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 whatare its dimensions?

Solution If we label the sides of the rectangle x and y then we are given

16 = xy⇒ y = 16x

This gives that the perimeter is

P = 2x + 2y = 2x + 32x

Differentiating we havedP

dx= 2 − 32

x2= 2x

2 − 32x2

Since x must be positive we have one critical point of x = 4. Testing the intervals we have thatthere is a relative minimum and thus an absolute minimum at x = 4. Solving for y we have

y = 164

= 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 ofcardboard by cutting congruent squares from the corners and folding up the sides. What are thedimensions 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 havedV

dx= 12x2 − 92x + 120 = 4(x − 6)(3x − 5)

This gives on critical point of x = 4 in the given domain.

V (0) = 0V (4) = 0

V (53) = 5

3(15 − 10

3)(8 − 10

3)

= 53⋅ 35

3⋅ 14

3

= 245027

So the max volume is V = 245027

in3 and the dimensions are given by

15 − 2(53) = 15 − 10

3= 45 − 10

3= 35

3

8 − 2(53) = 8 − 10

3= 24 − 10

3= 14

3

Thus the dimensions are5

3in by

14

3in by

35

3in.

2

Problem 7. A rectangular plot of farmland will be bounded on one side by a river and on the othertree sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largestarea you can enclose, and what are its dimensions?

Solution The picture is

River

x x

y

We are given2x + y = 800⇒ y = 800 − 2x

ThusA = 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 andthus 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 stainlesssteel plates together along their edges. As the production engineer, your job is to find dimensionsfor 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 = 500x2

We want to minimize the material so we want to minimize surface area which is given by

S = x2 + 4xy = x2 + 2000x

Differentiating we havedS

dx= 2x − 2000

x2= 2x

3 − 2000x2

Since x must be positive this gives one critical point of x = 10. Testing we have that there is arelative and thus absolute minimum at x = 10.

(a)

x = 10⇒ y = 500100

= 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 weightdepends 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 willminimize the amount of paper used?

Solution

We are given that

(x − 8)(y − 4) = 50⇒ y − 4 = 50x − 8 ⇒ y =

50

x − 8 + 4We want to minimize

A = xy = x( 50x − 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= 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 thereis 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. Whatvalue of θ will maximize the triangle’s area? (Hint: A = (1/2)ab sin θ)

Solution

A = 12ab sin θ⇒ dA

dθ= 1

2ab cos θ

Since θ is in a triangle we have that 0 ≤ θ ≤ π. Thus the only critical point is when θ = π/2 . Testingthe 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 willtake waste into account. There is no waste in cutting the aluminum for the side, but the top andbottom of radius r will be cut from squares that measure 2r units on a side. The total amount ofaluminum 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 mosteconomical can was 2 to 1. What is the ratio now?

Solution We are given that

V = 1000 = πr2h⇒ h = 1000πr2

This gives

A = 8r2 + 2πrh = 8r2 + 2�πr (1000�πr2

) = 8r2 + 2000r

Differentiating with respect to r we have

dA

dr= 16r − 2000

r2= 16r

3 − 2000r2

Solving for the critical point we have

16r3 = 2000⇒ r = 3√

2000

16= 5⇒ h = 1000

25π= 40π

Thus the ratio is

h

r= 40π⋅ 15

= 8π

Thus the ratio is 8 to π

6

Problem 19. Find the dimensions of a right circular cylinder of maximum volume that can beinscribed in a sphere of radius 10 cm. What is the maximum volume?

Solution The picture looks like

Using the picture we have that

(h2)2

+ r2 = 102 ⇒ r2 = 100 − h2

4⇒ V = π (100 − h

2

4)h = 100πh − πh

3

4

Differentiating with respect to h gives

dV

dh= 100π − 3πh

2

4= 400π − 3πh

2

4

Solving for the critical point we have

400π − 3πh2 = 0⇔ 3πh2 = 400π⇔ h2 = 4003⇔ h =

√400

3= 20√

3

Since the domain for h is [0,20] we have

V (0) = 0V (20) = 0

V ( 20√3) = 100π ( 20√

3) − π(20/

√3)3

4

= 2000π√3

− 8000π4 ⋅ 3√3

= 2000π√3

− 2000π3√

3

= 4000π3√

3

Thus the maximum volume is4000π

3√

3cm3

7

Problem 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangleis of clear glass, whereas the semicircle is of tinted glass that transmits only have as much light perunit area as clear glass does. The total perimeter is fixed. Find the proportions of the window thatwill admit the most light. Neglect the thickness of the frame.

20. a. The U.S. Postal Service will accept a box for domestic ship-ment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?

Square end

Girth = distancearound here

Length

b. Graph the volume of a 108-in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a).

21. (Continuation of Exercise 20.)

a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by w and the girth is 2h + 2w. What dimensions will give the box its largest volume now?

w

Girth

h

h

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.

T

T

10″

xx

x

x x

x

15″

Base Lid

x x

NO

T T

O S

CA

LE

a. Write a formula V(x) for the volume of the box.

b. Find the domain of V for the problem situation and graph V over this domain.

c. Use a graphical method to find the maximum volume and the value of x that gives it.

d. Confirm your result in part (c) analytically.

17. Designing a suitcase A 24-in.-by-36-in. sheet of cardboard is folded in half to form a 24-in.-by-18-in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid.

a. Write a formula V(x) for the volume of the box.

b. Find the domain of V for the problem situation and graph V over this domain.

c. Use a graphical method to find the maximum volume and the value of x that gives it.

d. Confirm your result in part (c) analytically.

e. Find a value of x that yields a volume of 1120 in3.

f. Write a paragraph describing the issues that arise in part (b).

24″

36″

x

24″

x

x x

x x

x x

18″

24″

36″

Base

The sheet is then unfolded.

18. A rectangle is to be inscribed under the arch of the curve y = 4 cos (0.5x) from x = -p to x = p. What are the dimen-sions of the rectangle with largest area, and what is the largest area?

19. Find the dimensions of a right circular cylinder of maximum vol-ume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?

T

256 Chapter 4: Applications of Derivatives

Solution Let P be the fixed perimeter. Then we have

P = 2r + 2h + πr⇒ 2h = P − 2r − πr⇒ h = 12(P − 2r − πr)

We want to maximize the light. If we related light to the area of the rectangle and circle we getthe following equation

L = 2rh + 12⋅ πr

2

2

= 2r (12(P − 2r − πr)) + πr

2

4

= Pr − 2r2 − πr2 + πr2

4

Differentiating with respect to r we get

dL

dr= P − 4r − 2πr + π

2r

= P + r (−4 − 2π + π2)

Solving for the critical point we have

dL

dr= 0⇔ r = P

4 + 2π − π2

= 2P8 + 4π − π =

2P

8 + 3π

Thus r = 2P8 + 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 togenerate a right circular cone. Find the radius, height, and volume of the cone of greatest volumethat 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 this way.

h

r

"

3

28. Find the point on the line xa +yb

= 1 that is closest to the origin.

29. Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible.

30. Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.

31. A wire b m long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a minimum, what is the length of each part?

32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.

33. Determine the dimensions of the rectangle of largest area that can be inscribed in the right tri-angle shown in the accompanying figure.

34. Determine the dimensions of the rect-angle of largest area that can be inscribed in a semicircle of radius 3. (See accompanying figure.)

35. What value of a makes ƒ(x) = x2 + (a>x) have a. a local minimum at x = 2? b. a point of inflection at x = 1?

36. What values of a and b make ƒ(x) = x3 + ax2 + bx have a. a local maximum at x = -1 and a local minimum at x = 3? b. a local minimum at x = 4 and a point of inflection at x = 1?

Physical Applications37. Vertical motion The height above ground 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 = 0.38. Quickest route Jane is 2 mi offshore in a boat and wishes to

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?

24. The trough in the figure is to be made to the dimensions shown. Only the angle u can be varied. What value of u will maximize the trough’s volume?

uu

20′

1′

1′

1′

25. Paper folding A rectangular sheet of 8.5-in.-by-11-in. paper is placed on a flat surface. One of the corners is placed on the oppo-site longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper.

a. Show that L2 = 2x3>(2x - 8.5). b. What value of x minimizes L2?

c. What is the minimum value of L?

Crease

D C

BPAx

x

L

R

Q (originally at A)"

L2 − x2

26. Constructing cylinders Compare the answers to the following two construction problems.

a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume?

b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?

x

y

y

(a)

Circumference = xy

x

(b)

r = 3

w

h

4

3

5w

h

4.6  Applied Optimization 257

Solution Using the pythagorean theorem we have

r2 + h2 = 3⇔ r2 = 3 − h2

We want to maximize volume and volume is given by

V = 13πr2h = 1

3π(3 − h2)h = πh − 1

3πh3

Differentiating with respect to h

dV

dh= π − πh2 = π(1 − h2)

This gives h = ±1 as critical points but since our domain is [0,√3] we have

V (0) = 0V (√3) = 0V (1) = π

3(2)

= 2π3

Thus the maximum volume is2π

3and occurs when h = 1 and r = √3 − 1 = √2

9

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 = 9632

= 3s(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

dt= 2 cos t − 2 cos(2t) = 2 cos t − 2 cos2 t + 1

Solving for the critical points we have

dD

dt= 0⇔ 2 cos t − 4 cos2 t + 2 = 0

⇔ 2 cos2 t − cos t − 1 = 0⇔ (2 cos t + 1)(cos t − 1) = 0

⇒ cos t = −12

or cos t = 1

⇒ t = 2π3+ 2πn, 4π

3+ 2πn,2πn

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 = 4π3+2πn.

Plugging this in we get

D (4π3

) = 2 sin(4π3

) − sin(8π3

)

= 2⎛⎝−√

3

2

⎞⎠ −

√3

2

= −3√

3

2

Thus the max distance is3√

3

2

12

Section 4.6

Problem 1. Use Newton’s method to estimate the solutions of the equation x2 + x − 1 = 0. Startwith x0 = −1 for the left-hand solution and with x0 = 1 for the solution on the right. Then, in eachcase, find x2.

Solutionf(x) = x2 + x − 1⇒ f ′(x) = 2x + 1

x1 = 0 − f(0)f ′(0)

= −−11

= 1

x2 = 1 − f(1)f ′(1)

= 1 − 13

= 23

13

Problem 3. Use Newton’s method to estimate the two zeros of the function f(x) = x4+x−3. Startwith x0 = −1 for the left-hand zero and with x0 = 1 for the zero on the right. Then, in each case,find x2.

Solutionf ′(x) = 4x3 + 1

Case x0 = −1:

x1 = −1 − f(−1)f ′(−1)

= −1 − −3−4= −1 − 1= −2

x2 = −2 − f(−2)f ′(−2)

= −2 + 5931

= − 331

Case x0 = 1:

x1 = 1 − f(1)f ′(1)

= 1 − −15

= 65

x2 = 65− f(6/5)f ′(6/5)

= 65− (1296/625) + (6/5) − 3(864/125) + 1

= 65− 1296 + 750 − 1875

4320 + 625= 6

5− 171

4945

= 57634945

14

Problem 9. Show that if h > 0, applying Newton’s method to

f(x) =⎧⎪⎪⎨⎪⎪⎩

√x, x ≥ 0√−x, x < 0

leads to x1 = −h if x0 = h and to x1 = h if x0 = −h. Draw a picture that shows whats going on.

Solution

f ′(x) = 12√x

for x ≥ 0 and f ′(x) = − 12√−x for x < 0

For x0 = h:

x1 = h − f(h)f ′(h)

= h −√h

1/2√h= h − 2h= −h

For x0 = −h:

x1 = −h − f(−h)f ′(−h)

= −h −√h

−1/2√h= −h + 2h= h

The picture looks like

15

Problem 10. Apply Newton’s method to f(x) = x1/3 with x0 = 1 and calculate x1, x2, x3, and x4.Find a formula for ∣xn∣. What happens to ∣xn∣ as n →∞? Draw a picture that shows what is goingon.

Solution

f ′(x) = 13x2/3

To make things easier we can simplify the formula for xn to

xn+1 = xn − (xn)1/3

1/3(xn)2/3 = xn − 3xn = −2xn

x1 = −2(1)= −2

x2 = −2(−2)= 4

x3 = −2(4)= −8

x4 = −2(−8)= 16

From this we see thatxn = (−1)n2n and lim

n→∞∣xn∣ =∞

The picture looks like

16

Problem 22. Find the approximate values of r1 through r4 in the factorization.

8x4 − 14x3 − 9x2 + 11x − 1 = 8(x − r1)(x − r2)(x − r3)(x − r4)

Root Finding 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.

2. Use Newton’s method to estimate the one real solution of x3 + 3x + 1 = 0. Start with x0 = 0 and then find x2.

3. Use Newton’s method to estimate the two zeros of the function ƒ(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.

4. Use Newton’s method to estimate the two zeros of the function ƒ(x) = 2x - x2 + 1. Start with x0 = 0 for the left-hand zero and with x0 = 2 for the zero on the right. Then, in each case, find x2.

5. Use Newton’s method to find the positive fourth root of 2 by solving the equation x4 - 2 = 0. Start with x0 = 1 and find x2.

6. Use Newton’s method to find the negative fourth root of 2 by solving the equation x4 - 2 = 0. Start with x0 = -1 and find x2.

7. Guessing a root Suppose that your first guess is lucky, in the sense that x0 is a root of ƒ(x) = 0. Assuming that ƒ′(x0) is defined and not 0, what happens to x1 and later approximations?

8. Estimating pi You plan to estimate p>2 to five decimal places by using Newton’s method to solve the equation cos x = 0. Does it matter what your starting value is? Give reasons for your answer.

Theory and Examples 9. Oscillation Show that if h 7 0, applying Newton’s method to

ƒ(x) = e2x, x Ú 02-x, x 6 0 leads to x1 = -h if x0 = h and to x1 = h if x0 = -h. Draw a

picture that shows what is going on.

10. Approximations that get worse and worse Apply Newton’s method to ƒ(x) = x1>3 with x0 = 1 and calculate x1, x2, x3, and x4. Find a formula for 0 xn 0 . What happens to 0 xn 0 as n S q? Draw a picture that shows what is going on.

11. Explain why the following four statements ask for the same information:

i) Find the roots of ƒ(x) = x3 - 3x - 1. ii) Find the x-coordinates of the intersections of the curve

y = x3 with the line y = 3x + 1. iii) Find the x-coordinates of the points where the curve

y = x3 - 3x crosses the horizontal line y = 1. iv) Find the values of x where the derivative of g(x) =

(1>4)x4 - (3>2)x2 - x + 5 equals zero. 12. Locating a planet To calculate a planet’s space coordinates, we

have to solve equations like x = 1 + 0.5 sin x. Graphing the func-tion ƒ(x) = x - 1 - 0.5 sin x suggests that the function has a root near x = 1.5. Use one application of Newton’s method to improve this estimate. That is, start with x0 = 1.5 and find x1. (The value of the root is 1.49870 to five decimal places.) Remember to use radians.

13. Intersecting curves The curve y = tan x crosses the line y = 2x between x = 0 and x = p>2. Use Newton’s method to find where.

T

14. Real solutions of a quartic Use Newton’s method to find the two real solutions of the equation x4 - 2x3 - x2 - 2x + 2 = 0.

15. a. How many solutions does the equation sin 3x = 0.99 - x2 have?

b. Use Newton’s method to find them.

16. Intersection of curves

a. Does cos 3x ever equal x? Give reasons for your answer.

b. Use Newton’s method to find where.

17. Find the four real zeros of the function ƒ(x) = 2x4 - 4x2 + 1. 18. Estimating pi Estimate p to as many decimal places as your

calculator will display by using Newton’s method to solve the equation tan x = 0 with x0 = 3.

19. Intersection of curves At what value(s) of x does cos x = 2x? 20. Intersection of curves At what value(s) of x does cos x = -x? 21. The graphs of y = x2(x + 1) and y = 1>x (x 7 0) intersect at

one point x = r. Use Newton’s method to estimate the value of r to four decimal places.

1

21−1 0

3

2

x

y

y = x1

y = x2(x + 1)

rr,1

a b

22. The graphs of y = 2x and y = 3 - x2 intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places.

23. Intersection of curves At what value(s) of x does e-x2 =x2 - x + 1?

24. Intersection of curves At what value(s) of x does ln (1 - x2) = x - 1?

25. Use the Intermediate Value Theorem from Section 2.5 to show that ƒ(x) = x3 + 2x - 4 has a root between x = 1 and x = 2. Then find the root to five decimal places.

26. Factoring a quartic Find the approximate values of r1 through r4 in the factorization

8x4 - 14x3 - 9x2 + 11x - 1 = 8(x - r1)(x - r2)(x - r3)(x - r4).

x

y

2

1−1 2

−4

−6

−2

−8

−10

−12

y = 8x4 − 14x3 − 9x2 + 11x − 1

T

T

T

Exercises 4.7

264 Chapter 4: Applications of Derivatives

Solution Let y = f(x). We want to do Newton’s Method with x0 = −1, x0 = 0, x0 = 1/2, andx0 = 2. We have

f ′(x) = 32x3 − 42x2 − 18x + 11

x0 = −1x1 = −1 − f(−1)

f ′(−1)≈ −0.978

x2 = −4445

− f(−44/45)f ′(−44/45)

≈ −0.977

x3 = −0.977 − f(−0.977)f ′(−0.977)

≈ −0.977

x0 = 0x1 = − f(0)

f ′(0)≈ 0.091

x2 ≈ 0.091 − f(0.091)f ′(0.091)

≈ 0.100

x3 ≈ 0.100 − f(0.1)f ′(0.1)

≈ 0.100

17

x0 = 12

x1 = 12− f(1/2)f ′(1/2)

≈ 0.722

x2 ≈ 0.722 − f(0.722)f ′(0.722)

≈ 0.651

x3 ≈ 0.651 − f(0.651)f ′(0.651)

≈ 0.643

x4 ≈ 0.643 − f(0.643)f ′(0.643)

≈ 0.643

x0 = 2x1 = 2 − f(2)

f ′(2)≈ 1.984

x2 ≈ 1.984 − f(1.984)f ′(1.984)

≈ 1.984

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