even number solutions

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(1)Answers to even-numbered problems in Mathematics for Economic Analysis Knut Sydsæter Peter Hammond.

(2) Preface Mathematics for Economic Analysis, Prentice Hall, 1995 has been out for a long time, and over the years we have had many request for supplying solutions to the even-numbered problems. (Answers to the odd-numbered problems are given in the main text.) This manual provides answers to all the even-numbered problems in the text. These answers are taken from the old Instructors Manual which is no longer available. (In fact, some of the old answers have been extended.) For many of the more interesting and/or difficult problems, detailed solutions are provided. For some of the simpler problems, only the final answer is presented. Appendix A in the main text reviews elementary algebra. This manual includes a Test I, designed for the students themselves to see if they need to review particular sections of Appendix A. Many students using our text will probably have some background in calculus. The accompanying Test II is designed to give information to the students about what they actually know about single variable calculus, and about what needs to be studied more closely, perhaps in Chapters 6 to 9 of the text. Oslo and Coventry, November 2010 Knut Sydsæter and Peter Hammond Contact addresses: [email protected] [email protected]. Version 1.0 07122010 1113 © Knut Sydsæter and Peter Hammond 2010.

(3) CHAPTER 1. INTRODUCTION. 1. Chapter 1 Introduction 1.3 2. (a) Put p/100 = x. Then the given expression becomes a + ax − (a + ax)x = a(1 − x 2 ), as required. (b) $2000 · 1.05 · 0.95 = $1995. (c) The result is precisely the formula in (a). (d) With the notation used in the answer to (a), a − ax + (a − ax)x = a(1 − x 2 ), which is the same expression as in (a). 4. (a) F = 32 yields C = 0; C = 100 yields F = 212. (b) F = 95 C + 32 (c) F = 40 for C ≈ 4.4, F = 80 for C ≈ 26.7. The assertion is meaningless. R(1 − p) − S(1 − q) pS − qR , y= (p − q)(1 − p − q) (p − q)(1 − p − q) (Use, for example, (A.39) in Section A.9.). 6. x =. if (p − q)(1 − p − q) = 0.. 1.4 2. (a) Correct. (b) Incorrect. (c) Incorrect. (d) Correct. (e) Incorrect. (f) Incorrect. (“Usually” the sum of two irrationals is irrational, but not always. For example, π and −π are both irrationals, but π + (−π) = 0, which is rational.) 4. (a) y ≤ 3 − 43 x. (c) y ≤ (m − px)/q. (b) y > 23 z. 6. |2 · 0 − 3| = | − 3| = 3, |2 ·. 1 2. − 3| = | − 2| = 2, |2 ·. 7 2. − 3| = |4| = 4. 8. (a) 3 − 2x = 5 or 3 − 2x = −5, so −2x = 2 or √ −8. Hence √ x = −1 or x = 4. (b) −2 ≤ x ≤ 2 (c) 1 ≤ x ≤ 3 √(d) −1/4 ≤ x ≤ 1 (e) x > 2 or x < − 2 √ (f) 1 ≤ x 2 ≤ 3, and so 1 ≤ x ≤ 3 or − 3 ≤ x ≤ −1. 1.5 2. (a) ⇒ right, ⇐ wrong (b) ⇒ wrong, ⇐ right (c) (d) ⇒ and ⇐ both right (e) ⇒ wrong (0 · 5 = 0 · 4, but 5 = 4), ⇐ right. ⇒ right, ⇐ wrong (f) ⇒ right, ⇐ wrong. 4. x = 2. (x = −1, 0 or 1 make the equation meaningless. Multiplying each term by the common denominator x(x − 1)(x + 1) yields (x + 1)3 + (x − 1)3 − 2x(3x + 1) = 0. Expanding and simplifying, 2x 3 − 6x 2 + 4x = 0, or 2x(x 2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 is the only solution.) √ √ 6. (a) No solutions. (Squaring each side yields x−4 = x+5−18 x + 5+81, which reduces to x + 5 = 5, with solution x = 20. But x√= 20 does not satisfy the given equation.) (b) Just as in part (a) we find that x must be a solution of x + 5 = 5, which has the solution x = 20. Inserting this value of x into the given equation we find that x = 20 is the solution. √ √ (1) (2) 8. (a) x + x + 4 = 2 ⇒ x + 4 = 2−x ⇒ x +4 = 4−4x +x 2 ⇒ x 2 −5x = 0 ⇒ x −5 = 0 ⇐ x = 5. Here implication (1) is incorrect (x 2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (2) is correct, but it breaks the chain of implications. (b) x = 0. (By modifying the argument we see that the given equation implies x = 5 or x = 0. But only x = 0 is a solution.) 10. No. (To study law it seems that you need a course in logic.) © Knut Sydsæter and Peter Hammond 2010.

(4) 2. CHAPTER 1. INTRODUCTION. 1.6 2. Logically the two statements are equivalent, but the second statement is still an expressive poetic reinforcement.. 1.7 2. (a) No, not necessarily.. (b) Yes (if anybody is both a painter and a poet).. 4. F ∩ B ∩ C is the set of all female biology students in the university choir; M ∩ F the female mathematics students; (M ∩ B) \ C \ T the students who study both mathematics and biology but neither play tennis nor belong to the university choir. 6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A = {1, 2}, B = {1}, C = {1, 3}. As for (e), note in particular that A ∪ B = A ∪ C = A whenever B and C are subsets of A, even if B = C.) 8. 50 − 35 = 15 liked only coffee, 40 − 35 = 5 liked only tea, 35 liked both, and 10 did not like either. In all there were 15 + 5 + 35 + 10 = 65 who responded. 10. (a) Consider Fig. 1.7.10(a). Both sets consist of the elements in (1) and in (3). (b) Consider Fig. 1.7.10(b): A B consists of (3), (4), (2), and (5). Then (A B) C consists of (4), (5), (6), and (7), which is described verbally in the problem.. B. A (5). (1) (4) (7). (2). (3) A. (1). (2). (3). B (8). (6) C. Figure 1.7.10(a). Figure 1.7.10(b). 12. (a) Consider Fig. 1.7.10(b), and let ni denote the number of people in the set marked (i), for i = 1, 2, . . . , 8. The responses reported imply that: n1 +n3 +n4 +n7 = 420; n1 +n2 +n5 +n7 = 316; n2 +n3 +n6 +n7 = 160; n1 + n7 = 116; n3 + n7 = 100; n2 + n7 = 30; n7 = 16. From these equations we easily find n1 = 100, n2 = 14, n3 = 84, n4 = 220, n5 = 186, n6 = 46, n7 = 16. (i) n3 + n4 = 304 had read A but not B; (ii) n6 = 46; (iii) 1000 − (n1 + n2 + n3 + n4 + n5 + n6 + n7 ) = 334. (b) (i) n(A \ B) (ii) n(C \ (A ∪ B)) (iii) n( \ (A ∪ B ∪ C)) = n() − n(A ∪ B ∪ C)) (c) The equality is n1 + n2 + n3 + n4 + n5 + n6 + n7 = (n1 + n3 + n4 + n7 ) + (n1 + n2 + n5 + n7 ) + (n2 + n3 + n6 + n7 ) − (n1 + n7 ) − (n3 + n7 ) − (n2 + n7 ) + n7 , which is easily verified. The last equality is a special case of n( \ D) = n() − n(D): The number of persons who are in , but not in D, is the number of persons in all of minus the number of those who are in D. © Knut Sydsæter and Peter Hammond 2010.

(5) CHAPTER 2. FUNCTIONS OF ONE VARIABLE: INTRODUCTION. 3. Chapter 2 Functions of One Variable: Introduction 2.2 2. F (0) = F (−3) = 10, F (a + h) − F (a) = 10 − 10 = 0 √ √ √ √ 2/3, f ( π ) = π /(1 + π ), 4. (a) f (−1/10) = −10/101, f (0) = 0, f (1/ 2) = f (2) = 2/5 (b) f (−x) = −x/(1 + (−x)2 ) = −x/(1 + x 2 ) = −f (x) and moreover, f (1/x) = (1/x)/[1 + (1/x)2 ] = (1/x) · x 2 /[1 + (1/x)2 ] · x 2 = x/(1 + x 2 ) = f (x) 6. F (0) = 2, F (−3) =. √ √ 19, F (t + 1) = t 2 + 3. 8. (a) b(0) = 0, b(50) = 100/11, b(100) = 200 (b) b(50 + h) − b(50) is the additional cost of removing h% more than 50% of the impurities. 10. (a) No. f (2 + 1) = f (3) = 2 · 32 = 18, whereas f (2) + f (1) = 2 ·√22 + 2 · 12 = 8 + 2 = 10. (b)Yes. f (2+1) = f (2)+f (1) = −9 (c) No. f (2+1) = f (3) = 3 ≈ 1.73, whereas f (2)+f (1) = √ 2 + 1 ≈ 2.41. 12. See Figs. 2.2.12(a) and 2.2.12(b).. 1. x·1. 1·1. 1 1. x. x2. 1·x. x. 1. Figure 2.2.12(a). Area = (x + 1)2 = x 2 + 2x + 1. x. x Figure 2.2.12(b) Area = x 2 +1. 14. (a) (−∞, 2) ∪ (2, ∞) (b) f (8) = 5 3x + 6 (c) f (x) = = 3 ⇐⇒ 3x + 6 = 3(x − 2) ⇐⇒ 6 = −6, which is absurd. x−2 16. (a) is invalid: |2 + (−2)| = 0, whereas |2| + | − 2| = 2 + 2 = 4. (b), called the triangle inequality, |x + y| ≤ |x| + |y|, is valid. In fact, if x and y have the same sign, then |x + y| = |x| + |y|. If x and y have opposite signs, then we see that |x + y| < |x| + |y|. (See Problem 12.4.8 for a generalization.) (c) is valid: |xy| = |x| · |y|. (Look at the 4 different sign combinations of x and y: If x is positive and y is negative, then xy is negative, |xy| = −xy and |x|·|y| = (−x)·y = −xy, and so on.) (d) is easily seen to be valid. (e) is not valid (put x = 1). (f) is easily seen (Note that −4)2 = −(−2) = 2.) to be valid. (g) is valid: | − 2x| = 2|x|. (h) is valid because |x| − |y| ≤ |x − y| ⇔ x 2 − 2|x||y| + y 2 ≤ x 2 − 2xy + y 2 ⇔ −2|x||y| ≤ −2xy ⇔ |x||y| ≥ x · y, which is true because (c) is valid. © Knut Sydsæter and Peter Hammond 2010.

(6) 4. CHAPTER 2. FUNCTIONS OF ONE VARIABLE: INTRODUCTION. 2.3 2. See Figs. 2.3.2(a)—2.3.2(f). y. y. y 4 3 2. 1. 1. 1. x. 1. x. 1. Figure 2.3.2(a). 1. Figure 2.3.2(b). y. 2. x. 3. Figure 2.3.2(c). y. y. 1. 1. 2 1 2. 1. 1. 2. x. x. 1. x. 1. 1. Figure 2.3.2(d). Figure 2.3.2(e). Figure 2.3.2(f). 4. See Figs. 2.3.4(a)–(c). y. y. y. 2. 2. 2. 1. 1. 1. −3 −2 −1 −1. 1. 2. −1. −3 −2 −1 −1. −2. −2. 3x. −2. Figure 2.3.4(a). 1. 2. 3. 4. 5. x. Figure 2.3.4(b). 1. 2. 3x. Figure 2.3.4(c). y y y A = (3, 2). 2 (2, 4) 2. x. 1. 2 P 2. 5. B = (5, −4). x. Figure 2.3.6. © Knut Sydsæter and Peter Hammond 2010. Figure 2.3.10. 1 Figure 2.3.12. x.

(7) CHAPTER 2. FUNCTIONS OF ONE VARIABLE: INTRODUCTION. 5. 6. (5 − 2)2 + (y − 4)2 = 13, or y 2 − 8y + 12 √ = 0, with solutions y = 2 and y = 6. Geometric explanation: The circle with center at (2, 4) and radius 13 intersects the line x = 5 at two points. If the radius were 2, the circle would not intersect the line x = 5. See Fig. 2.3.6. 8. (a) (x − 2)2 + (y − 3)2 = 16. (b) (x − 2)2 + (y − 5)2 = 13. 10. (x − 3)2 + (y − 2)2 = (x − 5)2 + (y + 4)2 , which reduces to x − 3y = 7. See Fig. 2.3.10 12. (x, y) = (1, 0), (0, 1), (1/4, 1/4) are all solutions. See Fig. 2.3.12. 14. The condition is: p + r x 2 + y 2 = p + s (x − a)2 + y 2 . Cancelling p, then squaring each side and rearranging yields (r 2 − s 2 )(x 2 + y 2 ) = s 2 (a 2 − 2ax). If r = s, then the markets are separated by the straight line x = 21 a. Otherwise, completing squares yields the equation [x + as 2 /(r 2 − s 2 )]2 + y 2 = [ars/(r 2 − s 2 )]2 . This is a circle with center at − as 2 /(r 2 − s 2 ), 0 and radius |ars/(r 2 − s 2 )|.. 2.4 2. (a) f (−5) = 0, f (−3) = −3, f (−2) = 0, f (0) = 2, f (3) = 4, f (4) = 0 (b) Df = [−5, 4], Vf = [−3, 4] 4. See Figs. 2.4.4(a)–(d). y. y. y. y. 1 1 1. 1. 1 1. x. 3. Figure 2.4.4(a). x. 1. Figure 2.4.4(b). x. 1. x. 1. Figure 2.4.4(c). Figure 2.4.4.4(d). 2.5 2. 0.78. 4. See Figs. 2.5.4(a)–(c). y. y. y. 1. 4 3. −1. 2. −2 −3. 1 1. 2. Figure 2.5.4(a). 3. 4. x. 4 1 2. 3 4. 5 6 7. 8. 10 x. 3 2 1. −4 −5. −1. Figure 2.5.4(b). Figure 2.5.4(c). 1. 2. 3. 4. 5 x. 6. (a) Assume F = aC + b. Then 32 = a · 0 + b and 212 = a · 100 + b. Therefore a = 180/100 = 9/5 and b = 32, so F = (9/5)C + 32. (b) If X = (9/5)X + 32, then X = −40. © Knut Sydsæter and Peter Hammond 2010.

(8) 6. CHAPTER 3. POLYNOMIALS, POWERS, AND EXPONENTIALS. 8. L: y = 3x − 2; M: y = − 43 x + 45 . P : (13/15, 3/5). N : y = − 43 x − 47 . See Fig. 2.5.8. 10. (a) 75 − 3P e = 20 + 2P e , and so P e = 11.. (b) P e = 90. 12. C = 45 y + 100. (The general equation is C = ay + b. Here 900 = a · 1000 + b, and a = 80/100 = 4/5, so b = 100.) 1 1 − x0 + h x0 . Multiplying both numerator and denominator by the product (x0 + h)x0 , then 14. The slope is x0 + h − x 0 simplifying, yields the given expression. 16. See Figs. 2.5.16(a)–(c). y. L. 1. y. y. y. 1. 1. 1. x. 1. x. 1. x. 1. 1. x. M N Figure 2.5.8. Figure 2.5.16(a). Figure 2.5.16(b). Figure 2.5.16(c). Chapter 3 Polynomials, Powers, and Exponentials 3.1 2. (a). −4. −3. −2. −1. 0. 1. 2. −2.5. 0. 1.5. 2. 1.5. 0. −2.5. x f (x). (b) See Fig. 3.1.2(b). (c) f (x) = (−1/2)(x + 1)2 + 2. Maximum point (−1, 2). (d) x = −3 and x = 1 (e) f (x) > 0 in the interval (−3, 1), while f (x) < 0 for x < −3 and for x > 1. y 3 2 f (x) = − 21 x 2 − x + 1 −4 −3 −2−1 −1 −2 −3 −4 Figure 3.1.2(b). © Knut Sydsæter and Peter Hammond 2010. 1 2 3 x. 3 2.

(9) CHAPTER 3. POLYNOMIALS, POWERS, AND EXPONENTIALS. 7. 4. (a) x(x + 4). Zeros 0 and −4. (b) Factorization not possible. No zeros. √ √ (c) −3(x − x1 )(x − x2 ), where the zeros are x1 = 5 + √15 and x2 = 5 − √ 15. (d) 9(x − x1 )(x − x2 ), where the zeros are x1 = 1/3 + 5 and x2 = 1/3 − 5. (e) −(x + 300)(x − 100), where the zeros are −300 and 100. (f) (x + 200)(x − 100), where the zeros are −200 and 100. 6. (a) If the other side is y, then 2x + 2y = L, and so y = L/2 − x. The area is then A(x) = x(L/2 − x) = Lx/2 − x 2 . The square with side L/4 gives the largest area. (b) Yes; the radius is L/2π so the area is L2 /4π > L2 /16. 8. (a) x = ± 1, x = ± 2 (b) (i) x 2 = 9 (x 2 < 0 is impossible), so x = ± 3 (ii) x 3 = 1 or x 3 = 8, so x = 1 or x = 2 10. y = 2x 2 + x − 6. ((1, −3) belongs to the graph if −3 = a + b + c, (0, −6) belongs to the graph if −6 = c, and (3, 15) belongs to the graph if 15 = 9a + 3b + c. It follows that a = 2, b = 1, and c = −6.) 12. (a) (i) 289 ≤ 290 (ii) 361 ≤ 377 (b) If B 2 − 4AC were > 0, then according to formula [3.2] the equation f (x) = 0 would have two distinct solutions, which is impossible when f (x) ≥ 0 for all x. We find that A = a12 + a22 + · · · + an2 , B = 2(a1 b1 + a2 b2 + · · · + an bn ), and C = b12 + b22 + · · · + bn2 , so the conclusion follows.. 3.3 2. (a) Integer roots must divide 6. Thus ±1, ±2, ±3, and ±6 are the only possible integer solutions. We find that −2, −1, 1, 3 all are roots, and since there can be no more than 4 roots in a polynomial equation of degree 4, we have found them all. (b) The same possible integer solutions. Only −6 and 1 are integer solutions. (The third root is −1/2.) (c) Neither 1 nor −1 satisfies the equation, so there are no integer roots. (d) First multiply the equation by 4 to have integer coefficients. Then ±1, ±2, and ±4 are seen to be the only possible integer solutions. In fact, 1, 2, −2 are all solutions. 4. (a) The answer is 2x 2 + 2x + 4 + 3/(x − 1), because + 2x − 1) ÷ (x − 1) = 2x 2 + 2x + 4 (2x 3 3 2 2x − 2x 2x 2 + 2x − 1 2x 2 − 2x 4x − 1 4x − 4 3 (b) The answer is. x2. remainder. + 1, because (x 4 + x 3 + x 2 + x) ÷ (x 2 + x) = x 2 + 1 x4 + x3 x2 + x x2 + x 0. © Knut Sydsæter and Peter Hammond 2010. no remainder.

(10) 8. CHAPTER 3. POLYNOMIALS, POWERS, AND EXPONENTIALS. (c) The answer is 3x 5 + 6x 3 − 3x 2 + 12x − 12 + (28x 2 − 36x + 13)/(x 3 − 2x + 1): x2. + 1) ÷ (x 3 − 2x + 1) = 3x 5 + 6x 3 − 3x 2 + 12x − 12. x2. + 1. − 3x 5 + 12x 4 − 6x 3 + x 2 + 6x 3 − 3x 2 − 3x 5. + 1. (3x 8 3x 8 − 6x 6 + 3x 5 6x 6 − 3x 5 + − 12x 4 + 6x 3 6x 6. 12x 4 − 12x 3 + 4x 2 + 1 − 24x 2 + 12x 12x 4 − 12x 3 + 28x 2 − 12x + 1 + 24x − 12 − 12x 3 28x 2 − 36x + 13. remainder. (d) The answer is x 3 − 4x 2 + 3x + 1 − 4x/(x 2 + x + 1), because (x 5 − 3x 4 x5 + x4 + x3. + 1) ÷ (x 2 + x + 1) = x 3 − 4x 2 + 3x + 1. − 4x 4 − x 3 − 4x 4 − 4x 3 − 4x 2. +1. 3x 3 + 4x 2 +1 3 2 3x + 3x + 3x x 2 − 3x + 1 x2 + x + 1 − 4x. remainder. 6. (a) p(x) = x(x 2 + x − 12) = x(x − 3)(x + 4), because x 2 + x − 12 = 0 for x = 3 and x = −4. (b) ±1, ±2, ±4, ±8 are the only possible integer zeros. By trial and error we find that q(2) = q(−4) = 0, so 2(x − 2)(x + 4) = 2x 2 + 4x − 16 is a factor for q(x). By polynomial division we find that q(x) ÷ (2x 2 + 4x − 16) = x − 1/2, so q(x) = 2(x − 2)(x + 4)(x − 1/2).. 3.4 2. (a) 2.511886 (b) 0.000098 (c) 0.530094 √ 16 100 50 (Find first the number that, when raised to the power of 100, yields 50. Then this 4. 500.16 = number is raised to the power of 16.) 1 6. (a) 23 = 8, so x = 3/2 (b) 81 = 3−4 , so 3x + 1 = −4, and then x = −5/3. 2 2 (c) x − 2x + 2 = 2, so x − 2x = 0, implying that x = 0 or x = 2.. 8. (a) Multiply by 5K 1/2 to obtain K 1/2 = 15L1/3 . Squaring each side, K = 225L2/3 . (b) abx0b−1 = p, so x0b−1 = p/ab. Now raise each side to the power 1/(b − 1). −1/ρ b (c) x = − 2a . (Multiply each side by (ax + b)2/3 .) (d) b = λ1/ρ c−ρ − (1 − λ)a −ρ 10. K = 57 315.86 for Y = 100, L = 6, t = 10. © Knut Sydsæter and Peter Hammond 2010.

(11) CHAPTER 3. POLYNOMIALS, POWERS, AND EXPONENTIALS. 9. 3.5 ∗. (b) The doubling time t ∗ is given by the equation (1.034)t = 2, and we. 2. (a) P = 1.22 · 1.034t million find t ∗ ≈ 20.7 (years). 4. (a) 210 = 1024. (b) f (t) = 2t. (c) f (20) = (210 )2 = (1024)2 > 10002 = 1 million.. 6. The graph is shown in Fig. 3.5.6. x 2. x2. −2. −1. 0. 1. 2. 16. 2. 1. 2. 16. 6 5. y. 4 y 6. 3 2 1. y. 4 −2 −1 −1. 1 2. y = x 2 2x. 3 x 2. −2 −3 −4. 1 1. x. Figure 3.5.6. −5 −4 −3 −2 −1. Figure 3.5.10. 1. 2 x. Figure 3.5.12. ∗. 8. If the initial time is t, the doubling time t ∗ is given by the equation Aa t+t = 2Aa t , which implies ∗ ∗ Aa t a t = 2Aa t , so a t = 2, independent of t. 10. (a) The graph is shown in Fig. 3.5.10.. x. −3. −2. −1. 0. 1. 2. 3. 4. 1 − 2−x. −7. −3. −1. 0. 1/2. 3/4. 7/8. 15/16. (b) 1 − 2−x gets close to 1 as x becomes very large, and becomes very large negative as x becomes large negative. (Formally, limx→∞ f (x) = 1, limx→−∞ f (x) = −∞. See Section 6.1.) 12. The graph is drawn in Fig. 3.5.12. x. −10. −5. −4. −3. −2. −1. 0. 1. 2. x 2 2x. 0.1. 0.8. 1.0. 1.1. 1.0. 0.5. 0. 2.0. 16. 14. I (t) = I0 (1/2)t/8 remains after t days. © Knut Sydsæter and Peter Hammond 2010.

(12) 10. CHAPTER 4. SINGLE VARIABLE DIFFERENTIATION. 3.6 2. The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today to his/her mother. (Though the youngest child of a mother with several children will have been different at different dates.) The function in (d) is one-to-one and has an inverse: the rule mapping the surface area to the volume. The function in (e) is one-to-one and has an inverse: the rule that maps (u, v) to (u − 3, v). The other functions are many-to-one, in general, and so have no inverses.. Chapter 4 Single Variable Differentiation 4.2 2. f (x) = 6x + 2, f (0) = 2, f (−2) = −10, f (3) = 20. The tangent equation is y = 2x − 1. 4.. f (x + h) − f (x) 1/(x + h) − 1/x x − (x + h) −h −1 1 = = = = → − 2 as h → 0, h h hx(x + h) hx(x + h) x(x + h) x which proves the implication.. 6. (a) f (x + h) − f (x) = a(x + h)2 + b(x + h) + c − (ax 2 + bx + c) = 2ahx + bh + ah2 , so [f (x + h) − f (x)]/ h = 2ax + b + ah → 2ax + b as h → 0. Thus f (x) = 2ax + b. (b) f (x) = 0 for x = −b/2a. The tangent is parallel to the x-axis at the minimum/maximum point. √ √ 8. (a) Use [A.12] in Section A.3. (b) and (c) [f (x + h) − f (x)]/ h = x + h − x)/ h. Multiplying both √ √ numerator and denominator by x + h + x , then using the identity in (a), yields the result. Letting √ √ h → 0, the formula follows. (d) x = x 1/2 and 1/ x = x −1/2 . 10. (a) [f (x + h) − f (x)]/ h = [(x + h)1/3 − x 1/3 ]/ h. Then use the hint. (b) Follows from (a) by letting h → 0.. 4.3 2. I is the fixed cost, whereas k is the marginal cost, and also the incremental cost of producing one additional unit. 4. T (y) = t, so the marginal tax rate is constant.. 4.4 2. (a) The following table seems to indicate that the limit is 9. x. 0.9. 0.99. 0.999. 1. 1.001. 1.01. 1.1. x 2 +7x−8 x−1. 8.9. 8.99. 8.999. ∗. 9.001. 9.01. 9.1. *not defined (b) x 2 + 7x − 8 = (x − 1)(x + 8), so for x = 1,. 4. (a) 22 + 3 · 2 − 5 = 5 6. (a) 4. (b) 5. (c) 6. (b) 1/5. (d) 2a + 2. © Knut Sydsæter and Peter Hammond 2010. (c) 1. x 2 + 7x − 8 = x + 8 → 9 as x → 1. x−1. (d) −2. (e) 2a + 2. (e) 3x 2. (f) 4a + 4. (f) h2 8. (a) 1/6. (b) 1/27. (c) n.

(13) CHAPTER 4. SINGLE VARIABLE DIFFERENTIATION. 11. 4.5 2. (a) 2g (x) (b) (−1/6)g (x) (c) (1/3)g (x) 6. Add an arbitrary constant to these answers:. (b) A(b + 1)y b (c) (−5/2)A−7/2 x a+1 (b) x 2 + 3x (c) a+1. 4. (a) 8π r (a) (1/3)x 3. 4.6 2. (a) (6/5)x − 14x 6 − (1/2)x −1/2. (b) 4x(3x 4 − x 2 − 1). (c) 10x 9 + 5x 4 + 4x 3 − x −2. 3 −x 4 + 5x 2 + 18x + 2 4. (a) √ √ (b) (c) −2(1+2x)x −3 (x 2 + 2)2 (x + 3)2 2 x( x + 1)2 2(x 3 − x 2 + x + 1) (f) 3x 3 (x + 1)2 √ √ √ √ 6. (a) x = 2 (b) x = − √ 3, x = 0, x = 3 (c) x = − 2, x = 2 √ (d) x = 0, x = −1/2 − 5/2, x = −1/2 + 5/2 8. (a). ad − bc (ct + d)2. (b) a(n + 1/2)t n−1/2 + nbt n−1. (c). (d). 4x (x 2 + 1)2. (e). −2x 2 + 2 (x 2 − x + 1)2. −(2at + b) (at 2 + bt + c)2. 10. Differentiating f (x) · f (x) = x gives f (x) · f (x) + f (x) · f (x) = 1, so 2f (x) · f (x) = 1. Hence, 1 1 f (x) = = √ . 2f (x) 2 x 12. The Newton quotient of F is f (x + h)/g(x + h) − f (x)/g(x) F (x + h) − F (x) = h h Multiplying the numerator and denominator by g(x)g(x+h) yields the expression for the Newton quotient given in the hint. Letting h → 0 yields the desired result.. 4.7 2. dy/dx = ax a−1 − ax −a−1 and d 2 y/dx 2 = a(a − 1)x a−2 + a(a + 1)x −a−2 4. g (t) = (t 2 − 2t)/(t − 1)2 , so g (t) = [(2t − 2)(t − 1)2 − (t 2 − 2t)2(t − 1)]/(t − 1)4 = 2 for t = 2. 6. For n = 1, the formula is true. Suppose it is true for n = k, so that y = x k ⇒ y (k) = k! . Then (d k+1 /dx k+1 )x k+1 = (d k /dx k )(d/dx)(x k+1 ) = (d k /dx k )(k +1)x k = (k +1)(d k /dx k )x k = (k +1)·k!, by the induction hypothesis. This is equal to (k + 1)!, so the given formula is also true for n = k + 1. By induction, the formula is true for all n.. © Knut Sydsæter and Peter Hammond 2010.

(14) 12. CHAPTER 5. MORE ON DIFFERENTIATION. Chapter 5 More on Differentiation 5.1 2. (a) 3(2x + 1)2 · 2 = 6(2x + 1)2 (b) −5(1 − x)4 (c) 2(x 2 − 2x + 2)(2x − 2) (d) (x + 1)4 (4x − 1)/x 2 (e) −21(3x − 4)−8 (f) −2(4x + 3)(2x 2 + 3x − 4)−3. −b(a + 1) at + b a 4. (a) −6at (at 2 + 1)−4 (b) na(at + b)n−1 (c) nt 2 nt 6. dx/dt = Ar(Ap + B)r−1 (2at + b) 8. Putting y = 1/(g(x))n implies that y = [0·(g(x))n −1·n(g(x))n−1 g (x)]/(g(x))2n = −n(g(x))−n−1 g (x).. k+1 = 10. For a = 1, the formula is true. Suppose the formula is true for a = k. Consider y = g(x). k g(x) · g(x). The product rule for differentiation and the induction hypothesis yield. k−1 . k. k y = k g(x) · g (x) · g(x) + g(x) · g (x) = (k + 1) g(x) · g (x), which is the given formula for a = k + 1. By induction, the formula is true for all natural numbers a.. 5.2 2. (a) dY /dt = (dY /dV )(dV /dt) = (−3)5(V + 1)4 t 2 = −15(t 3 /3 + 1)4 t 2 (b) dK/dt = (dK/dL)(dL/dt) = AaLa−1 b = Aab(bt + c)a−1 4. dY /dt = (dY /dK) · (dK/dt) = Y (K(t0 ))K (t0 ) √ 1 dx a √ ap − c = b − u, with u = ap − c. Then = − √ u = − √ . (The restriction dp 2 ap − c 2 u should be p > c/a, for x to be differentiable.) 8. (a) b(t) is the total fuel consumption after t hours. (b) b (t) = B s(t) s (t). So the rate of fuel consumption per hour is equal to the rate per kilometer multiplied by the speed in kph.. 6. x = b −. 10. f (f (x)) = f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. Then 9x + 28 = 100 for x = 8. 14. p (x) = 2(x − a)q(x) + (x − a)2 q (x), so p (a) = 0. 12. dC/dx = q(25 − 21 x)−1/2 16. F (x) = f x n g(x) nx n−1 g(x) + x n g (x). 5.3 2. By implicit differentiation, 4x + 6y + 6xy + 2yy = 0. So y = − at the point (1, 2).. 2x + 3y . In particular, y = −8/5 3x + y. 4. (a) 2x + 2yy = 0, and solve for y to get y = −x/y. √ √ √ (b) 1/2 x + y /2 y = 0, and solve for y to get y = − y/x. (c) 4x 3 − 4y 3 y = 2xy 3 + x 2 3y 2 y , and so y = 2x(2x 2 − y 3 )/y 2 (3x 2 + 4y). 6. (a) No relationship.. (b) f (x0 ) = g (x0 ). 8. Differentiating y n = x m w.r.t. x yields ny n−1 y = mx m−1 , so y = mx m−1 /ny n−1 = mx m−1 /n(x m/n )n−1 = (m/n)x m−1−(m/n)(n−1) = (m/n)x (m/n)−1 . © Knut Sydsæter and Peter Hammond 2010.

(15) CHAPTER 5. MORE ON DIFFERENTIATION. 13. 5.4 2. f (x) ≈ f (0) + f (0)x =. 1 9. −. 10 27 x. 4. F (1) = A, F (K) = αAK α−1 , so F (1) = αA. Then F (K) ≈ F (1) + F (1)(K − 1)= A + αA(K − 1) = A(1 + αA(K − 1)). 6. (a) (i) y = 0.61, dy = 0.6 (ii) y = 0.0601, dy = 0.06 (b) (i) y = 0.011494, dy = 0.011111 (ii) y = 0.001124, dy = 0.001111 (c) (i) y = 0.012461, dy = 0.0125 (ii) y = 0.002498, dy = 0.0025 8. g(0) = A − 1 and g (μ) = Aa/(1 + b) (1 + μ)[a/(1+b)]−1 , so g (0) = Aa/(1 + b). Hence, g(μ) ≈ aA g(0) + g (0)μ = A − 1 + μ 1+b. 5.5 2. Follows from formula [5.9] with f = U , a = y, x = y + M − s. 4. We find x(0) ˙ = 2[x(0)]2 = 2 · 1 = 2. Differentiating the expression for x(t) ˙ yields x(t) ¨ = x(t) + t x(t) ˙ + 2 = 4[x(t)]x(t), ˙ and so x(0) ¨ = x(0)+4[x(0)]x(0) ˙ = 1+4·1·2 = 9. Hence, x(t) ≈ x(0)+ x(0)t ˙ + 21 x(0)t ¨ 1 + 2t + 29 t 2 . (px p−1 − qx q−1 )(x p + x q ) − (x p − x q )(px p−1 + qx q−1 ) 2(p − q)x p+q−1 = , so h (1) = (x p + x q )2 (x p + x q )2 1 1 2 (p − q). Since h(1) = 0, h(x) ≈ h(1) + h (1)(x − 1) = 2 (p − q)(x − 1).. 6. h (x) =. 5.6 2. ElK T = 1.06. A 1% increase in expenditure on road building leads to an increase in the traffic volume of approx. 1.06 %. 4. (a) Elx f (x) = 0. (b) Elx f (x) = x/(x + 1) (c) Elx f (x) = −20x 2 /(1 − x 2 ) f (x)Elx f (x) 6. Elx Af (x) = Elx f (x), Elx A + f (x) = A + f (x) 1 + 2x 2x 2 30x 3 (c) 3 (d) Elx 5x 2 = 2, so Elx Elx 5x 2 = 0 (e) 1+x x +1 1 + x2 . x−1 x 5x 5 x Elx x x 5 Elx x 5 (f) Elx = Elx (x − 1) − Elx (x 5 + 1) = − 5 = − 5 (Some answers 5 x +1 x−1 x +1 x−1 x +1 are easier if the results of Problem 6 are used as well.). 8. (a) −3. (b). 10. (a) 6Elx y = 5, so Elx y = 5/6 (b) Using rules (c) and (b) in Problem 7, we obtain Elx y − Elx x = Elx (x + 1)a + Elx (y − 1)b . Here Elx x = 1, whereas Elx (x + 1)a = [x/(x + 1)a ]a(x + 1)a−1 = ax/(x + 1) Elx (y − 1)b = [x/(y − 1)b ]b(y − 1)b−1 y = [by/(y − 1)]Elx y Thus, Elx y − 1 = ax/(x + 1) + [by/(y − 1)]Elx y. Solving for Elx y yields (ax + x + 1)(y − 1) Elx y = (y − by − 1)(x + 1) © Knut Sydsæter and Peter Hammond 2010.

(16) 14. CHAPTER 6. LIMITS, CONTINUITY, AND SERIES. Chapter 6 Limits, Continuity, and Series 6.1 x−3 (1/x) − (3/x 2 ) 2. (a) 2 = −→ x +1 1 + (1/x 2 ) x→∞.

(17) (b). 2 + 3x = x−1. √ 3 + 2/x −→ 3 1 − 1/x x→∞. (c) a 2. 4. limx→∞ fi (x) = ∞ for i = 1, 2, 3; limx→∞ f4 (x) = 0. Then: (a) ∞ (b) 0 (c) −∞ (d) 1 (e) 0 (f) ∞ (g) 1 (h) ∞ 6. y = Ax + A(b − c) + d is an asymptote as x → ∞.. 6.2 2. (a) None of the six functions is continuous at a. (b) Only in (i) does f have a limit: limx→a f (x) = A (c) (i) limx→a − f (x) = limx→a + f (x) = limx→a f (x) = A, (ii) limx→a − f (x) = f (a), limx→a + f (x) = A; (iii) limx→a − f (x) = limx→a + f (x) = ∞; (iv) limx→a − f (x) = −∞, limx→a + f (x) = f (a); (v) limx→a − f (x) = ∞, limx→a + f (x) = f (a); (vi) limx→a − f (x) = A, limx→a + f (x) = B. (d) Only the function in (ii) is left-continuous at x = a. The functions in (iv), (v), and (vi) are right-continuous at x = a. (e) (v) limx→∞ f (x) = A; (vi) limx→∞ f (x) does not exist. 4. (a) Continuous for all x. (b) Continuous for all√ x = 1. (c) Continuous √ for all x < 2. (d) Continuous for all x. (e) Continuous for all x where x = 3 − 1 and x = − 3 − 1. (f) Continuous for all x where x ≤ −1 or x > 1. (g) Continuous for all x > 0. (h) Continuous for all x = 0. (i) Continuous for all x > 0. 6. See Fig. 6.2.6; y is discontinuous at x = a. In the case where y is the nearest point on the ground, it would be a continuous function. y. a. x. Figure 6.2.6. 8. Because limx→0− f (x) = −2 and limx→2+ f (x) = 3, we need f (0) = −2 and f (2) = 3. Because f is linear on [0, 2], put f (x) = (5/2)x − 2 for x ∈ [0, 2].. 6.3 2. f (0+ ) = 1, f (0− ) = −1. At x = 0, f is continuous, but not differentiable. © Knut Sydsæter and Peter Hammond 2010.

(18) CHAPTER 6. 4. The tax function is. ⎧ ⎨ 0.15x t (x) = 0.28x − 2 632.50 ⎩ 0.31x − 4 111.50. LIMITS, CONTINUITY, AND SERIES. 15. for x ∈ [0, 20 250] for x ∈ (20 250, 49 300] for x ∈ (49 300, ∞). In particular, t (22 000) = 3527.50 and t (50 000) = 11388.50.. 6.4 2. (a) Converges to 5.. (b) Diverges (to ∞).. (c) √. 3n 2n2 − 1. =. √ 3 3 2 −→ √ = 2 2 − 1/n2 n→∞ 2 3. 6.5 2. (a) Geometric series with quotient 1/8. Its sum is 8/(1 − 1/8) = 64/7. (b) Geometric with quotient −3. It diverges. (c) Geometric, with sum 21/3 /(1 − 2−1/3 ). (d) Not geometric. (One can show that the series is convergent with sum ln 2.) 4. Geometric series with quotient (1 + p/100)−1 . Its sum is b/[1 − (1 + p/100)−1 ] = b(1 + 100/p). 6. The general term does not approach 0 as n → ∞ in any of these three cases, so each of the series is divergent. 8. sn = n/(n + 1) = 1/(1 + 1/n) → 1 as n → ∞. The infinite series converges to 1.. 6.6 2. Offer (a) is best. The second offer has present value 4600. 1 − (1.06)−5 ≈ 20 540. 1 − (1.06)−1. 12 000 · 1.115 [1 − (1.115)−8 ] ≈ 67 644.42. 0.115 (1.115)12 − 1 = 66, 384.08. Thus the contract The present value of the contract in (c) is 22, 000 + 7000 0.115(1.115)12 in (c) is the best in any case. When the interest rate becomes 12.5 %, contracts (b) and (c) have present values equal to 65907.61 and 64374.33, respectively.. 4. Schedule (b) has present value. 6. This is a geometric series with first term a = D/(1 + r) and quotient k = (1 + g)/(1 + r). It converges D/(1 + r) D a = = . iff k < 1, i.e. iff 1 + g < 1 + r, or g < r. The sum is 1−k 1 − (1 + g)/(1 + r) r −g. 6.7 2. |(x + 1)3 − 1| = |x 3 + 3x 2 + 3x + 1 − 1| ≤ |x|3 + 3|x|2 + 3|x| ≤ |x| + 3 + 3|x| + 3|x| = 7|x|. Let > 0. Choose δ = /7. Then according to [6.24], limx→0 (x + 1)3 = 1. Because f (0) = 1, f is continuous at x = 0. 4x 2 − 100 4. (a) 4x 2 − 100 = 4(x 2 − 25) = 4(x + 5)(x − 5), so for x = 5, = 4(x + 5). Hence, for x = 5, x−5 4x 2 − 100 − 40 = |4(x + 5) − 40| = 4|x − 5|. Let > 0. Choose δ = /4. According to [6.24], the x−5 conclusion follows. (b) |x 2 − π 2 | = |(x + π )(x − π )|, so for x = −π , |(x 2 − π 2 )/(x + π ) + 2π | = |x − π + 2π| = |x + π |. Let > 0. Choose δ = . According to [6.24], the conclusion follows. © Knut Sydsæter and Peter Hammond 2010.

(19) 16. CHAPTER 7. IMPLICATIONS OF CONTINUITY AND DIFFERENTIABILITY. Chapter 7 Implications of Continuity and Differentiability 7.1 2. For x = 0, we have f (x) = x 3 + ax 2 + bx + c = x 3 (1 + a/x + b/x 2 + c/x 3 ), which shows that f (x) is (large) positive if x is sufficiently large, and that f (x) is (large) negative if x is sufficiently large and negative. Hence Theorem 7.2 implies that f (x) = 0 has a solution. A similar argument applies to the general case, when the order n of the polynomial is odd. If n is even, there may be no real roots. For instance, x 2 + 1 = 0 has no real roots. 4. Recall that, arbitrarily close to any given real number, there are rational as well as irrational numbers. The function f is continuous at a = 0, because |f (x) − f (0)| = |f (x) − 0| = |f (x)| ≤ |x| for any x, so f (x) → f (0) as x → 0. If a = 0 is rational, then |f (x) − f (a)| = |f (x)|, which is equal to |x| when x is irrational. But if a = 0 is irrational, then |f (x) − f (a)| = |f (a)| whenever x is rational. In either case, f (x) does not approach 0 as x approaches a. It follows that f is discontinuous for all x = 0. We prove that g is discontinuous by using the εδ definition of continuity. Let a be an arbitrary number and choose ε = 21 . For each positive number δ, there is a rational number x1 and an irrational number x2 in the interval (a−δ, a+δ). Then g(x1 ) = 1 and g(x2 ) = 0. If a is rational, then |g(x2 )−g(a)| = |0−1| = 1, and if a is irrational, then |g(x1 ) − g(a)| = |1 − 0| = 1. In both cases there is a number x arbitrarily close to a for which |g(x) − g(a)| = 1 > ε. Hence g is discontinuous at a.. 7.2 2. (a) No, because f (x) can get arbitrarily close to 1, by letting x be sufficiently close to 1. But there is no value of x for which f (x) = 1. Similarly, there is no minimum, because f (x) can get arbitrarily close to −1 by letting x be sufficiently close to −1, yet there is no value of x for which f (x) = −1. y (b) No, not at x = −1 and x = 1. 1. yx1 y. y 1. y x 2 3. 3. 1. y x 3. 2. 1. 1 1. 1. x. Figure 7.3.2(a). 1. 2. 3. 4. 5. Figure 7.3.2(b). 1. 2. x. x Figure 7.3.2(c). 7.3 2. In (a), f is not differentiable at x = 0; in (b), f is not differentiable at x = 3; in (c), f is not even defined at x = 1. See Fig. 7.3.2(a)–(c). 4. There is at least one point where you must be heading in the direction of the straight line joining A to B (even if that straight line hits the shore).. 7.4. √ 3 2 2. (a) 25 = 3(1 − 2/27)1/3 ≈ 3(1 − 13 27 − √ 5 1 1/5 (b) 33 = 2(1 + 1/32) ≈ 2(1 + 5·32 −. © Knut Sydsæter and Peter Hammond 2010. 1 4 9 272 ) ≈ 2.924 2 1 25 322 ) ≈ 2.0125.

(20) CHAPTER 7. IMPLICATIONS OF CONTINUITY AND DIFFERENTIABILITY. 17. 4. g(0) = g(x) = 0 follows immediately from [1] and [2]. Because f is thrice continuously differentiable, S(x) and g(x) are differentiable in (0, x). The formula for g (t) follows easily, and the condition g (c) = 0 yields S(x) = f (c). The conclusion follows.. 7.5 x 4 − 4x 3 + 6x 2 − 8x + 8 “0” 4x 3 − 12x 2 + 12x − 8 “0” = = lim = = x→2 x→2 x 3 − 3x 2 + 4 0 3x 2 − 6x 0 12x 2 − 24x + 12 12 lim = =2 x→2 6x − 6 6 1/2 2(1 + x) − 2 − x (1 + x)−1/2 − 1 “0” “0” = = lim = = (b) lim 2 −1/2 x→0 2(1 + x + x 2 )1/2 − 2 − x x→0 0 (1 + 2x)(1 + x + x ) −1 0. 2. (a) lim. lim. x→0. − 21 (1 + x)−3/2 2(1 + x + x 2 )−1/2 + (1 + 2x)2 (− 21 )(1 + x + x 2 )−3/2. =−. 1 3. 1 − (1 + v β )−γ γ (1 + v β )−γ −1 βv β−1 “0” = = lim+ . If β = 1, then v→0 v→0 v 0 1 G = γ . If β > 1, then G = 0, and if β < 1, then G = ∞.. 4. G = lim+. f (x) f (1/t) “0” f (1/t)(−1/t 2 ) f (1/t) f (x) = lim+ = = lim+ = lim+ = lim 2 x→∞ g(x) x→∞ g (x) t→0 g(1/t) t→0 g (1/t)(−1/t ) t→0 g (1/t) 0. 6. lim 8.. f (x) 1/g(x) “0” −1/(g(x))2 g (x) = lim = · = lim x→a g(x) x→a 1/f (x) x→a −1/(f (x))2 f (x) 0 (f (x))2 g (x) g (x) 1 2 = lim · lim = L = L2 lim x→a (g(x))2 f (x) x→a f (x) x→a f (x)/g (x). L = lim. The conclusion follows. (Here, we have ignored problems with “division by 0”, when either f (x) or g (x) tends to 0 as x tends to a.). 7.6 2. p = (157.8/D)10/3 4. f (x) = x 2 = (−x)2 is not one-to-one on (−∞, ∞), and therefore has no inverse. On [0, ∞), f is √ strictly increasing and has the inverse g(x) = x. 6. f (x) = 7x 6 + 25x 4 + 2 > 0 for all x, so f has an inverse g. g (−2) = 1/f (0) = 1/2. 8. (a) See Fig. 7.6.8(a).. (b) Triangles OBA and OBC in Fig. 7.6.8(b) are congruent.. 10. See Fig. 7.6.10. (This example shows that the commonly seen statement: “if the inverse function exists, the original and the inverse function must both be monotonic” is wrong. This claim is correct for continuous functions, however.) 12. Differentiating f (g(x)) = x yields (∗) f (g(x) g (x) = 1, so g (x) = −1/f (g(x) . Differentiating (∗) yields f g(x) g (x)g (x) + f g(x) g (x) = 0. Solving for g (x) gives 3 2 g (x) = −f g(x) g (x) /f (g(x) = −f g(x) / f g(x) . The conclusions follow. © Knut Sydsæter and Peter Hammond 2010.

(21) 18. CHAPTER 8. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. y. y y=x D. y y=x. C = (b, a). (3, 5) B (1, 3). (5, 3) A = (a, b). (3, 1) O. x Figure 7.6.8(a). E. x. x. Figure 7.6.8(b). Figure 7.6.10. −1/2 > 0 for all x > 1. Hence, f is strictly increasing in [1, ∞), and has an 14. f (x) = 21 (x+1)−1/2 + 21 (x−1) √ √ inverse g. Because f (1) = 2 and f (x) → ∞ as x → ∞, the range of f is [ 2, ∞). To find a formula √ √ √ √ for g, let (1) y = x + 1 + x − 1. Interchanging x and y, we have (2) x = y + 1 + y − 1. The √ √ √ next step is to solve (2) for y, with x ∈ [ 2, ∞) and y ∈ [1, ∞). We obtain x = y + 1 + y − 1 ⇐⇒ √ √ √ √ x − y + 1 = y − 1 ⇒ (x − y + 1 )2 = y − 1 ⇐⇒ x 2 − 2x y + 1 + y + 1 = y − 1 ⇐⇒ √ 2x y + 1 = x 2 + 2 ⇒ 4x 2 (y + 1) = x 4 + 4x 2 + 4 ⇐⇒ y = 41 x 2 + √ 1/x 2 . √ √ 2 2 Thus, x = y + 1 + y − 1 with y ≥ 1 implies y = x /4 + 1/x , x ≥ 2. We prove the converse by substitution: √ √ 2 2 2 2 4 2 +4 x 4 −4x 2 +4 y + 1 + y − 1 = x4 + x12 + 1 + x4 + x12 − 1 = x +4x + = x 2x+2 + x 2x−2 = x. 2 4x 4x 2 √ We conclude that g(x) = 41 x 2 + 1/x 2 , x ∈ [ 2, ∞) is the required inverse function.. Chapter 8 Exponential and Logarithmic Functions 8.1 x. 2. (a) ee ex = ee. x +x. (b) 21 (et/2 − e−t/2 ) (c) −. et − e−t (et + e−t )2. (d) z2 ez (ez − 1)−2/3 3. 3. 4. p (x) = kce−cx , p (x) = −kc2 e−cx . The graph is shown in Fig. 8.1.4. y 3 2 y = 21 (ex + e−x ). y = 21 (ex − e−x ). 1. p a+k. -3. -2. -1. 1 -1. p(x) = a + k(1 − e−cx ). -2 a. -3 x. Figure 8.1.4. © Knut Sydsæter and Peter Hammond 2010. Figure 8.1.6. 2. 3. x.

(22) CHAPTER 8. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. 19. 6. The graphs are in Fig. 8.1.6. Verifying the identities is straightforward. We prove only equality (a): cosh x cosh y + sinh x sinh y = 21 (ex + e−x ) 21 (ey + e−y ) + 21 (ex − e−x ) 21 (ey − e−y ) = 1 x+y + ex−y + e−x+y + e−x−y + ex+y − ex−y − e−x+y + e−x−y ) = 21 (ex+y + e−x−y ) = cosh(x + y) 4 (e 8. f (z + x) = a z+x = a z a x = f (z)f (x). Differentiating w.r.t. z while x is fixed gives f (z + x) = f (z)f (x). Putting z = 0 yields f (x) = f (0)f (x), which is [8.2].. 8.2 2. (a) ln 3x = x ln 3 = ln 8, so x = ln 8/ ln 3. (b) x = e3 (c) x 2 − 4x + 5 = 1, or√x 2 − 4x + 4 = 0, √ so (x − 2)2 = 0. Thus x = 2. (d) x(x − 2) = 1, so x = 1 − 2 or x = 1 + 2. (e) x = 0 or √ ln(x + 3) = 0, so x = 0 or x = −2. (f) x − 5 = 1, so x = 36. √ 4. (a) t = (ln x − b)/a (b) t = (ln 2)/a (c) t = ± ln(8/ 2π ) = ± 25 ln 2 − 21 ln π 6. (a) True (π e ≈ 22.5, eπ ≈ 23.1) (b) True because (a) gives (π e )1/eπ < (eπ )1/eπ . 8. (a) Wrong. (Put A = B = C = 1.) (b) Correct by rule [8.7] (b). (c) Correct. (Use [8.7] (b) twice.) (d) Wrong. (If A = e and p = 2, then the equality becomes 0 = ln 2.) (e) Correct by [8.7](c). (f) Wrong. (Put A = 2, B = C = 1.) 10. (a) 1/(x + 1). (b) 1/x. (c) ln x + 1. (d). ln x − 1 (ln x)2. 12. (a) (i) y = x − 1 (ii) y = 2x − 1 − ln 2 (iii) y = x/e (b) (i) y = x (ii) y = 2ex − e (iii) y = −e−2 x − 4e−2 14. (a) Let f (x) = ex − (1 + x + x 2 /2). Then f (0) = 0 and f (x) = ex − (1 + x), which is positive for all x > 0, as shown in the problem. Hence f (x) > 0 for all x > 0, and the inequality follows. (b) Let f (x) = ln(1 + x) − 21 x. Then f (0) = 0 and f (x) = 1/(x + 1) − 21 = (1 − x)/2(x + 1) > 0 in (0, 1), so ln(1 + x) > 21 x in (0, 1). To prove the other inequality, let g(x) = x − ln(1 + x). Then g(0) = 0 and g (x) = 1 − 1/(x + 1) = x/(1 + x) > 0 in (0, 1), so x > ln(1 + x) in (0, 1). (c) Let f (t) = ln[(1 + t)/(1 − t)] − 2t. Then f (0) = 0 and f (t) = 2/[(1 + t)(1 − t)] − 2 = 2t 2 /(1 − t 2 ) > 0 in (0, 1), so the required inequality follows. 16. (a) x − x = 0. (b) 4 ln x − x. (c) x 2 /y 2. 18. (a) x. (b) 1/ ln x. (c) x ln a. √ √ ln x + 2 20. (a) ln y = x ln x implies y /y = 21 x −1/2 ln x + x −1/2 so y = x x √ 2 x √ (b) 21 ( x )x (ln x + 1) (c) Differentiate ln ln y = x ln x + ln ln x to get. 1 x y /(y ln y) = ln x + 1 + 1/(x ln x) and so y = x x +x (ln x)2 + ln x + x . 22. First take logarithms to obtain the expression: ln a + (v/α)[α ln N + α ln K − ln(N α + bK α ) = B − (v/α) ln(N α + bK α ) for ln F (α), where B does not depend on α. Differentiation yields:. N α ln N + bK α ln K −2 α α F (α) = vα F (α) ln(N + bK ) − α N α + bK α. 8.3. 2. (a) x = 4. (b) x = e. (c) x = 1/27. © Knut Sydsæter and Peter Hammond 2010. (d) x = −1050 and x = 1050.

(23) 20. CHAPTER 8. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. 4. (a) ex+1−4/x = e1 , so x + 1 − 4/x = 1, hence x = ±2. (b) Putting u = ln(x + e), the equation becomes u3 − 4u2 − u + 4 = 0, so (u − 4)(u2 − 1) = 0, implying that u = ±1 or 4. Because x = eu − e, one has x = 0, x = −e + 1/e, or x = e4 − e. 6. (a) 1 (b) Does not exist. (c) 4/5 (d) 1 (x 1/x = (eln x )1/x = eln x/x , and [8.19].) (e) 0. (x ln x = ln x/(1/x), and then l’Hôpital’s rule.) (f) 1. (x x = (eln x )x = ex ln x , and then use (e).) 4 3 8. (a) xex ≈ x + x 2 + 21 x 3 (b) √e2x √ ≈ 1 + √2x + 2x 2 + 3x √ √ 1 9 2 1 3 1 3 1 1 + 2 x + 8 x + 48 x (d) ex + 1 ≈ 2 + 4 2x + 32 2x 2 + 384 (7 2)x 3. 1. (c) x 2 + e 2 x. ≈. 10. For x = 0, one has f (x) = x −3 2e−1/x , f (x) = x −6 (−6x 2 + 4)e−1/x , and therefore f (x) = 2 2 x −9 (24x 4 − 36x 2 + 8)e−1/x . Hence, f (k) = x −3k pk (x)e−1/x is correct for k = 1, 2, 3. The general formula is proved by induction. (Actually, all we need in the following is the fact that pk (x) is a polynomial. 2 Its degree is of no importance.) Now, f (0) = limx→0 [f (x) − f (0)]/x = limx→0 e−1/x /x = 0 √ 2 because, in general, (∗) e−1/x /x p → 0 as x → 0 for any natural number p. (Substitute x = 1/ t and use [8.18].) The general case is proved by induction: Suppose that f (k) (0) = 0 for an arbitrary natural number k. Then, by the definition of the derivative, f (k+1) (0) = limx→0 [f (k) (x) − f (k) (0)]/x = 2 2 limx→0 x −3k pk (x)e−1/x /x = limx→0 x −3k−1 pk (x)e−1/x = 0, using (∗) again. (Note that pk (x), as a polynomial, approaches a constant as x → 0.) 2. 2. 8.4 2. (a) S(t) = S0 e−at. (b) t = (ln 2)/a is the time it takes for sales to halve.. 4. k = 0.1 ln(705/641) ≈ 0.0095. P (15) ≈ 739, P (40) ≈ 938. 6. (a) In 1950 there were about 276 thousand. In the next 10 years the number increased by 155 thousand. (b) y → 479.36 as t → ∞. The graph is sketched in Fig. 8.4.6. lny. Tractors (in 1000). 8. 500 6 400 4. 300 200. 2 100 0 0 (1950). 5. 10 (1960). 15. Figure 8.4.6. 20 (1970). t. 2. 4. 6. 8. lnx. Figure 8.4.10. 8. h = −f K/f 2 = −rf (1 − f/K)K/f 2 = −r(K/f − 1) = −rh. Hence h = Ae−rt for some constant A. But then −1 + K/f = Ae−rt , and solving for f yields [8.24]. © Knut Sydsæter and Peter Hammond 2010.

(24) CHAPTER 9. SINGLE-VARIABLE OPTIMIZATION. 21. 10. (a) See Fig. 8.4.10 and the following table: ln x. 3.00. 3.69. 4.09. 4.38. 4.61. 4.79. 4.94. 5.08. 5.19. 5.30. ln y. 7.44. 6.31. 5.44. 4.79. 4.32. 4.09. 3.81. 3.56. 3.22. 3.00. (b) ln y = ln A+a ln x. We see from the graph in Fig. 8.4.10 that a ≈ −2. Assuming that the graph passes through (say) (ln x, ln y) = (5.19, 3.22), we obtain 3.22 = ln A + (−2)5.19, implying that ln A = 13.6, so A = e13.6 ≈ 806 129. The required formula is y = 806 129x −2 . 12. (a) y = 2.81x 0.83. (b) a =. ln y2 − ln y1 , A = y1 x1−a ln x2 − ln x1. 14. Take ln of each side and solve for t0 .. 8.5 2. (a) A = 4, k = 0.25. The price after 5 years is f (5) = 4e1.25 ≈ 14. (b) Price controls are needed after ≈ 6 years. The doubling time before price controls is 2.8 years, after price controls it is 7.3 years.. Chapter 9 Single-Variable Optimization 9.2. √ 3x)(2 + 3x) 2. = . Note that h(x) → 0 as x ± ∞. The function has a maximum at (3x 2 + 4)2 √ √ √ √ √ √ x = 2 3/3 and a minimum at x = −2 3/3, with h(2 3/3) = 2 3/3 and h(−2 3/3) = −2 3/3. h (x). 8(2 −. √. 4. (a) f (x) = [4x(x 4 + 1) − 2x 2 4x 3 ]/(x 4 + 1)2 , then simplify and factor. (b) f has maximum 1 at x = 1, because f (x) increases in [0, 1] and decreases in [1, ∞). We see that f (−x) = f (x) for all x, so f (x) is symmetric about the y-axis. On (−∞, ∞), f has maximum 1 at x = −1 and at x = 1. 6. (a) f (x) = −48x(3x 2 + 4)−2 , so f (x) > 0 for x < 0 and f (x) < 0 for x > 0. Hence, f has a maximum at x = 0. (b) g (x) = −2(x − 2), so g (x) > 0 for x < 2 and g (x) < 0 for x > 2. Hence, g has a maximum at x = 2. (c) h (x) = 20(x + 2)3 , so h (x) < 0 for x < −2 and h (x) > 0 for x > −2. Hence, h has a minimum at x = −2. (d) F (x) = 4x/(2 + x 2 )−2 , so F (x) < 0 for x < 0 and √ F (x) > 0 for x > 0. Hence, F has a minimum at x = 0. (e) G (x) = 1/2 1 − x > 0 for x < 1, so G has a maximum at x = 1. (f) H (x) = −4x 3 /(1 + x 4 )2 , so H (x) > 0 for x ∈ [−1, 0), and H (x) < 0 for x ∈ (0, 1]. Hence, H has a maximum at x = 0, and minima at x = ±1. 8. Here d (x) = 2(x − a1 ) + 2(x − a2 ) + · · · + 2(x − an ) = 2[nx − (a1 + a2 + · · · + an )]. So d (x) = 0 for x = x, ¯ where x¯ = n1 (a1 + a2 + · · · + an ), the arithmetic mean of a1 , a2 , . . . , an . If x > x, ¯ then d (x) > 0, and if x < x, ¯ then d (x) < 0. We conclude that x¯ minimizes d(x).. 9.3 2. In all cases the maximum and minimum exist by the extreme value theorem. Follow the recipe in [9.5]. (a) f (x) is strictly decreasing. Maximum −1 at x = 0, minimum −7 at x = 3. © Knut Sydsæter and Peter Hammond 2010.

(25) 22. CHAPTER 9. SINGLE-VARIABLE OPTIMIZATION. (b) Maximum 10 at x = −1 and x = 2. Minimum 6 at x = 1. (f (x) = 3x 2 − 3 = 0 at x = ±1.) (c) Maximum 2.5 at x = 1/2 and x = 2. Minimum 2 at x = f (x) = 1−1/x 2 = 0 √1. (f√(x) = x+1/x and 2 at x = ±1.) (d) Maximum 4 at x = −1. Minimum −6 3 at 3. (f (x) = 5x (x 2 − 3).) (e) Maximum 4.5 · 109 at x = 3000. Minimum 0 at x = 0. (f (x) = 3(x 2 − 3000x + 2 · 106 = 3(x − 500)(x − 2000).) 4. (a) Total costs when there are 61, 70, and 80 passengers are: $ 48,190; $ 49,000; $ 48,000. (b) C(x) = (60 + x)(800 − 10x) = 48, 000 + 200x − 10x 2 , x ∈ [0, 20] (c) Maximum cost is with 70 travelers (x = 10). √ √ √ 6. h (t) = 1/2 t − 21 = (1 − t)/2 t. We see that h (t) ≥ 0 in [0, 1] and h (t) ≤ 0 in [1, ∞), so t = 1 maximizes h(t). The plant is highest after 1 month, when it is 1/2 meter tall. 8. Because A (Q) = [C (Q)Q − C(Q)]/Q2 , the result follows from Example 4.20 in Section 4.6. 1/b . 10. A (Q) = a(b − 1)Qb−2 − c/Q2 , so there is a minimum at Q = c/a(b − 1). 9.4 2. (a) No local extreme points. (b) Local maximum 10 at x = −1. Local minimum 6 at x = 1. (c) Local maximum −2 √1. √ √ at x = −1.√Local minimum 2 at x = (d) Local maximum 6 3 at x = − 3. Local minimum −6 3 at x = 3. (e) No local maximum point. Local minimum 0.5 at x = 3. (f) Local maximum 2 at x = −2. Local minimum −2 at x = 0. 4. (a) (a) We find f (t) = 0.05(t + 5)(35 − t)e−t . Obviously, f (t) > 0 for t < 35 and f (t) < 0 for t > 35, so t = 35 maximizes f (with f (35) ≈ 278). (b) f (t) → 0 as t → ∞. The graph is shown in Fig. 9.4.4. y 10. y 5 200. √2 3. 100. 2. 1. 1 √2 2. 3. x. 5 30. 60. 90. Figure 9.4.4. t Figure 9.4.8. 6. (a) a > 0, b = 0 (f (0) = 0 ⇒ b = 0. f (0) ≥ 0 ⇒ a ≥ 0. If a = b = 0, then f (x) = x 3 + c has no loc. min. at x = 0.) (b) a = −6, b = 9. (f (1) = 0 and f (3) = 0 give 3 + 2a + b = 0 and 27 + 6a + b = 0.) 3 √ 2 − x2 (x) = 2x − 12x − 12 ; x = 8. (a) f (x) = ; f 2 is a local maximum point with 2 2 2 3 (x + 3x + (x + 3x + 2) √ √ 2) √ f ( 2) ≈ 0.17; x = − 2 is a local minimum point with f (− 2) ≈ 5.83. (b) There are no global extreme points. The graph is shown in Fig. 9.4.8. √ (c) ex = 2 and so x = 21 ln 2 gives a global maximum, but g has no global minimum. © Knut Sydsæter and Peter Hammond 2010.

(26) CHAPTER 10. INTEGRATION. 23. 10. f (x) = x 3 + ax + b → ∞ as x → ∞, and f (x) → −∞ as x → −∞. Thus f (x) has at least one real root. We have f (x) = 3x 2 + a. Thus, for a ≥ 0, f (x) > 0 for all x = 0, so f is strictly increasing, and there is only one real root. Note that for a ≥ 0, 4a 3 + 27b2 ≥ 0. Assume next that a < 0. Then f (x) = 0 √ √ √ √ for x = ± −a/3 = ± p, where p = −a/3 > 0. Then f has a local maximum at (− p, b + 2p p) √ √ and a local minimum at ( p, b − 2p p). If one of the local extreme values is 0, the equation has a double root, and this is the case iff 4p 3 = b2 , that is, iff 4a 3 + 27b2 = 0. The equation has three real roots iff the local maximum value is positive and the local minimum value is negative. This occurs iff √ |b| < 2p p or iff b2 < 4p3 or iff 4a 3 + 27b2 < 0.. 9.5 2. (a) f (x) = 3(x − 1)(x + 2), f (x) = 6x + 3 (b) Stationary points: x = −2 and x = 1. f increases in (−∞, −2) and in (1, ∞). (c) x = −1/2 is the only inflection point. 4. x = −2 and x = 4 are minimum points, whereas x = 2 is a maximum point. Moreover, x = 0, x = 1, x = 3, and x = 5 are inflection points. 6. a = −2/5, b = 3/5 (f (−1) = 1 gives −a + b = 1. Moreover, f (x) = 3ax 2 + 2bx and f (x) = 6ax + 2b, so f (1/2) = 0 yields 3a + 2b = 0.) 8. π(Q) is stationary at Q = (P /ab)1/(b−1) . Moreover, π (Q) = −ab(b − 1)Qb−2 < 0 for all Q > 0, so this is a maximum point.. 9.6 2. a ≥ 0, b arbitrary.. ¯ where y¯ = 4. (a) Since u(c) is concave, T1 Tt=1 u(ct ) ≤ u T1 Tt=1 ct ≤ u T1 Tt=1 yt = u(y), 1 T ¯ t = 1, 2, . . . , T , this holds with equality.) t=1 yt . Thus y¯ is optimal. (For ct = y, T T −t −t (b) Put λt = (1 + r) / t=1 (1 + r) . Then Tt=1 λt = 1 and T max Tt=1 (1 + r)−t u(ct ) = Tt=1 (1 + r)−t max Tt=1 λt u(ct ). Also, Tt=1 λt u(ct ) ≤ u t=1 λt ct ≤ T T u t=1 λt yt . So ct = c¯ = t=1 λt yt (t = 1, 2, . . . , T ) solves the problem.. Chapter 10 Integration 10.1 2. See Figs. 10.1.2(a) to 10.1.2(d). (a) y. y. 2. 1. 2 0. 3x 2 dx =. 2 0. x3 = 8. y. (b) 1/7. (c) e − 1/e. (d) 9/10. y 1. 2 1. 1. 2. x. Figure 10.1.2(a). 1. x. Figure 10.1.2(b). © Knut Sydsæter and Peter Hammond 2010. 1. 1. x. Figure 10.1.2(c). 1. Figure 10.1.2(d). 10. x.

(27) 24. CHAPTER 10. 4. A =. 10.2. 1 2. 1. −1 (e. x. INTEGRATION. + e−x ) dx =. 1 1 x 2 −1 (e. − e−x ) = e − e−1. . (t 3 + 2t − 3) dt = t 3 dt + 2t dt − 3 dt = 41 t 4 + t 2 − 3t + C d (b) (x −1)2 dx = (x 2 −2x +1) dx = 13 x 3 −x 2 +x +C. Alternative: Since (x −1)3 = 3(x −1)2 , dx we have (x − 1)2 dx = 13 (x − 1)3 + C1 . This agrees with the first answer, with C1 = C + 1/3. (c) (x − 1)(x + 2) dx = (x 2 + x − 2) dx = 13 x 3 + 21 x 2 − 2x + C. 2. (a). 3 3 2 (d) Either first evaluate (x + 2) = x + 6x + 12x + 8, to get (x + 2)3 dx = 41 x 4 + 2x 3 + 6x 2 + 8x + C, or: (x + 2)3 = 41 (x + 2)4 + C1 . (e) (e3x − e2x + ex ) dx = 31 e3x − 21 e2x + ex + C 3 4 x − 3x + 4 dx = x2 − 3 + dx = 13 x 3 − 3x + 4 ln |x| + C (f) x x 4. (a) Differentiate the right-hand side. √ 1 (2x + 1)5 + C (ii) 23 (x + 2)3/2 + C (iii) −2 4 − x + C (b) (i) 10 6. (a) F (x) = ( 21 − 2x) dx = 21 x − x 2 + C. F (0) = 21 implies C = 21 . 5 (b) F (x) = (x − x 3 ) dx = 21 x 2 − 41 x 4 + C. F (1) = 12 implies C = 16 . 1 4 x + Ax + B. If we require that 8. The general form for f is f (x) = 13 x 3 + A, so that for f is f (x) = 12 1 4 f (0) = 1 and f (0) = −1, then B = 1 and A = −1, so f (x) = 12 x − x + 1.. 10.3 2. (a) −12/5 4. (a) 6/5. (b) 41/2. (b) 26/3. 3 1 (c) + t dt = t −1 2. (c) α(eβ − 1)/β. (d) − ln 2. 3 2. 5 ln(t − 1) + 21 t 2 = ln 2 + 2 6. (a) 32/15 ≈ 2.133. See Fig. 10.3.6. y. y. y=. 2 2. √ x. 4/3 1. 1. 1 2 Figure 10.3.6. x. 1. x∗. 3. 4. x. Figure 10.3.10. b + c √ + d ln |b| (c) 2 2 − 3/2 (b) A b − 1 + (b − c) ln 1+c x 10. (a) Given F (x) = a f (t) dt, according to the mean-value theorem, there exists a number x ∗ ∈ (a, b) b such that [F (b) − F (a)]/(b − a) = F (x ∗ ). Here F (b) = a f (t) dt, F (a) = 0, and F (x ∗ ) = f (x ∗ ), so the conclusion follows. (b) f (x ∗ ) = 4/3 at x ∗ = 16/9. See Fig. 10.3.10, in which the two shaded areas are equal. 8. (a) 59/30. © Knut Sydsæter and Peter Hammond 2010.

(28) CHAPTER 11. FURTHER TOPICS IN INTEGRATION. 25. 10.4 2. (a) Let n be the total The number of individuals with income in the interval [b, 2b] 2b number of individuals. 2b 2b nB −2 −1 is then N = n . Their total income is M = n Br dr = n −Br = Br −2 r dr = 2b b b b 2b 2b Br −1 dr = n B ln r = nB ln 2. Hence the mean income is m = M/N = 2b ln 2. n b b 2b 2b 2b nD(p, r)f (r) dr = nAp γ r δ Br −2 dr = nABp γ r δ−2 dr = (b) Total demand is x(p) = b. nABp. b. . 15. 4. PDV =. b. r δ−1 2δ−1 − 1 = nABp γ bδ−1 . δ−1 δ−1. 2b γ. 0. b. 15 −1 −0.06t. 500. 500e−0.06t dt = 500 0 = e 1 − e−0.9 ≈ 4945.25. 0.06 0.06. FDV = e0.06·15 PDV = e0.9 PDV ≈ 2.4596 · 4945.25 ≈ 12163.3.. Chapter 11 Further Topics in Integration 11.1. 1. 1 1 1 x ln(x + 2) dx = −1 21 x 2 ln(x + 2) − −1 21 x 2 x+2 dx 1 1 1 4 3 = 2 ln 3 − 2 −1 x − 2 + x+2 dx = 2 − 2 ln 3 d x 2 = 2x ln 2, and therefore 2x / ln 2 is the indefinite integral of 2x . If follows that (b) Recall that dx 2 2 x 2 2 2 2x 2x 8 8 3 x2x dx = x = − dx = − − 2 ln 2 ln 2 (ln 2)2 0 0 ln 2 0 ln 2 0 (ln 2) (c) First use integration by parts on the indefinite integral: with f (x) = x 2 and g(x) = ex , (∗) x 2 ex dx = x 2 ex − 2xex dx. To evaluate the lastintegral we must use integration by parts once more: with f (x) = 2x and g(x) = ex , 2xex dx = 2xex − 2ex dx = 2xex −(2ex +C). Inserted into (∗) this gives x 2 ex dx = 1 1 x 2 ex −2xex +2ex +C, and hence, 0 x 2 ex dx = 0 (x 2 ex −2xex +2ex ) = (e−2e+2e)−(0−0+2) = e−2. Alternatively, more compactly using formula [11.2]: 1 1 1 1 1 1 2 x 2 x x x x e dx = x e − 2 xe dx = e − 2 xe − ex dx = e − 2[e − 0 ex ] = e − 2. 2. (a). −1. 0. 0. 0. 0. 0. 4. Using integration by parts, we have . t1. t1 F (t)μ(t) ˙ dt = F (t)μ(t) − t0. t0. t1. F˙ (t)μ(t) dt. t0. = F (t1 )μ(t1 ) − F (t0 )μ(t0 ) −. . t1. F˙ (t)μ(t) dt = −. t0. . t1. F˙ (t)μ(t) dt. t0. 6. Using [11.2], we have . T. U (C(t))e 0. −rt. T −rt dt = − U (C(t))(1/r)e + 0. which becomes the required expression when U (C(0)) = 0. © Knut Sydsæter and Peter Hammond 2010. T 0. (d/dt)U (C(t))(1/r)e−rt dt.

(29) 26. CHAPTER 11. FURTHER TOPICS IN INTEGRATION. 11.2 2. (a) 19 (x 2 + 1)9 + C. (Substitute u = x 2 + 1, du = 2x dx.) (b) With u = x 3 + 2, du = 3x 2 dx and 1 u 2 x 3 +2 1 u 1 x 3 +2 dx = + C. (c) First attempt: u = x + 2, which gives x e 3 e du = 3 e + C = 3 e ln(x + 2) ln u du = dx and dx = du. Not promising. A better idea: Substitute u = ln(x + 2). 2x + 4 2u dx ln(x + 2) 2 2 1 1 1 Then du = and dx = 2 u du = 4 (u) + C = 4 (ln(x + 2)) + C. x+2 2x + 4 √ √ (d) Attempt: u = 1 + x. Then, du = dx, and x 1 + x dx = (u − 1) u du = (u3/2 − u1/2 ) du = √ 2 5/2 − 23 u3/2 + C = 25 (1 + x)5/2 − 23(1 + x)3/2 + C. Second attempt: u = 1+ x. Then u2 = 1 + x 5u √ and 2udu = dx. Then the integral is x 1 + x dx = (u2 − 1)u2u du = (2u4 − 2u3 ) du e.t.c. Check that you get the same answer. Actually, even integration by parts works in this case. Put f (x) = x √ and g (x) = 1 + x, and choose g(x) = 23 (1 + x)3/2 . (The answer looks different, but is not.) x3 x2 · x 2 2 dx = dx = (e) With u = 1 + x , x = u − 1, and du = 2xdx, so (1 + x 2 )3 (1 + x 2 )3 u−1 −1 1 1 du = 21 (u−2 − u−3 ) du = − 21 u−1 + 41 u−2 + C = + C. + 2 2) 2 2 u3 2(1 + x 4(1 + x) √ (f) With u = 4 − x 3 , u2 = 4−x 3 , and 2udu = −3x 2 dx, so x 5 4 − x 3 dx = x 3 4 − x 3 x 2 dx = 2 5 2 (4−u2 ) u (− 23 )u du = (− 83 u2 + 23 u4 ) du = − 89 u3 + 15 u +C = − 89 (4−x 3 )3/2 + 15 (4−x 3 )5/2 +C . x −2x 2t − 2 ln u = ln 13 (x 2 − 2x). Thus, dt = 2 3 t − 2t 3 the equation implies that 13 (x 2 − 2x) = 23 x − 1. Hence, x 2 − 4x + 3 = 0, with solutions x = 1 and x = 3. Here x = 1 is impossible, because the integral is not defined when x = 1. So the solution is x = 3.. 4. Substituting u = t 2 − 2t and assuming x > 2 gives. x. 2. 6. Substitute z = x(t). Then dz = x (t)dt, and the result follows.. b+λ 8. (a) Introduce z = x − λ as a new variable in the right-hand side integral. Then dz = dx, and a+λ f (x − b λ) dx = a f (z) dz. Then replace the dummy variable z by x in the last integral. (b) Introduce z = x/λ as a new variable in the right-hand side integral. u8 1/6 du. Here u8 : (−u2 +1) = −u6 −u4 −u2 −1+1/(−u2 +1). 10. Substitute u = x . Then I = 6 1 − u2 It follows that I = − 67 x 7/6 − 65 x 5/6 − 2x 1/2 − 6x 1/6 − 3 ln |1 − x 1/6 | + 3 ln |1 + x 1/6 | + C. 11.3. . +∞. . b 1 1 1 dx = (b − a) = 1 x= b−a a b−a a b−a −∞ +∞ b b 1 1 1 1 (b) xf (x) dx = x dx = x2 = (b2 − a 2 ) = (a + b) b − a 2(b − a) 2(b − a) 2 −∞ a a b 1 b3 − a 3 1 1 2 3 2 x = (c) = (a + ab + b ) 3(b − a) a 3 b−a 3 b b 4. Divergence because 0 x/(1 + x 2 ) dx = 0 21 ln(1 + x 2 ) = 21 ln(1 + b2 ) → ∞ as b → ∞. But b b 1 2 2 −b x/(1 + x ) dx = −b 2 ln(1 + x ) = 0 for all b, so the limit is 0.. 2. (a). f (x) dx =. b. © Knut Sydsæter and Peter Hammond 2010.

(30) CHAPTER 12. 6.. 1 1+x 2. ≤. 1 x2. for x ≥ 1, and. b. dx =. 1 1 x2. LINEAR ALGEBRA: VECTORS AND MATRICES. b 1 (−1/x). 27. = 1 − 1/b → 1 as b → ∞, so by. Theorem 11.1 the given integral converges. 8. (a) z = rτ1 (1 − e−rτ ) (b) z = rτ2 1 − rτ1 (1 − e−rτ ) √ √ √ 10. Integrating by parts, ln x/ x dx = 2 x ln x − 4 x + C. Hence, 1 1 √ √ √ √ √ ln x/ x dx = (2 x ln x − 4 x) = −4 − (2 h ln h − 4 h) → −4 as h → 0+ , so the given h h √ integral converges to −4. ( h ln h → 0 by l’Hôpital’s rule.) ∞ ∞ p p p 12. See Fig. 11.3.12. If p > 1, then ∞ n=1 (1/n ) = 1 + n=2 (1/n ) is finite because n=2 (1/n ) is the sum of the shaded rectangles, and this sum is certainly less than the area under the curve y = 1/x p over p [1, ∞), which is equal to 1/(p − 1). If p ≤ 1, the sum ∞ n=1 (1/n ) is the sum of the larger rectangles in the figure, and this sum is larger than the area under the curve y = 1/x p over [1, ∞), which is unbounded p when p ≤ 1. Hence, ∞ n=1 (1/n ) diverges in this case. y. y = 1/x p. 1. 1. 2. 3. 4. x. Figure 11.3.12. 11.4 2. The Gini coefficients are approximately: 0.34 (US 1980), 0.37 (US 1990), 0.35 (Netherlands 1959), 0.27 (Netherlands 1985), 0.69 (World 1989). (These numbers are obtained by approximating the Lorenz curve with a broken line through the points corresponding to the observations, then computing the area between this broken line and the diagonal of the square. Note that this underestimates the Gini coefficients.). Chapter 12 Linear Algebra: Vectors and Matrices 12.1 2. (a) No sector delivers to itself. (b) The total amount of good i needed to produce one unit of each good. (c) This collection gives the number of units of each good which are needed to produce one unit of good j . (d) No economic interpretation. (The goods are usually measured in different units, so it is meaningless to add them together.) 4. The Leontief model for this three sector model is as follows: 0.9x1 − 0.2x2 − 0.1x3 = 85 −0.3x1 + 0.8x2 − 0.2x3 = 95 −0.2x1 − 0.2x2 + 0.9x3 = 20 which does have the claimed solution. © Knut Sydsæter and Peter Hammond 2010.

(31) 28. CHAPTER 12. LINEAR ALGEBRA: VECTORS AND MATRICES. 12.2 2. a + b + c = (−1, 6, −4), a − 2b + 2c = (−3, 10, 2), 3a + 2b − 3c = (9, −6, 9), −a − b − c = −(a + b + c) = (1, −6, 4) 4. 3(x, y, z) + 5(−1, 2, 3) = (3x − 5, 3y + 10, 3z + 15) = (4, 1, 3) provided that 3x − 5 = 4, 3y + 10 = 1, and 3z + 15 = 3. So x = 3, y = −3, z = −4. 6. (a). 3x − 2y = −1 −4x + 3y = 2. , x = 1, y = 2. (b). 2x + 4y = 1 −3x − 6y = 0. , which have no solution.. 8. We need to find numbers t and s such that t (2, −1)+s(1, 4) = (4, −11). This vector equation is equivalent to (2t + s, −t + 4s) = (4, −11), which in turn is equivalent to the equation system (i) 2t + s = 4 (ii) −t + 4s = −11. This system has the solution t = 3, s = −2, so (4, −11) = 3(2, −1) − 2(1, 4).. 12.3 2. (a) λ = 0 gives x = (3, 1), λ = 1/4 gives x = (2, 5/4), λ = 1/2 gives x = (1, 3/2), λ = 3/4 gives x = (0, 7/4), λ = 1 gives x = (−1, 2). See Fig. 12.3.2. (b) When λ runs through [0, 1], then x will run through all points on the line segment between a and b. 4. See Fig. 12.3.4. (The point R should be one unit further down.) z S (3,2, 4). y λ 1. b. λ 3/4. λ 1/2. λ 1/4. 1. 1. y. P. a 1. Q. λ 0 x. 2. 3. R. x. Figure 12.3.2. Figure 12.3.4. 12.4 2. (a) a · b = b · a = −6, (a + b) √· c = a · c + b · c = 9, a · (3b) = 3a · b = −18. (b) a = 3, b = 3, c = 29 , (c) |ab| = 6 ≤ ab = 9 4. The scalar product of the two vectors is x 2 +(x −1)x +3·3x = x 2 +x 2 −x +9x = 2x 2 +8x = 2x(x +4), which is 0 for x = 0 and x = −4. 6. x = (5, 7, 12), u = (20, 18, 25), x · u = 526 8. ||a + b||2 = (a + b) · (a + b) = a · a + 2a · b + b · b ≤ ||a||2 + 2||a||||b|| + ||b||2 = (||a|| + ||b||)2 . The conclusion follows because ||a + b|| and ||a|| + ||b|| are both nonnegative. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 8 2 6λ + 2 x1 10. (a) Total output is ⎝ x2 ⎠ = λ ⎝ 4 ⎠ + (1 − λ) ⎝ 6 ⎠ = ⎝ −2λ + 6 ⎠. 4 10 −6λ + 10 x3 © Knut Sydsæter and Peter Hammond 2010.

(32) CHAPTER 12. LINEAR ALGEBRA: VECTORS AND MATRICES. 29. To produce the output in (i) we can put λ = 1/2. To produce the output in (ii) would require a value of λ such that 6λ+2 = 7, −2λ+6 = 5, and −6λ+10 = 5, and the second of these equations gives a different value for λ from the other two. (b) In (a) we saw that (i) can be produced even without throwing away outputs. For (ii) to be possible if we were allowed to throw away output, there must exist a λ ∈ [0, 1] such that 6λ + 2 ≥ 7, −2λ + 6 ≥ 5, and −6λ + 10 ≥ 5. These inequalities reduce to λ ≥ 5/6, λ ≤ 1/2, λ ≤ 5/6, which are incompatible. (c) Revenue = R(λ) = p1 x1 + p2 x2 + p3 x3 = p1 (6λ + 2) + p2 (−2λ + 6) + p3 (−6λ + 10) = (6p1 − 2p2 − 6p3 )λ + 2p1 + 6p2 + 10p3 . If the constant slope 6p1 − 2p2 − 6p3 is > 0, then R(λ) is maximized at λ = 1; if 6p1 − 2p2 − 6p3 is < 0, then R(λ) is maximized at λ = 0. Only in the special case where 6p1 − 2p2 − 6p3 = 0 can the two plants both remain in use.. 12.5 2. (a) Direct verification. (To show that a lies on L, put t = 0.) (b) The direction of L is given by (−1, 2, 1). (c) The equation of plane is (−1)(x1 − 2) + 2(x2 − (−1)) + 1 · (x3 − 3) = 0, or −x1 + 2x2 + x3 = −1. (d) We must have 3(−t + 2) + 5(2t − 1) − (t + 3) = 6, and so t = 4/3. Thus P = (2/3, 5/3, 13/3). (b) x1 = −2 − t, x2 = 1 + 2t, x3 = −1 + 3t.. 4. (a) Direct verification.. 12.6 2. (a) A =. 12.7. . 2 3. 3 4. 4 5. (b) A =. 1 −1 −1 1. 1 −1. 4. A + B =. ⎛. ⎞ ⎛ ⎞ ⎛ 4 1 −1 −2 3 −5 5 2. A + B = ⎝ 9 2 7 ⎠, A − B = ⎝ 1 −2 −3 ⎠, AB = ⎝ 19 3 −1 4 −1 −1 −2 1 ⎛ ⎞ ⎛ ⎞ 0 4 −9 23 8 25 BA = ⎝ 19 3 −3 ⎠, (AB)C = A(BC) = ⎝ 92 −28 76 ⎠ 5 1 −3 4 − 8 −4 ⎛ ⎞ 0.2875 ⎝ 4. T(Ts) = 0.2250 ⎠ 0.4875. 12.8. 1 7. 0 0 , 3A = 5 6. 3 −5 −3. 3 9. ⎞ 3 16 ⎠, 0. ⎛. ⎞⎛ ⎞ ⎛ ⎞ a d e x ax + dy + ez 2. We start by performing the multiplication ⎝ d b f ⎠ ⎝ y ⎠ = ⎝ dx + by + f z ⎠. Next, e f c z ex + fy + cz ⎛ ⎞ ax + dy + ez (x, y, z) ⎝ dx + by + f z ⎠ = (ax 2 + by 2 + cz2 + 2dxy + 2exz + 2fyz) ex + fy + cz which is a 1 × 1 matrix.. 4. Equality in both [1] and [2] ⇔ AB = BA.. © Knut Sydsæter and Peter Hammond 2010. ⎞ 1 6. (a) x0 = ±(1/ 3) ⎝ −1 ⎠ 1 √. ⎛. (b) An x0 = x0 for all n..

(33) 30. CHAPTER 13. DETERMINANTS AND MATRIX INVERSION. a 2 + bc ab + bd 8. (a) Direct verification yields (1) = (a + d)A − (ad − bc)I2 = ac + cd bc + d 2 3 (b) Multiplying (1) by A and using A = 0 yields (2) (a + d)A2 = (ad − bc)A, and further, (3) 0 = (a + d)A3 = (ad − bc)A2 . If ad − bc = 0, (3) yields A2 = 0. If ad − bc = 0, 2 (2) yields (a + d)A2 = 0, and if a +d = 0, again A = 0. Finally, if ad − bc = a + d = 0, 1 1 then (1) implies A2 = 0. (c) Let A = . Then A2 = 0, and also (of course) A3 = 0. −1 −1 . A2. 12.9 2. A =. . AB =. . . −1 0 2 3 1 −6 , B = , (A + B) = , (αA) = 5 2 2 4 7 −4. . . 4 10 4 10 −2 4 , (AB) = = B A , A B = 10 8 10 8 10 14. 3 2 . 2 , −10. 4. Symmetry if a 2 − 1 = a + 1 and a 2 + 4 = 4a, which reduces to a = 2. 6. (a) (A1 A2 A3 ) = (A1 (A2 A3 )) = (A2 A3 ) A1 = (A3 A2 )A1 = A3 A2 A1 . (b) Easy induction proof. 8. (a) tr(A + B) = ni=1 (aii + bii ) = ni=1 aii + ni=1 bii = tr(A) + tr(B) (b) tr(cA) = ni=1 caii = c ni=1 aii = c tr(A) (c) tr(AB) = ni=1 jn=1 aij bj i , whereas tr(BA) = n n n n i=1 j =1 bij aj i = j =1 i=1 bj i aij , because the indices i and j can be interchanged. The two are equal because the order of multiplication and summation in each double sum is irrelevant. (If this is hard to understand, write out the sums in full for the cases when n = 2 and/or n = 3.) (d) tr(A ) = ni=1 aii = tr(A). Chapter 13 Determinants and Matrix Inversion 13.1. 3 2. See Fig. 13.1.2. The shaded area is 18 = 2. 0 . 6. (2, 6). (3, 0) Figure 13.1.2. a11 b11 + a12 b21 a11 b12 + a12 b22 , implying that a21 b11 + a22 b21 a21 b12 + a22 b22 |AB| = (a11 b11 + a12 b21 )(a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 )(a21 b11 + a22 b21 ). On the other hand, |A||B| = (a11 a22 − a12 a21 )(b11 b22 − b12 b21 ). A tedious computation shows that the two expressions are equal. . 4. The matrix product is AB =. © Knut Sydsæter and Peter Hammond 2010.

(34) CHAPTER 13. 6. We write the system as. Y − C = I0 + G0 −bY + C = a. I0 + G0 −1 a 1 Y = 1 −1 −b 1. . =. DETERMINANTS AND MATRIX INVERSION. 31. . Then Cramer’s rule yields. a + I 0 + G0 , 1−b. 1 I 0 + G0 −b a C= 1 −1 −b 1. . =. a + b(I0 + G0 ) 1−b. The expression for Y is most easily found by substituting the second equation into the first, and then solving for Y . Then use C = a + bY to find C. 8. |A(t)| = 2t (1 − t), which is 0 for t = 0 and t = 1.. 13.2 ⎛. −1 2. AB = ⎝ 7 5. ⎞ −1 −1 13 13 ⎠, |A| = −2, |B| = 3, |AB| = |A| · |B| = −6 9 10. 4. By Sarrus’ rule the determinant is (1 + a)(1 + b)(1 + c) + 1 + 1 − (1 + b) − (1 + a) − (1 + c), which reduces to the given expression. 6. Direct verification using [13.7] or Sarrus’ rule.. 13.3 2. +a12 a23 a35 a41 a54 (Four lines between pairs of elements rise as one goes to the right.). 13.4 ⎛. 2 ⎝ 2. A = 1 3. 1 0 1. ⎞ 1 2 ⎠, |A| = |A | = −2 5. 4. (a) The first and the second columns are proportional. (b) Add the second column to the third. Then the first and the third columns are proportional. (c) The term x − y is a common factor for each entry in the first row. If x = y, all elements in the first row are 0. If x − y = 0, we divide this row by x − y, and the resulting determinant has rows 1 and 2 identical, so it is 0. 6. |AB| = |A||B| = −12, 3|A| = 9, | − 2B| = (−2)3 (−4) = 32, |A| + |B| = −1, whereas |A + B| is not determined. 8. (a) Because A2 = In it follows from part 8 of Theorem 13.1 that |A|2 = 1, and so |A| = ±1. (b) Direct verification. (c) (In − A)(In + A) = In2 + In A − AIn − A2 = In − A2 = 0 ⇐⇒ A2 = In © Knut Sydsæter and Peter Hammond 2010.

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