Math 1050
Final Exam Review
The problems below are intended as a review of the topics that will be
covered on the final exam. The final exam will consist of a subset (about 50)
questions, which will be drawn from the types of questions in this sheet, with
different numbers. You should also be sure to study from the topics list,
which is posted separately.
Sets and Numbers
1. Put the following sets in order from smallest to largest: R, Z, N, Q.
Solution:
N⊆Z⊆Q⊆R
2. Decide whether each of the following statements is true or false:
(a) π ∈ Q
√
(b) 2 ∈
/R
(c) 0 ∈ Q
(d) 0 ∈ N
(e) N ⊆ Z − {0}
Solution:
(a) False, π is not a rational number.
√
(b) False, 2 is a real number.
(c) True, 0 is a rational number.
(d) False, 0 is not a natural number.
(e) True.
Rules for Numbers
3. Determine whether the following statements are true or false:
(a) 2 ∈ (2, ∞)
(b) π ∈ (3, 5)
(c) 27 ∈ (−∞, 3)
Solution:
(a) False, the interval (2, ∞) is the set of real numbers which are greater than 2,
and 2 is not in that set.
(b) True, the interval (3, 5) is the set of real numbers which are greater than 3 and
less than 5. π is a real number between 3 and 5.
(c) False, the interval (−∞, 3) is the set of all real numbers which are less than 3,
and 27 is larger than 3.
4. Determine whether each statement is true or false. State which rules of numbers (law of
associativity, commutativity, inverses, or identity, or the distributive law) are followed
or broken. If it is false, write a true statement that follows the rules.
(a) 3x − 3y = 3(x + y)
(b) 3 31 = 0
(c) x + 2x = x(1 + 2)
(d) 3x2 + 1 = 3(x2 + 1)
Solution:
(a) False, the distributive law is violated. A true statement would be 3x − 3y =
3(x − y) or 3x + 3y = 3(x + y).
(b) False, the law of inverses for multiplication is violated. A true statement would
be 3 31 = 1.
(c) True, it follows the distributive law: a(b + c) = ab + ac.
(d) False, the distributive law is violated. A true statement would be 3x2 + 1 =
3(x2 + 31 ) or 3x2 + 3 = 3(x2 + 1).
5. Write down the following rules for numbers:
(a) Law of Associativity for Addition
(b) Law of Commutativity for Addition
(c) Law of Inverses for Addition
(d) Law of Identity for Addition
(e) Law of Associativity for Multiplication
(f) Law of Commutativity for Multiplication
(g) Law of Inverses for Multiplication
Page 2
(h) Law of Identity for Multiplication
(i) Distributive Law
Solution:
(a) Law of Associativity for Addition: a + (b + c) = (a + b) + c
(b) Law of Commutativity for Addition: a + b = b + a
(c) Law of Inverses for Addition: a + (−a) = 0
(d) Law of Identity for Addition: a + 0 = a
(e) Law of Associativity for Multiplication: a(bc) = (ab)c
(f) Law of Commutativity for Multiplication: ab = ba
(g) Law of Inverses for Multiplication: a a1 = 1
(h) Law of Identity for Multiplication: a · 1 = a
(i) Distributive Law: a(b + c) = ab + ac
6. Write the following set as a set or an interval: “The set of real numbers which are greater
than or equal to -3.”
Solution:
[−3, ∞) or R − (−∞, −3)
7. Write the following set as a set or an interval: “The set of rational numbers, except for
2, −4, and 12 .
Solution:
1
Q − {2, −4, }
2
Solving Some Simple Equations
8. Solve for x if 3x − 2 = 4x − 5, using the rules from the previous section.
Page 3
Solution:
3x − 2 = 4x − 5
3x − 2 + 5 = 4x − 5 + 5
3x + 3 = 4x
−3x + 3x + 3 = −3x + 4x
3=x
(Law of Inverses for Addition)
(Simplify)
(Law of Inverses for Addition)
(Simplify)
Intro to Functions
9. Determine whether the following rule is a function:
“Assign to every real number x the positive real number y such that y 2 = x.”
Solution: This rule is not a function, because it “takes in” real numbers (including
negative numbers and 0) but there is no positive real number y whose square (y 2 ) is
either negative or zero. So the rule does not assign every object in the domain an
object in the target.
10. What is the target of the function g : N → [1, ∞) where g(x) = 2x + 21 ?
Solution: The target is [1, ∞).
11. If f (x) =
2x2 +1
,
x−3
find f (2). Write your answer in simplest form.
Solution:
2(2)2 + 1
2−3
2(4) + 1
=
−1
8+1
=
−1
9
=
−1
= −9
f (2) =
Page 4
Sequences
12. Decide whether each of the following sequences is arithmetic, geometric, or neither. If it
is arithmetic, find the common difference. If it is geometric, find the common ratio.
(a) 3, −6, 12, −24, . . .
(b) 3, 6, 9, 12, . . .
(c) 115, 108, 101, 94, . . .
(d) 8, −8, 8, −8, . . .
(e) 1, 4, 9, 16, . . .
Solution:
(a) The sequence is geometric with a common ratio of −2.
(b) The sequence is arithmetic with a common difference of 3.
(c) The sequence is arithmetic with a common difference of −7.
(d) The sequence is geometric with a common ratio of −1.
(e) The sequence is neither arithmetic or geometric.
13. Find the 56th term of the arithmetic sequence 12, 5, −2, −9, . . ..
Solution: We know that the sequence is arithmetic. The first term, a1 , is 12. We
can find the common difference, d, by subtracting the first term from the second
term:
5 − 12 = −7
To find the 56th term in the sequence, we need to use the formula for the nth term
in an arithmetic sequence:
an = a1 + (n − 1)d
In this case, n = 56, a1 = 12, d = −7, so
a56 = 12 + (56 − 1)(−7)
We could simplify further to get
a56 = 12 + 55(−7)
= 12 + −385
= −373
Page 5
14. Find the 33rd term in the geometric sequence 4, 2, 1, 12 , . . ..
Solution: We know that the sequence is geometric. The first term, a1 , is 4. The
common ratio can be found by dividing the second term by the first term:
2
1
=
4
2
To find the 33rd term in the sequence, we need to use the formula for the nth term
in a geometric sequence:
an = a1 (r)n−1
In this case, n = 33, a1 = 4, r = 12 , so
1
a33 = 4( )33−1
2
We can simplify a bit further to find that
1
a33 = 4( )3 2
2
2 −32
= 2 (2 )
= 22−32
= 2−30
=
1
1
or( )30
30
2
2
Sums and Series
15. Find the sum
3
X
(i2 + 1). Write your answer as an integer in simplest form.
i=0
Solution:
3
X
(i2 + 1) = (02 + 1) + (12 + 1) + (22 + 1) + (32 + 1)
i=0
= 1 + 2 + 5 + 10
= 18
16. Find the sum
100
X
(2i). Write your answer as an integer in simplest form.
i=1
Page 6
Solution:
!
100
100
X
X
(2i) = 2
(i)
i=1
i=1
= 2(1 + 2 + 3 + . . . + 99 + 100)
100
(1 + 100)
=2
2
= 100(101)
= 10100
17. Find the sum of the first 100 terms of the arithmetic sequence 4, −2, −8, −14, . . .. Write
your answer as a single term (you do not need to simplify further).
Solution: To find the sum of the first through k th terms of an arithmetic sequence,
we can use the formula:
X
k
i = 1k ai = (a1 + ak )
2
In this case k is 100 (the number of terms and the index of the last term. The first
term is 4, and we need to find the 100th term of the sequence:
a100 = 4 + (100 − 1)(−6) = 4 + 99(−6) = 4 − 594 = −590
So the sum is
X
i = 1100 ai =
100
(a1 + a100 ) = 50(4 − 590) = 50(−586)
2
18. Find the value of the series 6 + 2 + 32 +
(you do not need to simplify further).
2
9
+ . . . and write your answer as a single term
Solution: This is a geometric series (infinite sum of a geometric sequence) with
a1 = 6 and r = 13 . The formula for the value of a geometric series is
∞
X
ai =
i=1
a1
1−r
In this case the value of the sum will be
6
3
6
1 = 2 = 6( )
2
1− 3
3
Page 7
Counting I & II
19. There are 100 questions in a test bank, and a teacher wants to create a practice exam
with 50 questions. If the order of the questions doesn’t matter, and the teacher doesn’t
want to repeat any questions, how many different exams are possible?
Solution: In this case, we are choosing 50 things, from a pool of 100, without
caring
the order and without repeats. So the number of possible tests is
about
100
100!
=
.
(50!)(50!)
50
20. Suppose that you are on a scholarship committee at a university. You have chosen the
top 10 students, and you have 10 different scholarships to award. How many ways are
there to award one scholarship to each of the top students?
Solution: If we need to award one scholarship to each student, and all of the scholarships are different, then there are 10! different ways to make the awards.
21. You are teaching a high school English class. Your school has a list of 20 recommended
books, and you have time to read 8 of them over the course of the semester. If the order
in which you read the books matters, how many semester-long literary adventures can
you create for your students?
Solution: Since we are choosing 8 books from a pool of 20, and the order of the
20!
= 20!
.
books matters, the number of ways to do this is (20−8)!
12!
12
22. Write
as a natural number in simplest form.
9
Solution:
12!
12
=
9
(9!)(3!)
12 · 11 · 10 · (9!)
=
(9!)(3!)
12 · 11 · 10
=
3·2·1
= 2 · 11 · 10
= 220
Page 8
23. Use the Binomial Theorem
to rewrite (x − 2y)6 so that your answer doesn’t include any
numbers that look like nk .
Solution: The 6th row of Pascal’s Triangle is “1, 6, 15, 20, 15, 6, 1.” The Binomial
Theorem tells us that (x − 2y)6 can be rewritten as
„ «
„ «
„ «
„ «
„ «
„ «
„ «
6
6
6
6
6
6
6
6
0
5
1
4
2
3
3
2
4
1
5
0
6
x (−2y) +
x (−2y) +
x (−2y) +
x (−2y) +
x (−2y) +
x (−2y) +
x (−2y)
0
1
2
3
4
5
6
We can use Pascal’s Triangle to rewrite that as:
1x6 (−2y)0 + 6x5 (−2y)1 + 15x4 (−2y)2 + 20x3 (−2y)3 + 15x2 (−2y)4 + 6x1 (−2y)5 + 1x0 (−2y)6
The answer above or anything equivalent to it would be acceptable. Just out of
curiosity, let’s rewrite this in “simplest” form:
x6 − 12x5 y + 60x5 y 2 − 160x3 y 3 + 240x2 y 4 − 192xy 5 + 64y 6
More on Functions
24. If f (x) = x2 + 1 and g(x) = 4x, find f ◦ g(x).
Solution:
f ◦ g(x) = f (g(x))
= f (4x)
= (4x)2 + 1
= 16x2 + 1
25. Find the implied domain of f (x) =
3x2 +1
.
(x−2)(x+5)
Solution: The potential problem with this function is division by zero. The denominator is zero if x is 2 or -5 (those are the roots of the polynomial (x − 2)(x + 5)),
but we can plug in any other real number and the denominator will be non-zero. So
the implied domain is R − {2, −5}.
26. Find the implied domain of g(x) =
√
4
8 − 4x.
Page 9
Solution: The potential problem with this function is taking the 4th root of a negative number. So we need to make sure that 8 − 4x ≥ 0. We can solve this inequality
for x:
8 − 4x ≥ 0
8 − 4x + 4x ≥ 0 + 4x
8 ≥ 4x
1
1
(8) ≥ 4x(
4
4
2≥x
If x is any real number which is less than or equal to 2, then the function makes
sense. So the implied domain is the set of all real numbers which are less than or
equal to 2, which we write as
(−∞, 2]
27. Find the implied domain of h(x) =
√
5
x − 3.
√
Solution: The function 5 has domain R. So we can plug any real number into the
√
5
. Whenever x is a real number, x − 3 is also a real number, so we can plug any
real number into h(x) and get something that makes sense. So the implied domain
of h is R.
28. What is the domain of the function f : N → R?
Solution: The domain is N. This is what we are told in the problem.
29. What is the target of the function
√
2
as defined in class?
√
Solution: By definition, the 2 function is the inverse of the function x2 : [0, ∞) →
[0, ∞). Since the target of the inverse function is the domain of the original function,
√
the target of 2 is [0, ∞).
30. What is the range of the function f : R → R where f (x) = x2 ?
Page 10
Solution: Looking at the graph of f (x), we see a parabola, which sits on the x-axis
and opens upwards. If we look at the shadow of this graph on the y-axis, we should
see any positive real number or 0, and no negative real numbers. So the range is
[0, ∞)
Intro to Graphs
31. Graph the following functions. Label any x and y-intercepts, and at least one point on
the graph which is not an intercept. (You must use numbers to indicate scale. “Tick
marks” will not be assumed to have any “usual” meaning.)
(a) id(x)
(b) x2
(c) x5
(d)
(e)
1
x3
1
x4
(f) f (x) = 4
(g) g : {−2, 0, 3, 4} → R where g(x) = x + 2.
Solution:
(a)
Page 11
(b)
(c)
(d)
Page 12
(e)
(f)
(g)
32. Is the following graph the graph of a function? Explain why or why not.
Page 13
Solution: Yes. The graph passes the vertical line test so it is the graph of a function.
33. Given the graph of f (x) below, answer the following questions.
(a) What is f (2)?
(b) What is the domain of f ?
(c) What is the range of f ?
(d) What are the x-intercepts of the graph?
(e) What are the y-intercepts of the graph?
(f) Assuming that the target of f is equal to its range, determine whether f has an
inverse function. Explain your reasoning.
Page 14
Solution:
(a) f (2) = −4
(b) R − {3, 5}
(c) [−4, ∞]
(d) −2, 0, 1
(e) 0
(f) f does not have an inverse function because it is not one-to-one.
Graph Transformations
34. Graph the following functions. Label any x and y intercepts, vertical or horizontal
asymptotes.
√
(a) 5 32x − 2
(b) log 1 (−x)
3
x+3
(c) 2
(d) −2(x + 1)2 − 1 (In addition to the above instructions, label the vertex of the
parabola.)
√
(e) 2 −x + 1
(f)
1
x2 +2x+1
1
, and
xn
(Hint: Turn this function into a transformation of a function that looks like
then graph.)
Solution:
√
(a) 5 32x − 2
Page 15
(b) log 1 (−x)
3
(c) 2x+3
Page 16
(d) −2(x + 1)2 − 1
(e)
√
2
−x + 1
Page 17
(f)
1
x2 +2x+1
35. Given the graph of f (x) in question 33, graph the function −f (x).
Solution:
Page 18
Inverse Functions
36. If f (x) = 2x + 1 and g(x) = 12 x − 1, are f and g inverse functions?
Solution: To check whether f and g are inverses, we need to check whether f ◦ g
and g ◦ f are both the identity function:
f ◦ g(x) = f (g(x))
1
= f ( x − 1)
2
1
= 2( x − 1) + 1
2
=x−2+1
= x − 1 6= id(x)
Since f ◦ g(x) 6= id(x), we know that f and g are not inverse functions.
37. Find the inverse of the function h(x) =
3x−1
.
x+3
Solution: To find the inverse function, first set h(x) = y:
y=
3x − 1
x+3
Page 19
Next switch x and y:
x=
3y − 1
y+3
Now solve for y:
3y − 1
y+3
x(y + 3) = 3y − 1
xy + 3x = 3y − 1
1 + 3x = 3y − xy
3x + 1 = y(3 − x)
3x + 1
=y
3−x
x=
Finally, set f −1 (x) = y:
f −1 (x) =
3x + 1
3−x
You could check your work by checking that f ◦f −1 (x) = id(x) and f −1 ◦f (x) = id(x).
38. The graph of a function f : R → [−3, ∞) is shown below. Does f have an inverse?
Solution: The function does not pass the horizontal line test, so it is not one-to-one.
A function needs to be one-to-one in order to have an inverse, so this function does
not have an inverse.
Page 20
39. The graph of a function g : R → (0, ∞) is shown below. Does g have an inverse?
Solution: The function passes the horizontal line test so it is one-to-one. We can
see from the graph that the range of g is (0, ∞), which is the same as the target of
the function (the target is specified in the question) so the function is onto. Any
function which is one-to-one and onto has an inverse, so g does have an inverse.
40. The graph of an invertible function h : R → (0, ∞) is shown below, along with the graph
of the line y = x. Graph the inverse function h−1 (x).
Page 21
Solution:
nth Roots
41. What is the domain of the function
√
4
: R → R?
Solution: The 4th root function is defined as having domain [0, ∞).
**For implied domain questions, see the questions from ”More on Functions.”
√
42. Graph the function f (x) = 2 4 x + 1 and label any x and y intercepts.
Solution:
Page 22
43. Graph the function g(x) =
√
5
x − 1 and label any x and y intercepts.
Solution:
Basics of Polynomials
44. What is the leading coefficient of the polynomial 3(x − 2)(x + 2)(x2 + 1)?
Page 23
Solution: The leading term of a product of polynomials is the product of the leading
terms of the factors. So the leading term is
3(x)(x)(x2 ) = 3x4
45. If p(x) = 3x2 + 4x + 1 and q(x) = x4 − 2x + 3, find the polynomial p(x) − q(x).
Solution:
p(x) − q(x) =
=
(
3x2 + 4x + 1)
− (x4
− 2x + 3)
4
2
−x + 3x + 4x − 2
46. True or false: The polynomial 4x1 7 + 2x2 + 3 has 19 roots.
Solution: False. The degree of the polynomial is 17, which means that it has at
most 17 roots.
47. What is the degree 4 coefficient of the polynomial 4x1 7 + 2x2 + 3?
Solution: The degree 4 coefficient is 0; there is no x4 term.
48. If p(x) = 2x4 + 5 and q(x) = −x5 + 3x, find their product, p(x)q(x), and write it in
simplest form.
Solution:
p(x)q(x) = (2x4 + 5)(−x5 + 3x)
= (2x4 + 5)(−x5 ) + (2x4 + 5)(3x)
= (2x4 )(−x5 ) + 5(−x5 ) + (2x4 )(3x) + 5(3x)
= −2x9 − 5x5 + 6x5 + 15x
− −2x9 + x5 + 15x
Division
Page 24
49. Find the quotient:
2x4 + 3x2 − x + 1
.
x2 − 2
Solution:
2x2 + 7 +
50. Find the quotient
−x + 15
x2 − 2
3x5 + 2x + 5
.
x+1
Solution:
3x4 − 3x3 + 3x2 − 3x + 5
Roots and Factors
51. Is 3x4 − 3x3 + 3x2 − 3x + 5 a factor of the polynomial 3x5 + 2x + 5?
Solution: To determine whether one polynomial is a factor of another, we can divide
by the potential factor. If we get a remainder of 0 then it is a factor. Use long division
to find the quotient:
3x5 + 2x + 5
=x+1
3x4 − 3x3 + 3x2 − 3x + 5
Yes, 3x4 − 3x3 + 3x2 − 3x + 5 is a factor of 3x5 + 2x + 5.
52. Given that −2 is a root of the polynomial p(x) = 3x3 + 2x2 − 10x − 4, write p(x) as the
product of a linear polynomial and a quadratic polynomial.
Solution: Since −2 is a root, we know that x + 2 is the linear factor. Dividing p(x)
by x + 2, we can find the quadratic factor:
3x3 + 2x2 − 10x − 4
= 3x2 − 4x − 2
x+2
So p(x) = (x + 2)(3x2 − 4x − 2).
53. Given the graph of the polynomial p(x) below, list as many linear factors of p(x) as you
can.
Page 25
Solution: The x-intercepts of the graph are 2, −4, 0, and −1. So the roots of the
polynomial (the values for x which make p(x) = 0), are 2, −4, 0, and −1, and the
linear factors are (x − 2), (x + 4), (x), and (x + 1).
54. Is −3 a root of the polynomial p(x) = 2x4 − 9x2 + 2x + 6?
Solution: To determine whether −3 is a root of p(x), check whether p(−3) = 0:
p(−3) = 2(−3)4 − 18(−3)2 + 2(−3) + 6
= 2(81) − 18(9) − 6 + 6
= 162 − 162 + 6 − 6
=0
55. What is the degree of the polynomial (3x1 5 + 6)(x2 + 1)(x − 2)?
Solution: The degree is the sum of the degrees of the factors: 15 + 2 + 1 = 18
Constant and Linear Polynomials
56. Graph the function h(x) = −3 and label any x- and y-intercepts.
Page 26
Solution:
57. How many roots does the polynomial p(x) = 5 have?
Solution: p(x) is never 0, so it does not have any roots.
58. Graph the polynomial q(x) = 2x − 4 and label any x and y-intercepts.
Solution:
Page 27
59. Find all roots of the polynomial p(x) = 3x − 5.
Solution: A linear polynomial has exactly one root. We can find the root by either
completely factoring or by setting p(x) = 0 and solving:
3x − 5 = 0
3x = 5
5
x=
3
The root is 53 .
60. What is the slope of the line that passes through the points (5, −1) and (2, 3)?
Solution:
y2 − y1
x2 − x1
3 − (−1)
=
2−5
4
=
−3
4
=−
3
slope =
The slope of the line is − 34 .
Quadratic Polynomials
61. Complete the square: Write the polynomial p(x) = −2x2 −4x+3 in the form α(x+β)2 +γ
where α, β, γ ∈ R.
Solution:
p(x) = −2(x + 1)2 + 5
62. Graph the polynomial q(x) = −2(x − 1)2 + 4 and label the y-intercept.
Page 28
Solution:
63. How many roots does the polynomial −x2 − 3x + 1 have?
Solution: Check the discriminant to determine how many roots a quadratic polynomial has:
b2 − 4ac = (−3)2 − 4(−1)(1) = 9 + 4 = 13
Since the discriminant is greater than 0, the polynomial has 2 roots.
64. Find all roots of the polynomial x2 − 2x − 2.
Solution: First, double-check the discriminant to determine how many roots the
polynomial has:
b2 − 4ac = (−2)2 − 4(−2)(1) = 4 + 8 = 12
Since the discriminant is greater than 0, we are looking for two roots. They are:
√
√
1 + 3, and 1 − 3
Factoring Polynomials
65. Is the polynomial (2x2 + 1)(x − 2)(x + 3)(x − 1) completely factored?
Page 29
Solution: No, because the factor (2x2 + 1) is quadratic but not monic. For a polynomial to be completely factored, all quadratic factors must be monic, and have no
roots.
66. Completely factor the polynomial p(x) = 3x − 1.
Solution:
1
p(x) = 3(x − )
3
67. Completely factor the polynomial q(x) = 2x2 − 4x + 1.
Solution: First, check the discriminant to determine how many roots the polynomial
has:
b2 − 4ac = (−4)2 − 4(2)(1) = 16 − 8 = 8
Since the discriminant is greater than 0, the polynomial has two roots. Using the
quadratic equation to find them, we get
√
√
2+ 2
2− 2
and
2
2
So the completely factored form is
q(x) = 2 x −
√ !!
2+ 2
x−
2
√ !!
2− 2
2
68. Find a root of the polynomial p(x) = −2 − x − 3x2 − 2x3 + 2x4 .
Solution: Check factors of the degree 0 coefficient (−2 in this case) for roots. The
factors of −2 are 1, −1, 2, −2. Both p(−1) and p(2) are 0, so −1 and 2 are both
roots.
69. Completely factor the polynomial 3x3 − 5x2 − 11x − 3 given that 3 is a root.
Page 30
Solution:
1
3(x + )(x + 1)(x − 3)
3
70. Find all roots of the polynomial p(x) = −2(x − 1)(x + 2)(x − 1)(x)(x2 + 2).
Solution: Since p(x) is completely factored, the roots can be found from the linear
factors:
1, −2, 0
71. List all of the monic linear factors of q(x) that you know from its graph:
Solution:
(x + 1),
(x − 3)
Graphing Polynomials
72. Graph the polynomial −3(x − 2)(x + 4)(x + 1)(x2 + 1), and label the x- and y-intercept.
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Solution:
Rational Functions
73. Find all x-intercepts of the rational function
2(x − 2)(x + 1)(x2 + 4)
r(x) =
−2(x + 4)(x2 + 1)
Solution: x-intercepts are the roots of the numerator, which are 2, −1.
74. Find all vertical asymptotes of the rational function
2(x − 2)(x + 1)(x2 + 4)
r(x) =
−2(x + 4)(x2 + 1)
Solution: Vertical asymptotes are the roots of the denominator. In this case, the
only vertical asymptote is to the line x = −4.
75. Calculate the end behavior of the rational function
2(x − 2)(x + 1)(x2 + 4)
r(x) =
−2(x + 4)(x2 + 1)
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(In other words, find the function that r(x) looks like to the far left and far right of its
graph.
Solution: The end behavior is the ratio of leading terms:
2(x)(x)(x2 )
2x4
=
= −x
−2(x)(x2 )
−2x3
So the graph of r(x) looks like the graph of the function −x to the far left and far
right of its graph.
76. Graph the rational function
r(x) =
2(x − 2)(x + 1)(x2 + 4)
−2(x + 4)(x2 + 1)
Solution:
Exponential Functions
77. Rewrite the following expression as a single power of a:
r
a2
4
a−3
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Solution:
5
a4
78. Graph the function f (x) = −( 12 )x + 2, and label any x and y-intercepts.
Solution:
79. Suppose your credit card company charges you 3% interest each month. If you initially
owe $5,000, how much will you owe after 1 year (12 months) of not making payments?
Solution:
$5, 000(1 + 0.03)12
80. If
3x
9
= 3, solve for x.
Solution:
3x
=3
9
3x = 81
x = 4 since 34 = 81
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81. Why can’t an exponential function have base 1?
Solution: The function f (x) = 1x is a constant function (f (x) = 1), not an exponential function.
Logarithms
82. Graph the function g(x) = log3 (x + 3) and label any x and y-intercepts.
Solution:
83. Find the implied domain of the function h(x) = log 1 (3 − 2x) and write your answer as
2
a set or an interval, using the notation from Chapter 1.
Solution: In order for this function to make sense, we need to ensure that 3 − 2x is
positive (not negative or zero). Set up an inequality and solve to find the restriction
on x:
3 − 2x > 0
3 > 2x
3
>x
2
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The implied domain is the interval which consists of all real numbers less than 32 :
3
(−∞, )
2
2
84. Rewrite the expression loge ( xy3z ) as a sum of terms involving only loge (x), loge (y), and
loge (z).
Solution:
loge
x2 z
y3
= loge (x2 z) − loge (y 3 )
= loge (x2 ) + loge (z) − loge (y 3 )
= 2 loge (x) + loge (z) − 3 loge (y)
85. Rewrite the expression log2 (x) − 2 log2 (y) + 3 log2 (z) as a single logarithm.
Solution:
log2 (x) − 2 log2 (y) + 3 log2 (z) = log2 (x) − log2 (y 2 ) + log2 (z 3 )
x
= log2
+ log2 (x3 )
y2
3
xz
= log2
y2
86. Rewrite log2 15 in terms of logarithms with base 10.
Solution:
log2 15 =
log10 (15)
log10 (2)
87. What is the greatest integer which is less than log5 (59)?
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Solution: We know that log5 (25) = 2 and log5 (125) = 3. Since 25 < 59 < 125, we
know that 2 < log5 (59) < 3, so 2 is the greatest integer less than log5 (59).
88. Write log1000 (100) as a rational number. (Compute the exact value of log1000 (100).
Solution: First, use the change of base formula to change to a common base (in
this case, 10):
log10 (100)
log10 (1000)
Next, simplify log10 (100) = 2 and log10 (1000) = 3:
log1000 (100) =
2
3
89. If log4 (x) = 54 , solve for x.
Solution:
5
x = 44 =
√
4
45
Exponential and Logarithmic Functions
90. Solve for x if log5 (x + 1) + log5 (2x) = 3 log5 (2) + 1.
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Solution:
log5 (x + 1) + log5 (2x) = 3 log5 (2) + 1
log5 (2x(x + 1)) = log5 (8) + 1
log5 (2x(x + 1)) − log5 (8) = 1
2x(x + 1)
log5
=1
8
2x(x + 1)
5=
8
40 = 2x(x + 1)
40 = 2x2 + 2x
0 = 2x2 + 2x − 40
0 = 2(x2 + x − 20)
0 = 2(x + 5)(x − 4)
x = −5 or x = 4
Plugging back into the original equation: x = −5 is not in the domain of log5 (x + 1)
or log5 (2x). So x = 4 is the only solution.
91. Solve for x if 32(4x ) = 2x .
Solution:
32(4x ) = 2x
2x
32 = x
4 x
2
32 =
4
log 1 (32) = x
2
x = −5
Piecewise Defined Functions
92. Find p(1) if

2

x + 1 if x ∈ (−∞, 1)
p(x) = −2
if x = 1

 3
x
if x ∈ (1, ∞)
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Solution:
p(1) = −2
93. Graph the function:


−2 if x ∈ (−2, 0)
f (x) = 3
if x ∈ [0, 4)


−1 if x ∈ (4, ∞)
Solution:
94. The function g(x) = |x + 17| measures the distance from x to what number?
Solution: -17
95. Solve for x: |2 − 4x| < 5.
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Solution:
|2 − 4x| < 5
−5 < 2 − 4x < 5
−7 < −4x < 3
3
7
>x>−
4
4
3
7
− <x<
4
4
96. Compute the exact value of the following expression:
|2(3) − 4|| − 2| + 1
Solution: 5
97. State the Triangle Inequality, and briefly explain what it means.
Solution: The triangle inequality says that |x − y| ≤ |x − z| + |z − y|. It means that
the distance from x to y is no more than the distance from x to z plus the distance
from z to y.
Linear Equations in Two Variables
98. Is x = 1 and y = −1 a solution to the following system of equations?
(
2x + 2y = 0
2x − 2y = −4
Solution: x = 1 and y = −1 is not a solution to the second equation, so it is not a
solution to the system.
99. Graph the solution set of the linear equation 3x − y = −3.
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Solution:
100. Does the following system of equations have a unique solution?
(
x − 2y = −1
−2x + 4y = −2
Solution: The two lines have the same slope but different intercepts, so there are
no solutions to the system. No, the system does not have a unique solution.
Substitution
101. The following system of equations has a unique solution. Find the solution:
(
2x − 3y = 11
3x − y = −1
Solution: x = −2, y = −5
Linear Equations in Three Variables
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102. Is x = 1, y = −1, z = 0 a solution to the following system of equations?


2x − 3y + z = 5
3x + z = 1


−x − 3y + 6z = 2
Solution: No. It is not a solution to the second equation.
Vectors and Matrices
103. Compute the following matrices and vectors:

 

2
−3
(a)  −1  +  0 
17
10



2 3 −1
−1
(b)  −1 7 2   0 
0 4 −5
2


5
(c) 2  −1 
3


−1 0
5 3 −1 
2 0 
(d)
0 2 0
0 1
Solution:


−1
(a)  −1 
27


−4
(b)  5 
−10


10
(c)  −2 
6
1 −1
(d)
4 0
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104. Are the following matrices inverses?




−2 0
1
−1 −1 −1
 0 −1 −1  , and  1
1
2 
1
1
0
−1 −2 −2
Solution: Yes.
105. Write down the 2x2 identity matrix.
Solution:
1 0
0 1
Determinants and Inverse Matrices
106. Explain the relationship between determinants and inverses.
Solution: The determinant of a matrix tells you whether it has an inverse or not.
If the determinant is nonzero, the matrix has an inverse. If the determinant is zero,
the matrix does not have an inverse.
107. Compute det
−1 2
−2 1
.
Solution: −1 − (−4) = 3


−1 0 2
108. Compute det  −2 0 3 
1 1 1
Solution: 3 + 0 − 4 = −1


−1 0 2
109. Does the matrix  −2 0 3  have an inverse?
1 1 1
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Solution: Yes.
110. Find the inverse of the matrix:
Solution:
−1 0
−2 −3
−1
2
3
0
− 31
Matrix Equations
111. Solve the following system of equations
(
3x − y = 0
−2x + y = 2
3 −1
x
0
Solution: First, convert to a matrix equation:
=
Sec−2
1
y
2
3 −1
ond, find the determinant of the matrix
: 3 − 2 = 1. Since the deter−2 1
minant is nonzero, the matrix has an inverse.
−1 1 1
3 −1
=
2 3
−2 1
If we multiply both sides of the original matrix equation by the inverse from above,
we can find the solution:
x
1 1
0
2
=
=
y
2 3
2
6
So x = 2, y = 6.
112. Use the fact that
−1 

−1 0 2
3 −2 0
 −2 0 3  =  −5 3 1 
1 1 1
2 −1 0

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to solve the system of equations


−x + 2z = −1
−2x + 3z = 1


x+y+z =0
.
Solution: First, convert to a matrix equation:

  

−1 0 2
x
−1
 −2 0 3   y  =  1 
1 1 1
z
0
Then apply the inverse matrix to both sides of the equation to find the solution:
  

 

x
3 −2 0
−1
−5
 y  =  −5 3 1   1  =  8 
z
2 −1 0
0
−3
So x = −5, y = 8, z = −3 is the solution to the system.
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