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35. Formula Proof Problem I lf f(x) = k.g(x), then f'(x) = k.g'(x). Proof: r,(*) =;Tolllr*-lc) =litjij:@ h-ro h =limk.@ h+o h = k.g'(x), Q.E.D. 36. Formula Proof Problem ll lf f(x) = x5, then f'(c) = 56+. Proof: f'(c)= rim !0*nCI .. x"-c" =ltm- x--rc X - c ,,- (x : cXl4 + x3c .+ r3ql t-rs:l-g1) =llm- = jd o, + x3c + ,r", *f,i *""0; =&+c4+c4+c4+c4 = 5c4, Q.E.D. 37. Derivative of a Power Formula lf f(x) = xn, then f'(x) = ryn-1. Proof: lr-l=,1gf0aiFa 38. Derivative of a Sum of n Functions Problem llyn - u1 + u2 + u3+. . . + un, wherethe u; are differentiable functions of x, prove that Yn' = u1' + u2' + u3' +. . . + un' for a//integers n 22. Proof: Anchor: For n = 2, yz= u1 + u2. :. Y2' = u1' + u2' by the derivative of a sum of fwo functions property, thus anchoring the induction. lnduction Hvoothesis: Suppose that for n =k> 2, Yk'= ut'+ u2'+ u3'+. . . + u1'. Verification for n = k + 1: Letyl*1 =ul *U2+ u3+'.. + Uk+ uk+1. Thenyp*1 -(ur + u2+ u3+.. . + u1) + u1*1, which isa sum of tulo terms. .'. Yk+r'- (ur + u2 + u3 +. . . + up)'+ up*1' by the anchor = ul '+ u2'+ u3'+.. . + up' + u1*1', which completes the induction. .'. Yn'= u1'+ u2'r u3't . ' .* Un' for a//integers n ) 2' Q.E.D. 39. lntroduction to Antiderivatives a. f'(x) = 3x2- 10x + 5 + f(x) = xs- 5x2 + 5x b. g(x) = l(x) + 13 is also an answer to part a, because it has the same derivative as f(x). The derivative of a constant is zero. c. The name antiderivative is chosen because it is an inverse operation of taking the derivative. 3. x = -t3 + 13t2 - 351 + 27. Graph. The object starts out alx=27 ft when t = 0 sec. lt moves to the left to x = 0.16 ft when t = 1.7 sec. lt turns there and goes to the right to x = 70 ft when t = 7 sec. lt turns there and speeds up, going to the left for all higher values of t. -k. lim h+0 g(x+h)-s(x) h xn + m(rF-1h + hrrtlrgf + +tfl-f = lim h-+0 = lim (nxn-1 + ]n(n-1)xn-2h +. .. + hn-1 ) h-+0 =p1rFl+0+0+...+0 = nXr1, which is from the second term in the binomial expansion of (x + h)n, Q.E.D. Problem Set 3-5. poges 102 to 104 Displocement. Velocity. ond Accelerotion 45. (d/dx)(3x + 5) = 3 Q6. f(3) = 45 Q1. No values of t Q3. y'= -51x-a Q7. f'(3) = 30 Q9. Epsilon 1.y=sta-3F/+71 u=ff= 2gf -7.211.a * 2.y=0.3t4-5t u=ff= -1.2trs-5,a= Q2. dy/dx = 10x aa. f'(x) = 1.7x0.7 Q8. Limit = 45 Q10. Definite integral z,a=S= 60t2-10.08t0'4 #= u.u Turnsatt=7,x=76. Problem Set 3-5 Solutions Monuol 33

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+tfl-f b. g(x) = l(x) + 13 is also an answer to part a, because it has the same derivative as f(x). The derivative of a constant is zero. c. The name antiderivative is chosen because it is an inverse operation of taking the derivative. .'. Yn'= u1'+ u2'r u3't . ' .* Un' for a//integers n ) 2' Thenyp*1 -(ur + u2+ u3+.. . + u1) + u1*1, which isa =&+c4+c4+c4+c4 35. Formula Proof Problem I lf f(x) = k.g(x), then f'(x) = k.g'(x). The object starts out alx=27 ft when t = 0 sec. lt = 5c4, Q.E.D.

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Page 1: IMG_0007

35. Formula Proof Problem I

lf f(x) = k.g(x), then f'(x) = k.g'(x).Proof:

r,(*) =;Tolllr*-lc)=litjij:@h-ro h

=limk.@h+o h

= k.g'(x), Q.E.D.

36. Formula Proof Problem lllf f(x) = x5, then f'(c) = 56+.Proof:

f'(c)= rim !0*nCI.. x"-c"

=ltm-x--rc X - c,,- (x : cXl4 + x3c .+ r3ql t-rs:l-g1)

=llm-

= jd o, + x3c + ,r", *f,i *""0;

=&+c4+c4+c4+c4= 5c4, Q.E.D.

37. Derivative of a Power Formulalf f(x) = xn, then f'(x) = ryn-1.Proof:

lr-l=,1gf0aiFa

38. Derivative of a Sum of n Functions Problemllyn - u1 + u2 + u3+. . . + un, wherethe u; aredifferentiable functions of x, prove thatYn' = u1' + u2' + u3' +. . . + un' for a//integers n 22.Proof:Anchor: For n = 2, yz= u1 + u2.:. Y2' = u1' + u2' by the derivative of a sum of fwofunctions property, thus anchoring the induction.lnduction Hvoothesis:Suppose that for n =k> 2,

Yk'= ut'+ u2'+ u3'+. . . + u1'.Verification for n = k + 1:

Letyl*1 =ul *U2+ u3+'.. + Uk+ uk+1.Thenyp*1 -(ur + u2+ u3+.. . + u1) + u1*1, which isasum of tulo terms..'. Yk+r'- (ur + u2 + u3 +. . . + up)'+ up*1'

by the anchor= ul '+ u2'+ u3'+.. . + up' + u1*1',

which completes the induction..'. Yn'= u1'+ u2'r u3't . ' .* Un' for a//integers n ) 2'Q.E.D.

39. lntroduction to Antiderivativesa. f'(x) = 3x2- 10x + 5 + f(x) = xs- 5x2 + 5xb. g(x) = l(x) + 13 is also an answer to part a, because it

has the same derivative as f(x). The derivative of aconstant is zero.

c. The name antiderivative is chosen because it is aninverse operation of taking the derivative.

3. x = -t3 + 13t2 - 351 + 27. Graph.The object starts out alx=27 ft when t = 0 sec. ltmoves to the left to x = 0.16 ft when t = 1.7 sec. lt turnsthere and goes to the right to x = 70 ft when t = 7 sec. ltturns there and speeds up, going to the left for allhigher values of t.

-k. limh+0

g(x+h)-s(x)h

xn + m(rF-1h + hrrtlrgf + +tfl-f= lim

h-+0

= lim (nxn-1 + ]n(n-1)xn-2h +. .. + hn-1 )h-+0

=p1rFl+0+0+...+0= nXr1, which is from the second term in the

binomial expansion of (x + h)n, Q.E.D.

Problem Set 3-5. poges 102 to 104 Displocement. Velocity. ond Accelerotion

45. (d/dx)(3x + 5) = 3 Q6. f(3) = 45

Q1. No values of tQ3. y'= -51x-a

Q7. f'(3) = 30Q9. Epsilon

1.y=sta-3F/+71u=ff= 2gf -7.211.a *

2.y=0.3t4-5tu=ff= -1.2trs-5,a=

Q2. dy/dx = 10xaa. f'(x) = 1.7x0.7

Q8. Limit = 45Q10. Definite integral

z,a=S= 60t2-10.08t0'4

#= u.u

Turnsatt=7,x=76.

Problem Set 3-5 Solutions Monuol 33