# Lecture Notes on Calculus I (by Frederick T.-H (1). FONG).pdf

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Unformatted text preview: Calculus I Lecture Notes for MATH 1013 (Fall 2015) Frederick Tsz-Ho Fong Department of Mathematics Hong Kong University of Science and Technology Contents 1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Basic Notions 1 1.2 Compositions and Inverses 9 1.3 Logarithmic Functions 15 1.4 Inverse Trigonometric Functions 17 2 Limit and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.1 Limit of a Function 21 2.2 Continuity 31 2.3 Concept of Derivatives 35 3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.1 Basic Derivatives 39 3.2 Differentiability 43 3.3 Differentiation Rules 47 3.4 Derivatives of Trigonometric Functions 49 3.5 Chain Rule 51 3.6 Implicit Differentiation 54 4 Applications of Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 61 4.1 Rate of Change 61 4.2 Numerical Methods 65 4.3 l’Hospital’s Rule 69 4.4 Derivatives and Graphs 72 4.5 Optimization 80 4.6 Mean Value Theorem 82 5 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.1 Anti-Derivatives 85 5.2 Definite Integrals 88 5.3 Fundamental Theorem of Calculus 96 5.4 Integration by Substitutions 101 1 — Functions 1.1 Basic Notions 1.1.1 What is a function? 12 |||| CHAPTER 1 FUNCTIONS AND MODELS It’s helpful to think of a function as a ma the function f, then when x enters the machi produces an output f &x' according to the ru domain as the set of all possible inputs and t FIGURE 2 Theafter preprogrammed A function can be regarded as a machine give it ƒan input, then processing it functions in a calcu Machine diagram –foryou a function according to a defined rule, the machine gives you an output. We often denote For the example, input by athe square root key o machine. variable x, and the output by f ( x ). For instance, we can write: You press the key labeled s (or sx ) and e domain of this function; that is, x is not an a f ( x ) = x2 + 1 cate an error. If x # 0, then an approximatio your calculator to represent the “machine” that takes the input x and output the value + 1.onAs such, if we is not quite the sam sxx2key 2 input the value 2 into this function, it will output the value 2 + 1, orby equivalently, f &x' ! sx 5.. In short, we may simply write: Another way to picture a function is by an f (2) = 5. ƒ x connects an element of D to an element of E Likewise, it is easy to verify that fa(1) = 2, f (0) = 1, etc. with x, f &a' is associated with a, and so on. f(a) The most common method for visualizing Depending on the physical meaning, we can pick other suitable pair of letters to denote a i domain D, then its graph is the set of ordere function and its variable instead of the generic f and x. For instance, x (input) f ƒ (output) • A(r ) = pr2 , where r means the radius of a circle and A means the area of this circle. • h(t), where t means the time and fh means your height at time t. D !&x, f &x'' E (Notice that these are input-output pairs.) In FIGURE 3 1.1.2 Some terminology points &x, y' in the coordinate plane such tha Arrow for set ƒ of allowable inputs. For example, the function 1. The domain of a function f ( x )diagram means the The graph of a function f gives us a usef p defined by the expression f ( x ) = x has domain given by the interval: a function. Since the y-coordinate of any poin [0, •), or equivalently, {x : x 0} the value of f &x' from the graph as being th Figure 4). The graph of f also allows us to p 1 which means the set of all non-negative numbers. The functionrange g( x ) = domain on1the as in Figure 5. y-axis x has given by ( •, 1) [ (1, •) y { x, ƒ} which means the set of all real numbers except 1. r ƒ f (2) f(1) 0 1 2 x x Functions 2 i If every real number is an allowable input for a function f ( x ), then we can denote the domain of f ( x ) by ( •, •), or simply R. i Consider the function A(r ) = pr2 which is the area of a circle with radius r. Although it is mathematically legitimate to input a negative r into A(r ), a circle with negative or zero radius is not physically meaningful. Therefore, we can take (0, •) to be the domain of A(r ), instead of R. i If D is the domain of a function f ( x ), then any (non-empty) subset E of D can also taken to be the domain of f ( x ). When the domain of a function f ( x ) is not specified, the domain is usually taken as the largest set in which f ( x ) is defined. p Take f ( x ) = x as an example. Since [0, •) can be taken to be the domain of f ( x ), we can also declare that its subset [1, •)pto be the domain of f ( x ). However, without any declaration, the domain of f ( x ) = x is taken to be [0, •) by default, since it is the largest possible domain. 2. The codomain of a function f ( x ) is the set where the outputs belong to. In this course, the p output of a function is usually a real number, as in the examples f ( x ) = x, g( x ) = 1 1 x and A(r ) = pr2 we have seen so far. We may simply say R, or ( •, •), is the codomain of f ( x ), g( x ) and A(r ). i The codomain of a function simply indicates what kind of objects the outputs are, but it does not mean that everything in the codomain is a possible output. Take the function h( x ) = x2 + 1 as an example. Given any input x, the output x2 + 1 is always positive (in fact at least 1). However, it is perfectly fine to say that the codomain of h is R. 3. A function f ( x ) with domain D and codomain C is usually denoted by: f :D!C For example, we may indicate the domain and codomain of the function g( x ) = writing: g : ( •, 1) [ (1, •) ! R 1 1 x by 4. The range of a function f : D ! C is the set of all achievable outputs. It is usually denoted by f [ D ] where D is the domain of f . In set notations, the range of f is usually defined by: f [D] = { f (x) : x 2 D} For example, the range of h : R ! R defined by h( x ) = x2 + 1 is given by h[R ] = [1, •). i However, if we declare the domain of h to be a smaller set such as [2, 3], then the range of h will be smaller as it can only output values between 22 + 1 and 32 + 1. Therefore, the range of h : [2, 3] ! R becomes [5, 10]. i There is no universal way to find the range of a given function, although it is easy to do so in the above examples. One application of Calculus is to help us find the range of a given function. Another way to find the range is to look at the graph of the function which can be plotted by computer software. 1.1.3 Graph of a function The graph of a function is a geometric way of representing a function. It is a curve on the xy-coordinate plane which looks like the one in Figure 1.1. To read off the value of f (1) from the graph, simply locate the point on the curve with 1 as the x-coordinate, then the y-coordinate of that point will be the value f (1). Similar for f (2) and all other values f ( x ). Many computer softwares are able to plot the graph of a given function. With the graph, it is easy to find (or at least to estimate) the range of the function (see Example 1.1) FIGURE 3 points &x, y' in the coordinate plane such that y ! f &x' and x is in the domain of f . The graph of a function f gives us a useful picture of the behavior or “life history a function. Since the y-coordinate of any point &x, y' on the graph is y ! f &x', we can the value of f &x' from the graph as being the height of the graph above the point x Figure 4). The graph of f also allows us to picture the domain of f on the x-axis an 3 range on the y-axis as in Figure 5. Arrow diagram for ƒ 1.1 Basic Notions y y { x, ƒ} range ƒ f (2) f (1) 0 1 2 x x 0 FIGURE Figure4 1.1: graph of function y domain x FIGURE 5 EXAMPLE 1 The graph of a function f is shown in Figure 6. ⌅ (a) possible) Find the domain values ofoffthe and f &5'.defined by the expression: &1' function Example 1.1 Find the (largest (b) What are the s domain and range of f ? x2 x 2 x (a) We see from Figure36 that the point &1, 3' lies on the graph of f , so the value of SOLUTION f ( x ) = 1 0 y ! ƒ(x) 1 x at 1 is f &1' ! 3. (In other words, the point on the graph that lies above x ! 1 is 3 un By plotting the graph of f using computer softwares, estimate the range of f when the abovepossible. the x-axis.) domain is taken to be the largest When x ! 5, the graph lies about 0.7 unit below the x-axis, so we estimate that f &5' ( !0.7. FIGURE 6 ⌅ Solution First we observe that x 3 is a divisor so we need to exclude 3 from the domain. (b) We see that f &x' is defined when 0 "2 x " 7, so the domain of f is the closed int Then, in order for the square root to be meaningful, we need to ensure x x x3 2 is non-negative, so we solve the inequality: val \$0, 7%. Notice that f takes on all values from !2 to 4, so the range of f is x2 The notation for intervals is given in Appendix A. N 2 x x 3 " !y !2 " y " 4# ! \$!2, 4% 0. It is standard arithmetics that the quotient of two numbers is non-negative if and only if both numbers have the same sign (both 0, or both 0). Therefore, we can solve the inequality by dividing it into two cases: Case 1: x > 3 In this case the denominator x 3 is positive, so we need to have x2 x 2 0. x2 x ( x + 1)( x ( 2 0 2) 0 (factorization) x+1 0 x 2 0 ( x 1 x 2 x or 2 ( x+1 0 x 20 or ( or x x 1 x2 As x > 3 in this case, and ( x > 3) and ( x 2 or x in this case is x > 3 . Case 2: x < 3 In this case we need to have x2 x 2 0: x2 x ( x + 1)( x 20 2) 0 1 1) simply means x > 3, the solution Functions 4 ( x+1 0 x 2 0 ( x 1 x 2 or or 1x2 ( ( x+1 0 x 20 x 1 x2 Since the interval [ 1, 2] lies within x < 3, the solution in this case is simply Overall, the domain of f is given by inequalities 1 x 2 or x > 3, i.e. 1x2. [ 1, 2] [ (3, •). 6 4 2 -2 2 4 6 8 10 From the graph of f , it suggests that the range of f should be in the form of [0, a] [ [b, •). Later in the course, we will learn how to determine the exact values of a and b. In fact, the range of f can be shown to be [0, 1] [ [3, •). 1.1.4 Well-definedness When we define a function, it is important to keep in mind that: Each input in the domain must have exactly one output. It is the case for all examples we have seen so far. An expression with the property that every input in the domain has exactly one output is said to be well-defined. The expressions below are not well-defined and so they cannot be regarded as functions. 1. Let the domain be [0, •). If we define the output f ( x ) to be the number y such that y2 = x, then such an f ( x ) is not well-defined. Here is the reason. If we let the input to be 1, then f (1) is the number y such that y2 = 1. There are two possibilities, namely y = 1 or y = 1. Therefore, there are two possible values for f (1), and so such an f is not well-defined and cannot be regarded as a function. p 2. Let the domain be Q, the set of all rational numbers. Given any input q where p and q are integers, we define ✓ ◆ p f = p. q Given an input 12 , we have f ( 12 ) = 1, it seems that the output is simply 1. However, the input 12 has many other forms, such as 24 that gives f ( 24 ) = 2 as output. The fractions 12 and 24 are regarded as the same input just expressed in two different ways. Now that this input has at least two possible outputs, so f is not well-defined and cannot be regarded as a function. t!x" ! 1 1 ! The graph of a function is a curve in the xy-plane. But the question arises: Which curves x !x x!x ! 1" 2 in the xy-plane are graphs of functions? This is answered by the following test. and division by 0 is not allowed, we see that t!x" is not defined when x ! 0 or x ! 1. Thus the domain of t is THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of x if \$ 1.1 Basic Notions #x x " 0, xand " only 1% if no vertical line intersects the curve more than once. 5 Given a curve on the xy-plane, one can determine whether or not it is the graph of a which could also be writtenby in so-called interval notation as reason The for the truth of the Vertical Line Test can be seen in Figure 13. If each verfunction the vertical line test: tical line x ! a intersects a curve only once, at !a, b", then exactly one functional value (Vertical Line Test) A curve on the xy-plane is the graph of a function of x if and only if no M !!\$, 0" ! !0, ! !1,by\$"f !a" ! b. But if a line x ! a intersects is 1" defined the curve twice, at !a, b" and !a, c", vertical line intersects the curve at more than one point (see Figures 1.2 and 1.3). then the curve can’t represent a function because a function can’t assign two different valto a. But the question arises: Which curves The graph of a function is a curve in the xyues -plane. in the xy-plane are graphs of functions? This is answered by the following test. y y x=a (a, c) x=a THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once. (a, b) (a, b) x x a 13. If each a 0 be seen in Figure The reason for the truth of the Vertical Line Test can ver- 0 FIGURE 13 tical line x ! a intersects a curve only once, at !a, b", then exactly one functional value is defined by f !a" ! b. But if a line x ! a intersects the curve twice, at !a, b" and !a, c", Figure 1.2: pass the vertical line test, hence2a graph of function then the curve can’t represent a function because function the can’t assignxtwo val-in Figure 14(a) on the next page is not the Fora example, parabola ! ydifferent ! 2 shown ues to a. graph of a function of x because, as you can see, there are vertical lines that intersect the y parabola twice. The parabola, however, does contain the graphs of two functions of x. y Notice that the equation x ! y 2 ! 2 implies y 2 ! x " 2, so y ! #sx " 2 . Thus the x=a upper and lower halves of the(a, c) parabola are the graphs of the functions f !x" ! s x " 2 [from Example 6(a)] and t!x" ! !s x " 2 . [See Figures 14(b) and (c).] We observe that if we reverse the roles of x and y, then the equation x ! h!y" ! y 2 ! 2 does define x as a (a, b) function of y (with y as the independent variable and x as the dependent variable) and the parabola now appears as the graph of the function h. x=a (a, b) a 0 x 0 a x Figure 1.3: fail the vertical line test, hence not a graph of function For example, the parabola x ! y 2 ! 2 shown in Figure 14(a) on the next page is not the graph of a function of x because, as you can see, there are vertical lines that intersect the Although each input a function haveofexactly one output,ofit x is. perfectly fine for a parabola twice. The parabola, however, does ofcontain themust graphs two functions i 2 to give the same2output for two or more different inputs. For instance, consider the function Notice that the equation x ! y ! 2 implies y ! x " 2, so y ! #sx " 2 . Thus the function g : R ! R defined by upper and lower halves of the parabola are the graphs of the functions f !x" ! s x " 2 g( x ) = x2 + 1. [from Example 6(a)] and t!x" [See Figures 14(b) and (c).] We observe that ! ! x " 2 . Evidently,swe have both g(1) = 2 and g( 12) = 2. It is allowed. if we reverse the roles of x and y, then the equation x ! h!y" ! y ! 2 does define x as a function of y (with y as the independent variable and x as the dependent variable) and the 1.1.5 Piecewise Defined Functions parabola now appears as the graph of the function h. Examples of functions we have seen so far are defined using a simple expression such as f ( x ) = x2 + 1. However, in practice we sometimes define a function in a piecewise way, meaning that the function can have different expressions on different part of the domain. Let’s look at an example: Example 1.2 The monthly bill of a smartphone data plan is calculated as follows: At a monthly charge of HK\$168, you get 1024MB of data usage for that month. If you go over 1024MB, you will be additionally charged at the rate of HK\$0.5 per extra MB. Denote x to be your monthly usage. Express your month bill amount as a function f ( x ). ⌅ Solution First the domain is clearly [0, •) as your usage cannot be negative. When 0 x 1024, you will be charged at the standard monthly rate of HK\$168, so f ( x ) = 168 if 0 x 1024. When x > 1024, you have overused x 1024 (MB) and so you will be charged by extra 0.5( x 1024) dollars. Therefore, f ( x ) = 168 + 0.5( x 1024) in HKD if x > 1024. ⌅ Functions 6 We can write down the expression of this function in the following way: ( 168 if 0 x 1024 f (x) = 168 + 0.5( x 1024) if x > 1024 A classic example of piecewise-defined function is the absolute value function, defined piecewise by: ( x if x 0 |x| = x if x < 0 For example, we can easily see that |1| = 1 and | 3| = ( 3). In short, this function turns any negative number to its positive counterpart, and preserve all non-negative numbers. The 1 FUNCTIONS AND MODELS graph of the absolute value function18f ( x|||| ) = |CHAPTER x | is shown in Figure 1.4. EXAMPLE 8 Sketch the graph of the absolute valu y SOLUTION From the preceding discussion we know y=| x | \$x\$ ! 0 x FIGURE 16 Figure 1.4: graph of f ( x ) = | x |. • | xy| = | x | |y| and = x if \$x if Using the same method as in Example 7, we see line y ! x to the right of the y-axis and coincides y-axis (see Figure 16). EXAMPLE 9 Find a formula for the function f gra It is useful to keep in mind that x y # |x| ; |y| y • | x | = | x |; • | x + y| | x | + |y| and | x y| | x | + |y|. However, beware that generally we only have | x + y| | x | + |y|, and equality holds only when x and y are of the same sign, i.e. both 0, or both 0. FIGURE 17 Solving inequalities involving absolute values 1 0 1 To solve an inequality such as |2x 1| < 3, or |4 x | 2, etc., it is useful to keep in mind the SOLUTION The line through !0, 0" and !1, 1" has slo two facts: equation is y ! x. Thus, for the part of the graph If a is non-negative, we have: • |y| a if and only if a y a. f !x" ! x if 0 • |y| a if and only if y a or y a. If b is negative, we have: The line through !1, 1" and !2, 0" has slope m ! • |y| b is always false! Point-slope form of the equation of a line: • |y| b is always true! y \$ y1 ! m!x \$ x 1 " y \$ 0 ! !\$1"!x \$ 2" N See Appendix B. ⌅ So we have Example 1.3 Determine the (largest possible) domain of the function: f (x) = ⌅ q |3x + 1| 2 Solution Again, we need to take the square root, and so we need: |3x + 1| 2 |3x + 1| 0 2 f !x" ! 2 \$ x if We also see that the graph of f coincides with th tion together, we have the following three-piece f # x if f !x" ! 2 \$ x if 0 if EXAMPLE 10 In Example C at the beginning of t of mailing a first-class letter with weight w. In ef because, from the table of values, we have C 1 C!w" ! 0.39 0.63 0.87 1.11 ! ! ! if if if if 1.1 Basic Notions 7 From the rules mentioned above, we get: 3x + 1 x The domain of f is ( •, 1 or 2 or 3x + 1 2, and so: 1 . 3 x 1] [ [ 13 , •). 1.1.6 Various Types of Functions Definition 1.1 — Odd and Even Functions. Let D be a subset of R which is symmetric about zero. A function f : D ! R is said to be: • an even function if f ( x ) = f ( x ) for any x in D. • an odd function if f ( x ) = f ( x ) for any x in D. The graph of an even function is reflexively symmetric about the y-axis (see Figure 1.5). Important examples include: cos x, x2 , | x | , . . . y SYMMETRY f(_x) If a function f satisfies f !!x" " f !x" SECTION 1. even function. For instance, the func ƒ y0 _x SECTION x x SYMMETRY f !!x" If a function f satisfies f !!x" " f !x" fo f(_x) ƒ even function. For instance, the functio The geometric significance of an eve _x x x 0 to the y-axis (see Figure 19).f This !!x" me " Figure 1.5:19 graph of an even function we obtain the entire graph simply by FIGURE If f satisfies f !!x" " of !fan !x"even for fe An even function geometric significance The graph of an odd function is 180 -rotationally symmetric about The the origin (see Figure theThis function 1.6). Important examples include: tofunction. the y-axisFor (seeexample, Figure 19). mean y FIGURE 19 sin x, tan x, An even...
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