1
9. Limit of (sin x)/x Problem Prove that gbsrff = o See text proof. 'l 0. Derivative of the Sine Function See text proof. 1 1. Derivative of the Cosine Function See text proof. 12. Squeeze Theorem Problem a. See text statement of the theorem. b. See text proof of the theorem. c. See Figure 3-8c or 3-8d. 13. Group Discussion Problem 1 y=2+ (x-l)sin_ _ 1 a. Limit seems to equal 2. b. Graph. d. Prove that 1ir4 Y = 2. Proof: ,lgtt*1)=1+1=2. lg(3-x)=3-1=2' Forx<1,9(x)<y<h(x). .'. the squeeze theorem applies, and lim Y=2' x-J1- Forx>1,h(x)<y<g(x). .'. the squeeze theorem applies, and lim Y=2. x-J1+ Since both left- and right-hand limits equal 2, limy=2,Q.E.D. x+1 e. The word envelope (a noun) is used because the small window formed by the two lines "envelops" (a verb) the graph of the function. f. As lxl becomes large, (x - 1) ' sin *= - sin (J /(x -.1)) takes on the torm - 1/(x - 1) "i?-ggi:itt as the argument approaches zero. (argument) Thus the limit is 1, and y approaches 2 + 1, which equals 3. 14. Journal Problem Students should make the connection that they made a conjecture that the derivative of sine is cosine by examining the numerical derivative graph. Now they have proved it, using the definition of derivative and the limit properties. Graph in part (b). The lines have equations g(x)=x+1andh(x)=3-x. Q1. Squeeze theorem Q3. Y'= -51n ; Q5. No 07. xa8 Q9. GraPh. 1Y = lxl) Q2. Limit = 1 Q4.y=sinx+C 06. x8 Q8. log 32=5log2 Q10. Cube function 17. f'(x) = x4 and f(1) = 10 t1x;=lxs+C tO= *.15+C + C=9.8 t1x;=f5+s.e 18. f'(x) = x7 and f(-1) = 100 t1x1=lxa+C ,irur#*,"*. 1. f'(x) = 7x6 + f(x) = xz .' 6 2. f'(x) = 10xs + f(x) =;to a 6 3. f'(x) = x5 + f(x) = ]x6 + C 4. f'(x) = xa + f(x) =f x5+C 5. f'(x) = x-e + t1x; = -[r-s + C 6.f'(x) =x-1066 + f(x) =-f-s65r1065+C 7. f'(x) =cosx = f(x) = sin ; a 6 8. f'(x) = sin x =+ f(x) = -cos x + C g. f ixi = g6*7t2 -'f(x1 = 6*slz * a t o. f'(xi - 77x4t3 + f(x) - 33v7/3 a 6 11. f'(x) = sin 5x + f(x) = -0.2 cos 5x + C 12. f'(x) = cos 4x + l(x) = ] sin 4x + C 13. f'(x) = (4x + 5)7 + f(x) =$g^+ 5)B + C 1a. f'(x) - (8x + 3)5 + f(x) = *11et + 3)6 + C 15. f'(x) = x2 + 6x- 5 + f(x) =lx3 + 3x2 -5x + C 16.f'(x)=x2-1Ox +7 = f(x)=lx3 -5x2+7x+C 100 =*(-1)B + C + C = 99.875 f(x)=9.125*o+99.875 19. f'(x) = cos x and f(rt/2) = 5 f(x)=5i11 x16 5=sin(r/2) +C + C=4 f(x)-sinx+4 20. f'(x) = sin x and f(n) = 3 f(x)=-6esxa6 8={oSn +C + C=7 f(x)=assx*7 21. f'(x) = x2 - 8x + 3 and f(-2) = 13 f(x;=lP-4x2+3x+C 13=-9-16-6+c + C=113/3 f(x) = x3l3 - 4x2 +3x + 1 13/3 22. l'(x) = x2 + 12x- 7 and f(3) = -10 f(x;= ]1s+6x2-7x+C -10=9 +54-21 +C + C=12 f(x; = |s + 6x2 - 7x- 52 40 Colculus: Concepts ond Applicotions Problem Set 3-9

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g. f ixi = g6*7t2 -'f(x1 = 6*slz * a f(x)=-6esxa6 Q3. Y'= -51n ; a. Limit seems to equal 2. b. Graph. g(x)=x+1andh(x)=3-x. t1x;=f5+s.e f(x)-sinx+4 lim Y=2' 8. f'(x) = sin x =+ f(x) = -cos x + C e. The word envelope (a noun) is used because the small window formed by the two lines "envelops" (a verb) the graph of the function. Q8. log 32=5log2 lim Y=2. limy=2,Q.E.D. f(x)=5i11 x16 t1x;=lxs+C f(x)=9.125*o+99.875 22. l'(x) = x2 + 12x- 7 and f(3) = -10 Q1. Squeeze theorem Q9. GraPh. 1Y = lxl) 40 1

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Page 1: IMG_0014

9. Limit of (sin x)/x Problem

Prove that gbsrff = o

See text proof.'l 0. Derivative of the Sine Function

See text proof.

1 1. Derivative of the Cosine FunctionSee text proof.

12. Squeeze Theorem Problema. See text statement of the theorem.b. See text proof of the theorem.c. See Figure 3-8c or 3-8d.

13. Group Discussion Problem1y=2+ (x-l)sin_ _ 1

a. Limit seems to equal 2.b. Graph.

d. Prove that 1ir4 Y = 2.

Proof:

,lgtt*1)=1+1=2.lg(3-x)=3-1=2'Forx<1,9(x)<y<h(x)..'. the squeeze theorem applies, andlim Y=2'

x-J1-Forx>1,h(x)<y<g(x)..'. the squeeze theorem applies, andlim Y=2.

x-J1+Since both left- and right-hand limits equal 2,limy=2,Q.E.D.x+1

e. The word envelope (a noun) is used because thesmall window formed by the two lines "envelops" (a

verb) the graph of the function.

f. As lxl becomes large, (x - 1) ' sin *=- sin (J /(x -.1)) takes on the torm- 1/(x - 1)

"i?-ggi:itt as the argument approaches zero.(argument)

Thus the limit is 1, and y approaches 2 + 1, whichequals 3.

14. Journal ProblemStudents should make the connection that they madea conjecture that the derivative of sine is cosine byexamining the numerical derivative graph. Now theyhave proved it, using the definition of derivative andthe limit properties.

Graph in part (b). The lines have equationsg(x)=x+1andh(x)=3-x.

Q1. Squeeze theoremQ3. Y'= -51n ;Q5. No07. xa8Q9. GraPh. 1Y = lxl)

Q2. Limit = 1

Q4.y=sinx+C06. x8Q8. log 32=5log2

Q10. Cube function

17. f'(x) = x4 and f(1) = 10

t1x;=lxs+CtO= *.15+C + C=9.8

t1x;=f5+s.e18. f'(x) = x7 and f(-1) = 100

t1x1=lxa+C

,irur#*,"*.

1. f'(x) = 7x6 + f(x) = xz .' 62. f'(x) = 10xs + f(x) =;to a 63. f'(x) = x5 + f(x) = ]x6 + C

4. f'(x) = xa + f(x) =f x5+C

5. f'(x) = x-e + t1x; = -[r-s + C

6.f'(x) =x-1066 + f(x) =-f-s65r1065+C7. f'(x) =cosx = f(x) = sin ; a 68. f'(x) = sin x =+ f(x) = -cos x + Cg. f ixi = g6*7t2 -'f(x1 = 6*slz * a

t o. f'(xi - 77x4t3 + f(x) - 33v7/3 a 611. f'(x) = sin 5x + f(x) = -0.2 cos 5x + C

12. f'(x) = cos 4x + l(x) = ] sin 4x + C

13. f'(x) = (4x + 5)7 + f(x) =$g^+ 5)B + C

1a. f'(x) - (8x + 3)5 + f(x) = *11et + 3)6 + C

15. f'(x) = x2 + 6x- 5 + f(x) =lx3 + 3x2 -5x + C

16.f'(x)=x2-1Ox +7 = f(x)=lx3 -5x2+7x+C

100 =*(-1)B + C + C = 99.875

f(x)=9.125*o+99.87519. f'(x) = cos x and f(rt/2) = 5

f(x)=5i11 x165=sin(r/2) +C + C=4f(x)-sinx+4

20. f'(x) = sin x and f(n) = 3f(x)=-6esxa68={oSn +C + C=7f(x)=assx*7

21. f'(x) = x2 - 8x + 3 and f(-2) = 13

f(x;=lP-4x2+3x+C13=-9-16-6+c + C=113/3f(x) = x3l3 - 4x2 +3x + 1 13/3

22. l'(x) = x2 + 12x- 7 and f(3) = -10f(x;= ]1s+6x2-7x+C-10=9 +54-21 +C + C=12f(x; = |s + 6x2 - 7x- 52

40 Colculus: Concepts ond Applicotions Problem Set 3-9