Math 1050 Final Exam Review

The problems below are intended as a review of the topics that will becovered on the final exam. The final exam will consist of a subset (about 40)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 andless 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 ofassociativity, commutativity, inverses, or identity, or the distributive law) are followedor broken. If it is false, write a true statement that follows the rules.

(a) 3x− 3y = 3(x+ y)

(b) 313

= 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 wouldbe 31

3= 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 + 1

3) 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

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(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 1a

= 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 greaterthan 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 for2,−4, and 1

2.

Solution:

Q− {2,−4,1

2}

Solving Some Simple Equations

8. Solve for x if 3x− 2 = 4x− 5, using the rules from the previous section.

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Solution:

3x− 2 = 4x− 5

3x− 2 + 5 = 4x− 5 + 5 (Law of Inverses for Addition)

3x+ 3 = 4x (Simplify)

−3x+ 3x+ 3 = −3x+ 4x (Law of Inverses for Addition)

3 = x (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 y2 = x.”

Solution: This rule is not a function, because it “takes in” real numbers (includingnegative numbers and 0) but there is no positive real number y whose square (y2) iseither negative or zero. So the rule does not assign every object in the domain anobject in the target.

10. What is the target of the function g : N→ [1,∞) where g(x) = 2x+ 12?

Solution: The target is [1,∞).

11. If f(x) = 2x2+1x−3

, find f(2). Write your answer in simplest form.

Solution:

f(2) =2(2)2 + 1

2− 3

=2(4) + 1

−1

=8 + 1

−1

=9

−1

= −9

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Sequences

12. Decide whether each of the following sequences is arithmetic, geometric, or neither. If itis 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. Wecan find the common difference, d, by subtracting the first term from the secondterm:

5− 12 = −7

To find the 56th term in the sequence, we need to use the formula for the nth termin 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

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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. Thecommon ratio can be found by dividing the second term by the first term:

2

4=

1

2

To find the 33rd term in the sequence, we need to use the formula for the nth termin a geometric sequence:

an = a1(r)n−1

In this case, n = 33, a1 = 4, r = 12, so

a33 = 4(1

2)33−1

We can simplify a bit further to find that

a33 = 4(1

2)32

= 22(2−32)

= 22−32

= 2−30 =1

230or(

1

2)30

Sums and Series

15. Find the sum3∑

i=0

(i2 + 1). Write your answer as an integer in simplest form.

Solution:

3∑i=0

(i2 + 1) = (02 + 1) + (12 + 1) + (22 + 1) + (32 + 1)

= 1 + 2 + 5 + 10

= 18

16. Find the sum100∑i=1

(2i). Write your answer as an integer in simplest form.

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Solution:

100∑i=1

(2i) = 2

(100∑i=1

(i)

)= 2(1 + 2 + 3 + . . .+ 99 + 100)

= 2

(100

2(1 + 100)

)= 100(101)

= 10100

17. Find the sum of the first 100 terms of the arithmetic sequence 4,−2,−8,−14, . . .. Writeyour answer as a single term (you do not need to simplify further).

Solution: To find the sum of the first through kth terms of an arithmetic sequence,we can use the formula: ∑

i = 1kai =k

2(a1 + ak)

In this case k is 100 (the number of terms and the index of the last term. The firstterm 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∑i = 1100ai =

100

2(a1 + a100) = 50(4− 590) = 50(−586)

18. Find the value of the series 6 + 2 + 23

+ 29

+ . . . and write your answer as a single term(you do not need to simplify further).

Solution: This is a geometric series (infinite sum of a geometric sequence) witha1 = 6 and r = 1

3. The formula for the value of a geometric series is

∞∑i=1

ai =a1

1− r

In this case the value of the sum will be6

1− 13

=623

= 6(3

2)

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Counting I & II

19. There are 100 questions in a test bank, and a teacher wants to create a practice examwith 50 questions. If the order of the questions doesn’t matter, and the teacher doesn’twant to repeat any questions, how many different exams are possible?

Solution: In this case, we are choosing 50 things, from a pool of 100, withoutcaring about the order and without repeats. So the number of possible tests is(10050

)= 100!

(50!)(50!).

20. Suppose that you are on a scholarship committee at a university. You have chosen thetop 10 students, and you have 10 different scholarships to award. How many ways arethere 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 schol-arships 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 recommendedbooks, and you have time to read 8 of them over the course of the semester. If the orderin which you read the books matters, how many semester-long literary adventures canyou create for your students?

Solution: Since we are choosing 8 books from a pool of 20, and the order of thebooks matters, the number of ways to do this is 20!

(20−8)!= 20!

12!.

22. Write

(12

9

)as a natural number in simplest form.

Solution: (12

9

)=

12!

(9!)(3!)

=12 · 11 · 10 · (9!)

(9!)(3!)

=12 · 11 · 10

3 · 2 · 1= 2 · 11 · 10

= 220

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23. Use the Binomial Theorem to rewrite (x− 2y)6 so that your answer doesn’t include anynumbers that look like

(nk

).

Solution: The 6th row of Pascal’s Triangle is “1, 6, 15, 20, 15, 6, 1.” The BinomialTheorem tells us that (x− 2y)6 can be rewritten as

„6

0

«x6(−2y)

0+

„6

1

«x5(−2y)

1+

„6

2

«x4(−2y)

2+

„6

3

«x3(−2y)

3+

„6

4

«x2(−2y)

4+

„6

5

«x1(−2y)

5+

„6

6

«x0(−2y)

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 ofcuriosity, let’s rewrite this in “simplest” form:

x6 − 12x5y + 60x5y2 − 160x3y3 + 240x2y4 − 192xy5 + 64y6

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 denom-inator 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. Sothe implied domain is R− {2,−5}.

26. Find the implied domain of g(x) = 4√

8− 4x.

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Solution: The potential problem with this function is taking the 4th root of a nega-tive number. So we need to make sure that 8− 4x ≥ 0. We can solve this inequalityfor x:

8− 4x ≥ 0

8− 4x+ 4x ≥ 0 + 4x

8 ≥ 4x

1

4(8) ≥ 4x(

1

42 ≥ x

If x is any real number which is less than or equal to 2, then the function makessense. So the implied domain is the set of all real numbers which are less than orequal 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 the5√

. Whenever x is a real number, x − 3 is also a real number, so we can plug anyreal number into h(x) and get something that makes sense. So the implied domainof 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?

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Solution: Looking at the graph of f(x), we see a parabola, which sits on the x-axisand opens upwards. If we look at the shadow of this graph on the y-axis, we shouldsee 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 onthe graph which is not an intercept. (You must use numbers to indicate scale. “Tickmarks” will not be assumed to have any “usual” meaning.)

(a) id(x)

(b) x2

(c) x5

(d) 1x3

(e) 1x4

(f) f(x) = 4

(g) g : {−2, 0, 3, 4} → R where g(x) = x+ 2.

Solution:

(a)

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(b)

(c)

(d)

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(e)

(f)

(g)

32. Is the following graph the graph of a function? Explain why or why not.

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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 aninverse function. Explain your reasoning.

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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 horizontalasymptotes.

(a) 5√

32x− 2

(b) log 13(−x)

(c) 2x+3

(d) −2(x + 1)2 − 1 (In addition to the above instructions, label the vertex of theparabola.)

(e) 2√−x+ 1

(f) 1x2+2x+1

(Hint: Turn this function into a transformation of a function that looks like1

xn , and then graph.)

Solution:

35. Given the graph of f(x) in question 33, graph the function −f(x).

Solution:

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Inverse Functions

36. If f(x) = 2x+ 1 and g(x) = 12x− 1, are f and g inverse functions?

Solution: To check whether f and g are inverses, we need to check whether f ◦ gand g ◦ f are both the identity function:

f ◦ g(x) = f(g(x))

= f(1

2x− 1)

= 2(1

2x− 1) + 1

= 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−1x+3

.

Solution: To find the inverse function, first set h(x) = y:

y =3x− 1

x+ 3

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Next switch x and y:

x =3y − 1

y + 3

Now solve for y:

x =3y − 1

y + 3

x(y + 3) = 3y − 1

xy + 3x = 3y − 1

1 + 3x = 3y − xy3x+ 1 = y(3− x)

3x+ 1

3− x= y

Finally, set f−1(x) = y:

f−1(x) =3x+ 1

3− xYou 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 doesnot have an inverse.

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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 cansee from the graph that the range of g is (0,∞), which is the same as the target ofthe function (the target is specified in the question) so the function is onto. Anyfunction 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 graphof the line y = x. Graph the inverse function h−1(x).

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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:

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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)?

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Solution: The leading term of a product of polynomials is the product of the leadingterms 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)

= −x4 + 3x2 + 4x− 2

46. True or false: The polynomial 4x17 + 2x2 + 3 has 19 roots.

Solution: False. The degree of the polynomial is 17, which means that it has atmost 17 roots.

47. What is the degree 4 coefficient of the polynomial 4x17 + 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 insimplest 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

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49. Find the quotient:2x4 + 3x2 − x+ 1

x2 − 2.

Solution:

2x2 + 7 +−x+ 15

x2 − 2

50. Find the quotient3x5 + 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 divideby the potential factor. If we get a remainder of 0 then it is a factor. Use long divisionto find the quotient:

3x5 + 2x+ 5

3x4 − 3x3 + 3x2 − 3x+ 5= x+ 1

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 theproduct 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

x+ 2= 3x2 − 4x− 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 youcan.

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Solution: The x-intercepts of the graph are 2,−4, 0, and −1. So the roots of thepolynomial (the values for x which make p(x) = 0), are 2,−4, 0, and −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 (3x15 + 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.

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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:

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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 eithercompletely factoring or by setting p(x) = 0 and solving:

3x− 5 = 0

3x = 5

x =5

3

The root is 53.

60. What is the slope of the line that passes through the points (5,−1) and (2, 3)?

Solution:

slope =y2 − y1

x2 − x1

=3− (−1)

2− 5

=4

−3

= −4

3

The slope of the line is −43.

Quadratic Polynomials

61. Complete the square: Write the polynomial p(x) = −2x2−4x+3 in the form α(x+β)2+γwhere 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.

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Solution:

63. How many roots does the polynomial −x2 − 3x+ 1 have?

Solution: Check the discriminant to determine how many roots a quadratic poly-nomial 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 thepolynomial 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?

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Solution: No, because the factor (2x2 + 1) is quadratic but not monic. For a poly-nomial to be completely factored, all quadratic factors must be monic, and have noroots.

66. Completely factor the polynomial p(x) = 3x− 1.

Solution:

p(x) = 3(x− 1

3)

67. Completely factor the polynomial q(x) = 2x2 − 4x+ 1.

Solution: First, check the discriminant to determine how many roots the polynomialhas:

b2 − 4ac = (−4)2 − 4(2)(1) = 16− 8 = 8

Since the discriminant is greater than 0, the polynomial has two roots. Using thequadratic equation to find them, we get

2 +√

2

2and

2−√

2

2

So the completely factored form is

q(x) = 2

(x−

(2 +√

2

2

))(x−

(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. Thefactors of −2 are 1,−1, 2,−2. Both p(−1) and p(2) are 0, so −1 and 2 are bothroots.

69. Completely factor the polynomial 3x3 − 5x2 − 11x− 3 given that 3 is a root.

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Solution:

3(x+1

3)(x+ 1)(x− 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 linearfactors:

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

r(x) =2(x− 2)(x+ 1)(x2 + 4)

−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

r(x) =2(x− 2)(x+ 1)(x2 + 4)

−2(x+ 4)(x2 + 1)

Solution: Vertical asymptotes are the roots of the denominator. In this case, theonly vertical asymptote is to the line x = −4.

75. Calculate the end behavior of the rational function

r(x) =2(x− 2)(x+ 1)(x2 + 4)

−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 itsgraph.

Solution: The end behavior is the ratio of leading terms:

2(x)(x)(x2)

−2(x)(x2)=

2x4

−2x3= −x

So the graph of r(x) looks like the graph of the function −x to the far left and farright 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:

4

√a2

a−3

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Solution:a

54

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 initiallyowe $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

9= 3

3x = 81

x = 4 since 34 = 81

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81. Why can’t an exponential function have base 1?

Logarithms

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