4.1 Analysis of functions I: Increase, decrease and concavity

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1 4.1 Analysis of functions I: Increase, decrease and concavity Definition Let f be defined on an interval and let x 1 and x 2 denote points in that interval. a) f is said to be increasing on the interval if f (x 1 ) < f (x 2 ) whenever x 1 < x 2 b) f is said to be decreasing on the interval if f (x 1 ) > f (x 2 ) whenever x 1 < x 2 PR (FIU) MAC / 85

2 Theorem Let f be a function that is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) a) If f (x) > 0 for every value of x in (a, b), then f is increasing on [a, b]. b) If f (x) < 0 for every value of x in (a, b), then f is decreasing on [a, b] c) If f (x) = 0 for every value of x in (a, b), then f is constant on [a, b] PR (FIU) MAC / 85

3 Example: Find where f (x) = x 2 5x + 6 is increasing, decreasing or constant. f (x) = 2x 5 5 x 0 2 f (x) f(x) PR (FIU) MAC / 85

4 Example: Find where f (x) = x 2 5x + 6 is increasing, decreasing or constant. f (x) = 2x 5 5 x 0 2 f (x) f(x) f (x) = x 2 5x + 6 is decreasing on (, 5 2 ] and increasing on, + ). It is nowhere constant! [ 5 2 PR (FIU) MAC / 85

5 Concavity Concave up: The slope of tangent lines is increasing. Equivalently, the derivative function f (x) is increasing, that is (f ) (x) = f (x) > 0 PR (FIU) MAC / 85

6 Concavity Concave up: The slope of tangent lines is increasing. Equivalently, the derivative function f (x) is increasing, that is (f ) (x) = f (x) > 0 Concave down: The slope of tangent lines is decreasing. Equivalently, the derivative function f (x) is decreasing, that is (f ) (x) = f (x) < 0 PR (FIU) MAC / 85

7 Theorem a) If f (x) > 0 on an open interval (a, b), then the graph of f is concave up on (a, b). PR (FIU) MAC / 85

8 Theorem a) If f (x) > 0 on an open interval (a, b), then the graph of f is concave up on (a, b). b) If f (x) < 0 on an open interval (a, b), then the graph of f is concave down on (a, b). PR (FIU) MAC / 85

9 Example: Find where f (x) = 3x 3 4x + 3 is concave up or concave down. PR (FIU) MAC / 85

10 Example: Find where f (x) = 3x 3 4x + 3 is concave up or concave down. f (x) = 9x 2 4 f (x) = 18x x 0 f (x) f(x) PR (FIU) MAC / 85

11 The graph of f is concave up on (0, + ). It is concave down on (, 0). PR (FIU) MAC / 85

12 The graph of f is concave up on (0, + ). It is concave down on (, 0). Definition If f is continuous on an open interval containing x 0 and if the graph of f changes the direction of its concavity at x 0, then the point (x 0, f (x 0 )) on the graph of f is called an inflection point of f. PR (FIU) MAC / 85

13 Example: Find the largest intervals on which f (x) = 3x 4 4x 3 is increasing, decreasing; find the largest open intervals on which f is concave up, concave down and find the x coordinates of all inflection points. PR (FIU) MAC / 85

14 f (x) = 12x 2 (x 1) f (x) = 12x(3x 2) PR (FIU) MAC / 85

15 f (x) = 12x 2 (x 1) f (x) = 12x(3x 2) 2 x x x f (x) x x f (x) f(x) PR (FIU) MAC / 85

16 The function f is increasing on [1, + ), decreasing on (, 1]. Its graph is concave up on (, 0] and [ 2 3, + ) concave down on (0, 2 3 ). The graph has an inflection point at x = 0 and x = 2 3. PR (FIU) MAC / 85

17 4.2 Analysis of functions II: Relative extrema, graphing polynomials A critical point for a function f is any value of x in the domain of f at which f (x) = 0 or at which f is not differentiable. Recall an earlier statement that relative extrema of a function f can occur only at critical points. If f has a relative extremum at x 0, then either f (x 0 ) = 0 or f is not differentiable at x 0 ; in other words, x 0 is a critical point. PR (FIU) MAC / 85

18 Example 1. f (x) = x 2 5x + 6. Since f (x) = 2x 5 is defined everywhere and 2x 5 = 0 if and only if x = 5 2, we conclude that f can have a relative extremum only at x = 5 2 and nowhere else! PR (FIU) MAC / 85

19 Example 1. f (x) = x 2 5x + 6. Since f (x) = 2x 5 is defined everywhere and 2x 5 = 0 if and only if x = 5 2, we conclude that f can have a relative extremum only at x = 5 2 and nowhere else! Example 2. f (x) = { x if x 0 x if x < 0 f (x) = { 1 if x>0 1 if x<0 f (x) is undefined at x = 0 and it is nowhere zero, so f can have a relative extremum only at x = 0. PR (FIU) MAC / 85

20 First derivative test Suppose f is continuous at a critical point x 0. a) If f (x) > 0 on an open interval extending left from x 0 and f (x) < 0 on an open interval extending right from x 0, then f has a relative maximum at x 0. b) If f (x) < 0 on an open interval extending left from x 0 and f (x) > 0 on an open interval extending right from x 0, then f has a relative minimum at x 0. PR (FIU) MAC / 85

21 Example f (x) = 4x 2 + 2x 5 f (x) = 8x + 2 f (x) = 0 x = 1 4 x f (x) f(x) f (x) has a relative minimum at x = 1 4. PR (FIU) MAC / 85

22 Example: f (x) = x 3 + 3x 2 9x + 1 f (x) = 3(x 1)(x + 3) f (x) = 0 x = 1 or x = 3. PR (FIU) MAC / 85

23 x f (x) f(x) f (x) has a relative maximum at x = 3, a relative minimum at x = 1. PR (FIU) MAC / 85

24 Example f (x) = x 3 f (x) = 3x 2 f (x) = 0 x = 0 x 0 f (x) f(x) f (x) has no relative extremum. PR (FIU) MAC / 85

25 Second derivative test At a relative maximum, the graph of the function f is concave down, so as seen before, the second derivative should be negative at this point. Similarly, at a relative minimum, the graph is concave up, therefore the second derivative should be positive at such a point. Theorem Suppose f is twice differentiable at a stationary point x 0 (that is f (x 0 ) = 0. Then: a) If f (x 0 ) > 0, then f has a relative minimum at x 0. b) If f (x 0 ) < 0, then f has a relative maximum at x 0. PR (FIU) MAC / 85

26 Example f (x) = 2x 3 3x 2 3 f (x) = 6x 2 6x = 6x(x 1) f (x) = 0 x = 0 or x = 1 x = 0 and x = 1 are stationary points. f (x) = 12x 6 Since f (0) = 6 < 0, f (x) has a relative maximum at x = 0; and since f (1) = 6 > 0, f (x) has a relative minimum at x = 1. PR (FIU) MAC / 85

27 Simple curve sketching. Information to be included on the graphs X and Y intercepts. Critical points, relative extrema. Increasing, decreasing behavior. Concavity, inflection points. Behavior at ± PR (FIU) MAC / 85

28 Example Graph f (x) = x(x 2 1) 2. X intercepts: Solve 0 = x(x 2 1) 2 x=0, x=±1 Y intercept: Evaluate f (0) = 0. So y = 0. lim f (x) = + x + lim f (x) = x f (x) = (x 2 1)(5x 2 1) The critical points are x = ±1, x = ± 1 5 PR (FIU) MAC / 85

29 f 3 (x) = 0 x = 0, or x = ± 5. f (x) = 4x(5x 2 3) PR (FIU) MAC / 85

30 f 3 (x) = 0 x = 0, or x = ± 5. f (x) = 4x(5x 2 3) x ( 1 5 ) 1 2 ( 3 5 ) f f f PR (FIU) MAC / 85

31 lim x f (x) =, no global minimum. lim x + f (x) = +, no global maximum. 1 Local minima are f ( 5 ), and f (1). 1 Local maxima are f ( 1) and f ( 5 ). 3 Inflection points are ( 5, f ( 3 5 )), (0, f (0)) and ( 3 5, f ( 3 5 )). PR (FIU) MAC / 85

32 PR (FIU) MAC / 85

33 4.3 Analysis of functions III: Rational functions, cusps and vertical tangents Vertical asymptotes: Example: y = 1 (x+2) 3. 1 lim x 2 (x + 2) 3 = 1 lim x 2 + (x + 2) 3 = + The graph of y = 1 x = 2. (x+2) 3 will show a (two-sided) vertical asymptote at PR (FIU) MAC / 85

34 x = a is a vertical asymptote for the graph of y = f (x) if lim f (x) = ± x a PR (FIU) MAC / 85

35 x = a is a vertical asymptote for the graph of y = f (x) if lim f (x) = ± x a Horizontal asymptotes: Example: y = 1 x 2. lim x + lim x 1 x 2 = 0 1 x 2 = 0 The graph will show a (two-sided) horizontal asymptote y = 0 PR (FIU) MAC / 85

36 If then lim f (x) = d, x ± y = d is a horizontal asymptote for the graph of y = f (x). PR (FIU) MAC / 85

37 If then lim f (x) = d, x ± y = d is a horizontal asymptote for the graph of y = f (x). Slant asymptotes: y = mx + b m = lim f (x) x ± b = lim [f (x) mx] x ± PR (FIU) MAC / 85

38 Example: Find slant asymptotes for the graph of f (x) = x 2 +x 1 x 1. PR (FIU) MAC / 85

39 Example: Find slant asymptotes for the graph of f (x) = x 2 +x 1 x 1. f (x) = x 2 2x (x 1) 2 m = lim f (x) = 1 x ± b = 2 lim [x + x 1 2x 1 x] = lim x ± x 1 x ± x 1 = 2 y = x + 2 is a (two-sided) slant asymptote. PR (FIU) MAC / 85

40 Complete graph Graph f (x) = x 3 x 2 5 PR (FIU) MAC / 85

41 Complete graph Graph f (x) = x 3 x 2 5 x-intercept: x = 0 y-intercept: y = 0 lim x f (x) = lim x + f (x) = + f (x) = x 2 (x 2 15) (x 2 5) 2. Critical points are x = 0, x = ± 15, f (x) undefined at x = ± 5. f (x) = 10x(x 2 +15) (x 2 5) 3 f (x) = 0 at x = 0, undefined at x = ± 5 PR (FIU) MAC / 85

42 Vertical asymptote at x = ± 5 No horizontal asymptote. lim x ± f (x) = 1 x lim x ± [ 3 x] = 0. So x 2 5 y is a two-sided slant asymptote. = x PR (FIU) MAC / 85

43 x f (x) + 0 -? - 0 -? f (x) - -? + 0 -? + + f(x)? 0??? Local minimum value f ( 15) Local maximum value f ( 15) Inflection point at x = 0 PR (FIU) MAC / 85

44 PR (FIU) MAC / 85

45 Vertical tangent lines and cusps. Examples: f (x) = 6x x 4 3. Observe that f (0) is defined and f (x) = 2x 2/3 + 4x 1/3 lim f (x) = + x 0 The graph will have a vertical tangent line at x = 0 PR (FIU) MAC / 85

46 PR (FIU) MAC / 85

47 Another example: f (x) = (x 4) 2/3. Observe that f (4) is defined and f (x) = 2 3 (x 4) 1 3 lim f (x) = + x 4 + lim f (x) = x 4 The graph will show a cusp at x = 4. PR (FIU) MAC / 85

48 PR (FIU) MAC / 85

49 Graph f (x) = x 1/3 (2 x) 2/3. PR (FIU) MAC / 85

50 Graph f (x) = x 1/3 (2 x) 2/3. x intercepts: x = 0, x = 2 y intercept: y = 0 2 3x Computations show f (x) = 1 3. Critical points are x 2/3 (2 x) 1/3 x = 2/3, f (x) undefined at x = 0 and x = 2. Vertical tangent at x = 0, cusp at x = 2. Computations show f (x) = 8 9. Undefined at x = 0 x 5/3 (2 x) 4/3 and x = 2 1 PR (FIU) MAC / 85

51 x 0 2/3 2 f (x) +? + 0 -? + f (x) +? - -? - f(x) 0 PR (FIU) MAC / 85

52 x 0 2/3 2 f (x) +? + 0 -? + f (x) +? - -? - f(x) 0 A local minimum value is f (2). A local maximum value is f (2/3) Inflection point at x = 0 PR (FIU) MAC / 85

53 PR (FIU) MAC / 85

54 4.4 Absolute maxima and minima Recall: If f (x) is continuous on [a, b], then there exists numbers c and d in [a, b] such that f (c) is the minimum value and f (d) is the maximum value of f (x) on [a, b]. Moreover each of c and d is either critical points of f (x) or one of the endpoints a or b. PR (FIU) MAC / 85

55 Program for looking for extremal values of a continuous function f on a closed interval [a, b]. 1. Find all critical points of f (x) in [a, b] 2. Evaluate f (x) at those points and at the endpoints a and b 3. Identify the maximum and minimum values by comparison. PR (FIU) MAC / 85

56 Examples: Find the absolute extrema for: f (x) = 4x 3x on the interval [ 1, 1] 2 +1 PR (FIU) MAC / 85

57 Extrema values on open or unbounded intervals Examples State whether the given function attains a maximum value or a minimum value (or both). 1) f (x) = 1 on (, + ) x ) f (x) = 1 x(1 x) on (0, 1). Evaluation at endpoints is replaced by right and left limits as appropriate. PR (FIU) MAC / 85

58 4.5 Applied maximum and minimum problems Example A rectangle has its two lower corners on the X-axis and its two upper corners on the curve y = 16 x 2. For all such rectangles, what are the dimensions of the one with largest area? PR (FIU) MAC / 85

59 Strategy in solving applied maximum and minimum problems 1. Drawing a picture and labeling quantities. 2. Find a formula for the quantity to be maximized or minimized. 3. Using information from the text, express the quantity to be optimized as a function of one variable. 4. Find the interval of possible values for this variable. 5. Use the previous procedure to find the absolute extrema on the closed interval from 4. PR (FIU) MAC / 85

60 More applied extrema problems: Open or unbounded intervals Example: A closed rectangular container with a square base is to have a volume of 2000cm 3. It costs twice as much per square centimeter for the top and bottom as it does for the sides. Find the dimensions of the container of least cost. The procedure is similar, except evaluation at endpoints is replaced by right and left limits. PR (FIU) MAC / 85

61 Example 2 A lamp is suspended above the center of a round table of radius r. How high above the table should the lamp be placed to achieve maximum illumination at the edge of the table? Assuming that the illumination I is directly proportional to the cosine of the angle of incidence φ of the light rays and inversely proportional to the square of the distance l from the light source. PR (FIU) MAC / 85

62 Observation: A unique local extremum on an open interval is necessarily an absolute extremum! PR (FIU) MAC / 85

63 4.8 Rolle s Theorem; Mean Value Theorem Recall: (Rolle s Theorem) Let f be differentiable on (a, b) and continuous on [a, b]. If f (a) = 0 = f (b), then there is at least one point c in (a, b) where f (c) = 0. Recall: (Mean Value Theorem) Let f (x) be differentiable on (a, b) and continuous on [a, b]. Then there is at least one point c in (a, b) where f (c) = f (b) f (a). b a PR (FIU) MAC / 85

64 Independent proof of Mean Value Theorem The secant line joining (a, f (a)) and (b, f (b)) has equation y f (a) = f (b) f (a) (x a) b a PR (FIU) MAC / 85

65 Independent proof of Mean Value Theorem The secant line joining (a, f (a)) and (b, f (b)) has equation The function y f (a) = v(x) = f (x) f (a) f (b) f (a) (x a) b a satisfies the hypotheses in Rolle s Theorem. f (b) f (a) (x a) b a Indeed, v(x) is differentiable on (a, b), continuous on [a, b] and v(a) = 0 = v(b). PR (FIU) MAC / 85

66 So, by Rolle s Theorem, there is at least one c in (a, b) such that But v (c) = 0 is equivalent to v (c) = 0. f (c) f (b) f (a) b a = 0 That is f (c) = f (b) f (a). b a PR (FIU) MAC / 85

67 Exercise: Does f (x) = x + 1 x satisfy the Mean Value Theorem s hypotheses on [3, 4]. If it does, find all values c in ]3, 4[ that satisfy the conclusion of the Theorem. PR (FIU) MAC / 85

68 5.2 The indefinite integral The area Problem PR (FIU) MAC / 85

69 A(x) will denote the shaded area under the graph of y = f (x) and between fixed a and variable x. Claim: A (x) = f (x). PR (FIU) MAC / 85

70 Proof: assuming h > 0 for now: A (x) = lim h 0 A(x + h) A(x) h hf (x) A(x + h) A(x) hf (x + h) PR (FIU) MAC / 85

71 f (x) lim f (x) h 0 A(x+h) A(x) h f (x + h) lim h 0 A(x+h) A(x) h f (x) A (x) f (x) lim h 0 f (x + h) The same inequalities will be obtained assuming h < 0. So A (x) = f (x) PR (FIU) MAC / 85

72 Example: Find the area A(x) of the region under the graph of y = x 2, between x = 1 and x to the right of 1. PR (FIU) MAC / 85

73 Example: Find the area A(x) of the region under the graph of y = x 2, between x = 1 and x to the right of 1. A (x) = x 2. So A(x) = x C where C could be any constant. But since A(1) = 0, it follows that and hence C = = C A(x) = x PR (FIU) MAC / 85

74 Remark: To find an area to the left of 1, a negative sign needs to be introduced. For example the area between 1 and 1 2 is A( 1 2 ) = ( ) = PR (FIU) MAC / 85

75 A function F(x) such that F (x) = f (x) is called an antiderivative of f (x). If F(x) is an antiderivative of f (x), then for any constant C, F(x) + C is another antiderivative of f (x). It is true also that any two antiderivatives of f (x) differ by a (locally) constant function. PR (FIU) MAC / 85

76 The process of finding antiderivatives is called integration. We shall use the notation f (x)dx to represent all antiderivatives of f (x). That is f (x)dx = F(x) + C where F(x) is one of the antiderivatives of f (x). f (x)dx is also called the indefinite integral of f (x). PR (FIU) MAC / 85

77 Exercises: Find the integrals: 1) x 6 dx 2) x 2/3 dx 3) 1 2x 3 dx = PR (FIU) MAC / 85

78 (u 3 2u + 7)du = PR (FIU) MAC / 85

79 (u 3 2u + 7)du = We have used the following: u 4 4 u2 + 7u + C PR (FIU) MAC / 85

80 Properties of the indefinite integral Cf (x)dx = C f (x)dx (f (x) + g(x))dx = f (x)dx + g(x)dx PR (FIU) MAC / 85

81 (1 + x 2 )(2 x)dx = (t 2 cos t)dt = sin 2x cos x dx = x 2 sin x+2 sin x 2+x 2 dx = cot 2 xdx = Find f (x) such that f (x) + sin x = 0 and f (0) = 2. PR (FIU) MAC / 85

82 Some integration formulas Since d dx ( x r+1 r+1 ) = x r, for r 1, we deduce that x r dx = x r+1 r+1 + C. d dt tan t = sec2 t sec 2 tdt = tan t + C d dx ( cos x) = sin x sin xdx = cos x + C d dx sin x = cos x cos xdx = sin x + C PR (FIU) MAC / 85

83 5.3 Integration by substitution Basic integration by substitution formula: If f (x) = g(u) du dx, then f (x)dx = g(u) du dx dx = g(u)du Example: Evaluate (3x 1) 5 dx Put u(x) = 3x 1, then du dx = 3 Letting g(u) = u5 3, we see that (3x 1) 5 = g(u) du dx PR (FIU) MAC / 85

84 Therefore (3x 1) 5 dx = g(u) du dx dx u 5 = g(u)du = 3 du = 1 u C = 1 18 (3x 1)6 + C PR (FIU) MAC / 85

85 Exercises 1) sec 2 (4x + 1)dx PR (FIU) MAC / 85

86 Exercises 1) sec 2 (4x + 1)dx 2) sin (3x)dx PR (FIU) MAC / 85

87 Exercises 1) sec 2 (4x + 1)dx 2) sin (3x)dx 3) x cos (3x 2 )dx PR (FIU) MAC / 85

88 Exercises 1) sec 2 (4x + 1)dx 2) sin (3x)dx 3) x cos (3x 2 )dx 4) tan 3 (5x) sec 2 (5x)dx PR (FIU) MAC / 85

89 More exercises 1) 3x 4x 2 +5 dx 2) (4x x + 9) 2/3 dx 3) cos 4θ 2 sin 4θdθ 4) sin (5/x) dx x 2 5) x 2 sec 2 (x 3 )dx 6) cos 2 2t sin 2tdt PR (FIU) MAC / 85

90 10.1 Parametric equations; Tangent lines for parametric curves A parametric curve in the x, y plane is a pair of functions: x = f (t), y = g(t), t [a, b] Example 1: x = cos t, y = sin t, 0 t 2π PR (FIU) MAC / 85

91 The graph is the unit circle as we can see by eliminating t and obtaining Example 2: x 2 + y 2 = 1 x = 1 t2 1 + t 2, y = 2t, < t < t2 PR (FIU) MAC / 85

92 Eliminating t, we get x 2 + y 2 = 1 the unit circle again! Exercises: Eliminate the parameter and then sketch the curve: a) x = t 2 + 3t, y = t 2 b) x = e t, y = 4e 2t c) x = e t, y = e t d) x = 5 cos t, y = 5 sin t PR (FIU) MAC / 85

93 First eliminate the parameter and sketch the curve. Then describe the motion of the point (x(t), y(t)) as t varies in the given interval: 1) x = sin 2 πt, y = cos 2 πt; 0 t 2π 2) x = cos t, y = sin 2 t; π t π PR (FIU) MAC / 85

94 Tangent lines to parametric curves Given parametric equations of a curve: x = f (t), y = g(t) We wish to find dy dx at a given t = t 0. When dx dt 0, one has: dy dy dx = dt dx dt PR (FIU) MAC / 85

95 Example: x = t sin t, y = t cos t find the equation of the tangent line at t = π 2. At t = π 2, dy dx = π 2. dy dx = cos t t sin t sin t + t cos t The point corresponding to t = π 2 is ( π 2, 0). So the tangent line has equation: y = π 2 (x π 2 ) PR (FIU) MAC / 85

96 Exercises: 1. First write the equation of the tangent line to the given parametric curve at the point that corresponds to the given value of t, then calculate d 2 y to determine whether the curve is concave upward or dx 2 concave downward at this point. a) x = 2t 2 + 1, y = 3t 3 + 2; t = 1 b) x = t sin t, y = t cos t; t = π 2 PR (FIU) MAC / 85

97 Polar coordinates (r, θ) : 0 r < +, 0 θ < 2π x = r cos θ, y = r sin θ Conversion formulas: r 2 = x 2 + y 2 tan θ = y x for x 0 PR (FIU) MAC / 85

98 Exercises: Express the given rectangular equation in polar form. 1) xy = 1 2) x 2 y 2 = 1 3) x + y = 4 PR (FIU) MAC / 85

99 Express the given polar equation in rectangular form. 1) r = 3 2) θ = 3π 4 3) r = sin 2θ 4) r = 1 cos 2θ PR (FIU) MAC / 85

100 Graphing polar curves Polar curves are usually given by equations of the form r = f (θ) Example: Graph the circle r = 2 sin θ The period T = 2π will be divided into 4 quarters, θ = T 4, θ = 2 T 4, θ = 3 T 4, θ = 4 T 4. We will include any additional reference points θ satisfying r = 0, that is, 2 sin θ = 0 in this exemple. PR (FIU) MAC / 85

101 θ 0 π 2 π 3 π 2 2π r Observe that the point r = 2, θ = 3 π 2 will be plotted as r = 2, θ = 3 π 2 + π, that is, a reflection through the pole is applied. PR (FIU) MAC / 85

102 Cardioids are given by equations like: Example: Graph r = sin θ r = a(1 ± sin θ) r = a(1 ± cos θ) PR (FIU) MAC / 85

103 Rose curves are given by equations of the form: r = a cos nθ r = a sin nθ When n is even, the rose has 2n petals. When n is odd, the rose will have n petals. Examples: Graph r = 3 sin 3θ. Graph r = cos 2θ PR (FIU) MAC / 85

104 Polar curves as parametric curves From a polar curve r = f (θ) we get a parametric curve x = f (θ) cos θ, y = f (θ) sin θ In this case, the slope of tangent lines will be given by: dy dx = dr r cos θ + sin θ dθ r sin θ + cos θ dr dθ PR (FIU) MAC / 85

105 Example: Find the slope of the tangent line to the circle r = 4 cos θ at the point where θ = π 6. PR (FIU) MAC / 85