Calculus. Contents. Paul Sutcliffe. Office: CM212a.

Transcription

1 Calculus Paul Sutcliffe Office: CM212a. Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical sciences, Boas. Contents 1 Functions Functions, domain and range The graph of a function Graphs and transformations Even and odd functions Piecewise functions Operations with functions Inverse functions Elementary functions imits and continuity Definitions Results about limits and continuity imits as x Classification of discontinuities The intermediate value theorem

2 3 Differentiation Derivative as a limit Rules of differentiation Derivatives of higher order The chain rule Hopital s rule Implicit differentiation Boundedness and monotonicity Critical points Min-Max problems Rolle s theorem The mean value theorem The inverse function rule Partial derivatives Integration Indefinite integrals Definite integrals The fundamental theorem of calculus The logarithm and exponential function revisited Standard integrals Integration by parts Integration by substitution Integration of rational functions using partial fractions Integration using a recurrence relation Definite integrals using even and odd functions Differential equations First order separable ODE s First order homogeneous ODE s First order linear ODE s First order exact ODE s Bernoulli equations Second order linear constant coefficient ODE s The homogeneous case

3 5.6.2 The inhomogeneous case Initial and boundary value problems Systems of first order linear ODE s Taylor series Taylor s theorem and Taylor series agrange form for the remainder Calculating limits using Taylor series Fourier series Calculating the Fourier coefficients Parseval s theorem Half range Fourier series Complex Fourier series Functions of several variables Taylor s theorem for functions of two variables Chain rule Implicit differentiation Double integrals Rectangular regions Beyond rectangular regions Polar coordinates Change of variables and the Jacobian The Gaussian integral

4 1 Functions Notation: Symbol s.t. iff Denotes for every there exists such that if and only if Open interval, (a, b) = {x : a < x < b}. Closed interval, [a, b] = {x : a x b}. Half-open interval, (a, b] = {x : a < x b}, etc. Semi-infinite interval, (a, ) = {x : x > a}, etc. The set of real numbers, (, ), is denoted by R. For two sets A and B then A B is the union, A B is the intersection and A\B is the set of all elements in A that are not in B. 1.1 Functions, domain and range Defn: A function f is a correspondence between two sets D (called the domain) and C (called the codomain), that assigns to each element of D one and only one element of C. The notation to indicate the domain and codomain is f : D C. For x D we write f(x) to denote the assigned element in C, and call this the value of f at x, or the image of x under f, where x is called the argument of the function. Extending the above notation we write f : D C : x f(x) ie. f is a function from D to C that associates x in D to f(x) in C. The set of all images is called the range of f and our notation for the domain and range is Dom f and Ran f = {f(x) : x Dom f}. 4

5 Figure 1: Illustration of the domain, codomain and range of a function. In (almost all of) this course we shall only deal with real-valued functions of a real variable, meaning that our functions assign real numbers to real numbers ie. both the domain and codomain are subsets of R. There are various ways of representing a function, but perhaps the most familiar is through an explicit expression. Eg. f : R R given by f(x) = x 2, x R. In this case Dom f = R and Ran f = [0, ). We say that f maps the real line onto [0, ). Eg. f(x) = 2x + 4, x [0, 6]. Here Dom f = [0, 6] and Ran f = [2, 4]. If the domain of a function f is not explicitly given, then it is taken to be the maximal set of real numbers x for which f(x) is a real number. Eg. f(x) = 2x + 4. Here Dom f = [ 2, ) and Ran f = [0, ). Eg. f(x) = 1/(1 x). Here Dom f = R\{1} = (, 1) (1, ) and Ran f = R\{0}. In addition to explicit expressions, a function f may be represented by other means, for example, as a set of ordered pairs (x, y), where x Dom f and y Ran f. 5

6 Note: y(x) is another common notation for a function. The element x in the domain is called the independent variable and the element y in the range is called the dependent variable. This notation is often used if the function is defined by an equation in two variables, including differential equations (see later). The function f(x) = x 2 is defined by the equation y = x 2, as we have simply written y = f(x). However, not all equations will define a function. Eg. The equation y 2 = x does NOT define a function y(x), because for x < 0 there are no real solutions for y, and for x > 0 there are two solutions for y, whereas a function must assign only one element of the range for each element of the domain. 1.2 The graph of a function Defn: The graph of a function f is the set of all points (x, y) in the xy-plane with x Dom f and y = f(x), ie. graph f = {(x, y) : x Dom f and y = f(x)}. The graph of a function over the interval [a, b] is the portion of the graph where the argument is restricted to this interval. Note: If you are asked to graph a function, but no interval is given, then try to choose an appropriate interval that includes all the interesting behaviour eg. turning points. The graph of a function is a curve in the plane, but not every curve is the graph of a function. The following simple test determines whether a curve is the graph of a function. 6

7 Figure 2: The graph of f(x) = cos x over [ 2π, 2π]. The vertical line test. If any vertical line intersects the curve more than once then the curve is not the graph of a function, otherwise it is. The proof is obvious, given the defining property of a function that only one element in the range is associated with an element in the domain. Figure 3: The vertical line test applied to a cubic curve and a circle. In the above figure the first curve is the graph of a function; in fact the cubic function f(x) = x 3 x 2 1. The second curve is not the graph of a function as the vertical line shown intersects the curve twice. In fact the curve is given by the equation y 2 = 4 (x 1) 2 ie. the circle of radius 2 with centre (x, y) = (1, 0). Note that even when a curve is not the graph of a function, some sections of the curve may be graphs of functions. For the above circle example the section given by y 0 is the graph of the function f(x) = 4 (x 1) 2 with Dom f = [ 1, 3]. 7

8 1.3 Graphs and transformations The graph of a function f : R R can be shifted, reflected and rescaled by applying some simple transformations to the function and its argument, as follows: translation: f(x) + a shifts the graph of f(x) up by a units. f(x + b) shifts the graph of f(x) to the left by b units. reflection: f(x) reflects the graph of f(x) about the x-axis. f( x) reflects the graph of f(x) about the y-axis. f(x) reflects the portions of the graph of f(x) that are below the x-axis about the x-axis. scaling: λf(x) is a vertical stretch of the graph of f(x) if λ > 1 and a vertical contraction if 0 < λ < 1. f(µx) is a horizontal contraction of the graph of f(x) if µ > 1 and a horizontal stretch if 0 < µ < 1. Figure 4: Graph of f(x) = x 2 2x + 2, together with f(x) and f( x). Combining these transformations in turn is a way to create graphs from the graph of a simpler starting function. As an example, the figure shows how 8

9 Figure 5: Graph of f(x) = sin x, together with 2f(x) and f(2x). to obtain the graph of the function f(x) = (x 1) 2 1 over [ 1, 3] by starting from the function g(x) = x 2. Figure 6: Graphs of x 2, (x 1) 2, (x 1) 2 1, (x 1) 2 1, 2 (x 1) 2 1, 2+2 (x 1) 2 1. Here are some tips for graphing the reciprocal 1/f(x) from the graph of f(x). If f(x) is increasing (decreasing) then 1/f(x) is decreasing (increasing). The graphs of f(x) and 1/f(x) intersect iff f(x) = ± Even and odd functions Defn. A function f is even if f(x) = f( x) ± x Dom f. Defn. A function f is odd if f(x) = f( x) ± x Dom f. The graph of an even function is symmetric under a reflection about the y-axis. The graph of an odd function is symmetric under a rotation by 180 about the origin (equivalent to a reflection in the y-axis followed by a reflection in the x-axis). 9

10 Figure 7: Graph of f(x) = 1 + x 2, together with 1/f(x). Figure 8: Graphs of the even function x 2 and the odd function x 3. All functions f : R R can be written as the sum of an even and an odd function f(x) = f even (x) + f odd (x), where f even (x) = 1 2 f(x) f( x) and f odd(x) = 1 2 f(x) 1 2 f( x). This decomposition is often useful, as is the ability to spot an even or odd function as it can simplify some calculations. Eg. f(x) = (1 + x) sin x with f even (x) = x sin x and f odd (x) = sin x. 10

11 1.5 Piecewise functions Some functions are defined piecewise ie. different expressions are given for different intervals in the domain. Eg. The absolute value (or modulus) function { x if x 0 f(x) = x = x if x < 0. Figure 9: The graph of x over [ 1, 1]. Eg. f(x) = { x if 0 x < 1 x 1 if 1 x 2. In this example Dom f = [0, 2] but the function has a discontinuity at x = 1 (more about this later). Figure 10: The graph of a piecewise function. A step function is a piecewise function which is constant on each piece. An example is the Heaviside step function H(x) defined by { 0 if x < 0 H(x) = 1 if x > 0. Note that with this definition Dom H = R\{0}. It is sometimes convenient to extend the domain to R by defining the value of H(0) (some obvious candidates are 0, 1 2, 1). Figure 11: The Heaviside step function. 11

12 1.6 Operations with functions Given two functions f and g we can define the following: the sum is (f + g)(x) = f(x) + g(x), with domain Dom f Dom g. the difference is (f g)(x) = f(x) g(x), with domain Dom f Dom g. the product is (fg)(x) = f(x)g(x), with domain Dom f Dom g. the ratio is (f/g)(x) = f(x)/g(x), with domain (Dom f Dom g)\{x : g(x) = 0}. the composition is (f g)(x) = f(g(x)), with domain {x Dom g : g(x) Dom f}. Note that f g and g f are usually different functions. Eg. f(x) = sin x, g(x) = x 2 then (f g)(x) = sin(x 2 ) but (g f)(x) = sin 2 x. 1.7 Inverse functions Defn. A function f : D C is surjective (or onto) if Ran f = C, ie. if y C x D s.t. f(x) = y. Eg. f : R R given by f(x) = 2x + 1 is surjective. Eg. f : R R given by f(x) = x 2 is not surjective, since any negative number in the codomain is not the image of an element in the domain. Defn. A function f : D C is injective (or one-to-one) if x 1, x 2 D with x 1 x 2 then f(x 1 ) f(x 2 ). Eg. f(x) = 2x + 1 is injective since f(x 1 ) = f(x 2 ) implies x 1 = x 2. 12

13 Eg. f(x) = x 2 is not injective since f(x) = f( x) and x x if x 0. The following simple test can be applied to see if a function is injective from its graph. The horizontal line test. If no horizontal line intersects the graph of f more than once then f is injective, otherwise it is not. Eg. The function f(x) = x 3 3x is surjective as Ran f = R but it is not injective as the horizontal line y = 1 intersects the graph of f at 3 points (see Figure 12). Figure 12: Graph of f(x) = x 3 3x and the horizontal line y = 1. Defn. A function f : D C is bijective if it is both surjective and injective. Eg. From above we have seen that the function f(x) = 2x + 1 is bijective. Theorem of inverse functions. A bijective function f admits a unique inverse, denoted f 1, such that (f 1 f)(x) = x = (f f 1 )(x). It is clear from the definition that Dom f 1 = Ran f and Ran f 1 = Dom f. As the inverse undoes the effect of the function an equivalent definition is f(x) = y iff f 1 (y) = x. 13

14 Eg. We have seen that f(x) = 2x + 1 is bijective, so lets find its inverse. To simplify notation first write y = f 1 (x). Using the property x = f(f 1 (x)), we have x = f(y) = 2y + 1 and hence y = 1 2 (x 1) = f 1 (x). Note that given an injective function we can make a bijective function by taking the codomain equal to the range, and then an inverse exists. Eg. We have seen that f(x) = x 2 is not an injective function if Dom f = R, but it is injective if we take Dom f = [0, ). With this choice we can take the codomain equal to Ran f = [0, ) and f is now a bijective function. The inverse is f 1 (x) = x with Dom f 1 = Ran f = [0, ). Warning: Dont confuse the notation for the inverse with the same notation for the reciprocal. 1.8 Elementary functions Polynomials. inear functions are polynomials of degree one and take the general form f(x) = ax + b. Here Dom f = Ran f = R. The graph is a straight line with slope a and y-intercept (0, b), which is the point on the graph that intersects the y-axis. Quadratic functions are polynomials of degree two and take the general form f(x) = ax 2 + bx + c. Here Dom f = R, but the range is restricted. The graph is an upward parabola if a > 0 and a downward parabola if a < 0. By completing the square we can write f(x) = a(x + b 2a )2 + c b2 4a and see 2 that the vertex of the parabola is at ( b 2a, c b2 4a ). Hence, if a > 0 then the 2 range is given by Ran f = [c b2 4a, ) and if a < 0 then Ran f = (, c b2 2 4a ]. 2 Eg. f(x) = 2x 2 + 8x + 5 = 2(x 2) so Ran f = (, 13]. 14

15 Cubic functions are polynomials of degree three and take the general form f(x) = ax 3 + bx 2 + cx + d, with Dom f = Ran f = R. The graph of a typical example is shown in Figure 12 for the cubic function f(x) = x 3 3x. A cubic has at most two turning points (where the graph changes direction) and in the example these are at x = ±1. From the above examples it should be clear that for a polynomial function of arbitrary degree n, the domain is always R, and the range is also R if n is odd, but is restricted if n is even. Square root functions. These take the form f(x) = g(x) for some function g(x). By definition the square root is a non-negative number, so this can yield restrictions on the range. There can also be restrictions on the domain, since the square root is only defined if the argument g(x) is non-negative. Eg. f(x) = x has Dom f = Ran f = (, 0]. Rational functions. A rational function is a ratio of two polynomials f(x) = p(x)/q(x). The domain is given by Dom f = R\{x : q(x) = 0}. Eg. f(x) = 2x/(x 2 1) has Dom f = R\{±1}. The simplified form of a rational function is obtained by removing any common factors in the numerator and denominator. Eg. f(x) = 2x 4 x 2 4 = 2(x 2) (x + 2)(x 2) = 2 x + 2 if x 2. Note that we had to exclude the point x = 2 in the last equality to avoid a division by zero. Hence in this example the original rational function and its simplified form 2/(x + 2) are not quite the same function as they have different domains, being R\{±2} and R\{ 2} respectively. 15

16 If, in simplified form, q(a) = 0 then x = a is a vertical asymptote of the graph of f, which is a vertical line given by the equation x = a. Figure 13: Graph of f(x) = 2x/(x 2 1) and the vertical asymptotes x = ±1. There can also be horizontal asymptotes though a correct treatment requires the concept of a limit, which we have not yet studied. Roughly, a horizontal asymptote involves considering the behaviour of the function when x is large. For rational functions a horizontal asymptote exists only if the degree of the numerator is not larger than the degree of the denominator. In this case we can write f(x) = p 0 + p 1 x p n x n q 0 + q 1 x q n x n and the horizontal asymptote is the line y = p n /q n. In the example of Figure 13 we have n = 2 and p 2 = 0, q 2 = 1, so the horizontal asymptote is the line y = 0, which the graph approaches when x is large. A rational function is called a proper rational function if the degree of the numerator is less than the degree of the denominator. A rational function can always be written as the sum of a polynomial and a proper rational function by performing long division. 16

17 Figure 14: Graph of f(x) = 3x 2 /(2x 2 50), showing the vertical asymptotes x = ±5 and the horizontal asymptote y = 3/2. Eg. f(x) = 1+x6 x 3 x 2 is not a proper rational function. x 3 + x 2 + x + 1 x 3 x 2 x x 6 x 5 x x 5 x 4 x x 4 x 3 x x 3 x 2 x This yields the required decomposition 1 + x 6 x 3 x = 2 x3 + x 2 + x x2 + 1 x 3 x 2. 17

18 Exponential and logarithm functions. An exponential function is any function of the form f(x) = b x where b (called the base) is a positive real number not equal to 1. For all values of the base Dom f = R and Ran f = (0, ) and the graph is continuous and passes through the point (0, 1). However, there are two types of behaviour: If b > 1 the graph is increasing and has a horizontal asymptote y = 0 along the negative x-axis. If 0 < b < 1 the graph is decreasing and has a horizontal asymptote y = 0 along the positive x-axis. Figure 15: Graph of f(x) = 2 x and g(x) = ( 1 2 )x = 1 2 x = 1/f(x). Two important properties of exponential functions are b x b y = b x+y and (b x ) y = b xy. There is a special value of the base (called the natural base) b = e = , where e is called Euler s number and is transcendental (like π) ie. it is not the root of a polynomial equation with rational coefficients. One way to calculate e is via the sum e = ! + 1 2! + 1 3! + 1 4! +... The function f(x) = e x is usually referred to as the exponential function. 18

19 An exponential function f(x) = b x is injective and so admits an inverse. To determine this write y = f 1 (x) so that f(y) = x ie. b y = x. This is the defining relation of a logarithm function g(x) = log b x = f 1 (x) which is simply the name given to the inverse of an exponential function f(x) = b x. Explicitly, y = log b x iff b y = x. The domain of a logarithm function is the range of an exponential function and hence is (0, ). The range of a logarithm function is the domain of an exponential function and is thus R. Notation: I shall use the notation log x as the shorthand for the logarithm log e x, which is known as the natural logarithm. Warning: This notation is not universal as some people use ln x to denote the natural logarithm and instead use log x to denote log 10 x. A logarithm grows very slowly for x large and positive (more slowly than a linear function) and has a vertical asymptote at x = 0 (see Figure 16). Figure 16: The natural logarithm log x and the vertical asymptote at x = 0. Some important properties of logarithms are: log b 1 = 0 from b 0 = 1. log b b x = x from the definition as the inverse of an exponential. log b (αβ) = log b α + log b β from b α b β = b α+β. 19

20 log b (α/β) = log b α log b β from b α /b β = b α β. log b (x r ) = r log b x from (a x ) r = a xr. Trigonometric functions. Defn. A function f(x) is periodic if p > 0 s.t. f(x + p) = f(x) x. The smallest such value of p is called the period of f. The simplest periodic functions are the trigonometric functions sin x and cos x with period 2π. In both cases the domain is R and the range is [ 1, 1]. The function tan x = sin x/ cos x has period π and the domain is R\{x = ( n)π for integer n} with associated vertical asymptotes x = ( n)π. The range is R. Figure 17: The trigonometric functions sin x, cos x and tan x. It is expected that you know the values of the trigonometric functions at the particular values of the argument shown in the table, plus those related by the symmetries apparent in the plots shown in Figure 17. radians 0 π/6 π/4 π/3 π/2 degrees sin cos It is also expected that you know the important trigonometric identities: 20

21 sin 2 x + cos 2 x = 1. addition formulae sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y sin x sin y double angle formulae sin 2x = 2 sin x cos x cos 2x = 2 cos 2 x 1 = 1 2 sin 2 x half-angle formulae sin 2 x = cos 2x cos 2 x = cos 2x The reciprocals of the trigonometric functions have their own notation: csc x = 1/ sin x, sec x = 1/ cos x, cot x = 1/ tan x = cos x/ sin x. The above formulae lead to related formulae for these functions eg. sec 2 x = 1 + tan 2 x. The trigonometric functions are not injective over their full domains, but are injective over suitably restricted domains (given in the table), so that an inverse exists. f(x) Dom f Ran f f 1 (x) Dom f 1 Ran f 1 sin x [ π/2, π/2] [ 1, 1] sin 1 x or arcsin x [ 1, 1] [ π/2, π/2] cos x [0, π] [ 1, 1] cos 1 x or arccos x [ 1, 1] [0, π] tan x ( π/2, π/2) R tan 1 x or arctan x R ( π/2, π/2) 21

22 Figure 18: Graphs of the functions sin 1 x, cos 1 x and tan 1 x. Hyperbolic functions. Hyperbolic functions, with domain R, are defined in terms of the exponential function via sinh x = 1 2 (ex e x ), cosh x = 1 2 (ex + e x ), tanh x = sinh x/ cosh x. Figure 19: Graphs of the functions sinh x, cosh x and tanh x. From the properties of the exponential function it is easy to show (exercise) that some basic properties include sinh 0 = 0 and cosh 0 = 1. If x > 0 then 0 < sinh x < cosh x. sinh x is an odd function and cosh x is an even function. 22

23 Addition formulae sinh(x ± y) = sinh x cosh y ± cosh x sinh y cosh(x ± y) = cosh x cosh y ± sinh x sinh y. Double argument formulae sinh(2x) = 2 sinh x cosh x cosh(2x) = sinh 2 x + cosh 2 x. cosh 2 x sinh 2 x = 1. The reciprocals of hyperbolic functions have their own notation: cosech x = 1/sinh x, sech x = 1/cosh x, coth x = 1/tanh x. The above formulae lead to related formulae for these functions eg. sech 2 x = 1 tanh 2 x. 23

24 2 imits and continuity 2.1 Definitions Rough idea #1: A function f(x) has a limit at x = a if f(x) is close to whenever x is close to a. Rough idea #2: ets think about the function f(x) near x = a. If you approximate f(x) by a constant then you will make an error given by f(x). Suppose ε > 0 is the εrror at which you become unhappy with your approximation, that is, you are happy as long as f(x) < ε. The important issue for a limit is whether I can keep you happy simply by making sure that x stays within a δistance δ > 0 of a, that is by restricting to 0 < x a < δ. Note that we dont even care about what happens exactly at x = a. If I can always find a δistance δ that keeps you happy, no matter how small you choose your εrror ε, then we say that f(x) has a limit as x tends to a. Figure 20: The idea of a limit. 24

25 Defn: f(x) has a limit as x tends to a if We then write ε > 0 δ > 0 s.t f(x) < ε when 0 < x a < δ. lim f(x) = or equivalently f(x) as x a. x a If there is no such then we say that no limit exists. The limit does not require that f(a) is equal to or even that f(a) exists. This is because of the inequality 0 < x a in the definition. There is a lot more of this ε and δ stuff in Analysis I, where proofs concerning limits are considered. In this course we shall be less formal and deal only with methods to calculate limits or see that no limit exists. Rough idea #3: If f(a) exists then it would seem that this is a good candidate for the limit. This naive view turns out to be correct only if the graph of f(x) is a single unbroken curve with no holes or jumps, at least in a small region around x = a. Defn. A function f(x) is continuous at the point x = a if the following three properties all hold: f(a) exists. lim x a f(x) exists. lim x a f(x) is equal to f(a). Defn. A function f(x) is continuous on a subset S of its domain if it is continuous at every point in S. Defn. A function f(x) is continuous if it is continuous at every point in its domain. Eg. f(x) = x 2 is continuous and lim x 2 x 2 = f(2) = 4. 25

26 Eg. The following function has a discontinuity at x = 0 x 2 if 1 x < 0 f(x) = 1 if x = 0 x 2 if 0 < x 1 To indicate on a graph exactly which points are included we use a closed circle to denote an included point, so if it is at the end of a curve segment this denotes that this end is a closed interval. We use an open circle to denote an excluded point associated with an open interval. This notation is demonstrated in Figure 21 where the graph of the above function is presented. Figure 21: A graph illustrating the use of open and closed circles. The limit exists at x = 0. This function is not continuous at x = 0 but the limit exists at this point and lim x 0 f(x) = 0 1 = f(0). Eg. The following function also has a discontinuity at x = 0 (see Figure 23 for its graph) { 1 if x 0 f(x) = x 2 if x > 0 In this case no limit exists at x = 0. This is clear because if we consider x < 0 then to get arbitrarily close to the function for x arbitrarily close to 0 would require the limit to be 1. However, if we consider x > 0 then to get arbitrarily close to the function for x arbitrarily close to 0 would require the limit to be 0. These two requirements are incompatible and so no limit exists at x = 0. 26

27 Figure 22: A graph illustrating the use of open and closed circles. No limit exists at x = 0. Eg. For the function f(x) = sin(1/x) no limit exists at x = 0 because as x approaches zero this function oscillates between 1 and -1 over smaller and smaller intervals. In particular, in any given interval 0 < x < δ it is always possible to find values x 1 and x 2 s.t. f(x 1 ) = 1 and f(x 2 ) = 1, thus f(x) cannot remain close to any given constant. Figure 23: A graph of the function sin(1/x). 2.2 Results about limits and continuity There are some simple facts about limits that follow from the definition: The limit is unique. If f(x) = g(x) (except possibly at x = a) on some open interval containing a then lim x a f(x) = lim x a g(x). If f(x) K on either an interval (a, b) or an interval (c, a) 27

28 and if lim x a f(x) = then K. (A similar result holds by replacing all with ). If lim x a f(x) = and lim x a g(x) = M then (i) lim x a (f(x) + g(x)) = + M (ii) lim x a (f(x)g(x)) = M (iii) if M 0 then lim x a (f(x)/g(x)) = /M. Note: In Durham the results (i),(ii),(iii) are sometimes known as the Calculus Of imits Theorem (COT). However, this seems to be a rather grand name for these results and you will not find this name in any books. Pinching Theorem. If g(x) f(x) h(x) for all x a in some open interval containing a and lim x a g(x) = lim x a h(x) = then lim x a f(x) =. There are also some simple facts about continuous functions that follow from the definition and the above results: If f(x) and g(x) are continuous functions then so are f(x)+g(x), f(x)g(x) and f(x)/g(x). All polynomial, rational, trigonometric and hyperbolic functions are continuous. If f(x) is continuous then so is f(x). Eg. All the following functions are continuous: 2x 3 + x + 7, 3x/(x 1), (1 + x 2 )/ sin x, tan x. Note: Although tan x is continuous, it is not continuous on the interval [0, π], because this interval includes the point x = π/2 which is not in the domain of tan x. Eg. lim x π/2 x 2 sin x = (π/2) 2 sin(π/2) = π 2 /4. Eg. lim x 0 1 x does not exist. The value of f(x) can be made greater than any 28

29 finite constant by taking x sufficiently close to zero, so no limit exists. Eg. Calculate lim x 3 2x 2 18 x 3. 2x 2 18 x 3 = 2(x + 3)(x 3) x 3 = 2(x + 3) if x 3. The value of the function at x = 3 is irrelevant in defining the limit as x 3 so 2x 2 18 lim x 3 x 3 = lim x 3 2(x + 3) = 12. Eg. Calculate lim x 5 x 25 x 25. x 5 x 25 = ( x 5)( x + 5) (x 25)( x + 5) = x 25 (x 25)( x + 5) = 1 if x 25. x + 5 x 5 lim x 25 x 25 = lim 1 = 1 x 25 x Eg. Calculate lim x 0 x 2 sin( 1 x ). As 1 sin( 1 x ) 1 then x2 x 2 sin( 1 x ) x2. Also lim x 0 x 2 = 0 = lim x 0 x 2. Hence by the pinching theorem lim x 0 x 2 sin( 1 x ) = 0. Figure 24: Graphs of f(x) = x 2 sin(1/x) and the bounding functions g(x) = x 2 and h(x) = x 2. Two important trigonometric limits that you may assume to be true and use are: sin x 1 cos x lim = 1 and lim = 0. x 0 x x 0 x 29

30 The graphs in Figure 25 provide some evidence to support these results. They can be proved by applying the pinching theorem but it requires a bit of work. For example, the first result requires the fact that sin x < x < tan x for 0 < x < π/2. Figure 25: Graphs of sin x and 1 cos x. As x = 0 is not in the domain of either function there x x should really be open circles at x = 0 on both graphs. Eg. Calculate lim x 0 sin(2x) x. sin(2x) lim x 0 x = lim x 0 2 sin(2x) 2x = lim u 0 2 sin(u) u = 2 1 = 2. The above used the change of variable u = 2x and then the first of the two important trigonometric limits given above. Eg. Calculate lim x 0 tan x x. tan x lim x 0 x sin x = lim x 0 x cos x = Eg. Calculate lim x 0 1 cos x x 2. For π < x < π 1 cos x x 2 = ( lim x 0 (1 cos x)(1 + cos x) x 2 (1 + cos x) )( sin x lim x x 0 ) 1 = 1 1 = 1. cos x ( ) 2 sin x 1 = x 1 + cos x Hence by using the first important trigonometric limit we have that {( ) 2 } 1 cos x sin x 1 lim = lim = (1) 2 1 x 0 x 2 x 0 x 1 + cos x =

31 2.3 imits as x. So far we have only been concerned with the limit of a function f(x) as x approaches a finite point a. However, it is also possible to define a limit as x. Rough idea #4: A function f(x) has a limit as x if f(x) can be kept arbitrarily close to by making x sufficiently large. Defn: f(x) has a limit as x tends to if We then write ε > 0 S > 0 s.t f(x) < ε when x > S. lim f(x) = or equivalently f(x) as x. x If there is no such then we say that no limit exists. The corresponding definition associated with lim x f(x) = should be obvious and is given by Defn: f(x) has a limit as x tends to if ε > 0 S < 0 s.t f(x) < ε when x < S. 1 Eg. lim x x = 0. This should be clear because 1 x can be made as close to zero as required by making x sufficiently large. However, a better way to see this result is to first make the substitution u = 1/x as the limit x then becomes the limit u 0 that we are already familiar with ie. 1 lim x x = lim u 0 u = 0. Eg. Calculate lim x x cos(1/x)+2 x. By using the substitution u = 1/x we have that x cos(1/x) + 2 lim x x = lim u 0 1 u cos u u 31 = lim u 0 (cos u + 2u) = cos(0) + 0 = 1.

32 Our earlier discussion of a horizontal asymptote can now be made more precise, in that the graph of a function f(x) will have a horizontal asymptote y = if lim x f(x) =. 2x + 3 Eg. lim x x + 5 = lim x x x 1 Here we have used the result that lim x = = 2. x = 0, together with the earlier facts about limits. We see that calculating this limit shows that the graph of this rational function has the horizontal asymptote y = 2, reproducing our earlier observations about the horizontal asymptotes of rational functions. 2.4 Classification of discontinuities If a function f(x) is not continuous at a point x = a then we say it has a discontinuity at x = a. There are different types of discontinuity and these are best classified by considering how the function behaves on each side of the point x = a. This motivates the definition of the following one-sided limits. Defn: f(x) has a right-sided limit + = lim x a + f(x) as x tends to a from above if ε > 0 δ > 0 s.t f(x) + < ε when 0 < x a < δ. Note that the difference between this definition and the definition of the limit is the removal of the modulus sign in x a. This means that we only need to worry about points to the right of a when requiring the function to be close to +. Similarly, we have the definition Defn: f(x) has a left-sided limit = lim x a f(x) as x tends to a from below if ε > 0 δ > 0 s.t f(x) < ε when 0 < a x < δ. Here we only need to worry about points to the left of a when requiring the function to be close to. 32

33 It should be obvious that = lim x a f(x) exists iff + and both exist and are equal. In which case = + =. There are 3 types of discontinuity: (i) Removable discontinuity. In this case exists but f(a). The discontinuity can be removed to make the continuous function { f(x) if x a g(x) = if x = a. Eg. The following function (Figure 26i) has a removable discontinuity at x = 0, { x 2 if x 0 f(x) = 1 if x = 0. Removing this discontinuity yields the continuous function g(x) = x 2. Figure 26: 3 types of discontinuities at x = 0 : (i) removable; (ii) jump; (iii) infinite. (ii) Jump discontinuity. In this case both + and exist but +. Eg. The following function (Figure 26ii) has a jump discontinuity at x = 0, { 1 if x 0 f(x) = x 2 if x > 0. 33

34 In this example + = 0 1 =. Eg. The signum function 1 if x < 0 sgn(x) = 0 if x = 0 1 if x > 0 has a jump discontinuity at x = 0. In this example + = 1 1 =. (iii) Infinite discontinuity. In this case at least one of + or does not exist. Eg. The function f(x) = 1/x (Figure 26iii) has an infinite discontinuity at x = 0. In this example neither + nor exist. 2.5 The intermediate value theorem The intermediate value theorem states that if f(x) is continuous on [a, b] and u is any number between f(a) and f(b) (ie. either f(a) < u < f(b) or f(b) < u < f(a)) then (at least one) c (a, b) s.t. f(c) = u. Figure 27: An illustration of the intermediate value theorem. Eg. f(x) = sin x is continuous on [0, π/2] and f(0) = 0 < 1 2 < 1 = f(π/2) so by the intermediate value theorem there is (at least) one point x in (0, π/2) s.t. sin x =

35 It is important that f(x) is continuous throughout the interval, otherwise the theorem does not apply. Eg. f(x) = sgn(x) 1+x has f( 1) = < 1 5 < 1 2 = f(1) but there is no x in ( 1, 1) s.t. f(x) = 1 5. The intermediate value theorem does not apply because f(x) is not continuous at x = 0. There is a jump discontinuity at x = 0 with lim x 0 + f(x) = 1 1 = lim x 0 f(x). An application of the intermediate value theorem is locating the zeros of a function. If the function f(x) is continuous on [a, b] and we know that either f(a) < 0 < f(b) or f(b) < 0 < f(a) then by the intermediate value theorem the equation f(x) = 0 has at least one root between a and b. Eg. The function f(x) = x 2 2 is continuous on [1, 2] with f(1) = 1 < 0 and f(2) = 2 > 0, so there is at least one root of the equation x 2 2 = 0 in (1, 2) (of course the root is x = 2 = ). A repeated iterated application of this approach gives the bisection method, which can be used to locate the roots of a wide variety of equations to any desired accuracy. 35

36 3 Differentiation 3.1 Derivative as a limit Geometrically the derivative f (a) of a function f(x) at x = a is equal to the slope of the tangent to the graph of f(x) at x = a. Figure 28: The graph of a function f(x), the tangent at x = a and the secant through the points on the graph at x = a and x = a + h. A secant is a line that intersects a curve at two points (the part of the secant that is between the two intersection points is called a chord). Consider a secant that intersects the graph of f(x) at the two (x, y) points (a, f(a)) and (a+h, f(a+h)). The slope of this secant is given by the difference quotient (f(a + h) f(a))/h. As the two points approach each other ie. as h decreases, the secant approaches the tangent at x = a. The tangent is obtained in the limit as h 0 and the derivative at a is the slope of the secant in this limit ie. f (a) = lim h 0 f(a + h) f(a) h providing this limit exists. If this limit exists then we say that f(x) is differentiable at x = a. Eg. Use the limit definition of the derivative to calculate f (a) for f(x) = x 2. f (a) = lim h 0 f(a + h) f(a) h = lim h 0 (a + h) 2 a 2 h = lim h 0 2ah + h 2 h = lim h 0 (2a+h) = 2a. 36

37 This is, of course, the expected result from the knowledge that f (x) = 2x. Eg. Use the limit definition of the derivative to calculate f (π) for f(x) = sin x. f (π) = lim h 0 f(π + h) f(π) h = lim h 0 sin(π + h) sin π h = lim h 0 sin π cos h + cos π sin h h sin h = lim = 1 h 0 h where the final equality uses the earlier important trigonometric limit result. This is, of course, the expected answer from the knowledge that f (x) = cos x giving f (π) = cos π = 1. If f (a) exists for all a in Dom f we say that f(x) is differentiable and then f (a) defines a function f (x) called the derivative. Eg. Use the limit definition of the derivative to calculate the derivative of the function f(x) = x cos x. ( = lim h 0 f (x) = lim h 0 f(x + h) f(x) h x cos x cos h 1 h = lim h 0 (x + h) cos(x + h) x cos x h (x + h)(cos x cos h sin x sin h) x cos x = lim h 0 h x sin x sin h ) +cos x cos h sin x sin h = x sin x+cos x h where we have used both of the earlier important trigonometric limit results. The fact that the derivative is equal to the slope of the tangent of the graph allows us to determine a Cartesian equation for the tangent by calculating the derivative. Explicitly, the tangent line at the point (a, f(a)) is given by y = f(a) + f (a)(x a). Eg. For the function f(x) = 4x 3 x, find a Cartesian equation for the tangent to the graph at the point (1, 3). f (x) = 12x 2 1 so f (1) = 11. The tangent line is given by 37

38 y = f(1) + f (1)(x 1) = (x 1) = 11x 8 A necessary condition for a function to be differentiable at a point is that it is continuous at that point, but this is not a sufficient condition. Geometrically, there are two ways that a function can fail to be differentiable at a point where it is continuous: (i). The tangent line is vertical at that point. Eg. The function f(x) = x 1/3 is continuous in R but it is not differentiable at x = 0. The difference quotient at x = 0 is f(0 + h) f(0) h = h1/3 h = 1 h 2/3 and no limit exists as h 0 because the above grows without bound. (ii). There is no tangent line at that point. Eg. The function f(x) = x is continuous in R but it is not differentiable at x = 0. The difference quotient at x = 0 is { 1 if h < 0 f(0 + h) f(0) h = h h = 1 if h > 0 The left and right limits therefore both exist lim f(0+h) f(0) h 0 + h = 1 and lim f(0+h) f(0) f(0+h) f(0) h 0 h = 1 but these are not equal, so lim h 0 h does not exist. 3.2 Rules of differentiation The following rules of differentiation are easy to prove by considering the appropriate difference quotients. If f(x) and g(x) are differentiable at x then so are the following: 38

39 sum rule: f(x) + g(x) with derivative f (x) + g (x). product rule: f(x)g(x) with derivative f (x)g(x) + f(x)g (x). and providing g(x) 0 then so are the following: reciprocal rule: 1 g(x) with derivative g (x) (g(x)) 2. quotient rule: f(x) g(x) with derivative f (x)g(x) f(x)g (x) (g(x)) 2. Eg. Differentiate u(x) = (3x 2 1)/(x 4 + 3x + 1) u (x) = (3x2 1) (x 4 + 3x + 1) (3x 2 1)(x 4 + 3x + 1) (x 4 + 3x + 1) 2 = 6x(x4 + 3x + 1) (3x 2 1)(4x 3 + 3) = 6x5 + 18x 2 + 6x (12x 5 + 9x 2 4x 3 3) (x 4 + 3x + 1) 2 (x 4 + 3x + 1) Derivatives of higher order = 6x5 + 4x 3 + 9x 2 + 6x + 3 (x 4 + 3x + 1) 2. If f(x) is differentiable then its derivative f (x) may also be differentiable, in which case we denote its derivative by f (x) ie. the second derivative of f(x). Similarly the third derivative is f (x) and in general the n th derivative is f (n) (x) so that f (3) (x) = f. This notation is due to agrange ( ), but other notations are also common. f (x) = df dx, f (x) = d2 f dx 2, f (n) (x) = dn f dx n is due to eibniz ( ). f (x) = Df(x), f (x) = D 2 f(x), f (n) (x) = D n f(x) is due to Euler ( ). 39

40 Finally, if the independent variable represents time, say f(t), then f (t) = f and f (t) = f is due to Newton ( ). As we have seen, the derivative df dx is the slope of the tangent to the graph of f(x). This means that the derivative df dx measures the rate of change of the dependent variable f with respect to changes in the independent variable x. In the case that the independent variable is time t, the derivative measures the rate of change with time, so if the dependent variable, say X(t) represents the position of an object confined to a line then Ẋ is the velocity of the object, with Ẋ the speed, and Ẍ is the acceleration of the object. The product rule for differentiation is just the first case of the more general eibniz rule: If f(x) and g(x) are both differentiable n times then so is the product f(x)g(x) with D n (fg) = n k=0 ( ) n (D k f)(d n k g) k where ( ) n k = n! k!(n k)! is the binomial coefficient familiar from Pascal s triangle Figure 29: The binomial coefficients arranged into Pascal s triangle. Eg. Calculate D 3 (x 2 sin x). D 3 (fg) = f(d 3 g) + 3(Df)(D 2 g) + 3(D 2 f)(dg) + (D 3 f)g. f = x 2, Df = 2x, D 2 f = 2, D 3 f = 0. g = sin x, Dg = cos x, D 2 g = sin x, D 3 g = cos x. 40

41 D 3 (x 2 sin x) = x 2 cos x (3)2x sin x + (3)2 cos x + 0 sin x = (6 x 2 ) cos x 6x sin x. 3.4 The chain rule To handle the differentiation of composite functions (f g)(x) = f(g(x)) we turn to the following theorem: The chain rule theorem states that if g(x) is differentiable at x and f(x) is differentiable at g(x) then the composition (f g)(x) is differentiable at x with (f g) (x) = f (g(x))g (x). Recall that prime denotes differentiation with respect to the argument so in eibniz notation the above formula may be written more crudely as d df f(g(x)) = dx dg where we need to be aware that on the right hand side the argument of the first factor is g(x) and the argument of the second factor is x. This form is useful as a mnemonic (think of cancelling the dg factor) but it is no more than this. Eg. Calculate d (( x + 1 ) 3 ). dx x In terms of the above notation g(x) = x + 1 x and f(x) = x 3 giving g (x) = 1 1 x and f (x) = 3x 4 thus 2 (( d x + 1 ) 3 ) ( = f (g(x))g (x) = 3 x + 1 ) 4 (1 1x ). dx x x 2 dg dx Eg. d dt cos(t2 ) = 2t sin(t 2 ). 41

42 3.5 Hopital s rule Recall that the calculus of limits theorem states that if lim x a f(x) = and f(x) lim x a g(x) = M 0 then lim x a g(x) = M. Clearly this result is not applicable to the situation where = M = 0, which is called an indeterminant form. We have already seen how to deal with situations like this directly, but an alternative (and often easier) approach is sometimes available by making use of the following. Hopital s rule et f(x) and g(x) be differentiable on I = (a h, a) (a, a + h) for some h > 0, with lim x a f(x) = lim x a g(x) = 0. If lim x a f (x) g (x) exists and g (x) 0 x I then f(x) lim x a g(x) = lim f (x) x a g (x). Proof: We shall only consider the proof for the slightly easier situation in which f(x) and g(x) are both differentiable at x = a and g (a) 0. In this case f(x) lim x a g(x) = lim f(x) f(a) x a g(x) g(a) = lim x a f(x) f(a) x a g(x) g(a) x a = lim x a f(x) f(a) x a g(x) g(a) lim x a x a = f (a) g (a) = lim f (x) x a g (x). We can apply Hopital s rule to calculate the two important trigonometric limits from earlier. Eg. Calculate lim x 0 sin x x. f(x) = sin x and g(x) = x are both differentiable. Also lim x 0 sin x = sin 0 = 0 and lim x 0 x = 0. Furthermore, f (x) = cos x and g (x) = 1 0. Thus Hopital s rule applies to give sin x lim x 0 x = lim cos x x 0 1 = cos 0 = 1. 42

43 Eg. Calculate lim x 0 1 cos x x. f(x) = 1 cos x and g(x) = x are both differentiable. Also lim x 0 (1 cos x) = 1 cos 0 = 0 and lim x 0 x = 0. Furthermore, f (x) = sin x and g (x) = 1 0. Thus by Hopital s rule 1 cos x lim x 0 x = lim x 0 sin x 1 = sin 0 = 0. If applying Hopital s rule yields another indeterminant form then Hopital s rule can be reapplied to this form and so on. Eg. Calculate lim x 0 2 sin x sin(2x) x sin x. 2 sin x sin(2x) lim x 0 x sin x = lim x 0 2 cos x 2 cos(2x) 1 cos x = lim x 0 2 cos x + 8 cos(2x) cos x 3.6 Implicit differentiation = lim x 0 2 sin x + 4 sin(2x) sin x = = 6. So far we have been differentiating functions defined explicitly in terms of an independent variable. Even if an explicit expression is not available then we may still be able to obtain an expression for the derivative if we know that a differentiable function satisfies a particular equation. This is the process of implicit differentiation, which involves differentiating an equation and using the chain rule. Eg. Consider a function y(x) that satisfies the equation x 2 + y 2 = 1. Differentiating this equation with respect to x gives 2x + 2y dy dx = 0 and hence dy dx = x/y. In particular, the function y = 1 x 2 satisfies the given equation and the above result yields dy dx = x/ 1 x 2, which we could have obtained directly by differentiating the explicit expression. 43

44 Eg. Assume that x(t) is a differentiable function that satisfies the equation cos(t x) = (2t + 1) 3 x. Express dx dt in terms of x and t. Differentiating both sides of the equation with respect to t gives sin(t x)(1 dx dt ) = 3(2t + 1)2 2x + (2t + 1) 3 dx dt and hence dx dt = 6x(2t + 1)2 + sin(t x) sin(t x) (2t + 1) Boundedness and monotonicity The following definitions apply to a function f(x) defined in some interval I. Defn: If a constant k 1 s.t. f(x) k 1 x in I we say that f(x) is bounded above in I and we call k 1 an upper bound of f(x) in I. Furthermore, if a point x 1 in I s.t. f(x 1 ) = k 1 we say that the upper bound k 1 is attained and we call k 1 the global maximum value of f(x) in I. Similarly, we have the following. Defn: If a constant k 2 s.t. f(x) k 2 x in I we say that f(x) is bounded below in I and we call k 2 a lower bound of f(x) in I. Furthermore, if a point x 2 in I s.t. f(x 2 ) = k 2 we say that the lower bound k 2 is attained and we call k 2 the global minimum value of f(x) in I. Defn: f(x) is bounded in I if it is both bounded above and bounded below in I ie. if a constant k s.t. f(x) k x in I. If no interval I is specified then it is taken to be the domain of the function. Eg. cos x is bounded (in R) because cos x 1 x R. Both these upper and lower bounds are attained as cos 0 = 1 and cos π = 1. The global maximum value in R is 1 and the global minimum value in R is 1. Eg. Consider f(x) = sgn(x) 1+x 2 for 1 x 1. Then f(x) is bounded in [ 1, 1] because f(x) 1 for all x in [ 1, 1] but neither of the bounds is attained, so it has no global maximum value and no global minimum value in [ 1, 1]. 44

45 Eg. On [0, π/2) the function tan x is bounded below but not bounded above. A condition that guarantees the existence of both a global maximum and minimum value (called extreme values) is provided by the following: The extreme value theorem states that if f is a continuous function on a closed interval [a, b] then it is bounded on that interval and has upper and lower bounds that are attained ie. points x 1 and x 2 in [a, b] s.t. f(x 2 ) f(x) f(x 1 ) x [a, b]. Defn: f(x) is monotonic increasing in [a, b] if f(x 1 ) f(x 2 ) for all pairs x 1, x 2 with a x 1 < x 2 b. Defn: f(x) is strictly monotonic increasing in [a, b] if f(x 1 ) < f(x 2 ) for all pairs x 1, x 2 with a x 1 < x 2 b. Obvious definitions apply with increasing replaced by decreasing. Eg. sgn(x) is monotonic increasing in [ 1, 1] but not strictly. Eg. x 2 is strictly monotonic increasing in [0, b] for any b > Critical points Defn: We say that f(x) has a local maximum at x = a if f(a) f(x) x (a h, a + h). h > 0 s.t. Remark: If f(x) has a local maximum at x = a and is differentiable at this point then f (a) = 0. Proof: As f(x) is differentiable at a then f(x) f(a) lim x a x a = lim x a + f(x) f(a) x a = f (a). 45

46 f(x) has a local maximum at a implies f(x) f(a) x a 0 for x (a h, a) hence f f(x) f(a) (a) = lim 0. x a x a Similarly, f(x) has a local maximum at a also implies f(x) f(a) x a 0 for x (a, a + h) hence f f(x) f(a) (a) = lim 0. x a + x a Putting these two together gives f (a) 0 f (a) and thus f (a) = 0. Defn: We say that f(x) has a local minimum at x = a if f(x) f(a) x (a h, a + h). h > 0 s.t. Remark: If f(x) has a local minimum at x = a and is differentiable at this point then f (a) = 0. Defn: f(x) has a stationary point at x = a if it is differentiable at x = a with f (a) = 0. Defn: An interior point x = a of the domain of f(x) is called a critical point if either f (a) = 0 or f (a) does not exist. Every local maximum and local minimum of a differentiable function is a stationary point but there may be other stationary points too. Eg. f(x) = x 3 has f (0) = 0 so x = 0 is a stationary point but it is neither a local maximum nor a local minimum because f(x) > f(0) for x (0, h) but f(x) < f(0) for x ( h, 0). Defn: If f(x) is twice differentiable in an open interval around x = a with f (a) = 0 and if f (x) changes sign at x = a then we say that x = a is a 46

47 point of inflection. Figure 30: The graph of f(x) = x 3 indicating the point of inflection at x = 0. Eg. f(x) = x 3 has f (x) = 6x so f (0) = 0. As f (x) < 0 for x < 0 and f (x) > 0 for x > 0 then f (x) changes sign at x = 0 so it is a point of inflection. The first derivative test Suppose f(x) is continuous at a critical point x = a. (i). If h > 0 s.t. f (x) < 0 x (a h, a) and f (x) > 0 x (a, a + h) then x = a is a local minimum. (ii). If h > 0 s.t. f (x) > 0 x (a h, a) and f (x) < 0 x (a, a + h) then x = a is a local maximum. (iii). If h > 0 s.t. f (x) has a constant sign x a in (a h, a + h) then x = a is a not a local extreme value (minimum/maximum). Eg. f(x) = x is continuous but f (x) 0 so there are no stationary points. The derivative does not exist at x = 0 so this is a critical point. f (x) < 0 for x ( 1, 0) and f (x) > 0 for x (0, 1) so by the first derivative test x = 0 is a local minimum. In fact f(0) = 0 is a global minimum as x 0. Eg. f(x) = x 4 2x 3 has f (x) = 4x 3 6x 2 = 2x 2 (2x 3). The only critical points are x = 0, 3/2. Now f (x) < 0 for x < 3/2 and f (x) > 0 for x > 3/2. 47