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Luận văn, báo cáo, luận án, đồ án, tiểu luận, đề tài khoa học, đề tài nghiên cứu, đề tài báo cáo - Báo cáo khoa học, luận văn tiến sĩ, luận văn thạc sĩ, nghiên cứu - Cơ khí - Vật liệu Math 241 Homework 9 Solutions Section 4.5 Problem 1. What is the smallest perimeter for a possible rectangle whose area is 16 in2 , and what are its dimensions? Solution If we label the sides of the rectangle x and y then we are given 16 = xy ⇒ y = 16 x This gives that the perimeter is P = 2x + 2y = 2x + 32 x Differentiating we have dP dx = 2 − 32 x2 = 2x2 − 32 x2 Since x must be positive we have one critical point of x = 4. Testing the intervals we have that there is a relative minimum and thus an absolute minimum at x = 4. Solving for y we have y = 16 4 = 4 so the smallest perimeter is given by the dimensions 4 in by 4 in and the smallest perimeter is P = 8 + 8 = 16 in 1 Problem 5. You are planning to make an open rectangular box from an 8-in.-by-15-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume? Solution We have that 0 ≤ x ≤ 4 and V = x(15 − 2x)(8 − 2x) = 2x(2x2 − 23x + 60) = 4x3 − 46x2 + 120x Differentiating we have dV dx = 12x2 − 92x + 120 = 4(x − 6)(3x − 5) This gives on critical point of x = 4 in the given domain. V (0) = 0 V (4) = 0 V 5 3 = 5 3 15 − 10 3 8 − 10 3 = 5 3 ⋅ 35 3 ⋅ 14 3 = 2450 27 So the max volume is V = 2450 27 in3 and the dimensions are given by 15 − 2 5 3 = 15 − 10 3 = 45 − 10 3 = 35 3 8 − 2 5 3 = 8 − 10 3 = 24 − 10 3 = 14 3 Thus the dimensions are 5 3 in by 14 3 in by 35 3 in. 2 Problem 7. A rectangular plot of farmland will be bounded on one side by a river and on the other tree sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Solution The picture is River x x y We are given 2x + y = 800 ⇒ y = 800 − 2x Thus A = xy = x(800 − 2x) = 800x − 2x2 Differentiating with respect to x we have dA dx = 800 − 4x This gives one critical point of x = 200. Testing the intervals we have that there is a relative and thus absolute max at x = 200 ⇒ y = 800 − 400 = 400. So the dimensions are 200 m by 400 m A = 200(400) = 80000m2 3 Problem 9. Your iron works has contracted to design and build a 500 ft3 , square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. (a) What dimensions do you tell the shop to use? (b) Briefly describe how you took weight into account. Solution We have V = 500 = x2y ⇒ y = 500 x2 We want to minimize the material so we want to minimize surface area which is given by S = x2 + 4xy = x2 + 2000 x Differentiating we have dS dx = 2x − 2000 x2 = 2x3 − 2000 x2 Since x must be positive this gives one critical point of x = 10. Testing we have that there is a relative and thus absolute minimum at x = 10. (a) x = 10 ⇒ y = 500 100 = 5 Thus the dimensions are 10 ft by 10 ft by 5 ft (b) By minimizing the amount of material used, we minimize the weight used since the weight depends on the material. 4 Problem 11. You are designing a rectangular poster to contain 50 in.2 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used? Solution We are given that (x − 8)(y − 4) = 50 ⇒ y − 4 = 50 x − 8 ⇒ y = 50 x − 8 + 4 We want to minimize A = xy = x 50 x − 8 + 4 = 50 x x − 8 + 4x Since there is a 4 inch margin on the top and bottom, we have that x must be greater than 8. Differentiating with respect to x gives dA dx = 50(x − 8) − 50x (x − 8)2 + 4 = 4 − 400 (x − 8)2 = 4(x − 8)2 − 400 (x − 8)2 = 4(x2 − 16x − 36 ) (x − 8)2 = 4(x − 18)(x + 2 ) (x − 8)2 This gives one critical point in the given domain of x = 18. Testing the intervals we have that there is a relative and thus absolute minimum at x = 18 ⇒ y = 50 10 + 4 = 9. Thus the dimensions are 18 in by 9 in 5 Problem 13. Two sides of a triangle have lengths a and b, and the angle between them is θ . What value of θ will maximize the triangle’s area? (Hint: A = (1~2)ab sin θ) Solution A = 1 2 ab sin θ ⇒ dA dθ = 1 2 ab cos θ Since θ is in a triangle we have that 0 ≤ θ ≤ π. Thus the only critical point is when θ = π~2 . Testing the intervals we have that this is a relative and thus absolute maximum. Problem 15. You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be A = 8r2 + 2πrh rather than the A = 2πr2 + 2πrh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? Solution We are given that V = 1000 = πr2h ⇒ h = 1000 πr2 This gives A = 8r2 + 2πrh = 8r2 + 2πr 1000 πr2 = 8r2 + 2000 r Differentiating with respect to r we have dA dr = 16r − 2000 r2 = 16r3 − 2000 r2 Solving for the critical point we have 16r3 = 2000 ⇒ r = 3 ¾ 2000 16 = 5 ⇒ h = 1000 25π = 40 π Thus the ratio is h r = 40 π ⋅ 1 5 = 8 π Thus the ratio is 8 to π 6 Problem 19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume? Solution The picture looks like Using the picture we have that h 2 2 + r2 = 102 ⇒ r2 = 100 − h2 4 ⇒ V = π 100 − h2 4 h = 100πh − πh3 4 Differentiating with respect to h gives dV dh = 100π − 3πh2 4 = 400π − 3πh2 4 Solving for the critical point we have 400π − 3πh2 = 0 ⇔ 3πh2 = 400π ⇔ h2 = 400 3 ⇔ h = ¾ 400 3 = 20 √ 3 Since the domain for h is 0, 20 we have V (0) = 0 V (20) = 0 V 20 √3 = 100π 20 √3 − π(20~√3)3 4 = 2000π √3 − 8000π 4 ⋅ 3√3 = 2000π √3 − 2000π 3√3 = 4000π 3√ 3 Thus the maximum volume is 4000π 3√3 cm3 7 Problem 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only have as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame. 23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction. T24″ x x x x 18″ he arch of the curve What are the dimen- what is the largest area? nder of maximum vol- ius 10 cm. What is the Solution Let P be the fixed perimeter. Then we have P = 2r + 2h + πr ⇒ 2h = P − 2r − πr ⇒ h = 1 2 (P − 2r − πr) We want to maximize the light. If we related light to the area of the rectangle and circle we get the following equation L = 2rh + 1 2 ⋅ πr2 2 = 2r 1 2 (P − 2r − πr) + πr2 4 = P r − 2r2 − πr2 + πr2 4 Differentiating with respect to r we get dL dr = P − 4r − 2πr + π 2 r = P + r −4 − 2π + π 2 Solving for the critical point we have dL dr = 0 ⇔ r = P 4 + 2π − π 2 = 2P 8 + 4π − π = 2P 8 + 3π Thus r = 2P 8 + 3π , h = 1 2 (P − 4P 8 + 3π − 2πP 8 + 3π , give the maximum light. 8 Problem 27. A right triangle whose hypotenuse is √3 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.27. Constructing cones A right triangle whose hypotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made ...
Math 241 Homework 9 Solutions Section 4.5 Problem 1 What is the smallest perimeter for a possible rectangle whose area is 16 in2, and what are its dimensions? Solution If we label the sides of the rectangle x and y then we are given 16 = xy ⇒ y = 16 x This gives that the perimeter is P = 2x + 2y = 2x + 32 x Differentiating we have dP 32 2x2 − 32 =2− 2 = 2 dx x x Since x must be positive we have one critical point of x = 4 Testing the intervals we have that there is a relative minimum and thus an absolute minimum at x = 4 Solving for y we have y = 16 = 4 4 so the smallest perimeter is given by the dimensions 4 in by 4 in and the smallest perimeter is P = 8 + 8 = 16 in 1 Problem 5 You are planning to make an open rectangular box from an 8-in.-by-15-in piece of cardboard by cutting congruent squares from the corners and folding up the sides What are the dimensions of the box of largest volume you can make this way, and what is its volume? Solution We have that 0 ≤ x ≤ 4 and V = x(15 − 2x)(8 − 2x) = 2x(2x2 − 23x + 60) = 4x3 − 46x2 + 120x Differentiating we have dV = 12x2 − 92x + 120 = 4(x − 6)(3x − 5) dx This gives on critical point of x = 4 in the given domain V (0) = 0 V (4) = 0 V 5 = 5 15 − 10 8 − 10 33 3 3 = 5 ⋅ 35 ⋅ 14 33 3 = 2450 27 So the max volume is V = 2450 in3 and the dimensions are given by 27 15 − 2 5 = 15 − 10 = 45 − 10 = 35 3 3 3 3 8 − 2 5 = 8 − 10 = 24 − 10 = 14 3 3 3 3 5 14 35 Thus the dimensions are in by in by in 3 3 3 2 Problem 7 A rectangular plot of farmland will be bounded on one side by a river and on the other tree sides by a single-strand electric fence With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Solution The picture is River x x y We are given 2x + y = 800 ⇒ y = 800 − 2x Thus A = xy = x(800 − 2x) = 800x − 2x2 Differentiating with respect to x we have dA = 800 − 4x dx This gives one critical point of x = 200 Testing the intervals we have that there is a relative and thus absolute max at x = 200 ⇒ y = 800 − 400 = 400 So the dimensions are 200 m by 400 m A = 200(400) = 80000m2 3 Problem 9 Your iron works has contracted to design and build a 500 ft3, square-based, open-top, rectangular steel holding tank for a paper company The tank is to be made by welding thin stainless steel plates together along their edges As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible (a) What dimensions do you tell the shop to use? (b) Briefly describe how you took weight into account Solution We have V = 500 = x2y ⇒ y = x2 500 We want to minimize the material so we want to minimize surface area which is given by S = x2 + 4xy = x2 + 2000 x Differentiating we have dS 2000 2x3 − 2000 = 2x − 2 = dx x x 2 Since x must be positive this gives one critical point of x = 10 Testing we have that there is a relative and thus absolute minimum at x = 10 (a) x = 10 ⇒ y = 500 = 5 100 Thus the dimensions are 10 ft by 10 ft by 5 ft (b) By minimizing the amount of material used, we minimize the weight used since the weight depends on the material 4 Problem 11 You are designing a rectangular poster to contain 50 in.2 of printing with a 4-in margin at the top and bottom and a 2-in margin at each side What overall dimensions will minimize the amount of paper used? Solution We are given that (x − 8)(y − 4) = 50 ⇒ y − 4 = 50 x − 8 ⇒ y = 50 x − 8 + 4 We want to minimize A = xy = x 50 x − 8 + 4 = 50x x − 8 + 4x Since there is a 4 inch margin on the top and bottom, we have that x must be greater than 8 Differentiating with respect to x gives dA dx = (x − 8)2 50(x − 8) − 50x + 4 = 4 − (x − 8)2 400 = 4(x − 8)2 − 400 (x − 8) 2 = 4(x2 − 16x − 36) (x − 8) 2 = (x − 8)2 4(x − 18)(x + 2) This gives one critical point in the given domain of x = 18 Testing the intervals we have that there is a relative and thus absolute minimum at x = 18 ⇒ y = 50 + 4 = 9 Thus the dimensions are 10 18 in by 9 in 5 Problem 13 Two sides of a triangle have lengths a and b, and the angle between them is θ What value of θ will maximize the triangle’s area? (Hint: A = (1 2)ab sin θ) Solution A = 1 ab sin θ ⇒ dA = 1 ab cos θ 2 dθ 2 Since θ is in a triangle we have that 0 ≤ θ ≤ π Thus the only critical point is when θ = π 2 Testing the intervals we have that this is a relative and thus absolute maximum Problem 15 You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side The total amount of aluminum used up by the can will therefore be A = 8r2 + 2πrh rather than the A = 2πr2 + 2πrh in Example 2 In Example 2, the ratio of h to r for the most economical can was 2 to 1 What is the ratio now? Solution We are given that V = 1000 = πr2h ⇒ h = πr2 1000 This gives A = 8r2 + 2πrh = 8r2 + 2πr 2 1000 = 8r2 + 2000 πr r Differentiating with respect to r we have dA 2000 16r3 − 2000 = 16r − 2 = dr r r 2 Solving for the critical point we have 16r3 = 2000 ⇒ r = 3 2000 = 5 ⇒ h = 1000 = 40 16 25π π Thus the ratio is h = 40 ⋅ 1 r π5 =8 π Thus the ratio is 8 to π 6 Problem 19 Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm What is the maximum volume? Solution The picture looks like Using the picture we have that h 2 + r2 = 102 ⇒ r2 = 100 − h2 ⇒ V = π 100 − h2 h = 100πh − πh3 2 4 4 4 Differentiating with respect to h gives dV = 100π − 3πh2 = 400π − 3πh2 dh 4 4 Solving for the critical point we have 400π − 3πh2 = 0 ⇔ 3πh2 = 400π ⇔ h2 = 400 ⇔ h = 400 20 3 =√ 3 3 Since the domain for h is [0, 20] we have V (0) = 0 V (20) = 0 20 20 π(20 √3)3 V √ = 100π √ − 3 3 4 2000π 8000π =√ − √ 3 4⋅3 3 2000π 2000π =√ − √ 3 33 4000π =√ 33 4000π 3 Thus the maximum volume is √ cm 33 7 x x 24″ T b Graph the volume as a function of h and compare what you see with your answer in part (a) x x 22 A window is in the form of a rectangle surmounted by a semicircle The rectangle is of clear glass, whereas the semicircle is of tinted glass xx Problem 22 A window is in the form of a rectangle surmounted by a semicircle The rectangle 18″ is of clear glass, whtehraetatsratnhsemsietsmoincliyrchlealifsaosfmtiuncthedligghlatspsetrhuantittraarneasmasitcsleoanrlyglahsasvedoaess.much light per unit area as clear gTlahses tdootaels.peTrihmeettoetrailspfeixreimd.eFteinrdisthfiexepdro pFoirntidontsheofptrhoepowrtiniodnoswotfhtahte window that will admit the mostwliilglhatd.mNitetghleecmt othset ltighhict.kNneesgsleocft tthhee tfhriacmknee.ss of the frame 23 A silo (base not included) is to be constructed in the form of a he arch of the curve Let P bceyltihnedefirxseudrmpeoruimnteetderb yThaehnewmeishpahveere The cost of construction Solution What are the dimen- per square unit of surface area is twice as great for the hemisphere what is the largest area? asPit=is2rfo+r2thhe+cπyrli⇒nd2rihca=lPsid−e2wra−llπ rD⇒etehrm=i1ne(Pth−e2drim− eπnrs)ions to inder of maximum vol- be used if the volume is fixed and the cost of2construction is to be dius 10 cm WhW at eiswthaent to maximkizeepttthoealimghitn.imIfuwme Nreelagtleedctltihgehtthtiocktnheesasroefatohfe tshileoraencdtawngalseteanind circle we get the following equatcioonnstruction L = 2rh + 1 ⋅ πr2 + πr2 22 4 = 2r 1 (P − 2r − πr) 2 = P r − 2r2 − πr2 + πr2 4 Differentiating with respect to r we get dL = P − 4r − 2πr + π r dr 2 = P + r −4 − 2π + π 2 Solving for the critical point we have dL dr = 0 ⇔ r = 4 + 2π − π P = 2P 8 + 4π − π = 2P 8 + 3π 2 Thus r = 2P 8 + 3π , h = 12 (P − 4P 8 + 3π − 2πP 8 + 3π , give the maximum light 8 4.6 Applied Optimization 257 made to the dimensions shown 27 Constructing cones A right tr√iangle whose hypotenuse is Problem 27 A right triangle whose hypotenuse is 3 m long is revolved about one of its legs to What value of u will maximize 23 m long is revolved about one of its legs to generate a right generate a right circulcairrccuolanrec oFnein Fdinthdethreadraiudisu,s,hehiegighht,t,aanndd vvoolluummeeooffththeecocnoeneofof greatest volume that can be made this gwraeyat.est volume that can be made this way " h 3 20′ r xy 28 Find the point on the line a + = 1 that is closest to the origin b Solution Using 2th9e Fpiyntdhagpoorseiatinvetnhuemorbeemr fworewhhaivche the sum of it and its reciprocal eet of 8.5-in.-by-11-in paper is is the smallest (least) possible e corners is placed on the oppo- r2 + h2 = 3 ⇔ r2 = 3 − h2 30 Find a positive number for which the sum of its reciprocal and e figure, and held there as the four times its square is the smallest possible We want to maximize volume and volume is given by em is to make the length of the 31 A wire b m long is cut into two pieces One piece is bent into an e length L Try it with paper equilateral tria1ngle2and t1he other is2 bent into a c1ircle3 If the sum of V = πr h = π(3 − h )h = πh − πh 5) the areas enclo3sed by ea3ch part is a minimum3, what is the length of each part? Differentiating with respect to h 32 Answer Exercise 31 if one piece is bent into a L? square and the otdhVer into a circ2le w = π − πh = π(1 − h ) 2 5 4 C 33 Determine the didmhensions of the rectangle of largest area that can be inscribed in the right tri√- h This gives h = ±1 as critical points but since our domain is [0, 3] we have angle shown in the accompanying figure 3 34 Determine the dimensionVs(o0f)t=he0rect- angle of largest area t√hat can be w inscribed in a semicirVcl(e o3f )ra=di0us 3 Q (originally at A) (See accompanying figurVe.()1) = π (2) h x 35 What value of a makes ƒ(x) = x2 + (a>x) have 2π r = 3 3= B a a local minimum at x = 2? 3 √ √ b a point2oπf inflection at x = 1? re the answers to Tthheufsoltlhowe imngaximum volume is and occurs when h = 1 and r = 3 − 1 = 2 36 What value3s of a and b make ƒ(x) = x + ax + bx have32 er 36 cm and dimensions x cm a a local maximum at x = -1 and a local minimum at x = 3? ylinder as shown in part (a) of b a local minimum at x = 4 and a point of inflection at x = 1? d y give the largest volume? Physical Applications ed about one of the sides of 37 Vertical motion The height above ground of an object moving der as shown in part (b) of the give the largest volume? vertically is given by s = - 16t2 + 96t + 112, with s in feet and t in seconds Find a the object’s velocity when t = 0; b its maximum height and when it occurs; x c its velocity when s = 0 y 38 Quickest route Jane is 2 m9i offshore in a boat and wishes to (b) reach a coastal village 6 mi down a straight shoreline from the point nearest the boat She can row 2 mph and can walk 5 mph Where should she land her boat to reach the village in the least amount of time? Problem 31 The height of an object moving vertically is given by s = −16t2 + 96t + 112 with s in feet and t in seconds Find (a) the object’s velocity when t = 0 (b) its maximum height and when it occurs (c) its velocity when s = 0 Solution (a) v(t) = s′(t) = −32t + 96 ⇒ v(0) = 96 ft/sec (b) s′(t) = −32t + 96 = 0 ⇔ t = 96 = 3 32 s(3) = −16(9) + 96(3) + 112 = 256 feet Since our function is a parabola, the absolute max is 256 feet at t = 3 seconds (c) s = −16(t2 − 6t − 7) = −16(t − 7)(t + 1) Thus s is zero when t = −1, 7 since time is positive we have t = 7 v(7) = −32(7) + 96 = −128 ft/sec 10 Problem 38 Two masses hanging side by side from springs have positions s1 = 2 sin t and s2 = sin 2t, respectively (a) At what times in the interval 0 < t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t.) (b) When in the interval 0 ≤ t ≤ 2π is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos2 t − 1.) Solution (a) The masses will pass each other when their position functions are equal s1 = s2 ⇔ 2 sin t = sin(2t) ⇔ 2 sin t = 2 sin t cos t ⇔ 2 sin t − 2 sin t cos t = 0 ⇔ 2 sin t(1 − cos t) = 0 ⇒ t = 0 + πn where n is a nonnegative integer (b) The distance between the two masses is given by the absolute value of D = s1 − s2 = 2 sin t − sin(2t) Differentiating with respect to t we have dD = 2 cos t − 2 cos(2t) = 2 cos t − 2 cos2 t + 1 dt Solving for the critical points we have dD = 0 ⇔ 2 cos t − 4 cos2 t + 2 = 0 dt ⇔ 2 cos2 t − cos t − 1 = 0 ⇔ (2 cos t + 1)(cos t − 1) = 0 ⇒ cos t = − 1 or cos t = 1 2 ⇒ t = 2π + 2πn, 4π + 2πn, 2πn 3 3 where n is a nonnegative integer Testing the first few intervals gives 11 With the rest of the number line repeating Thus the distance is maximized when t = 4π +2πn 3 Plugging this in we get D 4π = 2 sin 4π − sin 8π 3 3 3 ⎛ √3 ⎞ √3 = 2 ⎝− 2 ⎠ − 2 √ = −3 3 2 √ 33 Thus the max distance is 2 12 Section 4.6 Problem 1 Use Newton’s method to estimate the solutions of the equation x2 + x − 1 = 0 Start with x0 = −1 for the left-hand solution and with x0 = 1 for the solution on the right Then, in each case, find x2 Solution f (x) = x2 + x − 1 ⇒ f ′(x) = 2x + 1 x1 = 0 − f ′f (0) (0) = − −1 1 =1 x2 = 1 − f ′f (1) (1) =1− 1 3 =2 3 13 Problem 3 Use Newton’s method to estimate the two zeros of the function f (x) = x4 + x − 3 Start with x0 = −1 for the left-hand zero and with x0 = 1 for the zero on the right Then, in each case, find x2 Solution f ′(x) = 4x3 + 1 Case x0 = −1: x1 = −1 − f ′f (−1) (−1) Case x0 = 1: = −1 − −3 −4 = −1 − 1 = −2 x2 = −2 − f ′f (−2) (−2) = −2 + 59 31 =− 3 31 x1 = 1 − f ′f (1) (1) = 1 − −1 5 =6 5 x2 = 65 − f ′f (6 5) (6 5) = 6 − (1296 625) + (6 5) − 3 5 (864 125) + 1 = 6 − 1296 + 750 − 1875 5 4320 + 625 = 6 − 171 5 4945 = 5763 4945 14 Problem 9 Show that if h > 0, applying Newton’s method to ⎧ f (x) = ⎪⎪√ ⎨√x, x≥0 ⎪⎪ −x, x2 Use Newton’s method to −12 Solution Let y = f (x) We want to do Newton’s Method with x0 = −1, x0 = 0, x0 = 1 2, and x0 = 2 We have f ′(x) = 32x3 − 42x2 − 18x + 11 x0 = −1 x1 = −1 − f ′f (−1) (−1) ≈ −0.978 x2 = − 4445 − f ′(−44 f (−44 45) 45) ≈ −0.977 x3 = −0.977 − f ′f (−0.977) (−0.977) ≈ −0.977 x0 = 0 x1 = − f ′f (0) (0) ≈ 0.091 x2 ≈ 0.091 − f ′f (0.091) (0.091) ≈ 0.100 x3 ≈ 0.100 − f ′f (0.1) (0.1) ≈ 0.100 17 x0 = 1 2 x1 = 12 − f ′(1 f (1 2)2) ≈ 0.722 x2 ≈ 0.722 − f ′f (0.722) (0.722) ≈ 0.651 x3 ≈ 0.651 − f ′f (0.651) (0.651) ≈ 0.643 x4 ≈ 0.643 − f ′f (0.643) (0.643) ≈ 0.643 x0 = 2 x1 = 2 − f ′f (2) (2) ≈ 1.984 x2 ≈ 1.984 − f ′f (1.984) (1.984) ≈ 1.984 18