 2: det( A ) = 9C21 + 8C22 + 0C23 ; M21 =
3 −7
= 16
1 3
M22 =
1 −7
= 10
1 3
C21 = ( −1)(16) = −16 and C22 = (1)(10) = 10 , so det(A) = 9(–16) + 8(10) = –64
31. C13 = ( 1 )M13 =
− 11 5
= 45
−9 0
−1 1
5 1
= 7 D x = det (A x ) =
= −8
3 2
6 2
D y 33
D
−8
⎛ − 8 33 ⎞
x= x =
y=
,
=
⎜
⎟
D
7
D
7
⎝ 7 7 ⎠
32. D = det( A ) =
33. D = det( A ) =
( )
Dy = det A y =
5 −1
= 33
3 6
7
−6
3
7
3 −6
= 0; Dx = det (A x ) =
= 0; Dy = det A y =
=0
− 6 12
− 14 12
− 6 − 14
( )
So, by Cramer’s Rule, the system is: dependent
1 1
−1
0 1
−1
1 0
−1
1 1 0
34. D = 2 1 − 2 = 1; Dx = 1 1 − 2 = 3; Dy = 2 1 − 2 = −1; Dz = 2 1 1 = 2
3 0 −4
1 0 −4
3 1 −4
3 0 1
*Use your calculator to compute the 3x3 determinants
x=
Dx 3
= =3
D
1
y=
Dy
D
=
D
2
−1
= −1 z = z = = 2
1
D
1
(3, –1, 2)
1⎤
⎡ 1 1⎤ ⎡ 7
⎡2 − 1⎤
35. a) Use calculator or by hand: 2⎢
=⎢
+ 3⎢
⎥
⎥
⎥
⎣− 2 1⎦ ⎣− 4 11⎦
⎣1 4 ⎦
1(2) + ( −3)(1)
1( −1) + ( −3)( 4)⎤ ⎡ − 1 − 13⎤
⎡
⎡ 1 − 3⎤
⎡2 − 1⎤ ⎢
⎥
⎢
1(2) + (2)(1)
b) ⎢ 1
2⎥ ⎢
=⎢
1( −1) + (2)( 4)⎥⎥ = ⎢⎢ 4
7⎥⎥
⎥
1 4⎦
⎢⎣ ( −2)(2) + ( −1)(1) ( −2)( −1) + ( −1)( 4)⎥⎦ ⎢⎣ − 5 − 2⎥⎦
⎢⎣− 2 − 1⎥⎦ ⎣
⎡− 3 6 ⎤
c)Use calculator: ⎢
⎥
⎣− 6 − 6⎦
60
⎡ 1 0 0⎤
⎡ 1 0 0⎤
⎥
⎢
36. Use your calculator to verify: AB = ⎢0 1 0⎥ and BA = ⎢⎢0 1 0⎥⎥ So, A and B are inverses!
⎢⎣0 0 1⎥⎦
⎢⎣0 0 1⎥⎦
⎡4
37. a) ⎢
⎣3
⎡
⎢1
⎢
⎢0
⎢⎣
⎡4 0 10 − 12⎤ 3R2 + R1
6 1 0⎤ ⎡4 6 1 0 ⎤
=⎢
=
⎥=⎢
⎥
⎥
5 0 1⎦ ⎣0 − 2 3 − 4⎦ 3R1 − 4R2 ⎣0 − 2 3 − 4 ⎦
⎤ 1
5
⎤
⎡ 5
− 3⎥
⎢
0 2 − 3⎥ 4 R1
So, A −1 = ⎢ 2
⎥ 1
⎥
3
3
1−
⎢−
2 ⎥ − R2
2⎥
⎥⎦ 2
2
⎦
⎣ 2
⎡1 − 1 1 1 0
⎢
b) ⎢0 2 − 1 0 1
⎢2 3 0 0 0
⎣
⎡1 − 1 0 1
⎢
= ⎢0 2 0 − 4
⎢0 0 − 1 4
⎣
⎡1 0 0 3
3
⎢
⎢0 1 0 − 2 − 2
⎢0 0 1 − 4 − 5
⎣
0⎤
⎥
0⎥ =
1⎥⎦
1
⎡1
⎢
⎢0
⎢0
⎣
0
⎡1 − 1 1 1 0 0 ⎤
0 0⎤
⎥
⎢
⎥
2 − 1 0 1 0⎥
= ⎢0 2 − 1 0 1 0 ⎥
5 − 2 − 2 0 1⎥⎦ R3 − 2R1 ⎢⎣0 0 − 1 4 5 − 2⎥⎦ 5R2 − 2R3
⎤ R1 + R3 ⎡ 1 − 1 0 1
1 0⎤
⎥
⎢
⎥
− 4 − 2⎥ R2 − R3 = ⎢0 1 0 − 2 − 2 1⎥ 0.5R2 =
⎢0 0 1 − 4 − 5 2⎥ − R3
5 − 2⎥⎦
⎣
⎦
− 1⎤ R1 − R2
3 − 1⎤
⎡3
⎥
⎢
A −1 = ⎢ − 2 − 2 1 ⎥⎥
1⎥
⎢⎣− 4 − 5 2 ⎥⎦
2 ⎥⎦
−1
1
1
38. a) 2 → 1 : 5 x − 4(5 x − 2) = −22 ⇒ −15 x + 8 = −22 ⇒ −15 x = −30 ⇒ x = 2 ⇒ y = 5(2) − 2 = 8
(2, 8)
Solve for x or y:
b) –10x + 6y = 4
∅
c) 5x – 3y = 6
⎛ 5 ⎞
⎜ x, x ⎟
10x – 6y = 3
–5x + 3y = –6
⎝ 3 ⎠
0=7
0 = 0 dependent
39.
⎡1
⎢
⎢3
⎢2
⎣
⎡1
⎢
⎢0
⎢0
⎣
2 − 1 − 1⎤ R1 ↔ R2 ⎡ 1
⎥
⎢
1 2 12 ⎥
= ⎢0
⎢0
2 3 10 ⎥⎦
⎣
⎡1
2 − 1 − 1⎤
⎥
⎢
1 − 1 − 3⎥
= ⎢0
0 3 6 ⎥⎦ R3 + 2R2 ⎢⎣0
2
−5
−2
2
1
0
⎡ 1 2 − 1 − 1⎤
− 1 − 1⎤
⎥
⎢
⎥
5 15 ⎥ R2 − 3R1 = ⎢0 1 − 1 − 3⎥ 1 5 R2 =
5 12 ⎥⎦ R3 − 2R1 ⎢⎣0 − 2 5 12 ⎥⎦
− 1 − 1⎤
⎥
− 1 − 3⎥
⇒ Now use back substitution!
1 2 ⎥⎦ 1 3 R3
z = 2 ; y – 2 = –3 so y = –1; x + 2(–1) – 2 = –1 so, x = 3: (3, –1, 2)
61
40.
41.
⎡3
⎢
⎢0
⎢0
⎣
⎡3
⎢
⎢0
⎢0
⎣
⎡3
⎢
⎢0
⎢0
⎣
⎡3 1
2 − 4⎤ ⎡3 1
2 −4 ⎤
2 − 4⎤
⎥ ⎢
⎥
⎢
⎥
− 3 − 2 − 5 ⎥ = ⎢0 − 3 − 2 − 5 ⎥
= ⎢0 − 3 − 2 − 5 ⎥
=
2
5 − 4⎥⎦ ⎢⎣0 0 11 − 22⎥⎦ 2R2 + 3R3 ⎢⎣0 0
1 − 2⎥⎦ 1 2 R3
⎡3 0 0 − 3⎤ R1 − R2
1 0 0 ⎤ R1 − 2R3 ⎡3 1 0 0 ⎤
⎥
⎢
⎥
⎢
⎥
− 3 0 − 9⎥ R2 + 2R3 = ⎢0 1 0 3 ⎥ − 1 3R2 = ⎢0 1 0 3 ⎥
=
⎢0 0 1 − 2⎥
⎢0 0 1 − 2⎥
0 1 − 2⎥⎦
⎣
⎦
⎣
⎦
0 0 − 3⎤ R1 − R2 ⎡ 1 0 0 − 1⎤ 1 3 R1
⎥
⎢
⎥
1 0 3⎥
= ⎢0 1 0 3 ⎥
⇒ (–1, 3, –2)
⎢0 0 1 − 2⎥
0 1 − 2⎥⎦
⎣
⎦
1
⎡2 − 5 3 ⎤
A = ⎢⎢3 2 − 1⎥⎥
⎢⎣ 1 − 3 − 4⎥⎦
⎡− 18⎤
B = ⎢⎢ − 12⎥⎥
⎢⎣ − 4 ⎥⎦
⎡− 5⎤
⎡x ⎤
⎥
⎢
−1
X = ⎢ y ⎥ ⇒ A B = ⎢⎢ 1 ⎥⎥ ⇒ ( −5, 1, − 1) Mult with calculator
⎢⎣ − 1⎥⎦
⎢⎣ z ⎥⎦
42. a) y 2 = 16 x ⇒ 4p = 16 ⇒ p = 4 Opens to the right. Vertex: (0, 0), Focus: (4, 0),
Directrix: x = – 4
b) (x − 2)2 = −8( y + 3) ⇒ 4p = −8 ⇒ p = −2 Vertex: (2, –3), Focus: (2, -5), Directrix: y = –1
y
y
x
x
a)
b)
43. a) x 2 + y 2 = 25 ; Center: (0, 0), r = 5
b) Complete the square to get: ( x − 1)2 + ( y + 4)2 = 20 ; Center: (1, -4), r =
y
20 = 2 5
y
a)
b)
x
x
44. a)
x2 y2
+
= 1 ⇒ a2 = b2 + c 2 ⇒ 25 = 4 + c 2 ⇒ c 2 = 21 ⇒ c = 21 ⇒ Center: (0, 0)
25 4
Vertices: (–5, 0) and (5, 0), Foci: ± 21, 0
(
)
( x + 2)2 ( y − 3)2
+
= 1, Center: (–2, 3), Vertices: (–2, 6) (–2, 0),
4
9
9 = 4 + c 2 ⇒ 5 = c 2 ⇒ c = 5 , Foci: − 2,3 ± 5
b) Complete the square to get:
(
)
62
a)
y
y
b)
x
x
45. a)
y2 x2
−
= 1 Center: (0, 0), Vertices: (0, –3) and (0, 3) c 2 = a2 + b2 ⇒ c 2 = 9 + 36 = 45
9 36
c = 45 = 3 5 ⇒ Foci : 0,−3 ± 3 5
(
)
( x + 2)2 ( y − 3)2
−
= 1 Center: (–2, 3), Vertices: (0, 3) and
4
36
(–4, 3), c 2 = a 2 + b 2 ⇒ c 2 = 4 + 36 = 40 ⇒ c = 40 = 2 10 ⇒ Foci : − 2 ± 2 10 , 3
y
y
a)
b)
b) Complete the square to get:
(
x
)
x
46. Info given tells that center is (2, 3), b = 4 and length of major axis is 10 or a = 5.
( x − 2)2 ( y − 3)2
+
=1
25
16
47. Info given tells major axis is parallel to x-axis and conjugate axis is parallel to y-axis, center
( x + 2) 2 ( y − 2 ) 2
−
=1
is (–2, 2) c 2 = a 2 + b 2 ⇒ 3 2 = a 2 + 2 2 ⇒ a 2 = 5 ⇒
5
4
48. a) By substitution: − 2x + 4 = x 2 + 1 ⇒ x 2 + 2x − 3 = 0 ⇒ ( x + 3)( x − 1) = 0 ⇒ x = −3, 1
y = (− 3 )2 + 1 = 10; y = 12 + 1 = 2 ⇒ (–3, 10) and(1, 2)
5 x 2 − y 2 = 20
b) By addition: − x 2 + y 2 = −4
So,
4 x 2 = 16 ⇒ x 2 = 4 ⇒ x = ±2
49. a)
y
b)
x
y 2 = x 2 − 4 ⇒ (± 2)2 − 4 = 0
( ±2,0)
y
x
63
2
3
⎛ 1 ⎞ 1
2 +1⎛ 2 ⎞
3 +1⎛ 3 ⎞
50. a1 = ( −1)1+1⎜
⎟ = ; a 2 = ( −1) ⎜
⎟ = − ; a 3 = ( −1) ⎜
⎟= ;
3
⎝ 1 + 1⎠ 2
⎝ 2 + 1⎠
⎝ 3 + 1⎠ 4
4
100
⎛ 4 ⎞
100 + 1⎛ 100 ⎞
a 4 = ( −1) 4 +1⎜
⎜
⎟=−
⎟ = − ; a100 = ( −1)
5
101
⎝ 100 + 1⎠
⎝ 4 + 1⎠
51. a2 = 2a1 + 1 = 2(1) + 1 = 3; a3 = 2a2 + 1 = 2(3) + 1 = 7; a4 = 2a3 + 1 = 2(7) + 1 = 15;
a5 = 2a4 + 1 = 2(15) + 1 = 31.
1, 3, 7, 15, 31
5
52.
∑ 2k −1 = 21−1 + 2 2 −1 + 23 −1 + 2 4 −1 + 25 −1 = 1 + 2 + 4 + 8 + 16 =
31
k =1
53. a n = 3 + (n − 1) 3 = 3 + 3 n − 3 ⇒
an = 3 n
a10 = 10 3
20
54.
∑ (1 − 2n) = (1 − 2(0)) + (1 − 2(1)) + (1 − 2(2)) + ... + (1 − 2(20)) = 1 − 1 − 3 − ... − 39
n=0
21
(1 + ( −39)) = − 399
2
55. a 5 = a1r 5 −1 ⇒ −7203 = −3r 4 ⇒ 2401 = r 4 ⇒ r = 4 2401 = 7 ⇒ a n = −3(7)n −1
S 21 =
a 3 = −3(7) 2 = −147
56. S10 = 4( −3)9 = −78732
n
∞
0
1
2
1
1 3
⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
= =
57. ∑ ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ... ⇒ S ∞ =
1 2 2
⎝3⎠
⎝3⎠ ⎝3⎠
n = 0⎝ 3 ⎠
1−
3 3
1
58. 1 + 4 + 7 + … + (3n – 2) = 2 n(3n – 1)
Proof
1
Prove true for n = 1: 1 = (1)(3(1) − 1) ⇒ 1 = 1
2
Assume true for n= k, prove true for n = k+1:
1
1 + 4 + 7 + … + (3k – 2)+(3(k + 1) –2) = 2 (k+1)(3(k + 1) – 1)
1
1 + 4 + 7 + … + (3k – 2)+(3k + 3 –2) = 2 (k+1)(3k + 3 – 1)
1
1
2 k(3k – 1) + (3k + 1) = 2 (k+1)(3k + 2)
3 2 1
1
3
5
1
k − k + 3k + 1 = (k + 1)(3k + 2) ⇒ k 2 + k + 1 = (k + 1)(3k + 2) ⇒
2
2
2
2
2
2
(
)
1
1
1
1
3k 2 + 5k + 2 = (k + 1)(3k + 2) ⇒ (k + 1)(3k + 2) = (k + 1)(3k + 2)
2
2
2
2
64
59. 42504
(use your calculator 24 [MATH][PRB]3:nCr 5 [Enter])
60. a) (2a + 5 )6 =
⎛ 6⎞
⎛ 6⎞
⎛ 6⎞
⎛ 6⎞
⎛ 6⎞
⎛ 6⎞
⎛ 6⎞
⎜⎜ ⎟⎟(2a )6 + ⎜⎜ ⎟⎟(2a )5 (5 ) + ⎜⎜ ⎟⎟(2a )4 (5 )2 + ⎜⎜ ⎟⎟(2a )3 (5 )3 + ⎜⎜ ⎟⎟(2a )2 (5 )4 + ⎜⎜ ⎟⎟(2a )(5 )5 + ⎜⎜ ⎟⎟(5 )6
⎝0⎠
⎝ 1⎠
⎝ 2⎠
⎝ 3⎠
⎝ 4⎠
⎝5⎠
⎝ 6⎠
(
) (
)
(
)
( )
( )
= 1 64a 6 + 6 32a 5 (5 ) + 15 16a 4 (25 ) + 20 8a 3 (125 ) + 15 4a 2 (625 ) + 6(2a )(3125 ) + 1(15625 )
= 64a 6 + 960a 5 + 6000a 4 + 20000a 3 + 37500a 2 + 3137a + 15625
(
)
5
b) x 2 − 3 y =
⎛5⎞ 2 5 ⎛5⎞ 2
⎜⎜ ⎟⎟ x
+ ⎜⎜ ⎟⎟ x
⎝0⎠
⎝ 1⎠
( )
( ) (− 3y ) + ⎛⎜⎜ 52⎞⎟⎟(x ) (− 3y )
( ) ( )
4
2 3
2
⎝ ⎠
( )( ) ( )(
( ) (− 3y )
⎛5⎞
+ ⎜⎜ ⎟⎟ x 2
⎝3⎠
2
) ( )(
3
( )
⎛5⎞
⎛5⎞
+ ⎜⎜ ⎟⎟ x 2 (− 3 y )4 + ⎜⎜ ⎟⎟(− 3 y )5
⎝ 4⎠
⎝5⎠
) (
= 1 x 10 + 5 x 8 (− 3 y ) + 10 x 6 9 y 2 + 10 x 4 − 27 y 3 + 5 x 2 81y 4 + 1 − 243 y 5
)
= x 10 − 15 x 8 y + 90 x 6 y 2 − 270 x 4 y 3 + 405 x 2 y 4 − 243 y 5
65