NOUN Elementary Mathematics II Course Guide PDF

NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 102 COURSE TITLE: ELEMENTARY MATHEMATICS II  NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 102 COURSE TITLE: ELEMENTARY MATHEMATICS II...

NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 102 COURSE TITLE: ELEMENTARY MATHEMATICS II  NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 102 COURSE TITLE: ELEMENTARY MATHEMATICS II COURSE GUIDE Course Developer/Writers Dr. Peter Ogedebe. and Ester Omonayin Base University, Abuja Content Editor Dr. Babatunde Disu And Babatunde J. Osho. Department of Mathematics, National Open University of Nigeria Course Coordinators Dr. Babatunde Disu And Babatunde J. Osho. Department of Mathematics, National Open University of Nigeria National Open University of Nigeria Headquarters 14/16 Ahmadu Bello Way Victoria Island Lagos Published in 2015 by the National Open University of Nigeria, 14/16 Ahmadu Bello Way, Victoria Island, Lagos, Nigeria © National Open University of Nigeria 2015 This publication is made available in Open Access under the Attribution-ShareAlike4.0 (CC-BY-SA 4.0) license (http://creativecommons.org/licenses/by-sa/4.0/). 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How to Reuse and Attribute this content Under this license, any user of this textbook or the textbook contents herein must provide proper attribution as follows: “First produced by the National Open University of Nigeria” and include the NOUN Logo and the cover of the publication.  If you use this course material as a bibliographic reference, then you should cite it as follows: “Course code: Course Title, National Open University of Nigeria, 2014 at http://www.nou.edu.ng/index.htm#  If you redistribute this textbook in a print format, in whole or part, then you must include on every physical page the following attribution: "Download for free at the National Open University of Nigeria Open Educational Resources Repository at http://www.nou.edu.ng/index.htm#).  If you electronically redistribute part of this textbook, in whole or part, then you must retain in every digital format page view (including but not limited to EPUB, PDF, and HTML) the following attribution: "Download for free, National Open University of Nigeria Open Educational Resources Repository athttp://www.nou.edu.ng/index.htm#)  Course Guide Course Code: MTH 102 Course Title: ELEMENTARY MATHEMATICS II Introduction MTH 102 - Elemetary Mathematics II is designed to teach you how differential and integral calculus could be used in solving problems in the contemporary business, technological and scientific world. Therefore, the course is structured to expose you to the skills required in other to attain a level of proficiency in Science, technology and Engineering Professions. What you will learn in this Course You will be taught the basis of mathematics required in solving s c i e n t i f i c problems. Course Aim There are ten study units in the course and each unit has its objectives. You should read the objectives of each unit and bear them in mind as you go through the unit. In addition to the objectives of each unit, the overall aims of this course include: (i) To introduce you to the words and concepts in Elementary mathematics (ii) To familiarize you with the peculiar characteristics in Elementary mathematics. (iii) To expose you to the need for and the demands of mathematics in the Science world. (iv) To prepare you for the contemporary Science world. Course Objectives The objectives of this course are:  To inculcate appropriate mathematical skills required in Science and Engineering.  Educate learners on how to use mathematical Techniques in solving real life problems.  Educate the learners on how to integrate mathematical models in Sciences and Engineering. Working through this Course { You have to work through all the study units in the course. There are two modules and ten study units in all. Course Materials Major components of the course are: 1. Course Guide 2. Study Units 3. Textbooks 4. CDs  5. Assignments File 6. Presentation Schedule Study Units The breakdown of the three modules and eight study units are as follows: MODULE ONE: UNIT 1: FUNCTION AND GRAPHS UNIT 2: LIMITS UNIT 3: IDEA OF CONTINUITY MODULE TWO: UNIT 1: THE DERIVATIVE AS LIMIT OF RATE OF CHANGE UNIT 2: DIFFERENTIATION TECHNIQUES UNIT 3:: INTEGRATION MODULE THREE: UNIT 1: DEFINITE INTEGRALS (Application to areas under curve and Volumes of solids) UNIT 2: VOLUME OF SOLIDS OF REVOLUTION BY DEFINATE INTEGRAL Recommended Texts * { Seymour, L.S. (1964). Outline Series: Theory and Problems of Set Theory and related topics, pp. 1 – 133. * Sunday, O.I. (1998). Introduction to Real Analysis (Real-valued functions of a real variable, Vol. 1) * Pure Mathematics for Advanced Level By B.D Bunday H Mulholland1970. * Introduction to Mathematical Economics By Edward T. Dowling. * Mathematics and Quantitative Methods for Business and Economics. By Stephen P. Shao. 1976. * Mathematics for Commerce & Economics By Qazi Zameeruddin & V.K. Khanne 1995. * Engineering Mathematics By K. A Stroad. * Business Mathematics and Information Technology. ACCA STUDY MANUAL By. Foulks Lynch. * Introduction to Mathematical Economics SCHAUM‟S Out lines Assignment File { In this file, you will find all the details of the work you must submit to your tutor for marking. The marks you obtain from these assignments will count towards the final  mark you obtain for this course. Further information on assignments will be found in the Assignment File itself and later in this Course Guide in the section on assessment. Presentation Schedule The Presentation Schedule included in your course materials gives you the important dates for the completion of tutor-marked assignments and attending tutorials. Remember, you are required to submit all your assignments by the due date. You should guard against falling behind in your work. Assessment Your assessment will be based on tutor-marked assignments (TMAs) and a final examination which you will write at the end of the course. } Exercises TMAS { Every unit contains at least one or two assignments. You are advised to work through all the assignments and submit them for assessment. Your tutor will assess the assignments and select four which will constitute the 30% of your final grade. The tutor-marked assignments may be presented to you in a separate file. Just know that for every unit there are some tutor-marked assignments for you. It is important you do them and submit for assessment. } Final Examination and Grading  { At the end of the course, you will write a final examination which will constitute 70% of your final grade. In the examination which shall last for two hours, you will be requested to answer three questions out of at least five questions. Course marking Scheme This table shows how the actual course marking as it is broken down. Assessment Marks Assignments Four assignments, Best three marks of the four count at 30% of course marks Final Examination 70% of overall course marks 100% of course marks Total How to Get the Most from This Course In distance learning, the study units replace the university lecture. This is one of the great advantages of distance learning; you can read and work through specially designed study materials at your own pace, and at a time and place that suits you best. Think of it as reading the lecture instead of listening to the lecturer. In the same way a lecturer might give you some reading to do, the study units tell you when to read, and which are your text materials or set books. You are provided exercises to do at appropriate points, just as a lecturer might give you an in-class exercise. Each of the study units follows a common format. The first item is an introduction to the subject matter of the unit, and how a particular unit is integrated with the other units and the course as a whole. Next to this is a set of learning objectives. These objectives let you know what you should be able to do by the time you have completed the unit. These learning objectives are meant to guide your study. The moment a unit is finished, you must go back and check whether you have achieved the objectives. If this is made a habit, then you will significantly improve your chances of passing the course. The main body of the unit guides you through the required reading from other sources. This will usually be either from your set books or from a Reading section. The following is a practical strategy for working through the course. If you run into any trouble, telephone your tutor. Remember that your tutor’s job is to help you. When you need assistance, do not hesitate to call and ask your tutor to provide it. In addition do the following: 1. Read this Course Guide thoroughly, it is your first assignment. 2. Organise a Study Schedule. Design a Course Overview ‟ to guide you through the Course”. Note the time you are expected to spend on each unit and how the assignments relate to the units. Important information, e.g. details of your tutorials, and the date of the first day of the Semester is available from the study centre. You need to gather all the information into one place, such as your diary or a wall calendar. Whatever method you choose to use, you should decide on and write in your own dates and schedule of work for each unit. 3. Once you have created your own study schedule, do everything to stay faithful to it. The major reason that students fail is that they get behind with their course work. If you get into difficulties with your schedule, please, let your tutor know before it is too late for help. 4. Turn to Unit 1, and read the introduction and the objectives for the unit. 5. Assemble the study materials. You will need your set books and the unit you are studying at any point in time. 6. Work through the unit. As you work through the unit, you will know what sources to consult for further information. 7. Keep in touch with your study centre. Up-to-date course information will be continuously available there. 8. Well before the relevant due dates (about 4 weeks before due dates), keep in mind that you will learn a lot by doing the assignment carefully. They have been designed to help you meet the objectives of the course and, therefore, will help you pass the examination. Submit all assignments not later than the due date. 9. Review the objectives for each study unit to confirm that you have achieved them. If you feel unsure about any of the objectives, review the study materials or consult your tutor. 10. When you are confident that you have achieved a unit’s objectives, you can start on the next unit. Proceed unit by unit through the course and try to pace your study so that you keep yourself on schedule. 11. When you have submitted an assignment to your tutor for marking, do not wait for its return before starting on the next unit. Keep to your schedule. When the Assignment is returned, pay particular attention to your tutor’s comments, both on the tutor-marked assignment form and also the written comments on the ordinary assignments. 12. After completing the last unit, review the course and prepare yourself for the final examination. Check that you have achieved the unit objectives (listed at the beginning of each unit) and the course objectives (listed in the Course Guide). Tutors and Tutorials The dates, times and locations of these tutorials will be made available to you, together with the name, telephone number and the address of your tutor. Each assignment will be marked by your tutor. Pay close attention to the comments your tutor might make on your assignments as these will help in your progress. Make sure that assignments reach your tutor on or before the due date. Your tutorials are important therefore try not to skip any. It is an opportunity to meet your tutor and your fellow students. It is also an opportunity to get the help of your tutor and discuss any difficulties encountered on your reading. Summary This course would train you on the concept of multimedia, production and utilization of it. Wish you the best of luck as you read through this course NATIONAL OPEN UNIVERSITY OF NIGERIA  SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 102 COURSE TITLE: ELEMENTARY MATHEMATICS II COURSE MATERIAL Course Developer/Writers Dr. Peter Ogedebe. and Ester Omonayin Base University, Abuja Content Editor Dr. Babatunde Disu And Babatunde J. Osho. Department of Mathematics, National Open University of Nigeria Course Coordinators Dr. Babatunde Disu And Babatunde J. Osho. Department of Mathematics, National Open University of Nigeria  MODULE ONE: UNIT 1: FUNCTION AND GRAPHS Content: 1.1 Introduction 1.2 Objectives 1.3 Functions 1.4 Function Notation 1.5 Graphs of function 1.6 Combination of function 1.7 Inverse Notation and Exercises 1.8 Conclusion and summary 1.9 References. 1.1 Introduction. In everyday life, many quantities depend on one or more changing variable. For example: a) Speed of a moving car or object depend on distance travelled and time taken b) The voltage of electrical devices depends on current and resistance. c) The volume of given mass of gas depends on the pressure at room temperature (I.e. temperature remain constant) A function is a phenomenon that relates how one variable or quantity depends on the other variables or quantities. For instance, in ohm’s law V I, mathematically V=IR where R is constant of proportionality. If I increase, so does the voltage V. If I decrease, so does the voltage. Hence, from this, we can say voltage is a function of current. 1.2 Objectives In this module, you will cover the following topics: Function(Definition of function) Function Notation Graphs of Function Combinations of Functions Inverse Function. 1.3 Function. Definition of a function A function is a relationship between two variables such that, to each value of the independent variable there is exactly one corresponding value of the dependent variable. The domain of the function is the set of all values of the independent variable for which the function is defined. The range of the function is the set of all values taken on by the dependent variable. OR A Function is a correspondence from a first set, called the domain, to a second set, called the range, such that each element in the domain corresponds to exactly one element in the range. domai n input Output Range  In fig.1, notice that you can think of a function as a machine that inputs values of the independent variable and inputs values of the dependent variable. Although function can be described by various means such as table, graphs and diagrams, they are most often specified by formulas or equation. For instance, the equation describes y as a function of. For this function, is the independent variable and y is the dependent variable. Example 1 Deciding whether relation are functions. Which of the equation below define y as a function of x? a) b) c) d) Solution To decide whether an equation defines a function, it is helpful to isolate the dependent variable on the left side. For instance, to decide whether the equation defines y as a function of x, write the equation in the form. From this form, you can see that for any value of , there is exactly one value of. So, is a function of. Original EquationRewritten EquationTest: Is a function of a. Yes, each value of determines exactly one value of y b. No, some values of x determine two values of. c. Yes, each value of x determine exactly one value of y. d. No, some value of x determine two values of y. Note that the equations that assign two values (+) to the dependent variable for a given value of the independent variable do not define functions of x. For instance, in part (b), when x=0, the equation indicates that or. Figure 1.12 shows the graphs of the four equations. y y 2 2 1 X2 + y2 x+y=1 1 2 1 2 1 -1 -2 x -1 -2 x -1 -1 -2 -2  y y 2 2 1 x2 + y = 1 1 X + y2 = 1 2 1 2 1 -1 -2 x -1 -2 x -1 -1 -2 -2 Fig 1.12 Checkpoint 1: Which of the equations below define y as a function of x? a. b. c. d. e. f. Example 2: Determine whether an Equation Represents a Function. Determine whether each equation defines as a function as a function of x: a. b. Solution Solve each equation for y in terms of x. If two or more values of y can be obtained for a given x, the equation is not a function. a. This is the given equation. Solve for y by subtracting x2 from both sides. Simplify. From this last equation we can see that for each value of x, there is one and only one value of y. For example, if x=1, then. The equation defines y as a function of x. b. This is the given equation. Isolate y2 by subtracting x2 from both sides. Simplify. Apply the square root properly: if u2=d then Then + in this last equation shows that for certain values of x (all values between -2 and 2), there are two values of y. For example, if x=1, then =. For this reason, the equation does not define y as a function of x. 1.4 Function Notation. When using an equation to define a function, you generally isolate the dependent variable on the left. For instance, writing the equation as indicates that y is the dependent variable. In function notation, this equation has the form The independent variable (input) is x and the name of the function is ‘’f’’. The symbol f(x) is read as ‘’f of x’’ or ‘’f at x’’ and it denotes the value of the independent variable (output) or the value of the function at the number x. For instance, the value of f when x 3 is The value f(3) is called a function value and lies in the range of f. This means that the point [3, f(3)] lies on the graph of f. One of the advantages of function notation is that it allows you to be less wordy. For instance, instead of asking ‘’what is the value of y when x=3?’’ you can ask ‘’what is f(3)? N.B: Study Tip: the notation f(x) does not mean ‘’f times x’’. The notation describes the value of the function at x. Example 3: Evaluating a function Find the value of the function When x= 1, 0, and 2. Is f one to one? Solution When x= 1, the value of is f(x) = 2x2 4x 1 This is the given function. f( 1) = 2( 1)2 4( 1) 1 Replace each occurrence of x with 1. = 2(1) 4( 1) 1 Evaluate the exponential expression.( 1)2= 1 =2 Perform the multiplications. =7 When x 0, the value of f is f(0) 2(0)2 4(0) 1 f(0) 0—0 1 1 When x 2, the value of f is f(2) 2(2)2 4(2) 1 2(4) 4(2) 1 8 8 1 1 Diagram F(x) = 2x2 – 4x + 1  Fig 1.21 Because two different values of x yield the same value of f(x), the function is not one to one, as shown in fig 1.21. Study tip: You can use the horizontal line to test or determine whether the function in Example 3 is one to one. Because the line y 1 intersects the graph of the function twice, the function is not one to one. Checkpoint Find the value of f(x) = x2 – 5x 1 when x is 0, 1 and 4? Answer: f (0) 1, f(1) , f(4). Example 4 If f (x) x2 5x 5, evaluate each of the following a. f (x 2) b. f ( x) Solution a. f (x 2) (x 2)2 5 (x 2) We find f(x 2) by substituting x 2 for x in the equation. f (x 2) (x 2)2 5(x 2) 5 Square x 2 using (A B)2= A2 2AB B2. x2 4x 4 5x 10 5 Distribute 5 throughout the Parentheses. x2 Combine like terms x2 9x 19 b. We find f( x) by substituting x for x in the equation f( x) = ( x)2 5( x) 5 = x2– 5x 5 Example 5 Let f(x) = x2 – 4x 7, find a. f(x+ x) b. f(x+ x)-f(x) x Solution a. To evaluate f(x+ x), substitute x+ x as x as shown in each set of parentheses as follows. f(x+ x) (x+ x)2 4(x+ x) x2 2x x ( x)2 4x 4 x 7 b. Using the result of part (a), you can write the following. f(x+ x)-f(x) x [(x+ x)2 4(x+ x) (x2 – 4x 7)] x [x2 2x x ( x)2 4x 4 x 7 x2 4x 7] x 2x x ( x)2 4 x x 2x x–4 as x 0 Checkpoint: 1. If , and find a. b. 2. If , evaluate each of the following a. b. c.  1.5 Graphs of Functions The graph of a function is the graph of its ordered pairs. For example, the graph of f(x) = 3x is the set of points (x, y) in the rectangular coordinates system satisfying y = 3x. When the graph of a function is sketched, the standard convention is to let the horizontal axis represents the independent variable. When this convention is used, the test described in Example 1 has vertical line test. This test states that if every vertical line intersects the graph of an equation at most once, then the equation defines y as a function of x. For instance, in figure 1.12, the graph in parts (a) and (c) pass the vertical line test but those in parts (b) and (d) do not The domain of a function may be described explicitly or it may be implied by an equation used to define the function. For example, the function given by y = 1/(x2 ) has an implied domain that consists of real x except x = 2. These two values (x2 – 4) excluded from the domain because division by zero is undefined. Another type of implied domain is that used to avoid even roots of negative number, as indicated in example below. Example 6:Finding the Domain and Range of a function Find the domain and range of each function a. b. Solution a. Because is not defined for X – 1 0 (that is for x 1), it follows that the domain of the function is the interval x 1 or [1, ]. To find the range, observe is never negative. Moreover, as x takes on the various values in the domain, y takes on all non negative values. So, the range is the interval y 0or [0, ]. The graph of the function, shown in figure 1.22 (a) confirms these results.  y Range: y≥0 y Range: y≥0 If x < 1 y= Y=1-x x Domain: x>1 Domain: all real x Because this function is defined for x and for x I, the domain is the entire set of real numbers. This function is called piecewise – definedfunction because it is defined by two or more equations over a specified domain. When x 1, the function behaves as in parts (a). For x I, the values of I x is positive, and therefore the range of the function is y or [0, ], as shown in figure 1.22 b. A function is one to one if to each value of the dependent variable in the range there corresponds exactly one value of the independent variable. For instance, the function in Example 6 a is one to one, where as the function in example 6b is not one- to -one Geometrically, a function is one- to- one if every horizontal line intersects the graph of the function at most once. This geometrical interpretation is the horizontal line test for one to one functions. So, a graph that represents a one- to –one test must satisfy both the vertical line test and the horizontal line test. Study Tips: The vertical line test for function: If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x. Example: Using the vertical line test  Use the vertical line test to identify graphs in which y is a function of x a. y B. y x x c. d. y y x x Solution a. B. y y x x Y is a function of x y is not a function of x. Two values of y corresponds to an x - value y y C. d. x x  Y is a function of x y is not a function of x. two values of y correspond to an x-value. Example7 Find the value of f(x) = x2 – 5x + 1 when x is 0, 1, and 4. Is f one to one? Solution When x = 0, the value of f is f (0) = 02 – 5(0) + 1 = 1 When x = 1, the value of f is f (1) = 12 – 5(1) + 1 =1–5+1=-5 When x = 4, the value of f(x) is = 42 – 5(4) + 1 =16 – 20 + 1 = - 3  f(x) (0,1 (4, -3 (0, 5 Checkpoint In the following exercise, use the vertical line test to identify graphs in which y is a function of x. i. y x ii. y x iii. y x iv. y y v. y x  1.4 combinations of functions (composite functions) Definition The function given by (f.g) (x) = f (g(x)) is the composite of f with g. The domain of (f.g) is the set of all x in the domain of g such that g(x) is the domain of f. Two functions can be combined in various ways to create new functions. For instance, if f (x) = 2x – 3 and g(x) = x2 + 1, you can form the following functions. f(x) + g(x) = (2x – 3) + (x2 + 1) = x2 + 2x – 2 Sum. f(x) – g(x) = (2x - 3) – (x2 + 1) = - x2 + 2x – 4 Difference. f(x) g(x) = (2x – 3) (x2 + 1) = 2x3 – 3x2 + 2x – 3Product. f(x)/g(x) = Quotient/ division Example Forming composite functions Let f(x) = 2x – 4 and g(x) =x2 3, and find a. f (g(x)) b. g(f(x)) Solution a. The composite of f with g is given by g (f(x)) = 2(g(x)) – 4 Evaluate f at g(x). = 2(x2 + 3) – 4 Substitute x2 3 for g(x). = 2x2 + 6 – 4 Simplify. = 2x2 – 2 = 2(x2 - 1) b. The composite of g with f is given by g (f(x)) = (f(x)) 2 + 1Evaluate g at f(x). = (2x – 4)2 + 1Substitute 2x – 4 for f(x). = 4x2 16x 16 1Simplify. = 4x2 16x 17 Checkpoint Let f(x) = 2x 1 and g(x) = x2 2, and find a) f (g(x)). b) g (f(x)). 1.7 INVERSE FUNCTION Definition of the inverse of a function. Let f and g be two functions such that: f (g(x)) = x for each x in the domain of g and g(f(x)) = x for each x in the domain of f Under these conditions, the function g is the inverse of the function f. The function is donated by , which is read as “f- inverse”. So, and The domain of f must be equal to the range of , and the range of f must be equal to the domain of. Example 8 Finding inverse functions Several functions and their inverse are shown below. In each case, note that the inverse function “undoes” the original function. For instance, to undo multiplication by 2, you should divide by 2. a) f(x) = 2x )= b) f(x) = /5x (x) = 5x c) f(x) = x + 8 (x) = x – 8 d) f(x) = 3x + 7 (x) = 1/3 (x + 7) e) f(x) = x3 (x) = f) f(x) = 1/x (x) = y = f(x) y=x (a,b) Y = f-1(x) (b,a) The graphs of f and f-1 are mirror images of each other (with respect to the line y = x), as in figure 1.23 Checkpoint Informally find the inverse function of each function a. f(x) = 1/5x b. f (x) = 3x + 2 Example 9: Finding the inverse function Find the inverse function of f(x) =. Solution Begin by substituting f(x) with y. Then interchange x and y and solve for y. Write original function. Replace f(x) with y. Interchange x and y. Squaring each side. Add 3 to each side. y Divide each side by 2. So, the inverse function has the form Using x as the independent variable, you can write as x 0. In the figure 1.24, note that the domain of coincides with the range of f. Diagram  f-1(x) = Y=x ( (1,2) (0, ) F(x) = (2,1) Figure 1.24 ( After you have found the inverse of a function, you should check your result. You can check your results graphically by observing that graphs of f and are reflections of each other in the line y =x. You can check your results algebraically by evaluating and – both should be equal to x. Check that Check that ) ). x, x Checkpoint Find the inverse of function of. NB: Not every function has an inverse function. In the fact, for a function to have an inverse function, it must be one- to- one. Example 10 A function that has no inverse function. Show that the function f(x) = x2 – 1 has no inverse function. (Assume that the domain of f is the set of all real numbers) Solution Begin by sketching the graph of f, as shown in figure 1.25 Note that f (x) = x2 – 1 f (2) = 22 – 1 = 3 And f( 2) = ( 2)2 1=3 So, f does not pass the horizontal line test, which implies that if is not one- to- one and therefore has no inverse function. The same conclusion can be obtained by trying to find the inverse of f algebraically. f (x) = x2 – 1 Write original function. 2 y=x –1 Replace f(x) with y 2 x=y –1 Interchange x and y x + 1 = y2 Add 1 to each side =y Take square root of each side The last equation does not define y as a function of x, and so f has no inverse function y F (x) = x2 – 1 write original function. Y = x2 – 1 Replace f(x) with y X = y2 – 1 Interchange x and y X + 1 = y2 Add1 to each side + Take square root of each side The last equation does not define y as a function of x, and so f has no inverse function  y (-2, 3) (2,3) y=3 F(x) = x2-1 Checkpoint: Show that the function f(x) = x2 +4 has no inverse function. 1.6 Exercises A. In the following exercises 1 - 4, decide whether the equation define y as a function of x. 1. x2 + y2 = 4 2. ½x – 6y = -3 3. x2 + y = 4 4. y2 = x2 - 1 B. In the following exercises 5 – 6, find the domain and range of the function. Use interval notation to write your result. 5. f(x) = x3 6. f(x) = 4 – x2  y y x x C. In exercises 7 – 8 evaluate the function at the specified the values of the independent variable. Simplify the result. 7. f(x) = 2x – 3 (a) f(0) (b). f(- 3) (c). f (x – 1) (d). f (1 8. g(x) = (a) g(2) (b). g(¼) (c). g( x ) (d). g( x D. In exercises 9 – 10, evaluate the difference quotient and simplify the result 9. f(x) = x3– x 10. g(x) 11. f(x) E.Find: (a) f(x) + g(x) (b) f(x).g(x) (c) (d) f(g(x)) and (e) g(f(x)) if defined. 12. f(x) , g(x) 13. f(x) , g(x) F. 14. Given that f(x) = and g (x) = x2 - 1, find the composite functions: a. f(g(1)) b. g(f(1)) c. g(f(0)) d. f(g( e. f(g(x)) f. g(f(x)) G. In exercises 15 – 16, show that f and g are inverse functions by showing that f(g(x)) = x and g (f(x)) = x. Then sketch the graphs of f and g on the same coordinate axes. 15. f(x) g(x) 16. f(x) g(x) , x 9 17. Find the inverse function of f. Then sketch the graph of f and on the same coordinate axis. f(x) , 0. H. In the exercises 18 and 19, use the vertical line test to determine whether y is a function of x. 18. 19. 1.8 Conclusion and Summary I. A function is a correspondence from a first set, called the domain to a second set, called the range such that each element in the domain corresponds to exactly one element in the range. II. A function is one -to -one if each value of the dependent variable in the range there corresponds exactly one value of the independent variable 1.9 References Calculus an Applied Approach Larson Edwards Sixth Edition Blitzer Algebra and Trigonometry Custom 4th Edition Engineering Mathematics by K.A Stroud. MODULE 2: LIMITS. Contents: 2.1 Introduction 2.2 Objectives 2.3 Limit (Definition) 2.4 Properties of Limit 2.5 The Limit of a Polynomial Function 2.6 Techniques for Evaluating Limits 2.7 Exercises 2.8 Summary 2.9 References. 2.1 Introduction: In everyday language, one always refers to limit one’s endurance, speed limit of a car, a wrestler’s weight limit or stretching a spring to its limit. These phrases all suggest that a limit is a bound, which on some instances may not be reached but on other instance may be reached or exceeded. Hooke’s law is a perfect illustration of limit which states provided an elastic limit of a spring is not exceeded; the extension (e) is directly proportional to the tension or force acting on it. That is a spring has a limit of extension when a load is suspended on it. If it exceeds the boundary, it will reach a point of plasticity or break without returning to its initial position. Limit is a concept/Fundamental to Calculus. 2.2 Objectives: In this Module, you will cover the following topics: 2.3 Limits (Definition) 2.4 Properties of Limits 2.5 The Limit of a Polynomial Function 2.6 Techniques for Evaluating Limits 2.1 Definition of Limit of a Function: If becomes arbitrarily close to a single number L as approaches c from either side, , then which is read as “the limit of as approaches c is L” 2.2 Properties of Limit: Many times, the limit of as approaches c is simply.Whenever the limit of as approaches c is.The limit can be evaluated by direct substitution. Properties of Limits: Let b and c be real numbers, and let n be a positive integer.  i. =b ii. =c n iii. = c n iv. = n In the property IV, if n is even, then c must be positive. By combining the properties of limits with the rules for operating with limits shown below, you can find limits for a wide variety of algebraic function. Operations with limits: Let b and c be real numbers let n be a positive integer, and let f and g be functions with the following limits. = L and =k I. Scalar multiple: =bL. II. Sum or Difference: = L k. III. Product: = L. k. IV. Quotient: = L/k provided that k 0. n n V. Power: =L VI. Radical: = In property VI, if n is even, then L must be positive. Example 1: Find the limit: 2 +1) Solution. Using direct substitution by substituting 1 for x 2 +1) = 12 1 = 2 Example 2: Find the limit: a. f(x) = b. f(x) = c. Solution a. = Factorizing the numerator by the difference of two square [ ]. =1+1=2 Substituting 1 for x. Therefore, b. = = =0 Substituting 1 for x. Therefore, does not exist. c. =1 2.3 The limit of a polynomial function If p is a polynomial function and c is any real number, then Example 3 Finding the limit of a polynomial function Find the limit: Solution: = + Applying property II. = Use direct substitution. 4+4–3=5 Simplify. =5 Note: Example 3 show or state that the limit of polynomial can be evaluated by direct substitution. Check point: 1. Find the limit: a. f(x) = b. f(x) = c. 2. Find the limit : 2.4 Techniques for Evaluating Limits. There are several techniques for calculating limits and these are based on the following important theorem. Basically, the theorem states that “if two functions agree at all but a single point c, then they have identical limit behavior at x = c.  The Replacement Theorem/Technique Let c be a real number and f(x) = g(x) for all x c. if the limit of g(x) exists as x→c, then the limit of f(x) also exists and To apply the Replacement Theorem, you can use a result from algebra which states that for a polynomial function p, p(c) = 0 if and only if (x-c) is a factor of p(x). Example 4 Finding the limit of a function Find the limit: Solution Note that the numerator and the denominator are zero when x=1.For the numerator = =0 the denominator. This implies or means that x - 1 is a factor of both and you can divide out this like factor using division of polynomial. = (x2 + x +1)(x -1) = = factor numerator. = Divide out factor. = ,x 1 Simplify. So, the rational function (x3 – 1)/(x – 1) and the polynomial function x2 + x + 1 agree for all value of x other than x = 1, and you can apply the Replacement theorem. = = + 1+1 =3 y y f(x) = g(x) = x2+x+1 x x  In fig2.1 illustrates this result graphically. Note that the two graphs are identical except that the graph of g contains the point (1, 3), whereas this point is missing on the graph of f. (in fig 2.1, the missing point is denoted by an open dot.) Checkpoint. Find the limit: Dividing Out Technique. Example 5 Find the limit: Solution. Using direct substitution will fails because both the numerator and thedenominator are zero when x = -3. Check: For the numerator = - + (-3)- 6 = 9 – 3 – 6 =0. Similarly for the denominator: (x + 3) = - 3+3= 0. Since the limits of both numerator and denominator are zero, you know that they have a common factor of x + 3 by factorizing the numerator. So, for all x ≠ 3, you can divide out this factor to obtain the following: = Factor numerator by factorization. = Divide out like factor. = Simplify. =-3–2 Substituting – 3 to be x. =-5 x  f(x) = (-3,-5) Fig 2.2: f is undefined when x = -3 This result is shown graphically above in fig 2.2. Note that the graph of f coincides with the graph of g(x) = x – 2, except that the graph of f has a hole at (-3, -5). Checkpoint Find the limit: Rationalizing The Numerator Technique. Example 6 Find the limit:. Solution Direct substitution fails because both the numerator and the denominator are zero when x = 0. In this case, you can rewrite the fraction by rationalizing the numerator by taking the conjugate of the numerator and using it to both the numerator and denominator. Taking the conjugate of the numerator of the numerator will be. Conjugate of is. = =  = = = , x Now, using the replacement theorem, you can evaluate the limit as follows: = = = = =½ Checkpoint Find the limit:. One Sided Limit One way in which a limit fails to exist is when a function approachesa different value from the left of c than it approaches from right of c. This type of behaviour can be described more concisely with the concept of a one-sided limit. Limit from the left. Limit from the right. The first of these two limits is read as “the limit of f(x) as x approaches c from the left is L”. The second is read as “limitf(x) as x approaches c from the right is L”. Example 7 Find the limit as x 0 from the left and the limit as x 0 from the right for the function: f(x) = Solution y f(x) = x Fig 2.3 From the graph of f, shown in fig 2.3, you can see that f(x)= -2 for all x 0, the limit from the right is: =2 Limit from the right. Unbounded Behaviour. A Limit can fail to exist when f(x) increases or decrease without bound as x approaches c.The equal sign in the statement does not mean that the limit exists. On the contrary, it tells you how the limit fails to exist by denoting the unbounded behaviour of f(x) as x approaches c. Example 8 Find the limit (if possible) : Solution = and = Because f is a unbounded as x approaches 2, the limit does not exist. Checkpoint: Find the limit (if possible): Solution. = = = 2.7 Exercises A. In exercise I and II, find the limit of (a) f(x) + g(x), (b) f(x).g(x) and (c) as x approaches c. I. II. B. In exercise III – XVI, find the limit: III. IV. V. VI. VII. IX. X. XI. XII. XIII. XIV. XV. XVI. In the following exercise XVII – XXX, find the limit (if it exists): XVII. XVIII. XIX. XX. XXI. XXII. XXIII. XXIV. XXV. XXVI. XXVII. , where f(x) = XXVIII. XXIX. XXX. 2.8 Summary. I. If f(x) becomes arbitrary close to a single number L as x approach c from either side, then = L which is as the limit of f(x) as x approaches c is L. II. If p is a polynomial function and c is any real number, then 2.9References: I. Engineering Mathematics by K. A Stroud. II. Blitzer Algebra and Trigonometry Custom 4th Edition. III. Calculus An Applied Approach Larson Edwards Sixth Edition.  MODULE ONE UNIT 2: LIMITS. Contents: 2.1 Introduction 2.2 Objectives 2.3 Limit (Definition) 2.4 Properties of Limit 2.5 The Limit of a Polynomial Function 2.6 Techniques for Evaluating Limits 2.7 Exercises 2.8 Summary 2.9 References. 2.1 Introduction: In everyday language, one always refers to limit one’s endurance, speed limit of a car, a wrestler’s weight limit or stretching a spring to its limit. These phrases all suggest that a limit is a bound, which on some instances may not be reached but on other instance may be reached or exceeded. Hooke’s law is a perfect illustration of limit which states provided an elastic limit of a spring is not exceeded; the extension (e) is directly proportional to the tension or force acting on it. That is a spring has a limit of extension when a load is suspended on it. If it exceeds the boundary, it will reach a point of plasticity or break without returning to its initial position. Limit is a concept/Fundamental to Calculus. 2.2 Objectives: In this Module, you will cover the following topics: 2.3 Limits (Definition) 2.4 Properties of Limits 2.5 The Limit of a Polynomial Function 2.6 Techniques for Evaluating Limits 2.1 Definition of Limit of a Function: If becomes arbitrarily close to a single number L as approaches c from either side, , then which is read as “the limit of as approaches c is L” 2.2 Properties of Limit: Many times, the limit of as approaches c is simply.Whenever the limit of as approaches c is.The limit can be evaluated by direct substitution. Properties of Limits: Let b and c be real numbers, and let n be a positive integer. v. =b vi. =c n vii. = c n viii. = n In the property IV, if n is even, then c must be positive. By combining the properties of limits with the rules for operating with limits shown below, you can find limits for a wide variety of algebraic function. Operations with limits: Let b and c be real numbers let n be a positive integer, and let f and g be functions with the following limits. = L and =k V. Scalar multiple: =bL. VI. Sum or Difference: = L k. VII. Product: = L. k. VIII. Quotient: = L/k provided that k 0. n V. Power: = Ln VI. Radical: = In property VI, if n is even, then L must be positive. Example 1: Find the limit: 2 +1) Solution. Using direct substitution by substituting 1 for x 2 +1) = 12 1 = 2 Example 2: Find the limit: a. f(x) = b. f(x) = c. Solution a. = Factorizing the numerator by the difference of two square [ ]. =1+1=2 Substituting 1 for x. Therefore, b. = = =0 Substituting 1 for x. Therefore, does not exist. c. =1 2.3 The limit of a polynomial function If p is a polynomial function and c is any real number, then Example 3 Finding the limit of a polynomial function Find the limit: Solution: = + Applying property II. = Use direct substitution. 4+4–3=5 Simplify. =5 Note: Example 3 show or state that the limit of polynomial can be evaluated by direct substitution. Check point: 3. Find the limit: b. f(x) = b. f(x) = c. 4. Find the limit : 2.4 Techniques for Evaluating Limits. There are several techniques for calculating limits and these are based on the following important theorem. Basically, the theorem states that “if two functions agree at all but a single point c, then they have identical limit behavior at x = c.  The Replacement Theorem/Technique Let c be a real number and f(x) = g(x) for all x c. if the limit of g(x) exists as x→c, then the limit of f(x) also exists and To apply the Replacement Theorem, you can use a result from algebra which states that for a polynomial function p, p(c) = 0 if and only if (x-c) is a factor of p(x). Example 4 Finding the limit of a function Find the limit: Solution Note that the numerator and the denominator are zero when x=1.For the numerator = =0 the denominator. This implies or means that x - 1 is a factor of both and you can divide out this like factor using division of polynomial. = (x2 + x +1)(x -1) = = factor numerator. = Divide out factor. = ,x 1 Simplify. So, the rational function (x3 – 1)/(x – 1) and the polynomial function x2 + x + 1 agree for all value of x other than x = 1, and you can apply the Replacement theorem. = = + 1+1 =3 y  y f(x) = g(x) = x2+x+1 x x In fig2.1 illustrates this result graphically. Note that the two graphs are identical except that the graph of g contains the point (1, 3), whereas this point is missing on the graph of f. (in fig 2.1, the missing point is denoted by an open dot.) Checkpoint. Find the limit: Dividing Out Technique. Example 5 Find the limit: Solution. Using direct substitution will fails because both the numerator and thedenominator are zero when x = -3. Check: For the numerator = - + (-3)- 6 = 9 – 3 – 6 =0. Similarly for the denominator: (x + 3) = - 3+3= 0. Since the limits of both numerator and denominator are zero, you know that they have a common factor of x + 3 by factorizing the numerator. So, for all x ≠ 3, you can divide out this factor to obtain the following:  = Factor numerator by factorization. = Divide out like factor. = Simplify. =-3–2 Substituting – 3 to be x. =-5 y x f(x) = (-3,-5) Fig 2.2: f is undefined when x = -3 This result is shown graphically above in fig 2.2. Note that the graph of f coincides with the graph of g(x) = x – 2, except that the graph of f has a hole at (-3, -5). Checkpoint Find the limit: Rationalizing The Numerator Technique. Example 6 Find the limit:. Solution Direct substitution fails because both the numerator and the denominator are zero when x = 0. In this case, you can rewrite the fraction by rationalizing the numerator by taking the conjugate of the numerator and using it to both the numerator and denominator. Taking the conjugate of the numerator of the numerator will be. Conjugate of is. = = = = = , x Now, using the replacement theorem, you can evaluate the limit as follows: = = = = =½ Checkpoint Find the limit:. One Sided Limit One way in which a limit fails to exist is when a function approachesa different value from the left of c than it approaches from right of c. This type of behaviour can be described more concisely with the concept of a one-sided limit. Limit from the left. Limit from the right. The first of these two limits is read as “the limit of f(x) as x approaches c from the left is L”. The second is read as “limitf(x) as x approaches c from the right is L”. Example 7 Find the limit as x 0 from the left and the limit as x 0 from the right for the function: f(x) = Solution y f(x) = x Fig 2.3 From the graph of f, shown in fig 2.3, you can see that f(x)= -2 for all x 0, the limit from the right is: =2 Limit from the right. Unbounded Behaviour. A Limit can fail to exist when f(x) increases or decrease without bound as x approaches c.The equal sign in the statement does not mean that the limit exists. On the contrary, it tells you how the limit fails to exist by denoting the unbounded behaviour of f(x) as x approaches c. Example 8 Find the limit (if possible) : Solution = and = Because f is a unbounded as x approaches 2, the limit does not exist. Checkpoint: Find the limit (if possible): Solution. = = = 2.7 Exercises A. In exercise I and II, find the limit of (a) f(x) + g(x), (b) f(x).g(x) and (c) as x approaches c. I. II. B. In exercise III – XVI, find the limit: III. IV. V. VI. VII. IX. X. XI. XII. XIII. XIV. XV. XVI. In the following exercise XVII – XXX, find the limit (if it exists): XVII. XVIII. XIX. XX. XXI. XXII. XXIII. XXIV. XXV. XXVI. XXVII. , where f(x) = XXVIII. XXIX. XXX. 2.8 Summary. I. If f(x) becomes arbitrary close to a single number L as x approach c from either side, then = L which is as the limit of f(x) as x approaches c is L. II. If p is a polynomial function and c is any real number, then 2.9References: I. Engineering Mathematics by K. A Stroud. II. Blitzer Algebra and Trigonometry Custom 4th Edition. III. Calculus An Applied Approach Larson Edwards Sixth Edition.  MODULE ONE UNIT 3: IDEA OF CONTINUITY 3.1 Introduction 3.2 Objective 3.3 Definition of Continuity 3.4 Continuity of Polynomial and Rational Functions 3.5 Exercises 3.6 References 3.1 Introduction: In mathematics, continuity means rigorous formulation of the intuitive concept of function that varies with no abrupt breaks or jumps. Continuity of a function is expressed some times by saying if the X values are closed together, then the y value of the function will also be close. 3.2 Objective In this module, you will cover the following topics 3.3 Definition of Continuity 3.4 Continuity of Polynomial and Rational functions 3.5 Continuity on a closed interval 3.3 Idea of Continuity 3.3.1 Definition of Continuity Let c be a number in the interval (a, b), and let f be a function whose domain contains the interval (a, b). The function f is continuous at the point cif the following conditions are true: i. F(c) is defined ii. exists iii. = f(c) If f is continuous at every point in the interval (a, b), then it is continuous on an open interval (a, b). N:B   Roughly, we can say that a function is said to be continuous on an interval if its graph on the interval can be traced using a pencil and paper without lifting the pencil from the paper (i.e. the graph of f is unbroken at c, and there are no holes, jumps, or gaps or to say that a function is continuous at x=c when there is no interruption in the graph of f at c.  Specifically, when direct substitution can be used to evaluate the limit of a function at c, then the function is continuous at c.The two types of functions that have this property are polynomial functions and rational functions. 3.4 Continuity of polynomial and Rational Functions. i. A polynomial function is continuous at every real number. iiA rational function is continuous at every number in its domain. Determining continuity of a polynomial function Example 1 Discuss the continuity of each function a. f(x) = x2 – 2x +3 b. f(x) = x3 – x Solution Each of these functions is a polynomial function. So, each is continuous on the entire real line. Both functions are continuous on (- , ). Determining continuity of each function Example 2 Discuss the continuity of each function (a) f (x) = (b) f(x) = (c) f(x) = Solution Each of these functions is a rational function and is therefore continuous at every number in its domain. (a) The domain of f(x) = consist of all real numbers except x = o. So, this function is continuous on the intervals (- , 0) and (0, ). (b) The domain of f(x) = (x2 – 1)/ (x – 1) consists of all real numbers except x = 1. So, this function is continuous on the intervals (- , 1) and (1, ). (c) The domain of f(x) = consists of all real numbers. So, this function is continuous on the entire real line. Class exercise Discuss the continuity of each function: (a) f(x) = (Answer: Continuous (- , 1) and (1, )) (b) f(x) = (Answer: Continuous (- , 2) and (2, - )) (c) f(x) = (Answer: Continuous on the entire real line) Consider an open interval I that contains a real number c. If a function f is defined on I (except possibly at c), and f is not continuous at c, then f is said to have a discontinuity at c. Discontinuities fall into two categories: removable and non-removable. A discontinuity is called removable if f can be made continuous by appropriately defining (or redefining) f(c). For instance, the function in Example 2b has a removable discontinuity at (1, 2). To remove the discontinuity, all you need to do is redefine the function so that F (1) = 2. A discontinuity at x = c is non removable if the function cannot be made continuous at x=c by defining or redefining the function at x=c. For instance, the function in Example 2a has a non removable discontinuity at x=0. 3.5 Continuity on a Closed Interval Definition Let f be defined on a closed interval [a, b]. If f is continuous on the open interval (a, b) and = f(a) and = f(b)  Then f is continuous on the closed interval [a, b]. Moreover, f is continuous from the right at a andcontinuous from the left at b. Similar definitions can be made to cover continuity on intervals of the form (a, b] and [a,b), or on infinite intervals. For example the function f(x) = is continuous on the infinite interval [0, ). Examining Continuity at an End point Example 3 Discuss the continuity of f(x) = Solution Notice that the domain of f is the set (- , 3]. Moreover, f is continuous from the left at x = 3 because: = =0 = f(3) For all x