Calculus I Course Syllabus PDF

Calculus I Han Li Associate Professor Office: Exley 639 Phone: 860-685-3221 Email: [email protected]  Course Syllabus Reference and Lecture Notes Our primary reference for this course is the open-access Calculus I textboo...

Calculus I Han Li Associate Professor Office: Exley 639 Phone: 860-685-3221 Email: [email protected]  Course Syllabus Reference and Lecture Notes Our primary reference for this course is the open-access Calculus I textbook from OpenStax, which you can download by clicking here. In addition to this, we will be developing our own lecture notes throughout the semester. Please be aware that the lectures may cover material beyond what is included in the OpenStax textbook. Course Description MATH121, together with MATH122, will cover both theoretical and practical aspects of limits, derivatives, and integrals; the calculus of exponential, logarithmic, trigonometric, and inverse trigonometric functions; techniques of integration; plane analytic geometry; various applica- tions of calculus; and sequences and series, including power series. Grading Policy In this course, your performance will be evaluated based on the following components: Homework: There will be a total of 9 homework sets, each worth 20 points. We will drop your 3 lowest homework grades. Your total homework grade will be out of 120 points. Midterm Exam: There will be a midterm exam scheduled for Monday, November 4, during the class time. The midterm exam is worth 80 points. Final Exam: The final exam will take place on Wednesday, December 11, from 9 AM to noon. i ii The final exam is worth 100 points. Course Attendance: Missing up to two lectures will not affect your course score. However, for each lecture missed beyond the third one and not backed up by your class Dean, you will lose 5 points. Now, let’s define some variables to calculate your course score: Let 𝐻 be your total homework grade. Let 𝑀 be your midterm exam grade. Let 𝐹 be your final exam grade. Let 𝐴 be the adjustment score based on your attendance (𝐴 = 0 if you miss no more than two lectures). Your course score will be calculated as the higher of the following two options: Option 1: 𝑆1 = 𝐻 + 𝑀 + 𝐹 − 𝐴 Option 2: 𝑆2 = 3𝐹. In both cases, your final course score will be out of a total of 300 points. Your final course grade will be determined as follows: If your course score is greater than or equal to 270 or ranked among the top 20% in the class, you will receive an A or above. If your course score is greater than or equal to 200 or ranked among the next 60% in the class, you will receive a B or above. If your course score is greater than or equal to 160 or ranked among the next 5% in the class, you will receive a C or above. If your course score is greater than or equal to 120 or ranked among the next 5% in the class, you will receive a D or above.  iii Policy on Makeup Exams Requests for makeup exams will be considered only if you are affected by a significant life event. Should that happen, you must have your class Dean send me a supporting email. With- out Dean’s support your request will be rejected and you will receive a zero exam grade. No makeup will be given before the original exam, no one will be allowed to retake any exam for any reason, and makeup exams may consist of different problems. Non-acceptable excuses for makeup exams include - but not limited to - “ I have a flight or trip scheduled and already bought an expansive ticket”, “I have to attend a social event or sport training”, “I have several exams or academic works on the same day or week”, “I am not ready for this exam and I want to study for a few more days”. Homework Submission Policy Homework assignments will be posted on Moodle, and it is your responsibility to submit them over Moodle in a single PDF file before the specified deadline. Please adhere to the following guidelines for homework submission: (a) Submission Deadline: Ensure that you submit your homework before the stated dead- line. Late submissions will not be accepted under any circumstances. Since the three lowest homework grades out of the nine total will be dropped, no late home- work will be accepted under any circumstances or for any reason. (b) Amendment Policy: After submitting your homework once, you are allowed one op- portunity to amend your submission by sending a new PDF file, but this must also be done before the deadline. (c) Originality: While it is acceptable to discuss homework problems with others, all writ- ten work you submit in this course must be your own. Submitting nearly identical solu- tions as other students may be considered plagiarism and will be referred to the Honor Board as a suspected violation of the Honor Code. (d) Neat and Organized: Homework submissions should be neat, clear, and well-organized. Ensure that your PDF file is arranged in the order of your homework pages, and be sure to print your name on the front page. (e) File Format: Submit your homework as a single PDF file. The graders may deduct points for submissions that do not adhere to this format, such as those consisting of multiple JPEG files or messy work. So please be diligent in following these instructions to ensure the smooth and fair evaluation of your homework assignments. If you have any questions or need clarification, don’t hesitate to reach out for assistance. iv Accommodation Statement Wesleyan University is committed to ensuring that all qualified students with disabilities are afforded an equal opportunity to participate in, and benefit from, its programs and services. Since accommodations may require early planning and generally are not provided retroac- tively, please contact Accessibility Services as soon as possible. If you have a disability, or think that you might have a disability, please contact Accessi- bility Services to arrange an appointment to discuss your needs and the process for requesting accommodations. Accessibility Services is located in North College, rooms 021/022, or can be reached by email ([email protected]) or phone (860-685-5581). Religious/Spiritual Observance Resources If you anticipate that your religious/spiritual observance may conflict with academic obliga- tions such as attending class, taking examinations, or submitting assignments, you can work directly with your professor to make reasonable arrangements. Should you require additional support or guidance, please feel free to reach out to Rabbi David Teva, Director of the Office of Religious and Spiritual Life at [email protected] or any of the chaplains in the Office of Religious and Spiritual Life at https://www.wesleyan.edu/orsl/index.html. For a list of religious holidays celebrated by members of the Wesleyan community, go to Wesleyan’s Multifaith calendar, which can be found at: https://www.wesleyan.edu/orsl/ multifaith-calendar.html. Title IX Resources If gender-based and/or sexual violence related trauma inhibits your ability to fully participate in class, please contact Debbie Colucci, Title IX Coordinator, at [email protected], or your class dean. Additionally, and if you are comfortable, you can work directly with your professor to make reasonable arrangements. You may also choose to talk with a confidential resource about all of your options for care and support. Confidential resources can be found in the Office of Counseling and Psychological Services (CAPS), WesWell, and the Office for Religious and Spiritual Life. Discrimination and Harassment Wesleyan University is committed to maintaining a positive learning, working, and living en- vironment and does not tolerate identity-based discriminatory harassment and/or sexual mis- conduct against students, faculty, staff, trustees, volunteers, and employees of any university contractors/agents. For purposes of this Wesleyan policy, identity refers to one’s race, color, religion, national or ethnic origin, age, disability, veteran status, sexual orientation, gender, gender identity, and gender expression. The Office for Equity and Inclusion serves students, faculty, administrators and develops policies and procedures regarding issues of diversity and  v equal opportunity/affirmative action. Individuals who believe they have been discriminated against should contact the Office for Equity and Inclusion at 860-685-4771. Honor Code All Wesleyan students are responsible for knowing and upholding the Honor Code. Feel free to contact me about any questions related to course expectations. If you have a question related specifically to the honor code, please contact Assistant Dean of Student/Director of Commu- nity Standards Kevin Butler - ([email protected]). AI Statement The use of AI tools (e.g., ChatGPT, Bing, Elicit, Google Translate, etc.) is prohibited for any part of this class, including, but not limited to, the generation of ideas, writing of text, or rewriting your own work. If you have any questions about a particular AI tool or use, please consult with me before using it. Unauthorized use of AI tools in this class will be considered a violation of Wesleyan’s Honor Code.  Contents vi Contents 1 1 Trigonometric Functions Functions 1 What are the Trig Functions? 1 Properties of Trig Functions 6 Inverse Functions 15 Inverse Trig Functions 20 25 2 Limits What are Limits? 25 Properties of Limits 31 Continuity 40 Bonus Examples 44 46 3 The Derivative Definitions and Basic Examples 46 Properties of Derivatives 52 Inverse Deriva- tives 59 Higher Derivatives 60 62 4 Applications of the Derivative Maxima and Minima 62 The Mean Value Theorem 65 Logarithms 66 71 Appendices Fractions 72 vi  1 CHAPTER Trigonometric Functions Functions Recall that a function can be expressed as 𝑦 = 𝑓 (𝑥). Functions are mathematical machines that take in an input (in this case x) and give a single output (in this case y). In this class we will be studying many of the properties that functions have. Some of these questions include: What is the minimum and maximum value that a function attains? Where does a function attain it’s minimum or maximum? What is the slope of any function at any point? Where is a function increasing or decreasing? etc. As a reminder, functions have operations which allow us to build new functions. (1.1) Example. If 𝑓 (𝑥) = 𝑥, then: 𝑓 (𝑥) + 𝑓 (𝑥) = 𝑥 + 𝑥 = 2𝑥, 𝑓 (𝑥) − 𝑓 (𝑥) = 𝑥 − 𝑥 = 0, 𝑓 (𝑥) ⋅ 𝑓 (𝑥) = 𝑥 ⋅ 𝑐 = 𝑥 2 𝑥 and 𝑓 (𝑥) ÷ 𝑓 (𝑥) = = 1. 𝑥 Using these operations, we can build new functions out of old ones. What are the Trig Functions? In this class, we will build the trig functions using the unit circle. We will also be using radians for our purposes. Why radians? If we try to use degrees, then certain properties which will 1 2 CHAPTER 1. TRIGONOMETRIC FUNCTIONS be discussed later will not be as nice to use. So let 𝛼 be any real number, that is −∞ < 𝛼 < ∞. Recall the unit circle: Unit𝑦 Circle 1.5 1 1 0.5 −1 𝑥 −1.5 −1 −0.5 0.5 11 1.5 −0.5 −1 −1 −1.5 Figure 1.1: Unit Circle When using the unit circle, we begin at the point (1, 0). Then 𝛼 will denote the angle (measured in radians) from this point and and point on the unit circle. When 𝛼 > 0 we move counterclockwise around the unit circle until we reach the point that results in angle measure 𝛼 and when 𝛼 < 0 we move clockwise around the circle. (1.2) Example. When 𝛼 = 𝜋/2 we get to the point (0, 1) as seen below: CHAPTER 1. TRIGONOMETRIC FUNCTIONS 3 𝑦 1.5 1 1 0.5 −1 𝑥 −1.5 −1 −0.5 0.5 11 1.5 −0.5 −1 −1 −1.5 Figure 1.2: Unit Circle (1.3) Definition. Let −∞ < 𝛼 < ∞ and (𝑥, 𝑦) be the point on the unit circle corresponding to the angle of 𝛼 radians. Then we define 𝑦 𝑥 1 1 sin 𝛼 = 𝑦, cos 𝛼 = 𝑥, tan 𝛼 = , cot 𝛼 = , csc 𝛼 = , sec 𝛼 =. 𝑥 𝑦 𝑦 𝑥 Using this definition, we can obtain some very important properties. (1.4) Theorem. sin2 𝛼 + cos2 𝛼 = 1 Proof. We apply the Pythagorean Theorem. To do this we create a triangle with hypotenuse one. 1 |y| |x| 4 CHAPTER 1. TRIGONOMETRIC FUNCTIONS Using the Pythagorean Theorem, we know that 𝑥 2 + 𝑦 2 = 1. Since every point on the unit circle is a point on the corner of a right triangle with hypotenuse length 1, using our definition for sin 𝛼 and cos 𝛼, we find that sin2 𝛼 + cos2 𝛼 = 1. The proof of the previous identity required the use of the Pythagorean Theorem. This begs the question: Why is the Pythagorean Theorem true? (1.5) Theorem. (Pythagorean Theorem) Let 𝑎 and 𝑏 represent the side-length of the legs of a right triangle and 𝑐 represent the length of the hypotenuse. Then 𝑎2 + 𝑏 2 = 𝑐 2. Proof. We first draw a square with side-length 𝑎 + 𝑏 and inscribe a second square with side- length c inside it. A diagram can be seen as follows: b a a c c b c c a b a b Then note that the area of the larger square can be written in two ways. It can first be written as (𝑎 + 𝑏)2 because the side-length of the square is 𝑎 + 𝑏. However, it can also be expressed as the sum of the area of the four triangles and the area of the inscribed square. When expressed in this second way, the area comes out to 𝑐 2 + 4 ⋅ 21 𝑎𝑏 = 𝑐 2 + 2𝑎𝑏. Since these areas are equal, we find that (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 = 𝑐 2 + 2𝑎𝑏. Subtracting 2𝑎𝑏 yields that 𝑎2 + 𝑏 2 = 𝑐 2 as desired. Now that we have established both the Pythagorean Theorem and the identity sin2 𝛼 + cos2 𝛼 = 1 we can establish some other identities. (1.6) Corollary. 1 + tan2 𝛼 = sec2 𝛼 Proof. sin2 𝛼 1 + tan2 𝛼 = 1 + cos2 𝛼 cos 𝛼 sin2 𝛼 2 = + cos2 𝛼 cos2 𝛼 cos2 𝛼 + sin2 𝛼 = cos2 𝛼 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 5 1 = cos2 𝛼 = sec2 𝛼 (1.7) Corollary. 1 + cot2 𝛼 = csc2 𝛼 Proof. cos2 𝛼 1 + cot2 𝛼 = 1 + sin2 𝛼 2 sin 𝛼 cos2 𝛼 = + sin2 𝛼 sin2 𝛼 cos2 𝛼 + sin2 𝛼 = sin2 𝛼 1 = sin2 𝛼 = csc2 𝛼 Now we can remind ourselves of what the trig functions evaluate to for certain special values as seen in the following table: 𝛼 sin 𝛼 cos 𝛼 tan 𝛼 csc 𝛼 sec 𝛼 cot 𝛼 0 0 1 0 undefined 1 undefined 𝜋/6 1/2 √ 3/2 1/√3 = √3/3 2 2/√3 = 2√3/3 √3 𝜋/4 √2/2 √2/2 1 2/√2 = √2 2/√2 = √2 1 𝜋/3 √3/2 1/2 √3 2/√3 = 2√3/3 2 1/√3 = √3/3 𝜋/2 1 0 undefined 1 undefined 0 Note that some of these values can also be obtained using the Pythagorean Theorem. (1.8) Proposition. sin(𝜋/6) = 12 Proof. Note the following picture of an equilateral triangle: Note that all the angles of an equilateral triangle are 𝜋/3 radians. When one draws a line bisecting one of the angles as shown above, the result is two right triangles, with one angle of 𝜋/6 radians and the other 𝜋/3 radians. Now suppose that the side length of the equilateral triangle is 2. Then the area 6 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 2 √3 1 of this triangle is √3. So knowing that the area of this triangle is also 1/2 ⋅ base ⋅ height, we obtain that √3 = 1/2 ⋅ 2 ⋅ height and further that height = √3. Now, since we know the height of the triangle and the base, we can obtain the length of each leg of the triangle using the Pythagorean theorem. In applying it we see that (√3)2 + 𝑏 2 = 22 and 𝑏 = √4 − 3 = 1. Finally, applying our definition of sin 𝛼, we obtain that sin 𝛼 = 1/2. (1.9) Proposition. sin 𝜋/4 = √2/2 Proof. Note the following figure: √2 1 1 When drawing a right triangle with angles 𝜋/4 radians, we know that the sides of the two legs must be the same. So suppose that the length of each leg is 1. Then by the Pythagorean Theorem, 12 + 12 = 2 = (hypotenuse length)2. Then applying the definition, sin 𝜋/4 = √2/2. Properties of Trig Functions Now that we understand the basics of what the trig functions are, we should try to understand some of their many properties. As always, we should try to understand why these properties are true instead of just memorizing them. (1.10) Proposition. sin (−𝛼) = − sin 𝛼 and cos −𝛼 = cos 𝛼 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 7 Proof. Let (𝑥, 𝑦) be the point on the unit circle corresponding to the angle 𝛼. Then we note that (𝑥, −𝑦) is the point on the unit circle corresponding to −𝛼. So by definition sin −𝛼 = −𝑦 = − sin 𝛼. Likewise, by applying the definition of cos 𝛼, we find that cos (−𝛼) = 𝑥 = cos 𝛼. (1.11) Proposition. sin (𝛼 + 2𝜋) = sin 𝛼 and cos 𝛼 + 2𝜋 = cos 𝛼 Proof. Note that 2𝜋 represents the circumference of the unit circle. So let (𝑥, 𝑦) represent the point on the unit circle corresponding to the angle 𝛼. Then rotating by 2𝜋 radians means moving one full rotation around the circle. So beginning at alpha, and then rotating by 2𝜋 radians, we land at the same point on the unit circle, that point being (𝑥, 𝑦). So by definition. (𝑥, 𝑦) represents the point on the unit circle corresponding to 𝛼 + 2𝜋. Therefore, by definition, sin (𝛼 + 2𝜋) = sin 𝛼 and cos 𝛼 + 2𝜋 = cos 𝛼 With this property, we can create graphs for both cos 𝛼 and sin 𝛼, which can be seen below: Graph of sin(𝑥) and cos(𝑥) 1 sin(x) cos(x) 0 𝑦 −1 −6 −4 −2 0 2 4 6 𝑥 But further, using the above proposition, we can begin to solve equations involving the trig functions. (1.12) Example. Find all 𝜃 such that cos 𝜃 = 12. To do this, we note that by the above property, both sin 𝜃 and cos 𝜃 are periodic with period 2𝜋. The result is that the above equation has infinitely many solutions since cos 𝜃 repeats itself after 2𝜋 radians. In order to then solve said equation, we can first restrict the domain to be [−𝜋, 𝜋] because this interval is the size of one period. We note that cos 𝜃 = 12 when 𝜃 = 𝜋/3 or 𝜃 = −𝜋/3. We can also note that these are the only two points in this period that result in the equation being true. Therefore, by applying the proposition, we find that 𝜃 = 𝜋/3 ± 2𝜋, 𝜃 = 𝜋/3 ± 4𝜋... as well as 𝜃 = −𝜋/3 ± 2𝜋, 𝜃 = 𝜋/3 ± 4𝜋... are all the solutions to the above 8 CHAPTER 1. TRIGONOMETRIC FUNCTIONS equation. If one wants to condense the solution, one could write that 𝜃 = 2𝑘𝜋 + 𝜋/3 and 𝜃 = 2𝑘𝜋 − 𝜋/3 where k is any whole number. So now we have a way of solving some equations with trig functions. However, it may be useful to be able to compute other values of sin 𝛼 and cos 𝛼. (1.13) Example. cos 75∘ ? As it stands, this is not a known value of cos 𝛼. So we need a new tool in order to compute it. In particular, we want to be able to use known values of sin 𝛼 and cos 𝛼 to compute new values. To do this, we have the following: (1.14) Proposition. cos (𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 Proof. We begin by noting the following geometric fact: Suppose 𝑝1 = (𝑥1 , 𝑦1 ) and 𝑝2 = (𝑥2 , 𝑦2 ) are points in the coordinate plane. Then we note that the distance between 𝑝1 and 𝑝2 is |𝑝1 𝑝2 | = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2. Why? This follows from the Pythagorean theorem. 𝑦 1.5 1 𝑝2 = (𝑥2 , 𝑦2 ) 2 2 𝑦2 − 𝑦1 √(𝑥2 − 𝑥1 ) + (𝑦2 − 𝑦1 ) 0.5 𝑥2 − 𝑥1 𝑝1 = (𝑥1 , 𝑦1 ) 𝑥 −1.5 −1 −0.5 0.5 1 1.5 −0.5 −1 −1.5 As seen in the diagram, one can form a right triangle between the points 𝑝1 and 𝑝2. In CHAPTER 1. TRIGONOMETRIC FUNCTIONS 9 doing so, one can find the lengths of each of the legs, and then applying the Pythagorean Theorem, one obtains the distance between the two points. Now that we have established this fact, let 𝑞1 = (cos 𝛼, sin 𝛼) and 𝑞2 = (cos 𝛽, sin 𝛽). Then |𝑞1 𝑞2 | = √(cos 𝛼 − cos 𝛽)2 + (sin 𝛼 − sin 𝛽)2. Now let 𝑞3 be the point on the unit circle corresponding to the angle 𝛼 − 𝛽, that is 𝑞3 = (cos (𝛼 − 𝛽), sin (𝛼 − 𝛽)). Finally, let 𝑃 = (1, 0). Then |𝑞3 𝑝| = √(cos (𝛼 − 𝛽) − 1)2 + sin2 (𝛼 − 𝛽). Now if 𝑂 represents the origin, then we can consider the triangles △𝑞1 𝑞2 𝑂 and △𝑝3 𝑃𝑂. 𝑞1 𝑞2 𝑞3 1 1 1 𝑂 1 𝑃 𝑂 We now note that the angle formed between now note that ∠𝑞1 𝑂𝑞2 is the same as ∠𝑃𝑂𝑞3 , and both have angle 𝛼 − 𝛽. The result is that we have two “Side Angle Side” triangles, both of which share the same angles and side lengths. Thus △𝑞1 𝑂𝑞2 ≅ △𝑞3 𝑂𝑃 by the “Side Angle Side” theorem from High School Geometry. Therefore, we know that |𝑞1 𝑞3 | = |𝑞3 𝑃| because the two triangles are equivalent. The result is that we have: 2 2 2 2 √[cos (𝛼 − 𝛽) − 1] + sin (𝛼 − 𝛽) = √(cos 𝛼 − cos 𝛽) + (sin 𝛼 − sin 𝛽) ⇒ [cos (𝛼 − 𝛽) − 1]2 + sin2 (𝛼 − 𝛽) = (cos 𝛼 − cos 𝛽)2 + (sin 𝛼 − sin 𝛽)2 Now expanding the left hand side of the equation yields: [cos (𝛼 − 𝛽) − 1]2 + sin2 (𝛼 − 𝛽) = cos2 (𝛼 − 𝛽) − 2 cos (𝛼 − 𝛽) + 1 + sin2 (𝛼 − 𝛽) = cos2 (𝛼 − 𝛽) + sin2 (𝛼 − 𝛽) − 2 cos (𝛼 − 𝛽) + 1 = 1 − 2 cos (𝛼 − 𝛽) + 1 = 2 − 2 cos (𝛼 − 𝛽) Expanding the right side yields: (cos 𝛼 − cos 𝛽)2 + (sin 𝛼 − sin 𝛽)2 = cos2 𝛼 − 2 cos 𝛼 cos 𝛽 + cos2 𝛽 + sin2 𝛼 − 2 sin 𝛼 sin 𝛽 + sin2 𝛽 = cos2 𝛼 + sin2 𝛼 + cos2 𝛽 + sin2 𝛽 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 = 1 + 1 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 = 2 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 So combining the left hand side and the right hand side yields: 10 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 2 − 2 cos (𝛼 − 𝛽) = 2 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 ⇒ −2 cos (𝛼 − 𝛽) = −2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 ⇒ cos (𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 Which is what we desired. Now that we have a formula for a difference in angles, it may be useful to have a formula for the sum of angles. We can use the difference in angles formula to prove a sum in angles formula. (1.15) Corollary. cos (𝛼 + 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽 Proof. cos (𝛼 + 𝛽) = cos (𝛼 − (−𝛽)) = cos 𝛼 cos (−𝛽) + sin 𝛼 sin (−𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 ⋅ (− sin 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽 Now that we have a sum of angles formula, we return to our example: (1.16) Example. cos 75∘ = cos 45∘ + 30∘ = cos 𝜋/4 + 𝜋/6 = cos 𝜋/4 cos 𝜋/6 − sin 𝜋/4 sin 𝜋/6 √2 √3 √2 1 = ⋅ − ⋅ 2 2 2 2 √6 − √2 = 2 So using the sum or difference of angles formula, we can now compute sin 𝛼 and cos 𝛼 for more values. However, note that the sum and difference of angles formulas both seem to show some relationship between sin and cos. One should ask: Are there useful relationships between sin and cos. In fact there are some useful relationships that one should know. (1.17) Proposition. cos (𝜋/2 − 𝛼) = sin 𝛼 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 11 Proof. We use the difference in angles formula. In doing so, we obtain: cos (𝜋/2 − 𝛼) = cos 𝜋/2 cos 𝛼 + sin 𝜋/2 sin 𝛼 = 0 ⋅ cos 𝛼 + 1 ⋅ sin 𝛼 = sin 𝛼 With this relationship in mind, we can obtain a sum and difference in angles formula for sin. (1.18) Proposition. sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 sin (𝛼 − 𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 Proof. sin (𝛼 + 𝛽) = cos (𝜋/2 − 𝛼 − 𝛽) = cos ([𝜋/2 − 𝛼] − 𝛽) = cos (𝜋/2 − 𝛼) cos 𝛽 + sin (𝜋/2 − 𝛼) sin 𝛽 = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 Now one may wonder why sin (𝜋/2 − 𝛼) = cos 𝛼. To see it note that if 𝛾 = 𝜋/2 − 𝛼 then sin (𝜋/2 − 𝛼) = sin 𝛾 = cos (𝜋/2 − 𝛾 ) = cos (𝜋/2 − [𝜋/2 − 𝛼]) = cos 𝛼. Now for the other sum of angles formula, we apply the same idea as before: sin (𝛼 − 𝛽 = sin (𝛼 + [−𝛽]) = sin 𝛼 cos −𝛽 + cos 𝛼 sin (−𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 Using our sum and difference of angles formulas, we can obtain the double angle identities. (1.19) Corollary. sin 2𝛼 = 2 sin 𝛼 cos 𝛼 Proof. sin 2𝛼 = sin (𝛼 + 𝛼) 12 CHAPTER 1. TRIGONOMETRIC FUNCTIONS = sin 𝛼 cos 𝛼 + sin 𝛼 cos 𝛼 = 2 sin 𝛼 cos 𝛼 (1.20) Corollary. cos 2𝛼 = cos2 𝛼 − sin2 𝛼 = 2 cos2 𝛼 − 1 = 1 − 2 sin2 𝛼 Proof. We show this in three parts: 1.) cos 2𝛼 = cos (𝛼 + 𝛼) = cos 𝛼 cos 𝛼 − sin 𝛼 sin 𝛼 = cos2 𝛼 − sin2 𝛼 2.) Now note that: cos 2𝛼 = cos2 𝛼 − sin2 𝛼 = (1 − sin2 𝛼) − sin2 𝛼 = 1 − 2 sin2 𝛼 3.) Likewise, for the last equality: cos 2𝛼 = cos2 𝛼 − sin2 𝛼 = cos2 𝛼 − (1 − cos2 𝛼) = 2 cos2 𝛼 − 1 The double-angle identities can come in handy and allow us to show what the sin and cos values are for even more angles. (1.21) Example. Let’s find cos 15∘. We can do this by noting that: √3 = cos 30∘ 2 = cos 2 ⋅ 15∘ = 2 cos2 15∘ − 1 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 13 Hence, we find that √23 = 2 cos2 15∘ − 1. Solving for cos 15∘ yields that √3 + 2 cos(15∘ ) = √ 4 Alternatively, one could identify that cos 15∘ = cos (45∘ − 30∘ ). Then applying the differ- ence in angles formula yields: cos 15∘ = cos (45∘ − 30∘ ) = cos (𝜋/2 − 𝜋/6) = cos 𝜋/2 cos 𝜋/6 + sin 𝜋/4 sin 𝜋/6 √2 √3 √2 1 = ⋅ + ⋅ 2 2 2 2 √6 + √2 = 4 Note that both methods are correct ways of computing cos 15∘. Since sin 𝛼 is periodic, we can obtain the following: (1.22) Proposition. sin (𝜋 − 𝑥) = sin 𝑥 Proof. We can solve this using the difference in angles formula. sin 𝜋 − 𝑥 = sin 𝜋 cos 𝑥 − cos 𝜋 sin 𝑥 = 0 ⋅ cos 𝑥 − (−1) sin 𝑥 = sin 𝑥 Now when proving the above, one may be tempted to write sin(𝜋 − 𝑥) = sin 𝜋 − sin 𝑥. This is not correct! We can not distribute sin or cos. One can check that this is false by noting sin 𝜋/3 − 𝜋/6 = sin 𝜋/6 = 1/2 but sin 𝜋/3 − sin 𝜋/6 = √3/2 − 1/2 ≠ 1/2. Using the above proposition can help us to solve equations involving sin 𝛼. (1.23) Example. Find all angles 𝜃 satisfying sin 𝜃 = −1 2. In order to solve for 𝜃, one should first remember the property that sin (−𝜃) = − sin 𝜃. Now note that sin 𝜋/6 = 1/2. Using this property, we know that sin −𝜋/6 = −1/2. Further, by the previous proposition, we know that −1/2 = sin −𝜋/6 = sin (𝜋 − (−𝜋/6)) = sin 7𝜋/6 = sin −5𝜋/6. So we have two values within the period of sin 𝜃 that satisfy the equation. Since sin 𝜃 attains each value (except for 1 and -1) twice on its period, we can obtain all solutions by writing 𝜃 = 2𝑘𝜋 + −𝜋 6 and 𝜃 = 2𝑘𝜋 + −5𝜋 6 where 𝑘 is a whole number. 14 CHAPTER 1. TRIGONOMETRIC FUNCTIONS So when solving equations involving sin 𝛼 = 𝑎 and cos 𝛼 = 𝑎 for −1 < 𝑎 < 1, the general strategy can be seen in the following chart: sin 𝛼 = 𝑎 1.) Find one value of 𝜃 for −𝜋 < 𝜃 < 𝜋 2.) Find 𝜋 − 𝜃 with −𝜋 < 𝜋 − 𝜃 < 𝜋 3.) The solution is 2𝑘𝜋 + 𝜃 and 2𝑘𝜋 + (𝜋 − 𝜃) where k is a whole number cos 𝛼 = 𝑎 1.) Find one 𝜃 and its respective −𝜃. 2.) The solution is 2𝑘𝜋 + 𝜃 and 2𝑘𝜋 − 𝜃 where 𝑘 is a whole number. Finally, using the properties we have, we can find solutions to some equations involving both sin 𝛼 and cos 𝛼. (1.24) Example. Find all 𝜃 satisfying 2 + 3 sin 𝜃 − cos (2𝜃) = 0. To solve, we will need to apply the double angle identity for cos (2𝜃). Note: 0 = 2 + 3 sin 𝜃 − 2 cos 2𝜃 = 2 + 3 sin 𝜃 − (1 − 2 sin2 𝜃) = 2 sin2 𝜃 + 3 sin 𝜃 + 1 = (2 sin 𝜃 + 1)(sin 𝜃 + 1) Thus, either 2 sin 𝜃 + 1 = 0 or sin 𝜃 + 1 = 0. Thus either sin 𝜃 = −1/2 or sin 𝜃 = −1. We solved sin 𝜃 = −1/2 in the previous example, so we will look at sin 𝜃 = −1. Note that sin 𝜃 = −1 only once within the period of sin 𝜃. The value of 𝜃 that results in sin 𝜃 = −1 is 𝜃 = −𝜋/2. So our final answer will be 𝜃 = 2𝑘𝜋 + −𝜋6 and 𝜃 = 2𝑘𝜋 + −5𝜋 6 and 2𝑘𝜋 − 𝜋/2 where 𝑘 is a whole number. Now note that in the above example, one may have also solved for sin 𝜃 by using the quadratic formula. By using the quadratic formula, one obtains: −3 ± √32 − 4 ⋅ 2 ⋅ 1 −3 ± 1 sin 𝜃 = = 2⋅2 4. However, in order to use the quadratic formula, one should know how to derive it. (1.25) Theorem. (Quadratic Formula) Let 𝑓 (𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 where 𝑎 ≠ 0. Then −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 15 Proof. To derive the formula we will “complete the square”. But first, since 𝑎 ≠ 0 we have that 𝑏 𝑎 and 𝑎𝑐 are defined. Thus, we have the following: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 𝑏 𝑐 ⇒𝑥 2 + 𝑥 + = 0 𝑎 𝑎 𝑏 𝑏 𝑏 𝑐 ⇒𝑥 2 + 𝑥 + ( )2 − ( )2 + = 0 𝑎 2𝑎 2𝑎 𝑎 𝑏 2 𝑏 2 𝑐 ⇒(𝑥 + ) − ( ) + = 0 2𝑎 2𝑎 𝑎 𝑏 2 𝑏 2 𝑐 ⇒(𝑥 + ) = ( ) − 2𝑎 2𝑎 𝑎 𝑏 𝑏 2 𝑐 ⇒(𝑥 + )2 = 2 − 2𝑎 4𝑎 𝑎 𝑏 𝑏2 4𝑎𝑐 ⇒(𝑥 + )2 = 2 − 2 2𝑎 4𝑎 4𝑎 𝑏 2 𝑏 2 − 4𝑎𝑐 ⇒(𝑥 + ) = 2𝑎 4𝑎2 𝑏 ±√𝑏 2 − 4𝑎𝑐 ⇒𝑥 + = 2𝑎 √4𝑎2 −𝑏 ±√𝑏 2 − 4𝑎𝑐 ⇒𝑥 = + 2𝑎 2𝑎 2 −𝑏 ± √𝑏 − 4𝑎𝑐 ⇒𝑥 = 2𝑎 Inverse Functions One should recall some of the terminology associated with functions. (1.26) Definition. The domain of a function is the collection of all the values of 𝑥 where the function is defined. When a function is presented as an algebraic expression, then it is in- terpreted that it is defined over the natural domain, i.e. all the x values for which the expression is defined. (1.27) Example. 𝑓 (𝑥) = 2𝑥 + 1 as written means that it’s domain is all of −∞ < 𝑥 < ∞ because it is defined for all x. However, if one sees 𝑔(𝑥) = 2𝑥 + 1; −1 ≤ 𝑥 ≤ 1, then the domain 16 CHAPTER 1. TRIGONOMETRIC FUNCTIONS of 𝑔(𝑥) is −1 ≤ 𝑥 ≤ 1. Now one may ask how to find the natural domain of a given function. To find the natural domain, generally one needs to look for where a function is not defined. This means avoiding situations where one has: √ negative number, 0𝑐 where c is a constant, or other restrictions on a given function. (1.28) Example. Find the natural domain of √1 − 𝑥 2. Since we want to avoid having the square root of a negative number, we must have that 1 − 𝑥 2 ≥ 0. But this happens when −1 ≤ 𝑥 ≤ 1. Thus our natural domain is −1 ≤ 𝑥1. 1 (1.29) Example. Find the natural domain of 2−𝑥 2. In this case, we want to avoid dividing by 0. So we need to find were 2 − 𝑥 2 = 0. But this happens when 𝑥 = ±√2. So those are the only points where the function is undefined. Therefore, the domain is when −∞ < 𝑥 < −√2, −√2 < 𝑥 < √2 and √2 < 𝑥 < ∞. Now that we have seen what the domain of a function is, it may be useful to see what values a function attains. (1.30) Definition. The range of a function is the collection of all values that are attained by said function. (1.31) Example. For the function 𝑦 = 2𝑥 + 1; −1 ≤ 𝑥 ≤ 1, the range of y is −1 ≤ 𝑦 ≤ 3 because the lowest value that y attains is at 𝑥 = −1, in which case 𝑦 = −1, and the highest value y attains is at 𝑥 = 1, in which case 𝑦 = 3. Note that y also attains all values in-between −1 and 3 because it is continuous (a concept that will be explained later). Therefore our range must be −1 ≤ 𝑦 ≤ 3. Note that if we were dealing with 𝑦 = 2𝑥 + 1 without the added restriction on the domain, then as x gets arbitrarily large, then so does y. Likewise, when x gets arbitrarily smaller, then so does y. The result is that the range is −∞ < 𝑦 < ∞. Finally, for a function like 𝑦 = √1 − 𝑥 2 , one may also find the range on the natural domain. In this case, the smallest value that y attains is 0 and the largest value that y attains is 1. Since y is continuous, we have that the range is 0 ≤ 𝑦 ≤ 1. Now that we have the ideas of the domain and range of a function, one may ask the question: For each 𝑦 in the range, what is the 𝑥 value in the domain satisfying 𝑦 = 𝑓 (𝑥)?. By answering this question, we obtain the inverse function. So if 𝑦 = 𝑓 (𝑥) then the inverse function is written 𝑥 = 𝑓 −1 (𝑦). CHAPTER 1. TRIGONOMETRIC FUNCTIONS 17 (1.32) Example. Find the inverse function of 𝑦 = 2𝑥 + 1. 𝑦−1 Step 1: Solve for 𝑥 in terms of 𝑦. In doing so in this case, 𝑥 = 2. Step2: Swap x and y. When doing so, one obtains the inverse function 𝑦 = 𝑥−1 2. As one can see, the general steps to find an inverse function are to: 1.) Find 𝑥 in terms of 𝑦; and; 2.) Switch x and y. Here’s one more example: (1.33) Example. Find the inverse function of 𝑦 = 𝑥 3 − 2. Step 1: 𝑦 = 𝑥 3 − 2 ⇒ 𝑥 = √ 3 𝑦 − 2. 3 Step 2: 𝑦 = √ 𝑥 − 2 is the inverse function. Now we pose a question: Does every function have an inverse? To answer this, let’s re- member that if 𝑦 = 𝑓 (𝑥) then 𝑥 = 𝑓 −1 (𝑦) if and only if (written iff) given 𝑦, 𝑥 is the unique value satisfying 𝑦 = 𝑓 (𝑥). We often will write 𝑦 = 𝑓 −1 (𝑥). The result of this is that if there exists a value 𝑦 and 𝑥1 , 𝑥2 , 𝑥1 ≠ 𝑥2 such that 𝑓 (𝑥1 ) = 𝑦 = 𝑓 (𝑥2 ), then 𝑓 has no inverse. To be able to see it a little more clearly, consider the following example: (1.34) Example. Consider the function 𝑦 = 𝑥 2. Given that 𝑦 = 4, one can see that there are two choices of 𝑥 that result in 𝑥 2 = 4, those values being 𝑥 = 2 and 𝑥 = −2. In this case, one can see that there is not a unique value of 𝑥 that results in 𝑦 = 𝑥 2. Therefore, this function does not have an inverse. One can also see it in the following graph: Here, one can see that the red line crossed the graph of 𝑦 = 𝑥 2 twice. Often this may be referred to the “Horizontal Line Test”. If a horizontal line ever meets multiple points of 𝑓 (𝑥) then 𝑓 (𝑥) will not have an inverse. Now that we have established that not all functions have an inverse, the next natural question should be: What functions do have an inverse. (1.35) Proposition. If a function is either strictly increasing (i.e. for all 𝑥1 < 𝑥2 in the domain of 𝑓 , we have 𝑓 (𝑥1 ) < 𝑓 (𝑥2 )) or strictly decreasing (i.e. for all 𝑥1 < 𝑥2 in the domain of 𝑓 , we have 𝑓 (𝑥1 ) > 𝑓 (𝑥2 )), then it has an inverse. Proof. Given y, if 𝑓 (𝑥) = 𝑦 then we have two cases: Case 1: If 𝑥1 > 𝑥, if 𝑓 is strictly increasing then 𝑓 (𝑥) < 𝑓 (𝑥1 ) = 𝑦. If 𝑓 is strictly decreasing then 𝑦 = 𝑓 (𝑥) > 𝑓 (𝑥1 ). So 𝑓 (𝑥1 ) ≠ 𝑦. 18 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 𝑦 6 5 4 3 2 1 𝑥 −2 −1 1 2 −1 −2 Case 2:If 𝑥1 < 𝑥, if 𝑓 is strictly increasing then 𝑦 = 𝑓 (𝑥) > 𝑓 (𝑥1 ). If 𝑓 is strictly decreasing then 𝑦 = 𝑓 (𝑥) < 𝑓 (𝑥1 ). So 𝑓 (𝑥1 ) ≠ 𝑦. This shows that for any given y in the range of 𝑓 , there is a unique 𝑥 satisfying 𝑓 (𝑥) = 𝑦. Therefore, 𝑓 has an inverse. Now that we have established some of the functions that have inverses, we can look at functions that do have inverses. (1.36) Example. 𝑦 = 𝑥 2 ; 𝑥 ≥ 0. Unlike before, when we restrict the domain of 𝑦 = 𝑥 2 we may obtain a function with an inverse because as written 𝑦 = 𝑥 2 ; 𝑥 ≥ 0 is strictly increasing. In this case we obtain that 𝑦 = √𝑥 is the inverse. One can see the plot below: We can also restrict the domain to 𝑥 ≤ 0 and obtain a new inverse. (1.37) Example. 𝑦 = 𝑥 2 ; 𝑥 ≤ 0. As written 𝑦 = 𝑥 2 ; 𝑥 ≥ 0 is strictly decreasing. In this case we obtain that 𝑦 = −√𝑥 is the inverse. Why −√𝑥? This occurs because the domain consists of the negative x values. So, when evaluating the inverse function, the range of the inverse function must also consist of the negative values. CHAPTER 1. TRIGONOMETRIC FUNCTIONS 19 √𝑥 𝑥2 4 3 2 1 0.5 1 1.5 2 Now that we have some inverse functions under our belt, we should also note that the graph of 𝑦 = 𝑓 (𝑥) and its inverse 𝑦 = 𝑓 −1 (𝑥) are symmetric with respect to the line 𝑦 = 𝑥. As of right now, we are not equipped to properly prove this fact. However, the idea behind the proof is that if (𝑎, 𝑏) is on the graph of 𝑓 (i.e. 𝑏 = 𝑓 (𝑎)) then (𝑏, 𝑎) is on the graph of 𝑓 −1. Then if one computes the midpoint between (𝑎, 𝑏) and (𝑏, 𝑎) is ( 𝑎+𝑏 2 , 𝑏+𝑎 2 ), which lies on the line 𝑦 = 𝑥. One can see this idea in the following graph: 20 CHAPTER 1. TRIGONOMETRIC FUNCTIONS 𝑓 (𝑥) 𝑓 −1 (𝑥) 6 4 2 −2 −1 1 2 −2 Inverse Trig Functions Now that we have the general idea of inverse functions and trig functions, one should ask whether trig functions have inverses. (1.38) Example. 𝑦 = sin 𝑥 Note that sin 𝑥 is not strictly increasing or decreasing. Further, multiple values of 𝑥 will result in the same value of y. For example sin 𝜋/2 = sin 5𝜋/2 = 1. So sin 𝑥 does not have an inverse. If we wish to obtain an inverse, we must restrict the domain. By restricting our domain to −𝜋/2 ≤ 𝑥 ≤ 𝜋/2. In doing so we obtain the “inverse” of sine with the restricted domain. We will write this function as arcsin 𝑥. (1.39) Definition. The arcsin function is the inverse of 𝑦 = sin 𝑥; −𝜋/2 ≤ 𝑥 ≤ 𝜋/2. The arccos function is the inverse of 𝑦 = cos 𝑥; 0 ≤ 𝑥 ≤ 𝜋. The graphs of each can be seen below. Now that we have the definitions of arcsin and arccos, we can now evaluate some values of them. In order to evaluate them we use the following general strategy: If 𝑦 = 𝑓 (𝑥) where 𝑥 is in the domain, then 𝑏 = 𝑓 −1 (𝑎) if and only if 𝑎 = 𝑓 (𝑏) and 𝑏 is in the domain of 𝑓. In terms of arcsin and arccos, this means the following: We have 𝑏 = arcsin (𝑥) if and only if 𝑥 = sin (𝑏) and −𝜋/2 ≤ 𝑏 ≤ 𝜋/2. We have 𝑏 = arccos (𝑥) if and only if 𝑥 = cos (𝑏) and 0 ≤ 𝑏 ≤ 𝜋. CHAPTER 1. TRIGONOMETRIC FUNCTIONS 21 Plot of arcsin(𝑥) arcsin(𝑥) 1 0 𝑦 −1 −1 −0.5 0 0.5 1 𝑥 Figure 1.3: Graphs of arcsin(𝑥) Plot of arccos(𝑥) arccos(𝑥) 3 2 𝑦 1 0 −1 −0.5 0 0.5 1 𝑥 Figure 1.4: Graphs of arcsin(𝑥) 22 CHAPTER 1. TRIGONOMETRIC FUNCTIONS (1.40) Example. arcsin (−1/2) = −𝜋/6 By the strategy above, we note that 𝑏 = arcsin −1/2 if and only if −1/2 = sin 𝑏 and −𝜋/2 ≤ 𝑏 ≤ 𝜋/2. Note that sin −𝜋/6 = −1/2 and −𝜋/2 ≤ −𝜋/6 ≤ 𝜋/2 (1.41) Example. arccos (√2/2) = 𝜋/4 By the strategy above, we note that 𝑏 = arccos (√2/2) if and only if √2/2 = cos (𝑏) and 0 ≤ 𝑏 ≤ 𝜋. Note that cos (𝜋/4) = √2/2 and −𝜋 ≤ 𝜋/4 ≤ 𝜋 (1.42) Example. sin (arccos (−1/2)) = √3/2 To do this, we first compute arccos (−1/2). Note that 𝑏 = arccos (−1/2) iff −1/2 = cos (𝑏) and 0 ≤ 𝑏 ≤ 𝜋. We note that 4𝜋/3 is the only value that satisfies both conditions. Thus sin (arccos (−1/2)) = sin 4𝜋/3 = √3/2. (1.43) Example. arccos (cos (1/2)). Note that 𝑏 = arccos (cos (1/2)) iff cos (𝑏) = cos (1/2) and 0 ≤ 𝑏 ≤ 𝜋. Noting that 𝑏 = 1/2 satisfies both conditions yields that arccos cos (1/2) = 1/2 (1.44) Example. arccos (cos (3𝜋)) Note that 𝑏 = arccos (cos (3𝜋)) iff cos 𝑏 = cos (3𝜋) and 0 ≤ 𝑏 ≤ 𝜋. Note that cos (𝜋) = cos (3𝜋) = −1 and 0 ≤ 𝜋 ≤ 𝜋. So arccos (cos (3𝜋)) = 𝜋. These last two examples tell us that we need to be careful to respect the range of arccos and arcsin when evaluating them. Now that we know how to evaluate these functions, we have two other functions that we can define. (1.45) Definition. The arctan function is the inverse of 𝑦 = tan (𝑥); −𝜋/2 < 𝑥 < 𝜋/2. The 𝑎𝑟𝑐𝑐𝑜𝑡 function is the inverse of 𝑦 = cot (𝑥); 0 < 𝑥 < 𝜋. The graph of arctan can be seen below: When evaluating these functions, we use a similar strategy as we did with arcsin and arccos: If 𝑏 = arctan 𝑎 iff 𝑎 = tan 𝑏 and −𝜋/2 < 𝑏 < 𝜋/2. 𝑏 = arccot(𝑎) iff 𝑎 = arccot(𝑏) and 0 < 𝑏 < 𝜋 One should note the notation arcsin, arccos and arctan is used to avoid ambiguity in the notation sin−1 𝑥. Note that sin −1𝑥 could be either the inverse sin function or it could mean (sin 𝑥)−1 = 1/ sin 𝑥 = csc 𝑥. So to avoid this ambiguity, we use the notation arcsin. Finally, one needs to be careful to remember that the arcsin, arccos, arctan and arccot functions are not inverses of the sin, cos, tan and cot functions. However, they are the inverses of these functions with the restricted domains. CHAPTER 1. TRIGONOMETRIC FUNCTIONS 23 Plot of arctan(𝑥) arctan(𝑥) 1 0.5 0 𝑦 −0.5 −1 −3 −2 −1 0 1 2 3 𝑥 Figure 1.5: Graph of arctan(𝑥) Plot of 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) 3 2 𝑦 1 0 −10 −5 0 5 10 𝑥 Figure 1.6: Graph of 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) 24 CHAPTER 1. TRIGONOMETRIC FUNCTIONS (1.46) Example. Compute arcsin (sin (7)) At first, one may be tempted to cancel sin and arcsin. However, if one inspects this a little bit more closely, one will realise that 7 is not the correct answer to this. Note that 𝑏 = arcsin (sin (7)) if and only if sin(𝑏) = sin(7) and −𝜋/2 ≤ 𝑏 ≤ 𝜋/2. Note that 7 does not lie in the acceptable range. However, note that sin (7 − 2𝜋) = sin (7) and 7 − 2𝜋 does lie within the acceptable range of values. Thus, arcsin (sin (7)) = 7 − 2𝜋 (1.47) Example. Compute sin (arcsin (7)). Once again, one may be tempted to try to cancel sin and arcsin. However, this is also not the case. Let’s begin by computing arcsin (7). We know that if 𝑏 = arcsin (7) then sin (𝑏) = 7 with −𝜋/2 ≤ 𝑏 ≤ 𝜋/2. However, the range of sin(𝑥) is −1 ≤ 𝑦 ≤ 1. So there is no such value that allows for sin (𝑏) = 7. Thus, this value does not exist. Therefore, sin (arcsin (7)) = 𝐷𝑁 𝐸 (Does not exist).  2 CHAPTER Limits What are Limits? This chapter will focus on looking at the value that a function approaches at certain points on a domain. Although the notion of “approach” will not be defined strictly, one can think of where a function is heading towards on certain points of the domain, or even ±∞. We begin by defining the notion of a left and right limit. (2.1) Definition. (Left Limit) Let −∞ < 𝑎 ≤ ∞ and −∞ ≤ 𝐿 ≤ ∞. Suppose 𝑓 (𝑥) is defined over the open interval (𝑐, 𝑎). If the values of 𝑓 (𝑥) approach 𝐿 as 𝑥 approaches 𝑎 within (𝑐, 𝑎) (from the left) then we say that lim− 𝑓 (𝑥) = 𝐿 𝑥→𝑎 (the limit as x approaches 𝑎 from the left is 𝐿. (2.2) Definition. (Right Limit) Let −∞ ≤ 𝑎 < ∞ and −∞ ≤ 𝐿 ≤ ∞. Suppose 𝑓 (𝑥) is defined over the open interval (𝑎, 𝑐). If the values of 𝑓 (𝑥) approach 𝐿 as 𝑥 approaches 𝑎 within (𝑎, 𝑐) (from the right) then we say that lim+ 𝑓 (𝑥) = 𝐿 𝑥→𝑎 (the limit as x approaches 𝑎 from the right is 𝐿. As a part of our convention, when 𝑎 = ±∞, we drop the 𝑎− or 𝑎+ notation because it is clear that one can only approach −∞ from the right and ∞ from the left. One of the first things that we should notice about the two definitions is that we are only looking at where a 25 26 CHAPTER 2. LIMITS function is approaching, and not the value the function actually attains. In fact, 𝑓 (𝑎) need not be defined, and yet the left and right limits may still exist. (2.3) Example. Consider 𝑦 = tan 𝑥, whose graph can be seen below: 𝑦 𝑥 −𝜋 − 𝜋2 𝜋 𝜋 2 Figure 2.1: Graph of tan(𝑥) Notice that tan(𝑥) is not defined at 𝑥 = 𝜋/2. However, we get that lim𝑥→(𝜋/2)+ = −∞ and lim𝑥→(𝜋/2)− = ∞ because if one looks to the right of 𝜋/2 and looks at where tan is going as it approached 𝜋/2, one sees that it drops to −∞. Likewise, when looking to the left of 𝜋/2 and looking where tan(𝑥) approaches, this time from the left side, one finds that tan(𝑥) approaches ∞ from the left. (2.4) Example. Now consider 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥), whose graph we can see below: Notice that as 𝑥 approaches ∞, 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) gets very close to 0, but not quite. In this case, we say that as x gets arbitrarily large, 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) gets close to zero. Therefore lim𝑥→∞ 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) = 0. Likewise, as 𝑥 gets very small (x gets closer to −∞) 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) gets very close to 𝜋 but never reaches it. We say in this case that lim𝑥→−∞ 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) = 𝜋. One should also notice that this definition of limit also respects the idea of the inverse trig functions. In particular, if 𝑏 = cot(𝑎) and 𝑎 approaches 0 from the right, one can notice that 𝑏 approaches ∞. So lim𝑎→0+ cot(𝑥) = ∞. But then we can notice that then 𝑎 = 𝑎𝑟𝑐𝑐𝑜𝑡(𝑏). So we find some consistency between the functions simply by looking at where 𝑎 approaches from one of the functions. Now that we have the ideas of a left and a right limit, we may now define the notion of a limit. CHAPTER 2. LIMITS 27 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) 3 2 𝑦 1 0 −10 −5 0 5 10 𝑥 Figure 2.2: Graph of 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) (2.5) Definition. Let −∞ < 𝑎 < ∞. We say that lim𝑥→𝑎 𝑓 (𝑥) (the limit as 𝑥 approaches a) exists if lim𝑥→𝑎− 𝑓 (𝑥) = lim𝑥→𝑎+ 𝑓 (𝑥) = 𝐿. In this case we write lim𝑥→𝑎 𝑓 (𝑥) = 𝐿. In short, the limit exists when the left and the right limit approach the same point. Now with this definition, we have a couple of warnings. 1.) 𝑓 (𝑥) does not need to be defined at 𝑎 in order to exist. 2.) lim𝑥→𝑎− 𝑓 (𝑥), lim𝑥→𝑎+ 𝑓 (𝑥) and lim𝑥→𝑎 𝑓 (𝑥) has nothing to do with 𝑓 (𝑎). One can see this in the following examples: 𝑥 +1 𝑥 0, we know that since 0 ≤ cos (𝑥) ≤ 1, we get 0 ≤ 𝑥 cos (𝑥) ≤ 𝑥. But then, as 𝑥 → 0+ , we find that lim𝑥→0+ 0 = 0 and lim𝑥→0+ 𝑥 = 0. Therefore, by applying the squeeze theorem, we obtain that lim𝑥→0+ 𝑥 cos (𝑥) = 0. 2.)lim𝑥→0− 𝑥 cos (𝑥). For the left limit, we look at points of 𝑥 close to zero but negative. But since 𝑥 < 0, we know that since −1 ≤ cos (𝑥) ≤ 0 on the left, we get −𝑥 ≤ 𝑥 cos (𝑥) ≤ 0. But then, as 𝑥 → 0− , we find that lim𝑥→0− 0 = 0 and lim𝑥→0− −𝑥 = 0. Therefore, by applying the squeeze theorem, we obtain that lim𝑥→0− 𝑥 cos (𝑥) = 0. Thus, applying the definition of a limit, we get lim𝑥→0 𝑥 cos (𝑥) = 0. The Squeeze Theorem can be a very powerful tool to help us to find limits. But in the previous example, we still went and used the definition of a limit to help us find the limit. Further, we still needed to check the sign of the function. So one should wonder whether there is an easier way to take this limit. In fact, the answer is yes, and can be seen in the following theorem. (2.23) Theorem. (Zero Limit Theorem) If lim𝑥→𝑎 |𝑓 (𝑥)| = 0 then lim𝑥→𝑎 𝑓 (𝑥) = 0 As it stands, we do not have the necessary tools to be able to prove this theorem. However, 38 CHAPTER 2. LIMITS this kind of theorem can be useful if a function oscillates between positive and negative values. It then allows to look at the absolute value of a function that oscillates, and conclude that if the limit of the absolute value of a function is 0, then the limit of the actual function must be zero. However, this will only work if the limit is 0. But let’s see how this theorem can help to make taking limits easier. (2.24) Example. lim 𝑥 cos (𝑥) 𝑥→0 This time, we use our new theorem. Notice that |𝑥 cos (𝑥)| is always positive. Therefore we know that 0 ≤ |𝑥 cos (𝑥)| ≤ 𝑥. Thus, since lim𝑥→0 0 = 0 and lim𝑥→0 𝑥 = 0, we know that by the squeeze theorem lim𝑥→0 |𝑥 cos (𝑥)| = 0 and thus, by the zero limit theorem, we know that lim𝑥→0 𝑥 cos (𝑥) = 0. Let’s look at another example. (2.25) Example. lim 𝑥 2 sin (1/𝑥) 𝑥→0 To solve this, let’s consider |𝑥 2 sin (1/𝑥)|. Note that since 0 ≤ | sin (1/𝑥) ≤ 1, we obtain that 0 ≤ |𝑥 2 sin (1/𝑥)| ≤ 𝑥 2. But then as 𝑥 → 0 we get that lim𝑥→0 0 = 0 and lim 𝑥 → 0𝑥 2 = 0. Therefore lim𝑥→0 |𝑥 2 sin (1/𝑥)| = 0 by squeeze theorem. Hence, lim𝑥→0 𝑥 2 sin (1/𝑥) = 0 by the zero limit theorem. One should note that the previous example can be done with the left and right limit, but it is much trickier because the statement 0 ≤ 𝑥 2 sin (1/𝑥) ≤ 𝑥 2 for positive x may not be true since sin (1/𝑥) because sin(1/𝑥) oscillates between positive and negative values. However, one could try to say that −𝑥 2 ≤ 𝑥 2 sin (1/𝑥) ≤ 𝑥 2 to obtain the result. Now we will show some limits which seem easy, but are a bit more difficult to justify using our methods. (2.26) Example. lim sin 𝑥 = 0 𝑥→0 Note that when 0 < 𝑥 < 𝜋/2, we know that 0 < sin 𝑥 < 𝑥. This can be taken as a fact, but if one wants to know why this is the case, one should note that if 0 < 𝑥 < 𝜋/2, then the arc length on the unit circle has length x. Then looking at the sin (𝑥) on the unit circle, we get that sin (𝑥) is bounded by 𝑥 because sin (𝑥) is bounded above by the distance between (1, 0) and the point on the unit circle defined by (cos(𝑥), sin(𝑥)). This, distance is in turn bounded by the arc of the circle at that point, namely 𝑥. CHAPTER 2. LIMITS 39 Now that we have this bound, we note that we can find lim𝑥→0+ sin(𝑥). Why? Notice that as 𝑥 approaches 0 from the right, we obtain lim𝑥→0+ 0 = 0 and lim𝑥→0+ 𝑥 = 0. Therefore, by squeeze theorem, lim𝑥→0+ sin(𝑥) = 0. Applying the same method to the left limit, we have that when −𝜋/2 < 𝑥 < 0, we get 𝑥 < sin(𝑥) < 0. Then as 𝑥 → 0 from the left, we get lim𝑥→0− 0 = 0 and lim𝑥→0− 𝑥 = 0. Therefore, applying the squeeze theorem yields that lim𝑥→0− sin (𝑥) = 0 Thus, since the left and right limits match, we obtain that lim𝑥→0 sin(𝑥) = 0 (2.27) Example. lim cos(𝑥) = 1 𝑥→0 Note that cos(𝑥) = ±√1 − sin2 (𝑥) by the Pythagorean Theorem. However, when −𝜋/2 < 𝑥 < 𝜋/2, we know that cos(𝑥) > 0. Thus, as 𝑥 → 0, we know that cos(𝑥) remains positive. Since sin(𝑥) → 0 as 𝑥 → 0, we obtain that lim𝑥→0 cos(𝑥) = lim𝑥→0 √1 − sin2 (𝑥) = √1 − 0 = 1 since sin(𝑥) → 0 as 𝑥 → 0. (2.28) Example. sin(𝑥) lim 𝑥→0 𝑥 This problem is a bit trickier and requires us to use the tangent function to help us. To begin, we note that when 0 < 𝑥 < 𝜋/2, we have that 0 < sin(𝑥) < 𝑥 < tan(𝑥). So sin 𝑥 < 𝑥 < sin(𝑥)/ cos(𝑥). Dividing by sin(𝑥) yields that 1 < 𝑥/ sin(𝑥) < 1/ cos(𝑥). But then as 𝑥 → 0+ 𝑥 we find that 1 → 1 and 1/ cos(𝑥) → 1. Thus, lim𝑥→0+ sin(𝑥) = 1. Then lim𝑥→0+ 1𝑥 = 1 = sin(𝑥) lim𝑥→0+ sin(𝑥) 𝑥. Now that we have the right limit, the left limit is obtained similarly.When −𝜋/2 < 𝑥 < 0, we obtain that tan(𝑥) < 𝑥 < sin(𝑥). Using the same process as before yields that lim𝑥→0− sin(𝑥) 𝑥 = 1. Therefore, lim𝑥→0 sin(𝑥) 𝑥 = 1. (2.29) Example. Analyze 𝑥 +2 lim 𝑥→−1 𝑥 +1 Notice that 𝑥+2 𝑥+1 1 = (𝑥 + 2) ⋅ 𝑥+1. Now we consider the left and right limits. For the left limit, 1 1 observe that lim𝑥→−1− (𝑥 + 2) ⋅ 𝑥+1 = −∞ since 𝑥+1 decreases toward −∞ while approaching from the left and (𝑥 + 2) is positive while approaching from the left. 1 1 We also can see that lim𝑥→−1+ (𝑥 + 2) ⋅ 𝑥+1 = ∞ since 𝑥+1 increases toward ∞ while approaching from the right and (𝑥 + 2) is positive while approaching from the right. Since the left limit is not the same as the right limit, we know that lim𝑥→−1 𝑥+2 𝑥+1 does not exist. 40 CHAPTER 2. LIMITS Continuity Now that we have some of the limit properties under our belt, we need to answer a very impor- tant question: What makes a function continuous? In heuristic terms, a function is continuous if it can be drawn without lifting a pencil from paper. But what does this mean mathemati- cally? (2.30) Definition. (Continuous Functions) We say that a function 𝑓 is continuous at a point 𝑎 in the domain if: (a) 𝑓 is defined at 𝑎; that is 𝑓 (𝑎) is defined. (b) lim𝑥→𝑎 𝑓 (𝑥) exists; that is lim𝑥→𝑎− 𝑓 (𝑥) = lim𝑥→𝑎+ 𝑓 (𝑥). (c) 𝑓 (𝑎) = lim𝑥→𝑎 𝑓 (𝑥) Let’s see some examples to see why we need these three conditions. 2 (2.31) Example. Is 𝑓 (𝑥) = 𝑥𝑥−2−4 continuous at 𝑥 = 2? Note that 𝑓 (𝑥) is not continuous at 𝑥 = 2 because 𝑓 (𝑥) is not even defined at 𝑥 = 2. In fact, there will be a hole at 𝑥 = 2. 𝑥 2 −4 𝑥≠2 (2.32) Example. Is 𝑓 (𝑥) = { 𝑥−2 continuous at 𝑥 = 2? 4 𝑥=2 Let’s check the three conditions of continuity. 1.) Unlike last time 𝑓 (2) = 4. So 𝑓 (2) is defined. 2.) We need to check that lim𝑥→2 𝑓 (𝑥) exists. But notice that lim𝑥→2 𝑓 (𝑥) = lim𝑥→2 (𝑥+2)(𝑥−2) (𝑥−2) = lim𝑥→2 𝑥 + 2 = 4. So the limit exists. 3.) Finally, we need to check that lim𝑥→2 𝑓 (𝑥) = 𝑓 (2). But this is true since lim𝑥→2 𝑓 (𝑥) = 4 = 𝑓 (2). Thus, 𝑓 (𝑥) is continuous at 𝑥 = 2 sin(𝑥) 𝑥≠0 (2.33) Example. Is 𝑓 (𝑥) = { 𝑥 continuous at 𝑥 = 0. 1 𝑥=0 We check the three conditions. 1.) Note that 𝑓 (0) = 1 by the way 𝑓 is defined. 2.) Note that lim𝑥→0 sin(𝑥) 𝑥 = 1 as previously shown. Recall that one should use the Squeeze Theorem in order to show this. 3.) Notice that 𝑓 (0) = lim𝑥→0 𝑓 (𝑥) = 1. Therefore, 𝑓 is continuous at 𝑥 = 0. CHAPTER 2. LIMITS 41 One should get into the habit of checking that functions are continuous by using the three conditions listed. Now, one should ask what functions are continuous. In particular, are the trig functions continuous? (2.34) Theorem. Trigonometric functions are continuous at any point in their natural domain. Proof. (For cos(𝑥)). We know that cos(𝑥) is defined everywhere. So we need to analyze lim𝑥→𝑎 cos(𝑥) to conclude that lim𝑥→𝑎 cos(𝑥) = cos(𝑎). Let −∞ < 𝑎 < ∞ be specified. Then, using the sum of angles formula, we obtain: lim cos(𝑥) = lim cos(𝑥 − 𝑎 + 𝑎) 𝑥→𝑎 𝑥→𝑎 = lim [cos(𝑥 − 𝑎) cos(𝑎) − sin(𝑥 − 𝑎) sin(𝑎)] 𝑥→𝑎 = cos(0) cos(𝑎) − sin(0) sin(𝑎) = cos(𝑎) Now that we have seen some examples of functions which are continuous and discontin- uous, we should label the types of discontinuities. (2.35) Definition. (a) We say that 𝑓 (𝑥) has a removable discontinuity at 𝑥 = 𝑎 if lim𝑥→𝑎 𝑓 (𝑥) exists but 𝑓 (𝑎) is undefined. (b) We say that 𝑓 (𝑥) has a jump discontinuity at 𝑥 = 𝑎 if lim𝑥→𝑎+ 𝑓 (𝑥) and lim𝑥→𝑎− 𝑓 (𝑥) are both numbers but not equal. (c) We say that 𝑓 (𝑥) has a infinite discontinuity if at least one of lim𝑥→𝑎+ 𝑓 (𝑥) or lim𝑥→𝑎− 𝑓 (𝑥) are ∞ or −∞ If 𝑓 (𝑥) has a discontinuity that does not fall into one of the three categories above, then 𝑓 has a different type of discontinuity. In this class we will not deal with those types of discon- tinuities. But let’s look at a couple of examples: 2 (2.36) Example. From before, we note that 𝑓 (𝑥) = 𝑥𝑥−2−4 has a jump discontinuity at 𝑥 = 2. The can be seen in the following graph: 4 − 𝑥2 𝑥 ≤ 3 (2.37) Example. Consider 𝑓 (𝑥) = { at 𝑥 = 3 4𝑥 − 8 𝑥 > 3 Note that 𝑓 (3) is defined, and in fact 𝑓 (3) = −5 42 CHAPTER 2. LIMITS 𝑦 6 4 2 𝑥 −4 −2 2 4 −2 Next, note that lim𝑥→3+ 𝑓 (𝑥) = lim𝑥→3− 4𝑥 − 8 = 4. Further lim𝑥→3− 𝑓 (𝑥) = lim𝑥→3− 4 − 𝑥 2 = −5. Since the left and right limits exist, but are not the same, we know that 𝑓 has a jump discontinuity at 𝑥 = 3 (2.38) Example. Consider 𝑓 (𝑥) = sin(𝑥)+cos(𝑥) tan(𝑥) at 𝑥 = 0 Note that 𝑓 is undefined at 𝑥 = 0 since tan(𝑥) is undefined at 𝑥 = 0. So 𝑓 (𝑥) is not con- tinuous. But let’s see what type of discontinuity 𝑓 (𝑥) has by considering the limit. Note that sin(𝑥)+cos(𝑥) tan(𝑥) = sin(𝑥)+cos(𝑥) sin(𝑥) 1 = cos(𝑥)(sin(𝑥)+cos(𝑥)) sin(𝑥). But note that: lim𝑥→0+ cos(𝑥)(sin(𝑥)+ cos(𝑥) 1 1 = ∞ and lim𝑥→0− cos(𝑥)(sin(𝑥)+cos(𝑥)) sin(𝑥) cos(𝑥)) sin(𝑥) = −∞. Therefore, 𝑓 has an infinite discontinuity at 𝑥 = 0. sin(𝑥) (2.39) Example. Consider 𝑓 (𝑥) = tan(𝑥) at the point 𝑥 = 0. Note that 𝑓 (𝑥) is not continuous because tan(𝑥) = 0 at 𝑥 = 0. Since tan(𝑥) is discontinuous, we can find out what kind of sin(𝑥) undefined tan(𝑥) = 0 discontinuity we have. Note that 𝑓 (𝑥) = tan(𝑥) = sin(𝑥) sin(𝑥) = {. Then cos(𝑥) cos(𝑥) otherwise notice that lim𝑥→0+ 𝑓 (𝑥) = lim𝑥→0 cos(𝑥) = lim𝑥→0+ 𝑓 (𝑥) = lim𝑥→0+ cos(𝑥) = 1. Therefore, 𝑓 has a removable discontinuity at 𝑥 = 0. We now move on to a very important theorem: (2.40) Theorem. (Intermediate Value Theorem) Let 𝑓 be a continuous function over a closed, bounded interval [𝑎, 𝑏]. If 𝜁 is any real number between 𝑓 (𝑎) and 𝑓 (𝑏) then there is a number 𝑐 in [𝑎, 𝑏] such that 𝑓 (𝑐) = 𝜁. To understand why this works, consider the following figure: CHAPTER 2. LIMITS 43 𝑓 (𝑥) 2 f(a) 𝑥 −1 a 1 2 b 3 4 f(b) −2 Notice that since 𝑓 (𝑥) is a continuous function, we can not lift the pencil from the paper when drawing it. Thus, if we want to reach 𝑓 (𝑏) from 𝑓 (𝑎) be must hit every point in-between 𝑓 (𝑎) and 𝑓 (𝑏). This theorem is incredibly useful if you want to show that functions attain certain values. (2.41) Example. Show that 𝑓 (𝑥) = 𝑥 − cos(𝑥) has at least one zero in [0, 𝜋/2] where a zero is a point 𝑐 in [0, 𝜋/2] such that 𝑓 (𝑐) = 0. Let’s use the Intermediate Value Theorem (IVT) to show this. Note that 𝑓 (𝑥) is continuous because 𝑥 and cos(𝑥) are continuous. Notice that 𝑓 (0) = 0 − cos(0) = −1 while 𝑓 (𝜋/2) = 𝜋/2 − cos(𝜋/2) = 𝜋/2 − 0 = 𝜋/2. By the IVT, we know that 𝑓 attains every value between −1 and 𝜋/2. Since −1 < 0 < 𝜋/2, there is a value 𝑐 in [0, 𝜋/2] such that 𝑓 (𝑐) = 0. Using the IVT, we may also obtain the following theorem: (2.42) Theorem. For 𝜁 > 0 and positive integer (whole number) 𝑛 > 0, there is a unique 𝑐 > 0 such that 𝑐 𝑛 = 𝜁. (This number is defined to be √ 𝑛 𝜁.) Proof. Once again, we employ the IVT. Let 𝑓 (𝑥) = 𝑥 𝑛 which is known to be continuous since it is a product of continuous functions. Let 𝜁 > 0 be given. Then note that 𝑓 (0) = 0. Then consider 𝜁 + 1. Note that 𝜁 + 1 > 1. Thus, 𝑓 (𝜁 + 1) = (𝜁 + 1)𝑛 > 𝜁 + 1 > 𝜁. Thus, since 𝑓 (0) < 𝜁 < 𝑓 (𝜁 + 1) the IVT tells us that there is a point 𝑐 in [0, 𝜁 + 1] such that 𝑓 (𝑐) = 𝑐 𝑛 = 𝜁. This value is unique because 𝑓 (𝑥) is strictly increasing. (2.43) Example. We now show an example of why continuity is needed in order for the IVT to apply. Consider the following graph: Notice that 𝑓 does not attain every value between 𝑓 (𝑎) and 𝑓 (𝑏) due to the jump discon- tinuity at 𝑥 = 0. Thus, we need the continuity assumption in order to use the IVT. 44 CHAPTER 2. LIMITS 𝑓 (𝑥) 2 f(b) 𝑥 −2 −1.5 −1 a −0.5 0.5 b 1 1.5 2 f(b) −2 Here’s one final example of how to use the IVT. (2.44) Example. Show that 𝑓 (𝑥) = 𝑥 3 − 𝑥 2 − 3𝑥 + 1 has a zero over [0, 1]. Note that 𝑓 (0) = 1 and 𝑓 (1) = −2. Applying the IVT, we know that for every −2 < 𝑎 < 1, there is a value 𝑐 in [0, 1] such that 𝑓 (𝑐) = 𝑎. Since −2 < 0 < 1, we know that there is a value 𝑐 such that 𝑓 (𝑐) = 0. Bonus Examples Here are some extra examples to help us see how to use some of the theorems in this chapter. (2.45) Example. Does lim𝑥→6 |𝑥−6| 𝑥−6 exist? 𝑥 𝑥≥0 Recall the definition of the absolute value function. |𝑥| = { −𝑥 𝑥𝑎 1 Notice that 𝑓 (𝑎) = 𝑎 and is therefore defined. However, lim𝑥→𝑎+ 𝑓 (𝑥) = lim𝑥→𝑎+ 𝑥−𝑎 = ∞. Therefore, 𝑓 (𝑥) has an infinite discontinuity at 𝑥 = 𝑎 despite being defined at 𝑥 = 𝑎.  3 CHAPTER The Derivative Definitions and Basic Examples 𝑥2 2𝑥 − 1 4 2 −2 −1 1 2 −2 −4 In this chapter, we will focus on the idea of finding the slope of the tangent line of any function at any given point. Recall that a tangent line will generally have the formula 𝑦 − 𝑦0 = 𝑘(𝑥 − 𝑥0 ). In general, we will be looking for what 𝑘 is. Now, to get some intuition for the derivative, consider the following graph: In this case, we want to find the slope of the graph at 𝑎. To do this, we begin by choosing another point, x, and we find the tangent line that goes through (𝑎, 𝑓 (𝑎)) and (𝑥, 𝑓 (𝑥)). We 46 CHAPTER 3. THE DERIVATIVE 47 4 2 (x,f(x)) (a,f(a)) −2 −1 1 2 −2 then find the slope of said tangent line. Then we keep moving 𝑥 closer and closer to the point 𝑎. This can be seen in the following: 4 2 −2 −1 1 2 −2 In doing so, we obtain the slopes of the tangent lines that get closer and closer to the 𝑓 (𝑥)−𝑓 (𝑎) tangent line at 𝑥 = 𝑎. Recall that the slope of each tangent line is 𝑥−𝑎 , which is known as the difference quotient. Thus, it is natural to define the slope of the tangent line of a point 𝑓 (𝑥)−𝑓 (𝑎) on a graph to be lim𝑥→𝑎 𝑥−𝑎. 𝑓 (𝑥)−𝑓 (𝑎) (3.1) Definition. Let 𝑓 (𝑥) be a function defined at a point 𝑎. If the limit lim𝑥→𝑎 𝑥−𝑎 = 𝐿 exists and is finite, we say that 𝑓 is differentiable at 𝑎 and define 𝑓 ′ (𝑎) = 𝐿 to be the derivative of 𝑓 at 𝑎. 48 CHAPTER 3. THE DERIVATIVE With this definition, we note the following facts: 1.) 𝑓 ′ (𝑎) is the slope of the tangent line of the graph of 𝑓 (𝑥) at the point 𝑎. 𝑓 (𝑥)−𝑓 (𝑎) 𝑓 (𝑎+ℎ)−𝑓 (𝑎) 2.) lim𝑥→𝑎 𝑥−𝑎 = limℎ→0 ℎ. This gives us another way to compute a deriva- tive at a particular point. This holds true because we can let ℎ = 𝑥 − 𝑎. If ℎ = 𝑥 − 𝑎 then when 𝑥 → 𝑎 we get ℎ → 0. Then, making a substitution yields the rest. But now that we know what a derivative is, let’s see some examples. (3.2) Example. Find the equation of the tangent line of 𝑓 (𝑥) = 1/𝑥 at 𝑥 = 2. Let’s use both methods to find the tangent line. We begin by computing the derivative of 𝑓 (𝑥) at 𝑥 = 2. Method 1: We apply the first definition: 1 1 𝑓 (𝑥) − 𝑓 (2) − lim = lim 𝑥 2 𝑥→2 𝑥 −2 𝑥→2 𝑥 − 2 2−𝑥 2𝑥 = lim 𝑥→2 𝑥 −2 2−𝑥 = lim 𝑥→2 −(𝑥 − 2)2𝑥 −1 = lim 𝑥→2 2𝑥 −1 = 4 Thus, we get that the slope of the tangent line at 𝑥 = 2 is −1 4. Then, note that to obtain the slope of the tangent line, we first compute 𝑓 (2) = 12. Finally, the equation is 𝑦 − 12 = −1 4 (𝑥 − 2). Method 2: We use the second formula: 1 𝑓 (2 + ℎ) − 𝑓 (2) 2+ℎ − 12 lim = lim ℎ→0 ℎ ℎ→0 ℎ 2 − (2 + ℎ) 1 = lim ⋅ ℎ→0 2(2 + ℎ) ℎ −1 = lim ℎ→0 2(2 + ℎ) −1 = 4 CHAPTER 3. THE DERIVATIVE 49 Thus, we get the same result as from the first method. (3.3) Example. Find the slope of the tangent line of 𝑓 (𝑥) = √𝑥 at 𝑥 = 4 We once again apply both methods. Method 1: 𝑓 (𝑥) − 𝑓 (4) √𝑥 − √4 lim = lim 𝑥→4 𝑥 −4 𝑥→4 𝑥 −4 √𝑥 − √4 √𝑥 + √4 = lim ⋅ 𝑥→4 𝑥 − 4 √𝑥 − √4 (𝑥 − 4) = lim 𝑥→4(𝑥 − 4)(√𝑥 + √4 1 = lim 𝑥→4 √𝑥 + √4 1 = 4 Method 2: 𝑓 (𝑎 + ℎ) − 𝑓 (𝑎) √4 + ℎ − √4 lim = lim ℎ→0 ℎ ℎ→0 ℎ √4 + ℎ − √4 √4 + ℎ + √4 = lim ⋅ ℎ→0 ℎ √4 + ℎ + √4 4+ℎ−4 = lim ℎ→0 ℎ(√4 + ℎ + √4) ℎ = lim ℎ→0 ℎ(√4 + ℎ + √4) 1 = lim ℎ→0 √4 + ℎ + √4 1 = 4 Notice that in order to use both methods, we must multiply by the conjugate of the nu- merator in order to help with the square root. In the future, both methods of finding the derivative will be acceptable. But also, we have only seen examples of finding a derivative at a particular point. What if we want a general formula for a derivative? 50 CHAPTER 3. THE DERIVATIVE (3.4) Example. Find the derivative of 𝑓 (𝑥) = √𝑥 at 𝑥 = 𝑎 where 𝑎 > 0. To find the derivative, we now use an arbitrary a. 𝑓 (𝑎 + ℎ) − 𝑓 (𝑎) √𝑎 + ℎ − √𝑎 √𝑎 + ℎ + √𝑎 lim = lim ⋅ ℎ→0 ℎ ℎ→0 ℎ √𝑎 + ℎ + √𝑎 𝑎+ℎ−𝑎 = lim ℎ→0 ℎ(√𝑎 + ℎ + √𝑎) ℎ = lim ℎ→0 ℎ(√𝑎 + ℎ + √𝑎) 1 = lim ℎ→0 √𝑎 + ℎ + √𝑎 1 = 2 √𝑎 (3.5) Example. Find the derivative of 𝑓 (𝑥) = 1/𝑥 at any 𝑥 = 𝑎 in the domain of 𝑓 (𝑥). 1 𝑓 (𝑎 + ℎ) − 𝑓 (𝑎) 𝑎+ℎ − 1𝑎 lim = lim ℎ→0 ℎ ℎ→0 ℎ 𝑎 − (𝑎 + ℎ) 1 = lim ⋅ ℎ→0 𝑎(𝑎 + ℎ) ℎ −1 = lim ℎ→0 𝑎(𝑎 + ℎ) −1 = 2 𝑎 Thus, we can say that 𝑓 ′ (𝑥) = (1/𝑥)′ = −1 𝑥2 Now that we have a couple of examples, we should see how this applies to trig functions. For instance, what is the derivative of sin(𝑥)? (3.6) Example. If 𝑓 (𝑥) = sin(𝑥) what is 𝑓 ′ (𝑥). To do this, we first need to compute a separate limit. cos(ℎ) − 1 cos(ℎ) − 1 cos(ℎ) + 1 lim = lim ⋅ ℎ→0 ℎ 𝑥→0 ℎ cos(ℎ) + 1 2 cos (𝑥) − 1 = lim ℎ→0 ℎ(cos(ℎ) + 1) − sin2 (ℎ) 1 = lim ⋅ ℎ→0 ℎ cos(ℎ) + 1 CHAPTER 3. THE DERIVATIVE 51 sin(ℎ) 1 = lim − sin(ℎ) ⋅ ⋅ ℎ→0 ℎ 1 + cos(ℎ) =0 Now that we understand this limit, note that: 𝑓 (𝑎 + ℎ) − 𝑓 (𝑎) sin(𝑎 + ℎ) − sin(𝑎) lim = lim ℎ→0 ℎ ℎ→0 ℎ sin(𝑎) cos(ℎ) + cos(𝑎) sin(ℎ) − sin(𝑎) = lim ℎ→0 ℎ sin(𝑎)(cos(ℎ) − 1) cos(𝑎) sin(ℎ) = lim + ℎ→0 ℎ ℎ cos(ℎ) − 1 sin(ℎ) = sin(𝑎) lim + cos(𝑎) lim ℎ→0 ℎ ℎ→0 ℎ = 0 + cos(𝑎) = cos(𝑎) Thus, we know that (sin(𝑥))′ = cos(𝑥). We can also do the same for cos(𝑥). (3.7) Example. Find 𝑓 ′ (𝑥) where 𝑓 (𝑥) = cos(𝑥). cos(𝑎 + ℎ) − cos(𝑎) 𝑓 ′ (𝑎) = lim ℎ→0 ℎ cos(𝑎) cos(ℎ) − sin(𝑎) sin(ℎ) − cos(𝑎) = lim ℎ→0 ℎ cos(ℎ) − 1 sin(ℎ) = lim cos(𝑎) − sin(𝑎) ℎ→0 ℎ ℎ = 0 − sin(𝑎) = − sin(𝑎) Thus, (cos(𝑥))′ = − sin(𝑎). What about basic functions like 𝑦 = 𝑥 or 𝑦 = 𝑐 for a constant c? (3.8) Example. Find 𝑓 ′ (𝑥) when 𝑓 (𝑥) = 𝑥. 52 CHAPTER 3. THE DERIVATIVE 𝑥 +ℎ−𝑥 𝑓 ′ (𝑥) = lim ℎ→0 ℎ ℎ = lim ℎ→0 ℎ = lim 1 ℎ→0 =1 Thus, 𝑓 ′ (𝑥) = (𝑥)′ = 1 for all x. (3.9) Example. Find 𝑓 ′ (𝑥) when 𝑓 (𝑥) = 𝑐 where 𝑐 is any constant. 𝑓 (𝑥 + ℎ) − 𝑓 (𝑥) 𝑓 ′ (𝑥) =