Math 527 Lecture Notes Topics in Calculus and Analysis Northern Illinois University Spring, Prof. Richard Blecksmith

Transcription

1 Math 527 Lecture Notes Topics in Calculus and Analysis Northern Illinois University Spring, 2014 Prof. Richard Blecksmith

2

3 Contents Module 4. Further Applications of Derivatives Direction of a Curve Optimization Optimization Exercises Due Saturday, Feb 5, Concavity 60 Module 5. Exponential Functions Definitions Derivatives of Exponential Functions The exponential function e x Exponential Function Exercises Due Saturday, March 5, iii

4

5 MODULE 4 Further Applications of Derivatives 1. Direction of a Curve 1.1. Finite Intervals. There are four kinds of finite intervals: Notation Condition Name [a,b] a x b closed (a,b) a < x < b open [a,b) a x < b half open (a,b] a < x b half open 1.2. Infinite Intervals. There are five kinds of infinite finite intervals: Notation Condition Name [a, ) x a closed ray (a, ) x > a open ray (,b] x b closed ray (,b) x < b open ray (, ) any x real number line infinite) 1.3. Increasing versus Decreasing. If I is any one of these nine intervals (finite or and f(x) is a function defined on I, We say f is increasing on I if for any two points x 1 and x 2 in I, 47

6 48 4. FURTHER APPLICATIONS OF DERIVATIVES if x 1 < x 2 then f(x 1 ) < f(x 2 ) In other words, as x values increase, the y values on the graph increase. We say f is decreasing on I if for any two points x 1 and x 2 in I, if x 1 < x 2 then f(x 1 ) > f(x 2 ) In other words, as x values increase, the y values on the graph decrease Relation of Deriv to Graph. function derivative increasing positive decreasing negative horizontal zero straight line constant steep rising large positive gradual rising small positive steep falling large negative gradual falling small negative 1.5. Major Theorem. Let f be a function which is conintuous on a closed interval [a,b] and differentiable of the open interval (a,b). If f (x) > 0 for every x in (a,b), then f(x) is increasing on [a,b]. If f (x) < 0 for every x in (a,b), then f(x) is decreasing on [a,b]. If f (x) = 0 for every x in (a,b), then f(x) is constant on [a,b].

7 1. DIRECTION OF A CURVE Derivatives. y = f(x) = x 3 6x 2 +9x+5 y = f (x) = 3x 2 12x+9 = 3(x 2 4x+3) = 3(x 1)(x 3) 1.7. Determining sign of f. f (x) = 3(x 1)(x 3) To determine the sign of f (x) use the following chart Interval (, 1) (1, 3) (3, ) x 1 neg pos pos x 3 neg neg pos (x 1)(x 3) pos neg pos Interpretation increasing decreasing increasing 1.8. Picture of Graph. (1,9) (3,5) f(x) = x 3 6x 2 +9x+5

8 50 4. FURTHER APPLICATIONS OF DERIVATIVES 1.9. Subset of the domain. Let f be a real valued function defined on a set S of real numbers. The set S is not necessarily the entire domain of f, although S must, of course, lie inside the domain of f. For example, if f(x) = x, then the domain of f is the set of real numbers 0. We could take S to be the closed interval [1,2]. 2. Optimization 2.1. Absolute Maximum&Minimum. The function f has an absolute maximum on the set S if there is at least one point c in S such that f(x) f(c) for all x in S. The number f(c) is called the absolute maximum value of f on S. We say f has an absolute minimum on the set S if there is at least one point c in S such that f(x) f(c) for all x in S Relative Maximum&Minimum. Let f be a real value function defined on a set S of real numbers. We say the function f has a relative maximum at a point c in S if there is an open interval I containing c such that f(x) f(c) for all x lying in both I and S.

9 2. OPTIMIZATION 51 The concept of relative minimum is similarly defined by requiring f(x) f(c) instead of f(x) f(c). A relative maximum/minimum is sometimes called a local maximum and minimum Recent Example. Rel Max (1,9) (3,5) Rel Min f(x) = x 3 6x 2 +9x Straight Line Example. Consider f(x) = x, S = [0,2). This function has absolute minimum on S when x = 0. No absolute maximum exists. This function fails to have an absolute maximum because x = 2 is not in the domain set S.

10 52 4. FURTHER APPLICATIONS OF DERIVATIVES 2.5. Reciprocal Example. Consider f(x) = 1 x, S = (0,2]. The reciprocal function has absolute minimum on S when x = 2. No absolute maximum exists. This function fails to have an absolute maximum because it is not continuous at x = 0. 1 that is, because lim x 0 + x = 2.6. Graph of 1/x. 1 lim x 0 + x = (2,.5) abs min 2.7. Vanishing Derivative Theorem. Assume f(x) is a continuous function defined on an open interval I. Assume that f(x) has a local maximum or minimum at a point c inside I. If f (c) exists, then f (c) = 0.

11 2. OPTIMIZATION First Warning. TheVanishingDerivativeTheoremdoesnotsaythatiff (c) = 0, then f has a maximum or a minimum at x = c. Example. Consider f(x) = x 3. Here f (x) = 3x 2 and so clearly f (0) = 0, yet x = 0 is neither a local maximum nor a local minimum Graph of x 3. no max / mins (0,0) horizontal tangent

12 54 4. FURTHER APPLICATIONS OF DERIVATIVES Second Warning. The Vanishing Derivative Theorem does not say that if f has a maximum or a minimum at x = c, then f (c) = 0. Example Consider f(x) = x. It is clear from the graph that the absolute value function has a relative (in fact, absolute) minimum at x = 0, yet f (0) is not even defined Graph of absolute value function. (0,0) absolute minimum no derivative Max-Min Theorem. Max-Min Theorem for Continuous Functions Assume f(x) is a continuous function defined on a set S and that f(x) has a local maximum or minimum at a point c in [a,b]. Then we have three possibilities: (i) f (c) = 0

13 2. OPTIMIZATION 55 (ii) f (c) is undefined or (iii) c is a boundary point of S Usually (iii) means that S is an endpoint of an interval. Points of type (i) or (ii) are called critical points Example 1. Find all local and absolute max/min points of f(x) = x 3 12x on the interval [0, 4]. f (x) = 3x 2 12 = 3(x 2 4) = 3(x 2)(x+2) The critical points are 2 and 2. Since 2 does not lie in the interval [0,4], we may ignore it. The only points that can possibly be relative max / mins are the critical point x = 2 and the endpoints 0 and Determining max / mins. f(x) = x 3 12x on the interval [0,4]. f (x) = 3(x 2)(x+2) x y = x 3 12x Result 0 0? = 16 Abs Min = 16 Abs Max Example 2. Find all local and absolute max/min points of f(x) = 3x 2/3 2x on the interval [ 1,2]. f (x) = x 1/3 2 = 2 x 1/3 2 f (x) = 0 2 x 1/3 2 = 0

14 56 4. FURTHER APPLICATIONS OF DERIVATIVES 2 x 1/3 = 2 1 x 1/3 = 1 x 1/3 = 1 x = 1 Note that the derivative is undefined when x = 0 Why? The critical points are 0 and 1 and the endpoints are 1 and Determining max / mins. f(x) = 3x 2/3 2x on the interval [ 1,2]. f (x) = 2 x 1/3 2 x y = 3x 2/3 2x Result = 5 Abs Max 0 0 Abs Min = 1? ? Example 1 Revisited. Find all relative max/min points of f(x) = x 3 12x on the interval [0, 4]. f (x) = 3x 2 12 = 3(x 2)(x+2) The critical points are 2 and 2. We ignore 2 since it is not in the interval [0,4]. The endpoints 0 and 4. We can use the sign of the derivative to determine when the function is going up and going down, therby isolating relative maximum and minimums.

15 2. OPTIMIZATION Finding Relative Extrema. Use the chart for f (x) = 3(x 2)(x+2) Interval (, 2) ( 2, 2) (2, ) x 2 neg neg pos x+2 neg pos pos (x 2)(x+2) pos neg pos Interpretation increasing decreasing increasing Conclusions: 0 is a relative max since the graph is decnreasing to the right of x = 0 2 is a relative min since the graph is decreasing to the left of 2 and increasing to the right of 2 4 is a relative max since the graph is increasing to the left of x = Graph of x 3 12x. (4, 16) Abs Max Rel Max (0,0) Abs Min (2, 16)

16 58 4. FURTHER APPLICATIONS OF DERIVATIVES First Derivative Test. Suppose f(x) is continuous on a closed interval [a, b] and c is a critical point of f(x) in the open interval (a,b) Interval (a, c) (c, b) Conclusion Case 1 y < 0 y > 0 MIN Case 2 y > 0 y < 0 MAX Case 3 y > 0 y > 0 INCR Case 4 y < 0 y < 0 DECR

17 3. OPTIMIZATION EXERCISES DUE SATURDAY, FEB 5, Optimization Exercises Due Saturday, Feb 5, 2011 (1) page 163: #2 and 5 optimization (2) page 163: #3 and 5 optimization (3) page 163: #4 and 5 optimization (4) page 164: #7 chocolate box (5) page 165: #9 oval track (6) page 166: #14 profit (7) Find the point(s) on the graph of y = 4 x 2 that are closest to the point (0,2). (8) A rectangular page is to contain 24 square inches of print. The margins at the top and bottom of the page are each inches wide, while the margins at the sides are each 1 inch wide. What are the dimensions of the page so that the least amount of paper is used? (9) The geometric mean of two positive numbers a and b is and the arithmetic mean is GM = ab AM = a+b 2. Use calculus to show that the geometric mean is the arithmetic mean. [Hint: Let P = ab. Among all positive numbers x and y with xy = P, when is the sum x+y the smallest?] You might experiment with different values of a and b. (10) The Road Less Paved [Courtesy of the Rose Hulman Institute of Technology and developed by Kimberly Foltz.] Four houses are located so that they form the vertices of a square which has sides of length 1 mile. These neighbors desire to connect their houses with roads in a manner that is as inexpensive as possible. The cost of building one unit of road is constant in the entire region. Construct this system of roads in the most cost efficient way possible. Determine the total length of this system of roads. Describe the path that is formed by the roads.

18 60 4. FURTHER APPLICATIONS OF DERIVATIVES 4. Concavity 4.1. Shape of a Curve. There are two ways that a curve can increase from point P to point Q Q concave down or concave up P

19 4. CONCAVITY Concave Up Graphs. When a graph is concave up, the slopes get progressively larger. Q slopes are increasing P 4.3. Concave Down Curve. There are two ways that a curve can increase from point P to point Q Q slopes are decreasing P

20 62 4. FURTHER APPLICATIONS OF DERIVATIVES 4.4. Concave Up Curve. y is increasing (y ) > 0 y > 0 Thus the graph is concave up if and only if y > Concave Down Curve. y is decreasing (y ) > 0 y < 0 Thus the graph is concave down if and only if y < 0

21 4. CONCAVITY 63 A point at which concavity changes is called an inflection point 4.6. Example - Revisited. y = f(x) = x 3 6x 2 +9x+5 y = f (x) = 3x 2 12x+9 = 3(x 2 4x+3) = 3(x 1)(x 3) Based on this factorization we saw earlier that Interval Behavior of graph (, 1) inceasing (1, 3) deceasing (3, ) inceasing 4.7. Determining Concavity. y = f(x) = x 3 6x 2 +9x+5 y = f (x) = 3x 2 12x+9 y = f (x) = 6x 12 = 6(x 2) To determine the sign of f (x) use the following chart Interval (, 2) (2, ) x 2 neg pos Sign of y y < 0 y > 0 Interpretation concave down concave up So the graph is concave down to the left of 2 and concave up to the right to 2. Since the concavity changes at x = 2, (2,f(2)) is an inflection point. Here, f(2) = = = 7

22 64 4. FURTHER APPLICATIONS OF DERIVATIVES 4.8. Picture of Graph. Rel Max (1,9) (2,7) Inflection (3,5) Rel Min f(x) = x 3 6x 2 +9x Two Faces. Happy Face Sad Face concave up mouth ++ eyes pos second deriv concave down mouth eyes neg second deriv

23 4. CONCAVITY Max versus Min. Concavity gives us a way to tell maximums from minimums when dealing with critical points Relative Min Relative Max concave up concave down y > 0 y < Second Derivative Test. Suppose f(x) is continuous on a closed interval [a, b] and c is a critical point of f(x) in the open interval (a,b) for which f (c) = 0 f (c) Conclusion pos Minimum neg Maximum zero Don t Know Test for x 3 12x. f(x) = x 3 12x f (x) = 3x 2 12 = 3(x 2)(x+2) Critical points: 2 and 2 f (x) = 6x x f(x) = x 3 12x f (x) = 6x Sign Concl = neg Max = pos Min

24 66 4. FURTHER APPLICATIONS OF DERIVATIVES Graph of x 3 12x. ( 2, 16) Rel Max Rel Min (2, 16) Warning. No conclusion can be made if f (c) = 0 as the following three examples show. For each of these three functions, f (0) = 0 and f (0) = 0 f(x) f (x) f (x) Behavior x 4 4x 3 12x 2 Min x 4 4x 3 12x 2 Max x 3 3x 2 6x Neither For f(x) = x 4, x = 0 is a minimum For f(x) = x 4, x = 0 is a maximum For f(x) = x 3, x = 0 is neither a max nor min

25 4. CONCAVITY Second Warning. The statement inflection points occur when the second derivative is zero is slightly misleading. course. Although this statement is true in practive for the functions you are likely to see in this Consider the function f(x) = x 4. Here f (x) = 12x 2 = 0 at x = 0 But x = 0 is not an inflection point, because This function is always concave up. It never changes concavity Strategy for Sketching Graphs. Step 1. [Derivatives] Find f (x) and f (x) Step 2. [Domain and symmetry] Note the domain of f(x) and look for possible symmetries: (a) f( x) = f(x) implies symmetry about y-axis (b) f( x) = f(x) implies symmetry about origin Step 3. [Critical points] Determine critical points where the derivative is zero or undefined. These are candidates for max/min points. Step 4. [Increasing/Decreasing] Determine the sign of the derivative in the intervals between successive critical points.

26 68 4. FURTHER APPLICATIONS OF DERIVATIVES (a) y > 0 means graph is increasing (b) y < 0 means graph is decreasing (c) Identify the maximum versus the minimums using the first derivative test Sketching Graphs Continued. Step 5. [Inflection points] Determine when f (x) = 0. These are candidates for infection points. Step 6. [Concavity] Determine the sign of the second derivative in the intervals between successive points where f (x) = 0. (a) y > 0 means graph is concave up (b) y < 0 means graph is concave down (c) Infection points occur when concavity changes Step 7. [Table of Points] Make a table of the y values of the critical and inflection points. Step 8. [Complete the sketch] Connect the point in you graph using the information about increasing/decreasing and concavity.

27 4. CONCAVITY 69 Graphing Exercises Due Saturday, Feb 19, 2011 Ten Step Graphing Outline: (1) Find f (x) (2) Find the critical points (3) On which intervals is the graph increasing? (4) On which intervals is the graph decreasing? (5) Which of the critical points are maximums / minimums? (6) Find f (x) (7) Find the inflection points (8) On which intervals is the graph concave up? (9) On which intervals is the graph concave down? (10) Sketch the graph, labelling each extreme point and each inflection point. Work the following 5 problems, applying the 10-step outline: (1) page 173: #11 f(x) = 4 3 x3 4x (2) page 173: #12 f(x) = x 3 +x 2 (3) page 189: #6 f(x) = x 4 2x 2 (4) not in book f(x) = x 3 6x 2 +9x+5 (5) not in book f(x) = x 4 2x 3 +2x

28

29 MODULE 5 Exponential Functions 1. Definitions 1.1. Exponential Functions. An exponential functions has the form f(x) = a x where the base number a is a positive real, not equal to 1. For a = 2 here is a table of values of a x : x x Defining 2 x. We know that if the exponent is a positive integer, say x = 5, then 2 5 = = 32 If x is a negative integer, say x = 5, then 2 5 = = 1 32 If x is a reciprocal, say x = 1 2, then = 2 71

30 72 5. EXPONENTIAL FUNCTIONS If x is a fraction, say x = 3 2, then = ( 2 ) 3 But what if x is not a fraction? How do we define 2 2? or 2 π? 1.3. Graph of y = 2 x. y = 2 x (0,1) (1,2) 1.4. Rules of exponents. (1) a m a n = a m+n (2) am a n = am n (3) ( a m) n = a mn (4) a 0 = 1 (5) a n = 1 a n (6) a 1/n = n a

31 1. DEFINITIONS Three Possible Answers. Things are not even clear when defining a m/n, where m/n is a fraction, at least when a is negative. For example, what is ( 8) 2 6? Answer 1: ( 8) 2 6 = 6 ( 8) 2 = 6 64 = 2 Answer 2: ( 8) 2 6 = ( 6 8 ) 2 which does not exist because you cannot take the sixth root of 8. Answer 3: ( 8) 2 6 = ( 8) 1 3 = 3 8 = 2 So which of these three answers is correct? 1.6. Thinking Like a Mathematician. The function satisfies the rule E(x) = a x E(x+y) = a x+y = a x a y = E(x)E(y) What could you prove about a function if all you knew about it was that it obeyed the rule E(x+y) = E(x)E(y) The function that is identically zero for all x satisfies this rule, but is very unintersting. So we ban function E(x) from being the zero function.

32 74 5. EXPONENTIAL FUNCTIONS 1.7. First Result. Assume and E(x) is not identically zero. E(x+y) = E(x)E(y) This means that for some real number b, E(b) 0. We have E(b) = E(b+0) = E(b) E(0) Dividing by E(b) which we know is not 0 gives Fact 1: E(0) = Second Result. The rule E(x+y) = E(x)E(y) implies E(x+ x) = E(x) E( x) But from Fact 1 we know E(x+ x) = E(0) = 1 Thus, E(x) E( x) = 1 or Fact 2: E( x) = 1/E(x)

33 1. DEFINITIONS Third Result. The rule E(x+y) = E(x)E(y) implies E(x+ y) = E(x) E( y) = E(x) 1 E(y) from Fact 2. Thus, Fact 3: E(x y) = E(x)/E(y) Application. Since the function E(x) = a x satisfies the rule E(x+y) = E(x)E(y) it must obey the first three Facts: Fact 1: a 0 = E(0) = 1 Fact 2: a x = E( x) = 1 E(x) = 1 a x Fact 3: a x y = E(x y) = E(x) E(y) = ax a y What are the advantages of this approach? What are its disadvantages? The Square root of a Million. To compute the square root of a million nobody can do that it their head! First write 1 million as 10 6

34 76 5. EXPONENTIAL FUNCTIONS Using the rule x = x 1/2 and the rule ( a m) n = a mn the square root of a million is ( 10 6 )1 2 = 10 (6 1 2 ) = 10 3 = 1000 The square root one million is one thousand. 2. Derivatives of Exponential Functions 2.1. The Derivative of y = a x. Let f(x) = a x and g(x) = f(x+b) = a x+b We will compute the derivative of g(x) in two ways. First Way: By the chain rule, g (x) = f (x+b) d dx (x+b) = f (x+b) 1 = f (x+b) Second Way: By the addition rule for exponents, g(x) = a x+b = a b a x = a b f(x) Since a b is a constant as far as the variable x is concerned, g (x) = a b f (x) Equating these two formula for g (x) gives us f (x+b) = a b f (x)

35 2. DERIVATIVES OF EXPONENTIAL FUNCTIONS The Derivative of y = a x. Now take f (x+b) = a b f (x) and plug in x = 0: f (0+b) = a b f (0) or f (b) = c a b where the constant c = f (0) Since we would like our variable to be x and not b, we just replace b by x in this formula: f (x) = c a x, c = f (0) 2.3. Summary. The derivative of an exponential function is that same exponential function times a constant. d dx ax = c a x The constant c is the derivative of a x at x = 0

36 78 5. EXPONENTIAL FUNCTIONS 2.4. The Constant. At this point you may be wondering, how do we know what the constant c is? Since c = f (0) we could ise the limit definition of derivative to evaluate c: For f(x) = a x c = f (0) = lim h 0 f(0+h) f(0) h = lim h 0 a h a 0 h = lim h 0 a h 1 h For a given value of a we could evaluate this limit, by taking values of h closer and closer to zero and examining the results The Constant when a = 2. 2 h 1 x h It looks like the constant when a = 2 is c = So d dx 2x = x

37 3. THE EXPONENTIAL FUNCTION e x Constant c for a = 2,...,10. a h 1 f(x) c = lim h 0 h 2 x x x x x x x x x Observe that for some value of a between 2 and 3, this constant will be exactly The exponential function e x 3.1. The function e x. c = 1. In calculus, the most commonly used base, by far, is the one for which the constant This base is so special, it has its own single-letter name: e Euler It was named in honor of the Swiss mathematician who disovered this number: Leonard The value of Euler s constant to twentyfive digits is e = The function e x is on on your scientific or graphing calculator.

38 80 5. EXPONENTIAL FUNCTIONS 3.2. What s important about e x. The single most important fact about e x is that e h 1 since c = lim = 1 h 0 h The derivative of y = e x is y = e x in other words, e x is its own derivative. What s the second derivative of e x? What s the third derivative of e x? What s the hundredth derivative of e x? 3.3. Working with e x. In computing derivatives involving the exponential function e x, just use the chain rule, product rule, and quotient rule. Here are some examples: 3.4. Example 1. Differentiate y = e 3x+1 By the chain rule, y = e 3x+1 = e 3x+1 3 d dx (3x+1) which is usually written 3e 3x+1

39 3. THE EXPONENTIAL FUNCTION e x Example 2. Differentiate y = e x2 2x+3 By the chain rule, y = e x2 2x+3 = e x2 2x+3 (2x 2) d dx (x2 2x+3) 3.6. Example 3. Differentiate y = x 2 e x By the product rule, y = d ( x 2 ) e x +x 2 d dx dx (ex ) = 2xe x +x 2 e x 3.7. Tangent Line. Find the tangent line to y = f(x) = e 2x 4 at (2,1) Differentiate: f (x) = 2e 2x 4 The slope of the tangent line is m = f (2) = 2e = 2e 0 = 2 By the point slope formula, the tangent line is y y 1 = m(x x 1 ) y 1 = 2(x 2) y 1 = 2x 4 y = 2x 3

40 82 5. EXPONENTIAL FUNCTIONS 4. Exponential Function Exercises Due Saturday, March 5, 2011 (1) page 182: #1 (2) page 182: #2 (3) page 182: #3 (4) page 182: #5 (5) page 182: #7 (6) page 182: #9 (7) page 182: #11 (8) page 182: #13 (9) page 182: #16 (10) page 189: #7 (11) page 189: #9 (12) page 189: #11 (13) page 189: #13 (14) page 189: #15 (15) page 189: #19