Assignment 1Solution From Monash University PDF

Title Assignment 1Solution From Monash University
Course Foundation Mathematics
Institution Monash University
Pages 4
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This document shows the solution of the assignment 1 in academic year 2020 at Monash University all right reserved and do not use....


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School of Mathematics Monash University Semester 1 2021

ENG1090 Assignment 1 ENG1090 FOUNDATION MATHEMATICS Assignment 1 Assignment instructions

Complete the following questions, upload and submit them in Moodle in a size A4 pdf file (see submission instructions below) in week 3 of the semester, no later than Friday March 19, 10 pm Melbourne time (7 pm Malaysia time). Submission Instructions: In Moodle, click on the following links: Grades, Asst1, and Add submission. Upload your assignment and click save changes. At this point, you have the options of editing, removing, or submitting your assignment. Make sure you submit your assignment and it is not left as ‘draft’, otherwise this will incur in a penalty of 0 marks. There is no need to submit a coversheet. Make sure you use a reliable program or way to produce pdf files (for instance you could use the university library photocopy machine to scan your assignment). If you write your assignment in a tablet, make sure the final pdf file is not editable, it should be flat. Do not include any special characters in your assignment as Moodle might not be able to process your assignment. If you take a picture of your assignment and then convert it to a pdf, make sure it is clear, has clear borders, the paper is not creased, and there is only a slight and well proportioned margin outside the paper. All steps in your working should be shown, as you must express a mathematical argument clearly in both sentences and correct mathematical notation. (See the ‘Guidelines for writing mathematics’ on Moodle for an indication of what is required). Marks are awarded both for many of the explanation steps (not just the final answer), and your mathematics communication and presentation skills (up to 3 marks). Complete and correct solutions to this assignment contribute up to 4% of the final unit mark for ENG1090. Special Consideration: If you cannot complete an in-semester assessment due to circumstances beyond your control, in the first instance you should submit a request for special consideration. Requests for special consideration have to be submitted no later than two university working days after the due date. For detailed information about how to apply for special consideration and what supporting documentation you need to submit please check the university’s special consideration web page. Short extensions: In addition to special consideration requests which are processed centrally by the University, unit coordinators may be able to grant an extension of up to two calendar days if you are experiencing short term exceptional circumstances such as carer responsibilities or a car accident. If you would like to ask for such a short extension, you have to let your unit coordinator (Clayton: Santiago, Malaysia: Lily) know via e–mail at least 12 hours before the due date. Any such request must be supported by a justification and, if at all possible, by documentation as in the case of a special consideration request. Extensions are not available for quizzes. Also, technical issues or workload in other units are not considered as a reasonable basis for extensions. If you miss the 12–hour cutoff, you can only apply for special consideration. If you request an extension, even a short one, your unit coordinator may ask you to submit a special consideration request instead. In the case of an extension request or a special consideration request, you may only hear back after the deadline. Unit coordinators will always try to get back to you about any extension request before the due date. However, during particularly busy times you may only hear back from us after two working days. The latest you can submit a special consideration request is two university working days after the due date. After you submitted such a request it may then take a few more days for your request to be processed (5 days is not unusual). At the same time, please keep in mind that assessment tasks submitted late without approval for an extension or special consideration will incur a 10% penalty per every 24 hour overdue. Assessment tasks submitted more than 7 calendar days late will receive a mark of 0. Therefore to avoid losing marks in case your request is denied please always: Submit your request for an extension or special consideration as soon as possible, preferably well before the due date. Even if you have submitted a request for an extension or special consideration, you should always aim to submit your assessment as soon as possible. Note that tutors are not authorised to approve late submission of assignments without penalty.

Assignment questions Question 1.1 (4 marks)

1 1 Consider the function f (x) = √ + defined on the maximal set of real numbers x for which f (x) x−2 x−3 is well defined. Identify the domain and range of f , giving analytical reasons for your answers. (Drawing the graph alone is not sufficient). √ S OLUTION : The square root function is only defined for non-negative arguments. Hence x − 2 is defined for 1 is defined for x > 2, x ≥ 2. The reciprocal function is only defined for non-zero arguments. Hence √ x−2 1 1 1 is defined for x 6= 3. In order for f (x) to be defined, both √ must be defined. and and x−3 x−3 x−2 Therefore Dom(f ) = (2, ∞) ∩ R \ {3} = (2, ∞) \ {3} 1 Consider the value of f (x) as x approaches 2 from above. √ increases to positive infinity without x−2 √ 1 1 approaches bound, since x − 2 approaches 0 yet remains positive. On the other hand = −1 . x−3 2−3 Hence f (x) approaches ∞ − 1 = ∞. 1 1 = 1. On the Now consider the value of f (x) as x approaches 3 from below. √ approaches √ 3−2 x−2 1 decreases to negative infinity without bound, since x − 3 approaches 0 yet remains negative. other hand x−3 Hence f (x) approaches 1 − ∞ = −∞. Since f (x) is well defined between 2 and −3, f (x) must take all values between +∞ and −∞ over the interval (2, 3). Therefore Ran(f ) = (−∞, ∞)

Question 1.2 (8 marks) 1 1 , where S is the last non–zero digit of your student ID Consider two functions f (x) = x S+1 and g(x) = x−S number, defined on the maximal set of real numbers x for which each formula is well defined. Write down your student ID number. Determine expressions (in terms of x) for the composite functions f ◦ g and g ◦ f and identify their domains and ranges, justifying your answer. S OLUTION : The composite function (f ◦ g)(x) is given by 

1 (f ◦ g)(x) = f (g(x)) = f x−S Last digit

0

f ◦g

1 x

5

Last digit f ◦g

1 



1 x−5

1 6



1 x−1

6 1 x−6



=



1 x−S

2 1



1



2

7

1 x−2

7 1 x−7

1  S+1

3 1



1



3

8

1 x−3

8 1 x−8

4 1



1



4

9

The composite function (g ◦ f )(x) is given by  1  (g ◦ f )(x) = g(f (x)) = g xS+1 =

1 x

1 S+1

−S

1 x−4

9

1 x−9

1 5

1

10

Last digit g◦f

Last digit g◦f

0

1

2

3

4

1 x

1

1

1

5 1 1 x6 −5

x 2 −1

x 3 −2

x 4 −3

1 1 x 5 −4

1 1 x 7 −6

1 1 x 8 −7

1 1 x 9 −8

1 1 x 10 −9

1

6

1

7

1

8

9

From lecture 2, the domain of the composite function (f ◦ g)(x) is Dom(f ◦ g) = Dom(g) ∩ {x : g(x) ∈ Dom(f )}. Note Dom(g) = R \ {S}, Dom(f ) = R if S is even, and Dom(f ) = [0, ∞) if S is odd. Hence when S is even, g(x) is always in Dom(f ); and when S is odd, g(x) is in Dom(f ) when x > S. It follows that Last digit 0 1 2 3 4 Dom(f ◦ g) R \ {0} (1, ∞) R \ {2} (3, ∞) R \ {4} Last digit 5 6 7 8 9 Dom(f ◦ g) (5, ∞) R \ {6} (7, ∞) R \ {8} (9, ∞) For the range of (f ◦ g)(x), note that the reciprocal function takes all values except 0, and if we restrict the reciprocal function to positive arguments, then its range is (0, ∞). Taking n-th roots does not change this range. Therefore we have Last digit 0 1 2 3 4 Ran(f ◦ g) R \ {0} (0, ∞) R \ {0} (0, ∞) R \ {0} Last digit 5 6 7 8 9 Ran(f ◦ g) (0, ∞) R \ {0} (0, ∞) R \ {0} (0, ∞) Similarly the domain of the composite function (g ◦ f )(x) is Dom(g ◦ f ) = Dom(f ) ∩ {x : f (x) ∈ Dom(g )}. Note f (x) is in Dom(g) when x 6= S S+1 . Therefore we have Last digit 0 1 2 3 4 Dom(g ◦ f ) R \ {0} [0, ∞) \ {1} R \ {23 } [0, ∞) \ {34 } R \ {45 } Last digit 5 6 7 8 9 Dom(g ◦ f ) [0, ∞) \ {56 } R \ {67 } [0, ∞) \ {78 } R \ {89 } [0, ∞) \ {91 0} For the range of (g ◦ f )(x), note that for for even S, g ◦ f takes all values of the reciprocal function, and for odd S, g ◦ f takes all values of the reciprocal function to the right of −S. Therefore we have Last digit 0 1 2 Ran(g ◦ f ) R \ {0} R \ (−1, 0] R \ {0} Last digit 5 6 7 R \ {0} R \ (−71, 0] Ran(g ◦ f ) R \ (− 51 , 0]

3 4 R \ {0} R \ (−31, 0] 8 9 R \ {0} R \ (− 91 , 0]

Question 1.3 (8 marks) √ The price of crude oil is x+100 dollars per gallon where x is the number of gallons in the order. The transx portation cost is a flat $100(1 + S) per order, where S is the last digit of your student ID number. (a) Determine the total cost function f (x) of an order of x gallons of crude oil. (b) What is the domain and range of f (x)? (c) Determine g(x), the amount of crude oil x dollars can purchase in an order. (d) What is the domain and range of g(x)?

S OLUTION :

(a) The total cost function is f (x) =

√ x+100 x

· x + 100(1 + S) =

√ x + 100 + 100(1 + S)

Last digit √ 0 1 2 3 4 √ √ √ √ f (x) x + 100 + 100 x + 100 + 200 x + 100 + 300 x + 100 + 400 x + 100 + 500 Last digit √ 5 6 7 8 9 √ √ √ √ f (x) x + 100 + 600 x + 100 + 700 x + 100 + 800 x + 100 + 900 x + 100 + 1000 (b) The purchase amount must be positive, hence x > 0. f (x) makes sense for all x > 0. Hence Dom(f ) = (0, ∞). The value of f (x) starts at 10 + 100(1 + S) and increases with x without bound. Therefore the range is (110 + 100S, ∞). Last digit 0 1 2 3 4 Ran(f ) (110, ∞) (210, ∞) (310, ∞) (410, ∞) (510, ∞) Last digit 5 6 7 8 9 Ran(f ) (610, ∞) (710, ∞) (810, ∞) (910, ∞) (1010, ∞) (c) Since f (x) is increasing, it is one-to-one. Therefore√g(x) = f −1 (x). To obtain the inverse function √ write x = f (y) = y + 100 + 100(1 + S). Then y + 100 = x − 100(1 + S), and g(x) = y = (x − 100(1 + S))2 − 100. Last digit 0 1 2 3 4 g(x) (x − 100)2 − 100 (x − 200)2 − 100 (x − 300)2 − 100 (x − 400)2 − 100 (x − 500)2 − 100 Last digit 5 6 7 8 9 g(x) (x − 600)2 − 100 (x − 700)2 − 100 (x − 800)2 − 100 (x − 900)2 − 100 (x − 1000)2 − 100 (d) Since g(x) is the inverse function of f (x), Dom(g) = Ran(f ) and Ran(g) = Dom(f ) = (0, ∞). Last digit 0 1 2 3 4 Dom(g) (110, ∞) (210, ∞) (310, ∞) (410, ∞) (510, ∞) Last digit 5 6 7 8 9 Dom(g) (610, ∞) (710, ∞) (810, ∞) (910, ∞) (1010, ∞) Enjoy learning!...


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