Solutions to Apostol

SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL.

ERNEST YEUNG

Fund Science! & Help Ernest finish his Physics Research! : quantum super-A-polynomials - a thesis by Ernest Yeung

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Ernest Yeung is supported by Mr. and Mrs. C.W. Yeung, Prof. Robert A. Rosenstone, Michael Drown, Arvid Kingl, Mr .and Mrs. Valerie Cheng, and the Foundation for Polish Sciences, Warsaw University.

SOLUTIONS TO VOLUME 1 One-Variable Calculus, with an Introduction to Linear Algebra

I 2.5 Exercises - Introduction to set theory, Notations for designating sets, Subsets, Unions, intersections, complements.Exercise 10. Distributive laws

Let X = A (B C), Y = (A B) (A C)Suppose x Xx A and x (B C) = x A and x is in at least B or in Cthen x is in at least either (A B) or (A C)x Y,X Y

Suppose y Yy is at least in either (A B) or A Cthen y A and either in B or Cy X,Y X

X = Y

Let X = A (B C), Y = (A B) (A C)Suppose x X

then x is at least either in A or in (B C)if x A, x Yif x (B C), x Y x Y,X Y

Suppose y Ythen y is at least in A or in B and y is at least in A or in Cif y A, then y Xif y A B or y A C, y X (various carvings out of A, simply )if y (B C), y X y X,Y X

X = Y1

Exercise 11. If x A A, then x is at least in A or in A. Then x A. So A A A. Of course A A A.If x A A, then x is in A and in A. Then x A. So A A A. Of course A A A.

Exercise 12. Let x A. y A B if y is at least in A or in B. x is in A so x A B. = A A B.Suppose b B and b / A. b A B but b / A. so A A B.

Exercise 13. Let x A , then x is at least in A or in . If x , then x is a null element (not an element at all). Thenactual elements must be in A. = A A.

Let x A. Then x A . A A . = A = A .Exercise 14. From distributivity, A (A B) = (A A) (A B) = A (A B).If x A (A B), x A and x A B, i.e. x A and x is at least in A or in B.= x is in A and is in B or is not in B. Then x A. = A (A B) A. Of course, A A (A B).= A (A B) = A (A B) = A.Exercise 15. a A, a C and b B, b C. Consider x A B. x is at least in A or in B. In either case, x C.= A B C.Exercise 16.

if C A and C B, then C A Bc C, c A and c Bx A B, x A and x B. Then c C, c A B. C A B

Exercise 17.

(1)if A B and B C thena A, a B.b B, b C.then since a B, a C,c C such that c / B.a A, a B so a 6= ca. = A C

(2) If A B,B C,A C since, a A, a B, b B, b C. Then since a B, a C. A C(3) A B and B C. B C or B = C. A B only. Then A C.(4) Yes, since a A, a B.(5) No, since x 6= A (sets as elements are different from elements)

Exercise 18. A (B C) = (AB) (A C)

Suppose x A (B C)then x A and x / B C = x / B Cthen x is not in even at least one B or C

= x (AB) (A C)Suppose x (AB) (A C)

then x is at least in (AB) or in (A C) = x is at least in A and not in B or in A and not in Cthen consider when one of the cases is true and when both cases are true = x A (B C)

Exercise 19.

Suppose x B AF

A

then x B, x /AF

A

x /AF

A = x / A,A F

since A F , x B, x / A, then x AF

(B A)2

Suppose x AF

(B A)

then x B A1 and x B A2 and . . .then A F , x B, x / Athen x / even at least one A F= x B

AF

A

Suppose x B AF

A

then x /AF

A

then at most x A for A F but onethen x is at least in one B A= x

AF

(B A)

Suppose x AF

(B A)

then x is at least in one B Athen for A F , x B and x / AConsider A F

= then x B AF

A

Exercise 20.

(1) (ii) is correct.

Suppose x (AB) Cthen x AB, x / Cthen x A and x / B and x / Cx / B and x / C = x / even at least B or Cx A (B C)

Suppose x A (B C)then x A, x / (B C)then x A and x / B and x / C= x (AB) C

To show that (i) is sometimes wrong,

Suppose y A (B C)y A and y / B Cy / B C

then y / B or y C or y / C(where does this lead to?)

Consider directly,Suppose x (AB) C

then x is at least in AB or in Cthen x is at least in A and / B or in C

Suppose x = c C and c / A3

(2)If C A,

A (B C) = (AB) CI 3.3 Exercises - The field axioms. The goal seems to be to abstract these so-called real numbers into just xs and ys thatare purely built upon these axioms.

Exercise 1. Thm. I.5. a(b c) = ab ac.Let y = ab ac;x = a(b c)Want: x = yac+ y = ab (by Thm. I.2, possibility of subtraction)

Note that by Thm. I.3, a(b c) = a(b+ (c)) = ab+ a(c) (by distributivity axiom)ac+ x = ac+ ab+ a(c) = a(c+ (c)) + ab = a(0 + b) = ab

But there exists exactly one y or x by Thm. I.2. x = y.Thm. I.6. 0 a = a 0 = 0.

0(a) = a(0) (by commutativity axiom)Given b R and 0 R, exactly one b s.t. b a = 0

0(a) = (b+ (b))a = ab ab = 0 (by Thm. I.5. and Thm. I.2)Thm. I.7.

ab = ac

By Axiom 4,y R s.t. ay = 1since products are uniquely determined, yab = yac = (ya)b = (ya)c = 1(b) = 1(c)

= b = cThm. I.8. Possibility of Division.

Given a, b, a 6= 0, choose y such that ay = 1.Let x = yb.

ax = ayb = 1(b) = b

Therefore, there exists at least one x such that ax = b. But by Thm. I.7, there exists only one x (since if az b, and sox = z).

Thm. I.9. If a 6= 0, then b/a = b(a1).

Let x =b

afor ax = b

y = a1 for ay = 1Want: x = by

Now b(1) = b, so ax = b = b(ay) = a(by)= x = by (by Thm. I.7)

Thm. I.10. If a 6= 0, then (a1)1 = a.Now ab = 1 for b = a1. But since b R and b 6= 0 (otherwise 1 = 0, contradiction), then using Thm. I.8 on b,ab = b(a) = 1; a = b1.

Thm.I.11. If ab = 0, a = 0 or b = 0.ab = 0 = a(0) = b = 0 or ab = ba = b(0) = a = 0. (we used Thm. I.7, cancellation law for multiplication)

Thm. I.12. Want: x = y if x = (a)b and y = (ab).ab+ y = 0

ab+ x = ab+ (a)b = b(a+ (a)) = b(a a) = b(0) = 00 is unique, so ab+ y = ab+ x implies x = y( by Thm. I.1 )

Thm. I.13. Want: x+ y = z, if a = bx, c = dy, (ad+ bc) = (bd)z.

(bd)(x+ y) = bdx+ bdy = ad+ bc = (bd)z

So using b, d 6= 0, which is given, and Thm. I.7, then x+ y = z.4

Thm. I.14. Want: xy = z for bx = a, dy = c, ac = (bd)z.

(bd)(xy) = (bx)(dy) = ac = (bd)z

b, d 6= 0, so by Thm. I.7, xy = z.Thm.I.15. Want: x = yz, if bx = a, dy = c, (bc)z = ad

(bc)z = b(dy)z = d(byz) = da

d 6= 0 so by Thm. I.7, by z = a, byz = abxb 6= 0 so by Thm. I.7, yz = x

Exercise 2. Consider 0 + z = 0. By Thm. I.2, there exists exactly one z, z = 0. By Axiom 4, z = 0. 0 = 0.Exercise 3. Consider 1(z)z(1) = 1. Then z = 11. But by Axiom 4, there exists distinct 1 such that z(1) = 1, so z = 1.

Exercise 4. Suppose there exists x such that 0x = 1, but 0x = 0 and 0 and 1 are distinct, so zero has no reciprocal .

Exercise 5. a+ (a) = 0, 0 + 0 = 0. Thena+ (a) + b+ (b) = (a+ b) + (a) + (b) = 0

(a+ b) = a+ (b) = a b

Exercise 6. a+ (a) = 0, b+ (b) = 0, soa+ (a) + b+ (b) = a+ (b) + (a) + b = (a b) + (a) + b = 0 + 0 = 0

(a b) = a+ b.Exercise 7.

(a b) + (b c) = a+ (b) + b+ (c) = a+ (b+ (b)) + (c) = a cExercise 8.

(ab)x = 1 (ab)1 = x

a(bx) = 1 a1 = bx

b(ax) = 1 b1 axa1b1 = (abx)x = 1(x) = (ab)1

Exercise 9. Want: x = y = z, if

z =a

b a = zt b+ t = 0

y =(a)b

by = u a+ u = 0

x = (ab

) (ab

)+ x = v + x = 0 vb = a

a+ (a) = vb+ by = b(v + y) = 0if b 6= 0, v + y = 0, but v + x = 0

by Thm. I.1 , x = y

b+ t = 0, then z(b+ t) = zb+ zt = zb+ a = z(0) = 0a+ zb = 0 = a = zb = bysince b 6= 0, z = y so x = y = z

Exercise 10. Since b, d 6= 0, Let

z =ad bcbd

(bd)z = ad bc by previous exercise or Thm. I.8, the possibility of division

x =a

bbx = a

t =cd

dt = c (By Thm. I.3, we know that b a = b+ (a) )5

dbx+ bdt = (bd)(x+ y) = ad bc = (bd)zb, d 6= 0, so x+ y = z

I 3.5 Exercises - The order axioms.

Theorem 1 (I.18). If a 0 then ac < bc

Theorem 2 (I.19). If a 0, then ac < bc

Theorem 3 (I.20). If a 6= 0, then a2 > 0Theorem 4 (I.21). 1 > 0

Theorem 5 (I.22). If a bc.

Theorem 6 (I.23). If a b. In particular, if a < 0, then a > 0.Theorem 7 (I.24). If ab > 0, then both a and b are positive or both are negative.

Theorem 8 (I.25). If a < c and b < d, then a+ b < c+ d.

Exercise 1.

(1) By Thm. I.19, c > 0a(c) < b(c) ac < bc

bc (ac) = ac bc > 0. Then ac > bc (by definition of > )(2)

a 0, then if b > 0, ab > 0(b) = 0. If b < 0, ab < 0(b) = 0. So if a > 0, then b > 0.

If a < 0, then if b > 0, ab < 0(b) = 0. If b < 0, ab > 0(b) = 0. So if a < 0, then b < 0.

(4)a < c so a+ b < c+ b = b+ c

b < d so b+ c < d+ cBy Transitive Law , a+ b < d+ c

Exercise 2. If x = 0, x2 = 0. 0 + 1 = 1 6= 0. So x 6= 0.

If x 6= 0, x2 > 0, and by Thm. I.21 , 1 > 0x2 + 1 > 0 + 0 = 0 x2 + 1 6= 0= @x R such that x2 + 1 = 0

Exercise 3.

a < 0, b < 0, a+ b < 0 + 0 = 0 ( By Thm. I.25)

Exercise 4. Consider ax = 1.

ax = 1 > 0. By Thm. I.24 , a, x are both positive or a, x are both negative

Exercise 5. Define x, y such that ax = 1, by = 1. We want x > y when b > a.

xb ax = xb 1 > 0 = bx > 1 = byb > 0 so x > y

Exercise 6.6

If a = b and b = c, then a = cIf a = b and b < c, then a < cIf a < b and b = c, then a < cIf a < b and b < c, then a < c (by transitivity of the inequality)= a c

Exercise 7. If a b and b c, then a c. If a = c, then by previous proof, a = b.Exercise 8. If a b and b c, then a c. If a = c, then by previous proof, a = b.Exercise 8. If a or b is zero, a2 or b2 = 0. By Thm. I.20, b2 0 or a2 0, respectively.

Otherwise, if neither are zero, by transitivity, a2 + b2 > 0.

Exercise 9. Suppose a x. Then a x 0.If a R so y R, such that a y = 0.

Consider y + 1 R (by closure under addition).a (y + 1) = a y 1 = 0 1 < 0 Contradiction that a y + 1

Exercise 10.If x = 0, done.

If x > 0, x is a positive real number. Let h =x

2.

= x2> x Contradiction.

I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line,Upper bound of a set, maximum element, least upper bound (supremum), The least-upper-bound axiom (completenessaxiom), The Archimedean property of the real-number system, Fundamental properties of the suprenum and infimum.We use Thm I.30, the Archimedean property of real numbers, alot.

Theorem 9 (I.30). If x > 0 and if y is an arbitrary real number, there exists a positive integer n such that nx > y.

We will use the least upper-bound axiom (completeness axiom) alot for continuity and differentiation theorems later.Apostol states it as an axiom; in real analysis, the existence of a sup for nonempty, bounded sets can be shown with analgorithm to zoom into a sup with monotonically increasing and monotonically decreasing sequence of guesses and showingits difference is a Cauchy sequence.

Axiom 1 (Least upper-bound axiom). Every nonempty set S of real numbers which is bounded above has a suprenum; thatis, theres a real number B s.t. B = supS.

Exercise 1. 0 < y x.= n(y x) > h > 0, n Z+, h arbitrary

y x > h/n = y > x+ h/n > xso let z = x+ h/n Done.

Exercise 2. x R so n Z+ such that n > x (Thm. I.29).Set of negative integers is unbounded below because

If m Z,x > m, then x is an upper bound on Z+. Contradiction of Thm. I.29. = m Z such that m < x < nExercise 3. Use Archimedian property.

x > 0 so for 1,n Z+ such that nx > 1, x > 1n .Exercise 4. x is an arbitrary real number. By Thm. I.29 and well-ordering principle, there exists a smallest n + 1 positive

integer such that x < n + 1 (consider the set of all m + 1 > x and so by well-ordering principle, there must be a smallestelement of this specific set of positive integers).

If x = n for some positive integer n, done.Otherwise, note that if x < n, then n+ 1 couldnt have been the smallest element such that m > x. x > n.

Exercise 5. If x = n, done. Otherwise, consider all m > x. By well-ordering principle, there exists a smallest element n such7

that n > x.If x+ 1 < n, then x < n 1, contradicting the fact that n is the smallest element such that x < n. Thus x+ 1 > n.

Exercise 6. y x > 0.n(y x) > h, h arbitrary , n Z+

y > x+ h/n = z > x

Since h was arbitrary, there are infinitely many numbers in between x, y.

Exercise 7. x = ab Q, y / Q.

x y = a byb

If a by was an integer, say m, then y = (amb

b

)which is rational. Contradiction.

xy =a

b

y

1=ay

b

If ay was an integer, ay = n, y =n

a, but y is irrational. = xy is irrational.

x

y

y is not an integer

Exercise 8. Proof by counterexamples. We want that the sum or product of 2 irrational numbers is not always irrational. If y

is irrational, y + 1 is irrational, otherwise, if y + 1 Q, y Q by closure under addition.= y + 1 y = 1

Likewise, y 1y = 1.

Exercise 9.

y x > 0 = n(y x) > k, n Z+, k arbitrary. Choose k to be irrational. Then k/n irrational.

y >k

n+ x > x. Let z = x+

k

n, z irrational .

Exercise 10.

(1) Suppose n = 2m1 and n+ 1 = 2m2.

2m1 + 1 = 2m2 2(m1 m2) = 1 m1 m2 = 12. But m1 m2 can only be an integer.

(2) By the well-ordering principle, if x Z+ is neither even and odd, consider the set of all x. There must exist asmallest element x0 of this set. But since x0 Z+, then there must exist a n < x such that n+ 1 = x0. n is even orodd since it doesnt belong in the above set. So x0 must be odd or even. Contradiction.

(3)(2m1)(2m2) = 2(2m1m2) even

2m1 + 2m2 = 2(m1 +m2) even

(2m1 + 1) + (2m2 + 1) = 2(m1 +m2 + 1) = sum of two odd numbers is even(n1 + 1)(n2 + 1) = n1n2 + n1 + n+ 2 + 1 = 2(2m1m2)

2(2m1m2) (n1 + n2) 1 odd, the product of two odd numbers n1, n2 is odd(4) If n2 even, n is even, since for n = 2m, (2m)2 = 4m2 = 2(2m2) is even.

a2 = 2b2. 2(b2) even. a2 even, so a even.

If a even a = 2n.a2 = 4n2

If b odd , b2 odd. b has no factors of 2 b2 6= 4n2

Thus b is even.8

(5) For pq , If p or q or both are odd, then were done.Else, when p, q are both even, p = 2lm, q = 2np,m, p odd.

p

q=

2lm

2np=

2lnmp

and at least m or p odd

Exercise 11. ab can be put into a form such that a or b at least is odd by the previous exercise.

However, a2 = 2b2, so a even, b even, by the previous exercise, part (d) or 4th part. Thus ab cannot be rational.

Exercise 12. The set of rational numbers satisfies the Archimedean property but not the least-upper-bound property.

Since pq Q R, np1q1 >p2q2

since if q1, q2 > 0,

np1q2q1q2

>q1p2q1q2

np1q2 > q1p2

n exists since (p1q2), (q1p2) R.The set of rational numbers does not satisfy the least-upper-bound property.

Consider a nonempty set of rational numbers S bounded above so that x = rs S, x < b.

Suppose x < b1, x < b2x S.r

s< b2 < nb1 but likewise

r

s< b1 < mb2, n,m Z+

So its possible that b1 > b2, but also b2 > b1.

I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, Thewell-ordering principle. Consider these 2 proofs.

N +N + +N = N2

(N 1) + (N 2) + + (N (N 1)) + (N N) = N2 Nj=1

j =

N1j=1

j

N2 +N = 2

Nj=1

j =Nj=1

j =N(N + 1)

2

An interesting property is that

S =

nj=m

j =

nj=m

(n+m j)

So thatNj=1

j =

Nj=m

j +

mj=1

j =

Nj=m

j +m(m+ 1)

2=N(N + 1)

2

Nj=m

j =N(N + 1)m(m+ 1)

2=

(N m)(N +m+ 1)2

Another way to show this is the following.

S = 1+ 2+ + (N 2)+ (N 1)+ Nbut S = N+ N 1+ + 3+ 2+ 1

2S = (N + 1)N S =N(N + 1)

29

Telescoping series will let you getNj=1 j

2 and other powers of j.

Nj=1

(2j 1) = 2N(N + 1)2

N = N2

Nj=1

(j2 (j 1)2) =Nj=1

(j2 (j2 2j + 1)) =Nj=1

(2j 1) = 2(N(N + 1)

2

)N = N2

Nj=1

(j3 (j 1)3) = N3 =Nj=1

(j3 (j3 3j2 + 3j 1)) =Nj=1

(3j2 3j + 1)

= 3Nj=1

j2 = 3N(N + 1)2

+N = N3 = 2N3 + 2N 3N2 3N

2=N(N + 1)(2N + 1)

6=

Nj=1

j2

Nj=1

j4 (j 1)4 = N4 =Nj=1

j4 (j4 4j3 + 6j2 4j + 1) =Nj=1

4j3 6j2 + 4j 1 =

= 4

Nj=1

j3 6N(N + 1)(2N + 1)6

+ 4N(N + 1)

2N = N4

=Nj=1

j3 =1

4(N4 +N(N + 1)(2N + 1) 2N(N + 1) +N) = 1

4(N4 + (2N)N(N + 1)N(N + 1) +N)

=1

4(N4 + 2N3 + 2N2 N2 N +N) = 1

4N2(N2 + 2N + 1) =

1

4

(N(N + 1))2

2

Exercise 1. Induction proof.

1(1 + 1)

2

N+1j=1

j =

nj=1

j + n+ 1 =n(n+ 1)

2+ n+ 1 =

n(n+ 1) + 2(n+ 1)

2=

(n+ 2)(n+ 1)

2

Exercise 6.

(1)

A(k + 1) = A(k) + k + 1 =1

8(2k + 1)2 + k + 1 =

1

8(4k2 + 4k + 1) +

8k + 8

8=

(2k + 3)2

8(2) The n = 1 case isnt true.(3)

1 + 2 + + n = (n+ 1)n2

=n2 + n

2 1 + 2x+ 2x2

1 + 2x+ x2 > 1 + 2x+ 2x2

0 > x2 = Impossible(1 + x)3 = 1 + 3x+ 3x2 + x3 > 1 + 3x+ 3x2

= x3 > 0By well-ordering principle, we could argue that n = 3 must be the smallest number such that (1 + x)n > 1 + 2x+ 2x2. Orwe could find, explicitly

(1 + x)n =

nj=0

(n

j

)xj = 1 + nx+

n(n 1)2

x2 +

nj=3

(n

j

)xj

10

andn(n 1)

2> n

n2 n > 2nn2 > 3n

n > 3

Exercise 8.a2 ca1, a3 ca2 c2a1an+1 can ca1cn1 = a1cn

Exercise 9.

n = 1,

1 = 112 + 12 =

2

(

2)2 + 12 =

3(n)2 + 12 =

n+ 1

Exercise 10.1 = qb+ r

q = 0, b = 1, r = 1

2 = qb+ r, q = 0, r = 2, b = 1, 2 or r = 0, q = 2; q = 1, r = 0

Assume n = qb+ r; 0 r < b; b Z+, b fixedn+ 1 = qb+ r + 1 = qb+ 1 + r = qb+ 1 + b 1 = (q + 1)b+ 0

Exercise 11. For n > 1, n = 2, 3 are prime. n = 4 = 2(2), a product of primes.

Assume the k 1th case. Consider kj , 1 j k.If kj Z+, only for j = 1, j = k, then k prime.If kj Z+, for some 1 < j < k, kj = c Z+. c, j < k.

Thus k = cj. c, j are products of primes or are primes, by induction hypothesis. Thus k is a product of primes.

Exercise 12. n = 2. G1, G2 are blonde. G1 has blue eyes. Consider G2. G2 may not have blue eyes. Then G1, G2 are not all

blue-eyed.

I 4.7 Exercises - Proof of the well-ordering principle, The summation notation. Exercise 1.

(1) n(n+1)2 =4k=1 k = 10

(2)5n=2 2

n2 =3n=0 2

n = 1 + 14 = 15

(3) 23r=0 2

2r = 23r=0 4

r = 170

(4)4j=1 j

j = 1 + 4 + 27 + 44 = 288

(5)5j=0(2j + 1) = 2

5(6)2 + 6(1) = 36

(6)

1k(k+1) =

nk=1

(1k 1k+1

)= 1 1n+1 = nn+1

Exercise 9.

n = 1(1)(3) + 5 = 2 = 2nn = 2(1)(3) + 5 + (1)7 + 9 = 4 = 2n

n

2nk=1

(1)k(2k + 1) = 2n

n+ 1

2(n+1)k=1

(1)k(2k + 1) =2nk=1

(1)k(2k + 1) + (1)2n+1(4n+ 3) + (1)2n+2(4n+ 5) =

= 2n+ 2 = 2(n+ 1)

Exercise 10.11

(1) am + am+1 + + am+n(2)

n = 11

2=

1

1 1

2=

1

2

n+ 1

2(n+1)k=n+2

1

k=

2nm=1

(1)m+1m

1n+ 1

+1

2n+ 1+

1

2n+ 2=

2nm=1

(1)m+1m

+ 12(n+ 1)

+(1)2n+1+1

(2n+ 1)

=

2(n+1)m=1

(1)m+1m

Exercise 13.

n = 12(

2 1) < 1 < 2 since 12>

2 1

n case (n+ 1n)(n+ 1 +n) = n+ 1 n = 1 1

So then, using the telescoping property,n1n=1

2(n+ 1n) = 2(m 1) x

x3 > x2 > x

xn+1 = xnx > x2 > x

0 < x < 1

x2 < x

X3 < x2 < x

xn+1 = xnx < x2 < x = xn+1 < x

Exercise 11. Let S = {n Z+|2n < n!}.By well-ordering principle, smallest n0 S. Now24 = 16, 4! = 24. So S starts at n = 4.

Exercise 12.

13

(1) (1 +

1

n

)n=

nj=0

(n

k

)(1

n

)j=

nk=0

n!

(n k)!k!(

1

n

)kk1r=0

(1 r

n

)=

k1r=0

(n rn

)=

(1

nk

)n!

(n k)!nk=1

1

k!

k1r=0

(1 r

n

)=

(1

nk

)n!

(n k)!(2)

(1 +1

n)n = 1 +

nk=1

(1

k!

k1r=0

(1 rn

)

)< 1 +

nk=1

1

k!< 1 +

nk=1

1

2k= 1 +

12

(12

)n+112

= 1 + (1(

1

2

)n)

< 3

The first inequality obtained from the fact that if 0 < x < 1, xn < x < 1. The second inequality came from theprevious exercise, that 1k! > 2

Exercise 13.

(1)

S =

p1k=0

(b

a

)k=

1 ( ba)p1 ba

p1k=0

bkap1k = ap11 ( ba)p

1 ba=bp apb a

(2)(3) Given

np

> (p+ 2) + 2(2p+1 (p+ 1)) + 2p(2) = p+ 2p+2 + 2p+1 > 2p+2

So the n = 2 case is true for all p Z+.Assume nth case is true. We now prove the n+ 1 case.

nk=1

kp =

n1k=1

kp + np x )If x is an integer [x] = [x]

(3) Let x = q1 + r1, y = q2 + r2; 0 r1, r2 < 1.

= [q1 + q2 + r1 + r2] =

{q1 + q2

q1 + q2 + 1 if r1 + r2 1[x] + [y] = q1 + q2 [x] + [y] + 1 = q1 + q2 + 1

(4)

If x is an integer , [2x] = 2x = [x] + [x+1

2] = [x] + [x] = 2x

[x] + [x+1

2] = q +

{q if r < 122q + 1 if r > 12

[2x] = [2(q + r)] = [2q + 2r] =

{2q if r < 122q + 1 if r > 12

(5)

[x] + [x+1

3] + [x+

2

3] = q +

{q if r < 23q + 1 if r > 23

+

{q if r < 13q + 1 if r > 13

=

3q if r < 133q + 1 if 13 < r 23

[3x] = [3(q + r)] = [3q + 3r] =

3q if r < 133q + 1 if 13 < r 23

Exercise 5. Direct proof.

[nx] = [n(q + r)] =

nq if r < 1nnq + 1 if 1n < r n1nExercise 6.

a(R) = hk = IR +1

2BR 1

bn=a

[f(n)] = [f(a)] + [f(a+ 1)] + + [f(b)]

[f(n)] = g f(n), g Z, so that if f(n) is an integer,g = f(n), and if f(n) is not an integer, g is the largest integer suchthat g < f(n), so that all lattice points included and less than g are included.

Exercise 7.

19

(1) Consider a right triangle with lattice points as vertices. Consider b+ 1 lattice points as the base with b length.Start from the vertex and move across the base by increments of 1.The main insight is that the slope of the hypotenuse of the right triangle is ab so as we move 1 along the base, the

hypotenuse (or the y-value, if you will) goes up by ab . Now

(5)[nab

]= number of interior points at x = n and below the hypotenuse line of the right triangle of sides a, b,

including points on the hypotenuse

b1n=1

[nab

]+

1

2((a+ 1) + b) 1 = ab

2

Now(a 1)(b 1)

2=ab

2 a

2 b

2+

1

2

=b1n=1

[nab

]=

(a 1)(b 1)2

(2) a, b Z+b1n=1

[nab

]=

b1n=1

[a(b n)

b

](reverses order of summation)

b1n=1

[a an

b

]=

{b1n=1 [anb a] if anb a4 is an integer (but a (nb 1) cant be!)b1n=1 ([anb a] 1) otherwise

= b1n=1

([anb a] 1)

= b1n=1

([anb

] a) (b 1) =

= b1n=1

[anb

]+ a(b 1) (b 1)

b1n=1

[nab

]=

(a 1)(b 1)2

Exercise 8. Recall that for the step function f = f(x), theres a partition P = {x0, x1, . . . , xn} of [a, b] such that f(x) = ckif x Ik.

Given that s(x) =

{1 x S0 x / S .

If x [a, b], then x must only lie in one open subinterval Ij , since real numbers obey transitivity.nk=1

ckIk(x) = cj for x Ij =nk=1

ckIk(x) = f(x)x [a, b]

1.15 Exercises - The definition of the integral for step functions, Properties of the integral of a step function, Othernotations for integrals. Exercise 1.

(1) 3

1[x]dx = (1) + 1 + (2) = 2

(2) 31[x+

12 ]dx =

7/21/2[x]dx = (1) 12 + (1)(1) + (2)(1) + 123 = 4

(3) 31([x] + [x+

12 ])dx = 6

(4) 31 2[x]dx = 4

(5) 31[2x]dx =

12

62[x]dx =

12 ((2)1 + (1) + (1) + 2 + 3 + 4 + 5) = 6

(6) 31[x]dx =

31

[x]dx = 13[x]dx = 3 +2 +1 = 6

Exercise 2.

s =

{5/2 if 0 < x < 21 if 2 < x < 5

20

Exercise 3. [x] = y x so y x.

y 1 x, otherwise if y 1 x, y + 1 x and so y wouldnt be the largest integer x.= [x] + [x] = y y 1 = 1

Or use Exercise 4(c), pp. 64. ba

([x] + [x])dx = ba

[x x]dx = ba

(1)dx = a b

Exercise 4.

(1) n Z+, n0

[t]dt =n1t=0 t =

(n1)(n1+1)2 =

(n1)n2

(2)

Exercise 5.

(1) 2

0[t2]dt =

21

[t2]dt = 1(

2 1) + 2(32) + 3(23) = 523(2)

33[t

2]dt = 3

0[t2]dt+

03[t

2]dt = 3

0[t2]dt+ 0

3[t2]dt = 2

30

[t2]dt 32

[t2]dt = 4(

5 2) + 5(

6

5) + 6(

7

6) + 7(

8

7) + 8(3

8)

16

5

6

7

8 20

[t2]dt+

32

[t2]dt = 21 3

2

3

5

6

7

8

= 33

[t2]dt = 42 2(3

2 +

3 +

5 +

6 +

7)

Exercise 6.

(1) n

0[t]2dt =

n1

[t]2dt =n1j=1 j

2 = (n1)n(2n1)6(2)

x0

[t]2dt =[x1]j=1 j

2 + q2r where x = q + r, q Z+, 0 r < 1. x0

[t]2dt =q(q 1)(2q 1)

6+ q2r = 2(x 1) = 2(q + r 1)

= q(q 1)(2q 1) + 6q2r = 12q + 12r 12= x = 1, x = 5/2

Exercise 7.

(1) 90

[t]dt =

91

[t]dt = 3(1) + 5(2) = 13 1

0

6[t]dt = 3(1) + 5(2) + 7(3) = 34 =

(4)(3)(17)

6

Assume n2

0

[t]dt =

n(n 1)(4n+ 1)6 (n+1)2

0

[t]dt =

n20

[t]dt+

(n+1)2n2

[t]dt =

n(n 1)(4n+ 1)6

+ n((n+ 1)2 n2) =

=(n2 n)(4n+ 1) + 6n(2n+ 1)

6=

4n3 + n2 4n2 n+ 12n2 + 6n6

=4n3 + 9n2 + 5n

6indeed ,

(n+ 1)(n)(4(n+ 1) + 1)

6=

(n2 + n)(4n+ 5)

6=

4n3 + 5n2 + 4n2 + 5n

621

Exercise 8. b+ca+c

f(x)dx = b+cca+cc f(x (c))dx =

baf(x+ c)dx

Exercise 9. kbkaf(x)dx = 11

k

(kb)/k(ka)/k

f(

x1/k

)dx = k

baf(kx)dx

Exercise 10. Given s(x) = (1)nn if n x < n+ 1;n = 0, 1, 2, . . . p 1; s(p) = 0, p Z+. f(p) = p0s(x)dx.

So for f(3) = 3

0s(x)dx, we need to consider n = 0, 1, 2.

s(0 x < 1) = 0s(1 x < 2) = (1)(1)s(2 x < 3) = 2;s(3 x < 4) = 3

So thenf(3) = (1)(1) + 2(1) = 1f(4) = 1 + (3)(1) = 2f(f(3)) = f(1) = 0

We obtain this formula

f(p) =

{p2 (1)p+1 p evenp1

2 (1)p+1 p evensince

f(p+ 1) = f(p) +

p+1p

s(x)dx ={p1

2 (1)p+1 p even + (1)pp

=

{p2 p evenp1

2 p odd+

{p

p ={p2p1

2

=

=

{ (p+1)2 if p+ 1 evenp2 if p+ 1 odd

Thus, p = 14, p = 15.

Exercise 11.

(1) ba

s(x)dx =

nk=1

s3k(xk xk1) ba

s+

cb

s =

n1k=1

s2k(xk xk1) +n2k=n1

s3k(xk xk1) =n2k=1

s3k(xk xk1) = ca

s(x)dx

(2) ba

(s+ t) =n3k=1(s+ t)

3k(xk xk1) 6=

bas+

bat

(3) bacs =

nk=1(cs)

3(xk xk1) 6= c bas

(4) Consider these facts that are true, that xk1 < x < xk, s(x) = sk; x0 = a+ c, xn = b+ c,xk1 c < x c < xl c = yk1 < y < yk so then s(y + c) = sk.

nk=1

s3k(xk xk1) =kk=1

s3k(xk c (xk1 c)) =

=

nk=1

s3k(yk yk1) = ba

s(y + c)dy

(5) s < t, bas =

nk=1 s

3k(xk xk1).

if 0 < s, s3 < s2t < st2 < t3

if s < 0t, s3 < and t3 > 0

if s < t < 0, s3 < s2t, s(st) < t(ts) = t2sts > t2

t2s < t3

s3 < s2t < t2s < t3

22

Then bas 0, basdx =

nk=1 sk(x

2k x2k1) 0) ba

s(x)dx =

nk=1

sk(xk xk1) =n3l=1

sl(zl zl1) 0). So cx3 < x2 (since x2 > 0).

f g =x2 cx3 =

(1

3x3 c

4x4)1/c

0

=1

12c3f g = 2

3=

1

12c3; c =

1

2

2

Exercise 16. f = x(1 x), g = ax.f g =

1a0

x x2 ax =(

(1 a)12x2 1

3x3)1a

0

= (1 a)3 16

= 9/2 = a = 2

Exercise 17. pi = 2 11

1 x2dx(1) 3

3

9 x2dx = 3

33

1

(x3

)2= 3(3)

11

1 x2 = 9pi

2

Now kbka

f(xk

)dx = k

ba

fdx

(2) 20

1 1

4x2dx = 2

10

1 x2dx = 2pi

4=

pi

2

(3) 22(x 3)

4 x2dx 2

2x

4 x2dx = (1) 2

2

x

4 x2 = 2 22x

4 x2 = 0

3 22

2

1

(x2

)2dx = (6)(2)

11

1 x2 = 6pi

Exercise 18. Consider a circle of radius 1 and a twelve-sided dodecagon inscribed in it. Divide the dodecagon by isosceles

triangle pie slices. The interior angle that is the vertex angle of these triangles is 360/12 = 30 degrees.

Then the length of the bottom side of each triangle is given by the law of cosines:

c2 = 1 + 1 2(1)(1) cos 30 = 2(

1

3

2

)= c =

2

1

3

2

25

The height is given also by the law of cosines

h = 1 cos 15 =

1 + cos 30

2=

1 +

3

2

2

The area of the dodecagon is given by adding up twelve of those isosceles triangles

(12)1

2

1 + 32

(12

)2

1

3

2

= 3So 3 < pi.

Now consider a dodecagon thats circumscribing the circle of radius 1.

(12)1

2

21 32

1 +

32

(1) = 12(2 32

)> pi

Exercise 19.

(1) (x, y) E if x = ax1, y = by1 such that x21 + y21 1= (xa)2 + (yb )2 = 1

(2)

y = b

1

(xa

)22

aab

1

(xa

)2= 2ba

11

1 x2 = bapi

2(2) = piba

Exercise 20. Let f be nonnegative and integrable on [a, b] and let S be its ordinate set.

Suppose x and y coordinates of S were expanded in different ways x = k1x1, y = k2y1.If f(x1) = y1, g(x) = k2f

(xk1

)= k2y1 = y.

integrating g on [k1a, k1b], k1bk1a

g(x)dx =

k1bk1a

k2f

(x

k1

)dx = k2k1

ba

f(x)dx = k2k1A

2.8 Exercises - The trigonometric functions, Integration formulas for the sine and cosine, A geometric description ofthe sine and cosine functions. Exercise 1.

(1) sinpi = sin 0 = 0. sine is periodic by 2pi, so by induction, sinnpi = 0.

sin 2(n+ 1)pi = sin 2pin+ 2pi = sin 2pin = 0

sin (2(n+ 1) + 1)pi = sin (2n+ 3)pi = sin ((2n+ 1)pi + 2pi) = sin (2n+ 1)pi = 0

(2) cospi/2 = cospi/2 = 0by induction, cospi/2 + 2pij = cospi/2(1 + 4j)

cospi/2 + 2pij = cos (4j 1)pi/2, j Z+

Exercise 2.

(1) sinpi/2 = 1, sinpi/2(1 + 4j) = 1, j Z+.(2) cosx = 1, cos 0 = 1, cos 2pij = 1

Exercise 3.

sinx+ pi = sinx+ pi/2 + pi/2 = cosx+ pi/2 = sinxcosx+ pi = cosx+ pi/2 + pi/2 = sinx+ pi/2 = cosx

Exercise 4.26

sin 3x = sin 2x cosx+ sinx cos 2x = 2 sinx cos2 x+ sinx(cos2 x sin2 x) = 3 cos2 x sinx sin3 x == 3(1 sin2 x) sinx sin3 x = 3 sinx 4 sin3 x

cos 3x = cos 2x cosx sin 2x sinx = (cos2 x sin2 x) cosx (2 sinx cosx) sinx = cosx 4 sin2 x cosxcos 3x = 3 cosx+ 4 cos3 x

Exercise 5.

(1) This is the most direct solution. Using results from Exercise 4 (and it really helps to choose the cosine relationship,not the sine relationship),

cos 3x = 4 cos3 x 3 cosxx = pi/6

cos 3pi/6 = 0 = 4 cos3 pi/6 3 cospi/6 = cospi/6(4 cos2 pi/6 3) = 0= cospi/6 =

3/2, sinpi/6 = 1/2( by Pythagorean theorem )

(2) sin 2pi/6 = 2 cospi/6 sinpi/6 =

32, cospi/3 = 1/2 (by Pythagorean theorem)(3) cos2pi/4 = 0 = 2 cospi/4 1, cospi/4 = 1/2 = sinpi/4

Note that the most general way to solve a cubic is to use this formula. For x3 + bx2 + cx+ d = 0,

R =9bc 27d 2b3

54

Q =3c b2

9

S = (R+Q3 +R2)1/3

T = (RQ3 +R2)1/3

x1 = S + T b/3x2 = 1/2(S + T ) b/3 + 1/2

3(S T )x3 = 1/2(S + T ) b/3 1/2

3(S T )Exercise 6.

tanx y = sinx ycosx y =

sinx cos y sin y cosxcosx cos y + sinx sin y

(1

cos x cos y1

cos x cos y

)=

tanx tan y1 + tanx tan y

if tanx tan y 6= 1Similarly,

tanx+ y =sinx+ y

cosx+ y=

sinx cos y + sin y cosx

cosx cos y sinx sin y =tanx+ tan y

1 tanx tan y , tanx tan y 6= 1

cotx+ y =cosx+ y

sinx+ y=

cosx cos y sinx sin ysinx cos y + sin y cosx

=cotx cot y 1cot y + cotx

Exercise 7. 3 sinx+ pi/3 = A sinx+B cosx = 3(sinx 12 +

32 cosx) =

32 sinx+

3

32 cosx

Exercise 8.C sinx+ = C(sinx cos+ cosx sin) = C cos sinx+ C sin cosx

A = C cos,B = C sin

Exercise 9. If A = 0, B cosx = B sinpi/2 + x = C sinx+ so C = B, = pi/2 if A = 0.

If A 6= 0,A sinx+B cosx = A(sinx+

B

Acosx) == A(sinx+ tan cosx)

=A

cos(cos sinx+ sin cosx) =

A

cos(sinx+ )

where pi/2 < < pi/4, B/A = tan, C = AcosExercise 10. C sinx+ = C sinx cos+ C cosx sin.

C cos = 2, C sin = 2, C = 2

2, = pi/4

Exercise 11. If A = 0, C = B, = 0. If B = 0, A = C, = pi/2. Otherwise,27

A sinx+B cosx = B(cosx+A

Bsinx) =

B

cos(cosx cos + sin sinx) = C cosx+

where AB = tan, = , C = Bcos .Exercise 12.

sinx = cosx =

1 cos2 x = cosx = 1/

2 = x = pi4

Try 5pi/4. sin 5pi/4 = cos 3pi/4 = sinpi/4 = 1/2.cos 5pi/4 = sin 3pi/4 = cospi/4 = 1/2. So sin 5pi/4 = cos 5pi/4. x = 5pi/4 must be the other root.

So = pi/4 + pin (by periodicity of sine and cosine).

Exercise 13.

sinx cosx = 1 =

1 cos2 x = 1 + cosx= 1 cos2 x = 1 + 2 cosx+ cos2 x = 0 = 2 cosx(1 + cosx)

cosx = 1, x = pi/2 + 2pinExercise 14.

cosx y + cosx+ y = cosx cos y + sinx sin y + cosx cos y sinx sin y = 2 cosx cos ycosx y cosx+ y = sinx cos y sin y cosx+ sinx cos y + sin y cosx = 2 sinx cos ysinx y + sinx+ y = sinx cos y sin y cosx+ sinx cos y + sin y cosx = 2 sinx cos y

Exercise 15.

sinx+ h sinxh

=sin (x+ h/2) cosh/2 + cos (x+ h/2) sinh/2 sin (x+ h) cosh/2 cosx+ h/2 sinh/2

h

=sinh/2

h/2cos (x+ h/2)

cosx+ h cosxh

=cos (x+ h/2) cosh/2 sin (x+ h/2) sinh/2 (cos (x+ h/2) cosh/2 + sin (x+ h/2) sinh/2)

h

= sinh/2h/2

sin (x+ h/2)

Exercise 16.

(1)sin 2x = 2 sinx cosx

if sin 2x = 2 sinx and x 6= 0, x 6= pin, cosx = 1 but x 6= pin = x = 2pin(2) cosx+ y = cosx cos y sinx sin y = cosx+ cos y.

cosx cos y cosx cos y = sin y

1 cos2 xLetting A = cosx,B = cos y,

A2B2 +A2 +B2 2A2B 2AB2 + 2AB = 1A2 B2 +A2B2A2 +B2 A2B AB2 +AB = 1/2

B2(1A) +B(AA2) +A2 1/2 = 0

B =A(1A)A2(1A)2 4(1A)(A2 1/2)

1A = A1

1A (A2(1A) 4(A2 1/2))1/2 =

= A 11A (3A

2 A3 + 2)1/2

Note that 1 B 1, but for |A| 1.Solve for the roots of 3A2 A3 + 2, A0 = 1,1 +

3,1 3. So suppose cosx = 9/10. Then there is

no real number for y such that cos y would be real and satisfy the above equation.(3) sinx+ y = sinx cos y + sin y cosx = sinx+ sin y

= sin y(1 cosx) + sin y + cosx sin y = 0,= y = 2pinChecking our result, we find that sin (2pin+ y) = sin 2pin+ sin y(1)

28

(4) y0

sinxdx = cosx|y0 = (cos y 1) = 1 cos y = sin y

= 1 cos y =

1 cos2 y

1 2 cos y + cos2 y = 1 cos2 y = cos y(cos y 1) = 0; y = 2(j + 1)pi2

, 2pin

Exercise 17. ba

sinxdx = cosx|ba = cos b+ cos a

(1)

32 + 1

(2)

22 + 1

(3) 12(4) 1(5) 2(6) 0 We were integrating over one period, over one positive semicircle and over one negative semicircle.(7) 0 We had integrated over two equal parts, though it only shaded in up to x = 1.(8)

2

2 +

32

Exercise 18. pi

0(x+ sinx)dx = ( 12x

2 cosx)pi0

= pi2

2 (1 1) = pi2

2 + 2

Exercise 19. pi/2

0(x2 + cosx)dx = ( 13x

3 + sinx)pi/20

= 13 (pi/2)3 + 1

Exercise 20. pi/2

0(sinx cosx)dx = ( cosx sinx)|pi/20 = 1 (1) = 0

Exercise 21. pi/2

0| sinx cosx|dx = ( by symmetry )2 pi/4

0(cosx sinx)dx = 2(sinx+ cosx)|pi/40 = 2(

2 1)

Exercise 22. pi

0( 12 + cos t)dt = (

12 t+ sin t)

pi0

= pi2

Exercise 23. 2pi/30

(1

2+ cos t)dt+

pi2pi/3

(12

+ cos t)dt = (t

2+ sin t)

2pi/30

+ (t

2+ sin t)

2pi/3pi

= 2(pi

3+

3

2) pi

2=pi

6+

3

Exercise 24. If pi < x 2pi3 , xpi(1

2+ cos t)dt =

pix

(1

2+ cos t)dt =

(t

2+ sin t

)pix

= pi2 x

2 sinx

If 2pi/3 x 2pi/3, 2pi/3pi

(12

+ cos t)dt+

x2pi/3

(1

2+ cos t)dt =

pi/6+

3/2 + (t/2 + sin t)|x2pi/3

= x/2 + sinx pi/3

3/2 +

3/2 pi/6 = x2

+ sinx pi/3If 2pi/3 x pi,

3/2 +

x2pi/3

(1/2 + cos t)dt =

3/2 + (t/2 + sin t)|2pi/3x = pi/3 +

3 x/2 sinx

Exercise 25. x2x

(t2 + sin t)dt = ( 13 t3 + cos t) = x6x33 + cosx cosx2

Exercise 26. pi/2

0sin 2xdx =

( cos (2x)

2

)pi/20

= (1/2)(1 1) = 1

Exercise 27. pi/3

0cosx/2dx = 2 sinx/2|pi/30 = 2 12 = 1

Exercise 28.29

x0

cos (a+ bt)dt =

x0

(cos a cos bt sin a sin bt)dt =(cos a

bsin bt sin a( cos bt/b)

)x0

=

=cos a

bsin bx+

sin a

b(cos bx 1) = 1

bsin a+ bx sin a/b x

0

sin (a+ bt)dt =

x0

(sin a cos bt+ sin bt cos a)dt =

(sin a

bsin bt cos a

bcos bt

)x0

=

=1

b(cos bx+ a+ cos a)

Exercise 29.

(1) x0

sin3 tdt =

x0

3 sin t sin 3t4

dt =

(3

4cos t+ cos 3t/12

)x0

= 3/4(cosx 1) + cos 3x 112

=

=1

3 3

4cosx+

1

12(cos 2x cosx sin 2x sinx) = 2/3 1/3 cosx(2 + sin2 x)

(2) x0

cos3 tdt =

x0

1

4(cos 3t+ 3 cos t)dt =

(1

4

sin 3t

3+

3

4sin t

)x0

=

=1

12(sin 2x cosx+ sinx cos 2x) +

3

4sinx =

1

12(2 sinx cosx+ sinx(2 cos2 x 1)) =

=sinx cos2 x+ 2 sinx

3

Exercise 30. Now using the definition of a periodic function,

f(x) = f(x+ p); f(x+ (n+ 1)p) = f(x+ np+ p) = f(x+ np) = f(x)

and knowing that we could write any real number in the following form,

a = np+ r; 0 < p, r R;n Zthen a+p

a

f(x)dx =

r+pr

f(x+ np)dx =

r+pr

f(x)dx =

pr

f +

r+pp

f(x)dx =

=

pr

f +

r0

f(x p)dx = pr

f +

r0

f =

p0

f

Exercise 31.

(1) 2pi0

sinnxdx =1

n

2pin0

sinxdx =1

n( cosx)

2pin0

= 1n

(1 1) = 0 2pi0

cosnxdx =1

n

2pin0

cosxdx =1

nsinx

2pin0

= 0

(2) 2pi0

sinnx cosmxdx =

2pi0

1

2(sin (n+m)x+ sin (nm)x)dx = 0 + 0 = 0 2pi

0

sinnx sinmxdx =

2pi0

1

2(cos (nm)x+ cos (n+m)x)dx = 0 + 0 = 0 2pi

0

cosnx cosmxdx =

2pi0

1

2(cos (nm)x+ cos (n+m)x)dx = 0 + 0 = 0

While 2pi0

sin2 nxdx =

2pi0

1 cos 2nx2

dx = pi 2pi0

cos2 nxdx =

2pi0

1 + cos 2nx

2dx = pi

30

Exercise 32. Given that x 6= 2pin; sinx/2 6= 0,nk=1

2 sinx/2 cos kx = 2 sinx/2

nk=1

cos kx =

nk=1

sin (2k + 1)x

2 sin (2k 1)x

2= sin (2n+ 1)

x

2 sinx/2

= sinnx cosx/2 + sinx/2 cosnx sinx/2 == 2 sinnx/2 cosnx/2 cosx/2 + sinx/2(1 2 sin2 nx/2) sinx/2 == 2(sinnx/2)(cos (n+ 1)x/2)

Exercise 33. Recall that

cos (2k + 1)x/2 cos (2k 1)x/2 = cos kx+ x/2 cos kx x/2 == cos kx cosx/2 sin kx sinx/2 (cos kx cosx/2 + sin kx sinx/2) == 2 sin kx sinx/2

2 sinx/2nk=1

sin kx =

nk=1

(cos (2k + 1)x/2 cos (2k 1)x/2) = cos (2n+ 1)x/2 cosx/2 =

= cosnx+ x/2 cosx/2Now

sinnx/2 sinnx/2 + x/2 = sinnx/2(sinnx/2 cosx/2 + sinx/2 cosnx/2) =

= sin2 nx/2 cosx/2 + sinx/2 cosnx/2 sinnx/2 =

=

(1 cosnx

2

)cosx/2 +

sinnx

2sinx/2 =

=1

2(cosx/2 cosx/2 cosnx+ sinnx sinx/2) = 1

2(cosx/2 cos (nxx/2)

Then

2 sinx/2nk=1

sin kx = 2 sinnx/2 sin 12

(n+ 1)x

nk=1

sin kx =sinnx/2 sin 12 (n+ 1)x

sinx/2

Exercise 34. Using triangle OAP, not the right triangle, if 0 < x < pi/2

1

2cosx sinx pi/2, sinx < 0,| sinx| = sinx = sinx = sin |x| < |x|

2.17 Exercises - Average value of a function. Exercise 1. 1bax2dx = 13 (b

2 + ab+ a2)

Exercise 2. 110x2 + x3 = 712

Exercise 3. 140x1/2 = 43

Exercise 4. 181x1/3 = 4528

Exercise 5. 1pi/20 pi/2

0sinx = 2pi

Exercise 6. 1pi/2pi/2

cosx = 2/pi

Exercise 7. 1pi/20

sin 2x = 1/pi(1 1) = 2/pi

Exercise 8. 1pi/40

sinx cosx = 1pi31

Exercise 9. 1pi/20

sin2 x = 1pi (x sin 2x/2)pi0

= 12

Exercise 10. 1pi0

cos2 x = 12

Exercise 11.

(1) 1a0x2 = a2/3 = c2 = c = a/3

(2) 1a0xn = 1a

1n+1x

n+1a0

= an

n+1 = cn = c = a

(n+1)1/n

Exercise 12.

A =

wf/

w

wx2 = k

x

x3 =1

4x4 = k

1

2x2; k =

1

2, w = x

x4 =1

5x5 = k

1

3x3; k =

3

5, w = x2

x5 =1

6x6 = k

1

4x4; k =

2

3, w = x3

Exercise 13.

A(f + g) =1

b af + g =

1

b af +

1

b ag = A(f) +A(g)

A(cf) =1

b acf = c

(1

b a)

f

A(f) =1

b af 1

b ag = A(g)

Exercise 14.

A(c1f + c2g) =

w(c1f + c2g)

w=c1wfw

+c2wgw

= c1A(f) + c2A(g)

f g w > 0( nonnegative ),= wf wg

Exercise 15.

Aba(f) =1

b a ba

f =1

b a

( ca

f +

bc

f

)=

(c ab a

)( caf

c a

)+b a (c a)

b a

baf

b ca < c 2.

1

4x4 1

2x2 =

1

3

(1

2t2 1

4t4)x

2

=1

6x2 1

12x4

= 13x4 2

3x2 = 0 = x = 0, x =

2 1

0(t3 t)dt+ 2

1(t3 t)dt cancel each other out.

Exercise 18. f(x) = x [x] 12 if x is not an integer; f(x) = 0 if x Z.For any real number, x = q + r, 0 r < 1, q Z. So then

x [x] = rf(x) = r 1

2

(1) To show the periodicity, consider

f(x+ 1) = x+ 1 [x+ 1] 12

= r 12

= f(x) sincex+ 1 = q + 1 + r, [x+ 1] = q + 1

x+ 1 [x+ 1] = r 12

(2) P (x) = x

0f(t)dt =

x0

(t 12 ) = 12x2 12x because given 0 < x 1, then q = 0 for x, so we can use r = t.To show periodicity,

P (x+ 1) =

x+10

f(t)dt =

10

f(t)dt+

x+11

f(t)dt = 0 +

x0

f(t+ 1)dt =

x0

f(t)dt = P (x)

since 1

0

f(t)dt =1

2(x2 x)

10

= 0

(3) Since P itself is periodic by 1, then we can consider 0 x < 1 only. Now x [x] = r and P (x) = 12 (r2 r). SoP (x) = 12 ((x [x])2 (x [x])).

(4) 10

(P (t) + c)dt = 0 = 1

0

P (t)dt = c

0 t 1 so P (t) = 12

(t2 t)

= 1

0

P (t)dt =1

2

(1

3t3 1

2t2)1

0

=1

2

16

= c = 112

(5) Q(x) = x

0(P (t) + c)dt

Q(x+ 1) =

x+10

(P (t) + c)dt =

10

(P (t) + c)dt+

x+11

(P (t) + c)dt =

= 0 +

x0

(P (t+ 1) + c)dt =

x0

(P (t) + c)dt = Q(x)

so without loss of generality, consider 0 x < 1

= Q(x) x

0

1

2(t2 t) + 1

12=

1

6x3 1

4x2 +

x

12

35

Exercise 19. g(2n) = 2pi

0f(t)dt

Consider 11f(t)dt =

10

f(t)dt+

01f(t)dt =

10

f(t)dt+1

1 0

1

f(1t)dt =

=

10

f +

01

f(t)dt = 0

Consider that 3

1f(t)dt =

11 f(t+ 2)dt =

11 f(t)dt = 0. Then, by induction, 2n+1

1

f =

2n11

f +

2n+12n1

f(t)dt = 0 +

11f(t+ 2n)dt =

11f(t)dt = 0

(1)

g(2n) =

10

f +

2n11

f +

2n2n1

f =

10

f +

01f(t)dt =

10

f + 0

1

f(t)dt

=

10

f +

01

f = 0

(2)

g(x) = x

0

f = x

0

f(t)dt = x

0

f(t)dt = g(x)

g(x+ 2) =

x+20

f(t)dt =

20

f +

x+22

f =

x0

f(t+ 2)dt =

x0

f(t)dt = g(x)

Exercise 20.

(1) g is odd since

g(x) = x

0

f(t)dt = x

0

f(t)dt = x

0

f(t)dt = g(x)Now

g(x+ 2) =

x+20

f =

20

f +

x+22

f = g(2) +

x0

f(t+ 2)dt = g(2) +

x0

f(t)dt = g(2) + g(x)

= g(x+ 2) g(x) = g(2)(2)

g(2) =

20

f =

21

f +

10

f =

21

f +A =

01f(t+ 2)dt+A =

01f(t)dt+A =

0

1

f(t)dt+A = 2Ag(5) g(3) = g(2)

g(3) = g(2) +

32

f(t)dt = 2A+

10

f(t+ 2)dt = 2A+A = 3A

= g(5) = 3A+ 2A = 5A(3) The key observation is to see that g must repeat itself by a change of 2 in the argument. To make g(1) = g(3) = g(5),

theyre different, unless A = 0!

Exercise 21. From the given, we can derive

g(x) = f(x+ 5), f(x) =

x0

g(t)dt

= f(5) = 5

0

g(t)dt = g(0) = 7

(1) The key insight I uncovered was, when stuck, one of the things you can do, is to think geometrically and drawa picture.

g(x) = f(x+ 5) = g(x) = f(x 5)= g(x) = f(x 5)

36

(2) 50

f(t)dt =

05f(t+ 5)dt =

05g(t)dt =

50

g(t)dt =

50

g(t)dt = 5

0

g(t)dt = f(5) = 7

(3) x0

f(t)dt =

x55

f(t+ 5)dt =

x55

g(t)dt =

x50

g +

05g = f(x 5) +

50

g(t)dt =

f(x 5) + 5

0

g(t)dt = f(x 5) + f(5) = g(x) + g(0)

where weve used f(x 5) = g(x) in the second and third to the last step.

3.6 Exercises - Informal description of continuity, The definition of the limit of a function, The definition of continuityof a function, The basic limit theorems. More examples of continuous functions, Proofs of the basic limit theorems.Polynomials are continuous.

Exercise 1. limx2 1x2 =1

limx2 x2= 14

Exercise 2. limx0(25x3+2)

limx0(75x72) = 1

Exercise 3. limx2(x2)(x+2)

(x2) = 4

Exercise 4. limx1(2x1)(x1)

x1 = 1

Exercise 5. limh0 t2+2th+h2t2

h = 2t

Exercise 6. limx0(xa)(x+a)

(x+a)2 = 1

Exercise 7. lima0(xa)(x+a)

(x+a)2 = 1

Exercise 8. limxa(xa)(x+a)

(x+a)2 = 0

Exercise 9. limt0 tan t = limt0 sin tlimx0 cos t =01 = 0

Exercise 10. limt0(sin 2t+ t2 cos 5t) = limt0 sin 2t+ limt0 t2 limt0 cos 5t = 0 + 0 = 0

Exercise 11. limx0+|x|x = 1

Exercise 12. limx0|x|x = 1

Exercise 13. limx0+x2

x = +1

Exercise 14. limx0x2

x = 1

Exercise 15. limx0 2 sin x cos xx = 2

Exercise 16. limx0 2 sin x cos xcos 2x sin x = 2

Exercise 17. limx0 sin x cos 4x+sin 4x cos xsin x = 1 + limx02 sin 2x cos 2x

sin x = 1 + 2(limx0 2 sin x cos x cos 2xsin x

)= 5 Exercise 18.

limx0 5 sin 5x5x limx0 3 sin 3x3x = 5 3 = 2 Exercise 19.37

limx0

sin(x+a

2 +xa

2

) sin (x+a2 (xa2 ))x a =

= limx0

(sin x+a2 cos

xa2 + sin

xa2 cos

x+a2

(sin x+a2 cos

xa2 sin xa2 cos x+a2

)x a

)=

= limxa

2 sin xa2 cosx+a

2

x a = cos a

Exercise 20. limx02 sin2 x/24(x/2)2 =

12

(limx0

sin x/2x/2

)= 12

Exercise 21. limx0 1

1x2x2

(1+

1x21+

1x2)

= limx01(1x2)

x2(1+

1x2) =1

2

Exercise 22. b, c are given.

sin c = ac+ b, a = sin cbc , c 6= 0.if c = 0, then b = 0, a R.Exercise 23. b, c are given.

2 cos c = ac2 + b, a = 2 cos cbc2 , c 6= 0.If c = 0, then b = 2, a R.Exercise 24.

tangent is continuous for x / (2n+ 1)pi/2cotangent is continuous for x / 2npi

Exercise 25. limx0 f(x) =. No f(0) cannot be defined.

Exercise 26.

(1) | sinx 0| = | sinx| < |x|. Choose = for a given .Then > 0, > 0 such that | sinx 0| < when |x| < .

(2)

| cosx 1| = | 2 sin2 x/2| = 2| sinx/2|2 < 2|x2|2 = |x|

2

2< 2/2 =

If we had chosen 0 =

2 for a given . |x 0| < = 2.(3)

| sinx(cosh 1) + cosx sinh| | sinx|| cosh 1|+ | cosh|| sinh| < 2

+

2=

| cosx+ h cosx| = | cosx cosh sinx sinh cosx| = | cosx(cosh 1) sinx sinh| | cosx|| cosh 1|+ | sinx|| sinh| <

2+

2=

since > 01, 2 > 0 such that | cosh 1| < 0; | sinh| < whenever |h| < min (1, 2)Choose 3 such that if |h| < 3; | cosh 1| <

2; | sinh| <

2

Exercise 27. f(x)A = sin 1x A.Let x = 1npi .

|f(x)A| = | sinnpi A| > || sinnpi| |A|| > |1 |A||Consider |x 0| = |x| = 1npi (n). Consider 0 = |1|A||2 . Then suppose a (n) |x 0| but |f(x) A| > 0. Thus,contradiction.

Exercise 28. Consider x 1n , n Z+, n > M(n) (n is a given constant)

f(x) =

[1

x

]= [n] = n, for m > M(n), x =

1

mf(x) > M(n)

so > 0, we cannot find = 1n such that |f(x)A| < for x < .So f(x) as x 0+.

38

Consider 1n x > 0, n Z; n > M(n).

f(x) =

[1

x

]= [n] = n < M(n)

Since integers are unbounded, we can consider n < A, so that

|f(x)A| > ||f | |A|| = n |A| > M(n) |A|. Choose n such that M(n) |A| > 0

Exercise 29.

|f A| = |(1)[1/x] A| ||(1)[1/x]| |A|| = |1 |A||Choose < |1 |A||. Then > 0 ( such that |x| < ), |f A| > . Thus theres no value for f(0) we could choose tomake this function continuous at 0.

Exercise 30. Since

|f(x)| = |x||(1)[1/x]| = |x|So , let = .Exercise 31. f continuous at x0.

Choose some 0, 0 < 0 < min (b x0, x0 a). Then 0 = (x0, 0).Consider 1 = 02 and 1 = (x0, 1)Consider x1 (x0 1, x0 + 1), so that |f(x1) f(x0)| < 1.

Proceed to construct a for x1, some (x1; 0)

|x x1| = |x x0 + x0 x1| < |x x0|+ |x0 x1|Without loss of generality, we can specify x1 such that |x0 x1| < 12 . Also, pick only the xs such that

|x x0| < 12< 1

= |x x1| < 12

+12

= 1

Thus, for these xs

|f(x) f(x1)| = |f(x) f(x0) + f(x0) f(x1)| < |f(x) f(x0)|+ |f(x1) f(x0)| < 1 + 1 = 0So 0,1 for x1. f is continuous at x1 (a, b). Thus, there must be infinitely many points that are continuous in (a, b),and at the very least, some or all are clustered around some neighborhood about the one point given to make f continuous.

Exercise 32. Given = 1n , |f(x)| = |x sin 1x | = |x|| sin 1/x| < |x|(1).Let = (n) = 1n , so that |x| < 1n .= |f(x)| < 1nExercise 33.

(1) Consider x0 [a, b].Choose some 0, 0 < 0 < min (b x0, x0 a) 6= 0 , ( x0 could be a or b )

Consider, without loss of generality, only xs such that x [a, b].|f(x) f(x0)| |x x0|

Let 0 = (0, x0) = 0 = |f(x) f(x0)| < 0.Since we didnt specify x0,x0 [a, b], f is continuous at x0.

(2) ba

f(x)dx (b a)f(a) =

ba

(f(x) f(a))dx

ba

|f(x) f(a)|dx

ba

|x a|dx = (12x2 ax)

ba

=1

2(b a)(b+ a) a(b a) = (b a)

2

2

39

(3) ba

f(x)dx (b a)f(c) =

ba

(f(x) f(c))dx

ba

|f(x) f(c)|dx ba

|x c|dx =

=

ca

(c x)dx+ bc

(x c)dx = c(c a) 12

(c a)(c+ a) + 12

(b c)(b+ c) c(b c) =

=1

2((c a)2 + (b c)2)

Draw a figure for clear, geometric reasoning.Consider a square of length (b a) and a 45 45 right triangle inside. From the figure, its obvious that right

triangles of c a length and (b c) length lie within the (b a) right triangle.Compare the trapezoid of c a, b a bases with the b a right triangle.

1

2(b c)(b a+ c a) = 1

2(b c)(b c+ 2(c a)) > 1

2(b c)2

Indeed, the trapezoid and c a right triangle equals the b a trapezoid since1

2(b c)(b a+ c a) + 1

2(c a)2 = 1

2(b2 c2 2ab+ 2ac+ c2 2ca+ a2) = 1

2(b a)2

= 12

(b a)2 > 12

(b c)2 + 12

(c a)2

so then

ba

f(x)dx (b a)f(c) (b a)22

3.11 Exercises - Bolzanos theorem for continuous functions, The intermediate-value theorem for continuous func-tions. These theorems form the foundation for continuity and will be valuable for differentiation later.

Theorem 10 (Bolzanos Theorem).Let f be cont. at x [a, b].Assume f(a), f(b) have opposite signs.Then at least one c (a, b) s.t. f(c) = 0.Proof. Let f(a) < 0, f(b) > 0.Want: Fine one value c (a, b) s.t. f(c) = 0Strategy: find the largest c.Let S = { all x [a, b] s.t. f(x) 0 }.S is nonempty since f(a) < 0. S is bounded since all S [a, b].= S has a suprenum.Let c = supS.

If f(c) > 0, (c , c+ ) s.t. f > 0c is an upper bound on S

but c is a least upper bound on S. Contradiction.If f(c) < 0, (c , c+ ) s.t. f < 0c+ is an upper bound on S

but c is an upper bound on S. Contradiction.

Theorem 11 (Sign-preserving Property of Continuous functions).Let f be cont. at c and suppose that f(c) 6= 0.then (c , c+ ) s.t. f be on (c , c+ ) has the same sign as f(c).Proof. Suppose f(c) > 0. > 0, > 0 s.t. f(c) < f(x) < f(c) + if c < x < c+ (by continuity).Choose for = f(c)2 . Then

f(c)

2< f(x) 0

By Bolzano, c (x1, x2) s.t. g(c) = 0 = f(c) = k

Exercise 1. f(0) = c0. f(0) 0.

Since limx ckxk

ck1xk1= limx ckck1x = M > 0 such that |cnMn| > |

n1k=0 ckM

k. So then

f(M) = cnMn +

n1k=0

ckMk cn

By Bolzanos theorem b (0,M) such that f(b) = 0.Exercise 2. Try alot of values systematically. I also cheated by taking the derivatives and feeling out where the function

changed direction.

(1) If P (x) = 3x4 2x3 36x2 + 36x 8, P (4) = 168, P (3) = 143, P (0) = 8, P ( 12 ) = 1516 , P (1) = 7,P (3) = 35, P (4) = 200

(2) If P (x) = 2x4 14x2 + 14x 1, P (4) = 231, P (3) = 7, P (0) = 1, P ( 12 ) = 18 , P ( 32 ) = 118 , P (2) = 2(3) If P (x) = x4 + 4x3 + x2 6x + 2, P (3) = 2, P ( 52 ) = 316 , P (2) = 2, P ( 13 ) = 2281 , P ( 12 ) = 316 , P ( 23 ) = 1481 , P (1) = 2.

Exercise 3. . Consider f(x) = x2j+1 a. f(0) = a > 0.

Since a is a constant, choose M < 0 such that M2j+1 a < 0. f(M) < 0.By Bolzanos theorem, there is at least one b (M, 0) such that f(b) = b2j+1 a = 0.

Since x2j+1 a is monotonically increasing, there is exactly one b.Exercise 4. tanx is not continuous at x = pi/2.

Exercise 5. Consider g(x) = f(x) x. Then g(x) is continuous on [0, 1] since f is.Since 0 f(x) 1 for each x [0, 1], consider g(1) = f(1) 1, so that 1 g(1) 0. Likewise 0 g(0) 1.If g(1) = 0 or g(0) = 0, were done (g(0) = f(0) 0 = 0. f(0) = 0. Or g(1) = f(1) 1 = 0, f(1) = 1 ).Otherwise, if 1 g(1) < 0 and 0 < g(0) 1, then by Bolzanos theorem, at least one c such thatg(c) = 0 (g(c) = f(c) c = 0. f(c) = c).

Exercise 6. Given f(a) a, f(b) b,

Consider g(x) = f(x) x 0. Then g(a) = f(a) a 0, g(b) = f(b) b 0.Since f is continuous on [a, b] (so is g) and since g(a), g(b) are of opposite signs, by Bolzanos theorem, at least one c

such that g(c) = 0, so that f(c) = c.41

3.15 Exercises - The process of inversion, Properties of functions preserved by inversion, Inverses of piecewise mono-tonic functions. Exercise 1. D = R, g(y) = y 1

Exercise 2. D = R, g(y) = 12 (y 5)

Exercise 3. D = R, g(y) = 1 y

Exercise 4. D = R, g(y) = y1/3

Exercise 5. D = R,

g(y) =

y if y < 1y if 1 y 16(y8

)2if y > 16

Exercise 6. f(Mf ) = f(f1(

1n

ni=1 f(ai)

)) = 1n

ni=1 f(ai)

Exercise 7. f(a1) 1nni=1 f(ai) f(an). Since f is strictly monotonic.

g preserves monotonicity.= a1 Mf an

Exercise 8. h(x) = af(x) + b, a 6= 0

Mh = H

(1

n

ni=1

h(ai)

)= H

(1

n

ni=1

(af(ai) + b)

)= H

(a

1

n

ni=1

f(ai) + b

)The inverse for h is g

(hba

)= H(h) = h1. So then

Mh = g

(1

n

ni=1

f(ai)

)= Mf

The average is invariant under translation and expansion in ordinate values.

3.20 Exercises - The extreme-value theorem for continuous functions, The small-span theorem for continuous func-tions (uniform continuity), The integrability theorem for continuous functions.

Since for c [a, b], m = minx[a,b] f f(c) maxx[a,b] f = M

and baf(x)g(x)dx bag(x)dx

= f(c)

Exercise 1.

g = x9 > 0 for x [0, 1]; f = 11 + x

m =12,M = 1 1

0

x9 =1

10x1010

=1

10

1

10

2 1

0

x91 + x

dx 110

Exercise 2. 1 x2 = 1 x

2

1 x2 . f =

11 x2 g = (1 x

2)M =23,m = 1 1/2

0

(1 x2)dx = (x 13x3)

1/20

=11

24

11

24 1/2

0

1 x2dx 11

24

4

3

Exercise 3.42

f =1

1 + x6g = 1 x2 + x4

a0

1 x2 + x4 =(x 1

3x3 +

1

5x5)a

0

= a a3

3+a5

5

m =1

1 + a6M = 1

1

1 + a6

(a a

3

3+a5

5

) a

0

1

1 + x2dx

(a a

3

3+a5

5

)So if a = 110 , (a a3/3 + a5/5) = a 0.333 . . . a3 + 0.2a5 = 0.099669

Exercise 4. (b) is wrong, since it had chosen g = sin t, but g needed to be nonnegative.

Exercise 5. At worst, we could have utilized the fundamental theorem of calculus.sin t2dt =

(1

2t

)(2t sin t2)dt =

1

2c( cos t2)(n+1)pi

npi=

=12c

((1)n+1 (1)n) = 1c

(1)n

Exercise 6. ba

(f)(1) = f(c) ba

1 = f(c)(b a). Then f(c) = baf

ba = 0 for some c [a, b] by Mean-value theorem forintegrals.

Exercise 7. f nonnegative. Consider f at a point of continuity c, and suppose f(c) > 0. Then 12f(c) > 0.

|f(x) f(c)| < = f(c) < f(x) < f(c) + Let =

1

2f(c) > 0 for = 1

2f(c) c+

cf(x)dx >

1

2f(c)(2) = f(c) > 0

But baf(x)dx = 0 and f is nonnegative. f(c) = 0.

Exercise 8.

m

g

fg M

g = m

g 0 M

g g

m 0 M forg = 1 but also

m 0 M = m 0M 0 forg = 1

So because of this contradiction, m = M = 0. By intermediate value theorem, f = 0, x [a, b].

4.6 Exercises - Historical introduction, A problem involving velocity, The derivative of a function, Examples of deriva-tives, The algebra of derivatives. Exercise 1. f = 1 2x, f (0) = 1, f (1/2) = 0, f (1) = 1, f (10) = 19Exercise 2. f = x2 + x 2

(1) f = 0, x = 1,2(2) f (x) = 2, x = 0,1(3) f = 10, x = 4, 3

Exercise 3. f = 2x+ 3

Exercise 4. f = 4x3 + cosx

Exercise 5. f = 4x3 sinx+ x4 cosx

Exercise 6. f = 1(x+1)2

Exercise 7. f = 1(x2+1)2 (2x) + 5x4 cosx+ x5( sinx)

Exercise 8. f = x1(x)(x1)2 =1

(x1)243

Exercise 9. f = 1(2+cos x)2 ( sinx) = sin x(2+cos x)2

Exercise 10.

(2x+ 3)(x4 + x2 + 1) (4x3 + 2x)(x2 + 3x+ 2)(x4 + x2 + 1)2

=2x5 9x4 + 12x3 3x2 2x+ 3

(x4 + x2 + 1)2

Exercise 11.

f =( cosx)(2 cosx) (sinx)(2 sinx)

(2 cosx)2 =2 cosx 2 sinx+ 1

(2 cosx)2

Exercise 12.

f =(sinx+ x cosx)(1 + x2) 2x(x sinx)

(1 + x2)2=

sinx+ x cosx+ x3 cosx x2 sinx(1 + x2)2

Exercise 13.

(1)f(t+ h) f(t)

h=v0h 32th 16h2

h= v0 + 32t 16h

f (t) = v0 32t(2) t = v032(3) v0(4) T = v016 , v0 = 16 for 1sec. v0 = 160 for 10sec.

v016 for Tsec.

(5) f = 32(6) h = 20t2

Exercise 14. V = s3, dVdS = 3s2

Exercise 15.

(1) dAdr = 2pir = C(2) dVdr = 4pir

2 = A

Exercise 16. f = 12x

Exercise 17. f = 1(1+x)2

(1

2x

)Exercise 18. f = 32x

1/2

Exercise 19. 32 x5/2

Exercise 20. f = 12x1/2 + 13x

2/3 + 14x3/4 x > 0

Exercise 21. f = 12x3/2 + 13x4/3 14x5/4

Exercise 22. f =12x1/2(1+x)x

(1+x)2 =1

2x(1+x)2

Exercise 23. f =(1+x)x 12 1x

(1+x)2

=1+ 12x

(1+x)2

Exercise 24.44

g = f1f2

g = f 1f2 + f1f2

g

g=f 1f1

+f 2f2

g = f1f2 . . . fnfn+1

g = (f1f2 . . . fn)fn+1 + (f1f2 . . . fn)f n+1;

g

g=

(f1f2 . . . fn)

f1f2 . . . fn+f n+1fn+1

=f 1f1

+f 2f2

+ + fn

fn+f n+1fn+1

Exercise 25.

(tanx) =(cosx

sinx

)=

cos2 x ( sinx) sinxcos2 x

= sec2 x

(cotx) =(cosx

sinx

)= sinx sinx cosx cosx

sin2 x= csc2 x

(secx) =1

cos2 x( sinx) = tanx secx

(cscx) =1

sin2 xcosx = cotx cscx

Exercise 35.

f =(2ax+ b)(sinx+ cosx) (cosx sinx)(ax2 + bx+ c)

(sinx+ cosx)2=

=(2ax+ b)(sinx+ cosx) (cosx sinx)(ax2 + bx+ c)

(sinx+ cosx)2

Exercise 36.

f = a sinx+ (ax+ b) cosx+ c cosx+ (cx+ d)( sinx) = ax cosx+ (b+ c) cosx+ (a d) sinx cx sinxSo then a = 1, d = 1, b = d, c = 0.

Exercise 37.

g = (2ax+ b) sinx+ (ax2 + bx+ c) cosx+ (2dx+ e) cosx+ (dx2 + ex+ f)( sinx) == ax2 cosx dx2 sinx+ (2a e)x sinx+ (b+ 2d)x cosx+ (b f) sinx+ (c+ e) cosx

g = x2 sinx. So d = 1, b = 2, f = 2, a = 0, e = 0, c = 0.Exercise 38. 1 + x+ x2 + + xn = xn+11x1

(1)

(1 + x+ x2 + + xn) = 1 + 2x+ + nxn1 = (n+ 1)xn(x 1) (1)(xn+1 1)

(x 1)2

=(n+ 1)(xn+1 xn) xn+1 + 1

(x 1)2 =nxn+1 (n1)xn + 1

(x 1)2

x(1 + 2x+ + nxn1) = x+ 2x2 + + nxn = nxn+2 (n+ 1)xn+1 + x

(x 1)2(2)

(x+ 2x2 + + nxn) = (1 + 22x1 + + n2xn1) =

=(n(n+ 2)xn+1 (n+ 1)2xn + 1)(x 1)2 2(x 1)(nxn+2 (n+ 1)xn+1 + x)

(x 1)4

x+ 22x2 + + n2xn = (n(n+ 2)xn+2 (n+ 1)2xn+1 + x)(x 1) 2(nxn+3 (n+ 1)xn+2 + x2)

(x 1)3 =

=n2xn+3 + (2n2 2n+ 1)xn+2 + (n+ 1)2xn+1 x2 x

(x 1)345

Exercise 39.

f(x+ h) f(x)h

=(x+ h)n xn

h

(x+ h)n =

nj=0

(n

j

)xnjhj

(x+ h)n xnh

=

nj=1

(nj

)xnjhj

h=

nj=1

xnjhj1(n

j

)

limh0

(x+ h)n xnh

=

(n

1

)xn1 = nxn1

4.9 Exercises - Geometric interpretation of the derivative as a slope, Other notations for derivatives. Exercise 6.

(1)f = x2 + ax+ b

f(x1) = x21 + ax1 + b

f(x2) = x22 + ax2 + b

f(x2) f(x1)x2 x1 =

x22 x21 + a(x2 x1)x2 x1

= x2 + x1 + a

(2)f = 2x+ a

m = x2 + x1 + a = 2x+ a x =x2 + x1

2

Exercise 7. The line y = x as slope 1.

y = x3 6x2 + 8x y = 3x2 12x+ 93x2 12x+ 8 = 1 = x = 3, 1

The line and the curve meet under the condition

x = x3 6x2 + 8x = x = 3; f(3) = 3At x = 0, the line and the curve also meet.

Exercise 8. f = x(1 x2). f = 1 3x2.

f (1) = 2 = y = 2x 2For the other line,

f (a) = 1 3a2= y(1) = 0 = (1 3a2)(1) + b = b = 1 3a2

Now f(a) = a(1 a2) = a a3 at this point. The line and the curve must meet at this point.y(a) = (1 3a2)a+ (1 3a2) =

= a 3a3 + 1 3a2 = a a3

= 2a3 + 1 3a2 = 0 = a3 12

+3

2a2

The answer could probably be guessed at, but lets review some tricks for solving cubics.First, do a translation in the x direction to center the origin on the point of inflection. Find the point of inflection by taking

the second derivative.

f = 6a+ 3 = a = 12

So

a = x 12

= (x 12

)3 +3

2(x 1

2)2 1

2= x3 =

3

4x 1

4= 0

46

Then recall this neat trigonometric fact:

cos 3x = cos 2x cosx sin 2x sinx = 4 cos3 x 3 cosx= cos3 x = 3

4cosx cos 3x

4= 0

Particularly for this problem, we have cos 3x = 1. So x = 0, 2pi/3, 4pi/3. cosx = 1, 12 . Plugging cosx x back intowhat we have for a, a = 1, which we already have in the previous part, and a = 12 . So

f

(1

2

)=

3

8

y(x) =

(1

4x

)+

1

4

Exercise 9.

f(x) =

{x2 if x cax+ b if x > c

f (x) =

{2x if x ca if x > c

a = 2c; b = c2

Exercise 10.

f(x) =

{1|x| if |x| > ca+ bx2 if |x| c

Note that c 0 since |x| c, for the second condition.

f (x) =

1x2 if x > c1x2 if x < c2bx if |x| c

So b = 12c3

, a =3

2c.

Exercise 11.

f =

{cosx if x ca if a > c

Exercise 12. f(x) =(

121+

2

)= 1A1+A

A =xA = a =

1

2x1/2 =

1

2A;A = 1

4x3/2 = 1

4A3

f =A(1 +A)A(1A)

(1 +A)2=2A

(1 +A)2=

1x(1 +

x)2

f =1

(A(1 +A2))2(A(1 +A)2 +A(2)(1 +A)A) =

3x+ 1

2x3/2(1 +x)3

f =1

2

( 1A2A

(A2(1 +A)3) (2AA(1 +A)3 + 3A2(1 +A)2A)(3 + 1A )(A2(1 +A)3)2

)=34

( 1A + 4 + 5A

A4(1 +A4)

)= 3

4

(1 + 4x+ 5x)

x(x+x)4

Exercise 13.47

P = ax3 + bx2 + cx+ d

P = 3ax2 + 2bx+ c

P = 6ax+ 2b

P (0) = 2b = 10 = b = 5P (0) = c = 1 P (0) = d = 2

P (1) = a+ 5 +1 +2 = a+ 2 = 2 = a = 4

Exercise 14.

fg = 2,f

g= 2

f

g= 4

g

g= 2 f =

1

2, g = 4

(1)

h =f g gf

g2=f

g g

g

f

g= 4 2(1

8) =

15

4

(2)

k = f g + fg = 4g2 + f2g = 64 + 4 = 68

(3)

limx0

g(x)f (x)

=limx0 g(x)limx0 f (x)

=1

2

Exercise 15.

(1) True, by definition of f (a).(2)

limh0

f(a) f(a h)h

= limh0

f(a h) f(a)h

= limh0

f(a h) f(a)h = f

(a)

True, by definition of f (a).(3)

limt0

f(a+ 2t) f(a)t

= 2 lim2t0

f(a+ 2t) f(a)2t

= 2f (a)

False.(4)

limt0

f(a+ 2t) f(a) + f(a) f(a+ t)2t

=

lim2t0

f(a+ 2t) f(a)2t

+12

limt0

f(a+ t) f(a)t

=

f (a) 12f (a) =

1

2f (a)

False.

Exercise 16.

(1)

D(f + g) = limh0

(f(x+ h) + g(x+ h))2 (f(x) + g(x))2h

= limh0

(F +G)2 (f + g)2h

=

= Df +Dg + limh0

2FG 2fgh

limh0

2FG 2fgh

= limh0

(2(FG) 2fG)(F + f)(F + f)h

+(2fG 2fg)(G+ g)

(g +G)h=

= limh0

2g

F + hlimh0

F 2 f2h

+ limh0

2f

G+ glimh0

G2 g2h

=

=g

fDf +

f

gDg

48

D(f g) = limh0

(f(x+ h) g(x+ h))2 (f(x) g(x))2h

=

= limh0

(F G)2 (f g)2h

=

= Df +Dg + limh0

2FG 2fgh

= Df +Dg gfDf +

f

gDg

D(fg) = limh0

((fg)(x+ h))2 ((fg)(x))2h

=

= limh0

(f2(x+ h))(g2(x+ h)) f2(x)g2(x+ h) + (g2(x+ h) g2(x))f2(x)h

=

= g2Df + f2Dg

D(f/g) = limh0

f2(x+h)g2(x+h) f

2(x)g2(x)

h= limh0

f2(x+h)f2(x)g2(x+h) +

f2(x)g2(x+h) f

2(x)g2(x)

h=

=Dfg2

+f2

g4(Dg) when g(x) 6= 0

(2)(3)

4.12 Exercises - The chain rule for differentiating composite functions, Applications of the chain rule. Related ratesand implicit differentiation. Exercise 1. 2 sin 2x 2 cosxExercise 2. x

1+x2

Exercise 3. 2x cosx2 + 2x(x2 2) sinx2 + 2 sinx3 + 6x3 cosx3

Exercise 4.

f = cos (cos2 x)(2 cosx sinx) cos (sin2 x) + sin (cos2 x) sin (sin2 x)(2 sinx cosx) == sin 2x(cos (cos 2x))

Exercise 5.

f = n sinn1 x cosx cosnx+n sinnx sinn xExercise 6.

f = cos (sin (sinx))(cos (sinx))(cosx)

Exercise 7.

f =2 sinx cosx sinx2 2x cosx2 sin2 x

sin2 x2=

sin 2x sinx2 2x sin2 x cosx2sin2 x2

Exercise 8. f = 12 sec2 x

2 +12 csc

2 x2

Exercise 9. f = 2 sec2 x tanx+2 csc2 x cotx

Exercise 10. f =

1 + x2 + x2

1+x2= 1+2x

21+x2

Exercise 11. f = 4(4x2)3/2

Exercise 12.

f =1

3

(1 + x3

1 x3)2/3(

3x2(2)

(1 x3)2)

=2x2

(1 x3)2(

1 + x3

1 x3)2/3

Exercise 13. This exercise is important. It shows a neat integration trick.49

f(x) =1

1 + x2(x+

1 + x2)=

11 + x2(x+

1 + x2)

(x1 + x2x1 + x2

)=

=x1 + x21 + x2 = 1

x1 + x2

f =

1 + x2 x2

1+x2

1 + x2=

1

(1 + x2)3/2

Exercise 14.

1

2(x+

x+x)1/2(1 +

1

2(x+

x)1/2

(1 +

1

2x

))

Exercise 15.

f = (2 + x2)1/2(3 + x3)1/3 + (1 + x)x(2 + x2)1/2(3 + x2)1/3 + (1 + x)(2 + x2)1/2(3 + x3)2/3x2

Exercise 16.

f =1(

1 + 1x)2 (1x2

)=

1

(x+ 1)2g =

1(1 + 1f

)2 (1f2)f =

f

(f + 1)2

g =(x+ 1)2(xx+1 + 1

)2 = 1(2x+ 1)2Exercise 17. h = f g

x h h k k

0 f(2) = 0 2(5) = 10 g(1) = 0 1(5) = 51 f(0) = 1 5(1) = 5 g(3) = 1 6(2) = 122 f(3) = 2 4(1) = 4 g(0) = 2 5(2) = 103 f(1) = 3 2(6) = 12 g(2) = 3 1(4) = 4

Exercise 18.

g(x) = xf(x2)

g(x) = f(x2) + x(2x)f (x2) = f(x2) + 2x2f (x2)

g(x) = 2xf (x2) + 4xf (x2) + 2x2(2x)f (x2) = 6xf (x2) + 4x3f (x2)

x g(x) g(x) g(x)0 0 0 01 1 3 102 12 6 + 8(3) = 30 12(3) + 32(0) = 36

Exercise 19.

(1)

g =df(x2)

dx22x = 2xf

(2)g = 2 sinx cosxf 2 cosx sinxf = (sin 2x)(f (sin2 x) f (cos2 x))

(3)

g =df(f(x))

d(f(x))f

(4)

g =df(f(f(x)))

d(f(f(x)))

d(f(f(x)))

d(f(x))

df

dx

50

Exercise 20. V = s3, s = s(t) dVdt = 3s2 dsdt .

s = 5cm 75cm3/sec

s = 10cm 300cm3/sec

s = xcm 3x2cm3/sec

Exercise 21.

l =x2 + h2

dl

dt=

1

lxdx

dtdx

dt=l

x

dl

dt=

10mi

102 82 (4mi/sec) =20

3

mi

sec

(3600sec

1hr

)Exercise 22.

l2 = x2 + s2

2ldl

dt= 2x

dx

dt

dl

dt=x

l

dx

dtdl

dt

(x =

s

2

)= 20

5

dl

dt(x = s) = 50

2

Exercise 23.

dl

dt=x

l

dx

dt=

3

512 =

36

5mi/hr

Exercise 24. Given the preliminary information

r

h=

2

5= , V =

1

3pir2h =

1

3pi2h3

(1)

V =pir2

h2(h2y hy2 + 1

3y3)

dV

dt=pir2

h2(h2 2hy + y2)dy

dtdy

dt=

h2

pir2

(1

h2 2hy + y2)dV

dt=

102

pi42

(1

102 2(10)5 + 25)

5 =5

4pi

(2)dV

dt= pi2h2

dh

dt,dh

dt=

1

pi2h2dV

dt=

5

4pi

Exercise 25.

=r

h=

3

2dV

dt= pi2h2

dh

dt

c 1pi 94

(22)4 = 36pi = c = 36pi + 1

Exercise 26. The constraint equation, using Pythagorean theorem on the geometry of a bottom hemisphere, is

r2 = R2 (R h)2 = 2Rh h2So then

rdr

dt= (R h)dh

dt

V =

pir2dh = dV

dh= pir2 = pi(2Rh h2)

= dVdh

= pi(2(10(5) 25)) = 50pi51

dV

dt=dV

dh

dh

dt, = dh

dt=dV

dt

(1

pi(2Rh h2))

(r

R h)dr

dt=dV

dt

(1

pi(2Rh h2))

=dr

dt=dV

dt

(R h

rpi(2Rh h2))

=

= (5

3)

(10 5

pi(2(10)5 25)3/2)

=1

15pi

Exercise 27. I suppose the area of the triangle is 0 at t = 0.

Now the point on vertex B moves up along the y axis according to y = 1 + 2t. y(

72

)= 8.

A =1

2

(y 1)36

7y

dA

dt=

1

2

(36

y

1

2

1y 1y +

(y 1)36

7

)dy

dt=

=1

2

(6

2(7)8 + 6

)(2) =

66

7

Exercise 28. From the given information, h = 3r + 3. The volume formula is V = piR2

3 H . So then

V = pi/3r2(3r + 3) = pir3 + pir2

dV

dr= pir(3r + 2)

dr

dt

With the given information, we getdr

dt=

1

pi(6)(20)

Using this, we can plug this back in for the different case:

dV

dt= n = pi(36)(110)/(120pi) = 33

Exercise 29.

(1) dydt = 2xdxdt ; when x =

12 , y =

14 ,

dydt =

dxdt

(2) t =pi

6

Exercise 30.

(1) 3x2 + 3y2y = 0 = x2 + y2y = 0(2)

2x+ 2yy2 + y2y = 0 = y2y = 2(x+ yy2)

= y = 2(xy4 + yx4

y6

)= 2xy5

Exercise 31.

1

2

1x

+1

2yy = 0 y =

yx

< 0

Exercise 32.

12 3x24

52

6x+ 8yy = 0 = y = 3x4y

3 + 4(y2 + yy) = 0

y =(3

4 y2

)1

y=94y3

Exercise 33.

sinxy + x cos2 xy(y + xy) + 4x = 0

yx2 cosxy + xy cosxy + sinxy + 4x = 0

Exercise 34. y = x4. yn = xm.

yn = xm, ynyn1 = mxm1; y =mxm1

nyn1=m

n

xm1

xm(11/n)=

y =m

nxm/n1

4.15 Exercises - Applications of differentiation to extreme values of functions, The mean-value theorem for derivatives.Lets recap what was shown in the past two sections:

Theorem 13 (Theorem 4.3).Let f be defined on I .Assume f has a rel. extrema at an int. pt. c I .If f (c), f (c) = 0; the converse is not true.Proof. Q(x) = f(x)f(c)xc if x 6= c, Q(c) = f (c)f (c), so Q(x) Q(c) as x c so Q is continuous at c.If Q(c) > 0, f(x)f(c)xc > 0. For x c 0, f(x) f(c), thus contradicting the rel. max or rel. min. (no neighborhoodabout c exists for one!)If Q(c) < 0, f(x)f(c)xc < 0. For x c 0, f(x) f(c), thus contradicting the rel. max or rel. min. (no neighborhoodabout c exists for one!)

Converse is not true: e.g. saddle points.

Theorem 14 (Rolles Theorem).Let f be cont. on [a, b], f (x) x (a, b) and let

f(a) = f(b)

then at least one c (a, b), such that f (c) = 0.Proof. Suppose f (x) 6= 0 x (a, b).By extreme value theorem, abs. max (min) M, m somewhere on [a, b].M,m on endpoints a, b (Thm 4.3).F (a) = f(b), so m = M . f constant on [a, b]. Contradict f (x) 6= 0 Theorem 15 (Mean-value theorem for Derivatives). Assume f is cont. everywhere on [a, b], f (x) x (a, b). at least one c (a, b) such that(6) f(b) f(a) = f (c)(b a)Proof.

h(x) = f(x)(b a) x(f(b) f(a))h(a) = f(a)b f(a)a af(b) + af(a)h(b) = f(b)(b a) b(f(b) f(a)) = bf(a) af(b) = h(a)

= c (a, b), such that h(c) = 0 = f (c)(b a) (f(b) f(a))

Theorem 16 (Cauchys Mean-Value Formula). Let f, g cont. on [a, b], f , g x (a, b)Then c (a, b). x(7) f (c)(g(b) g(a)) = g(c)(f(b) f(a)) (note how its symmetrical)

53

Proof.h(x) = f(x)(g(b) g(a)) g(x)(f(b) f(a))h(a) = f(a)(g(b) g(a)) g(a)(f(b) f(a)) = f(a)g(b) g(a)f(b)h(b) = f(b)(g(b) g(a)) g(b)(f(b) f(a))

= h(c) = f (c)(g(b) g(a)) g(c)(f(b) f(a)) = 0 (by Rolles Thm.)

Exercise 1. For any quadratic polynomial y = y(x) = Ax2 +Bx+ C,

y(a) = Aa2 +Ba+ C

y(b) = Ab2 +Bb+ C

y(b) y(a)b a =

A(b a)(b+ a) +B(b a)b a = A(b+ a) +B

y = 2Ax+B

y(a+ b

2

)= A(a+ b) +B

Thus the chord joining a and b has the same slope as the tangent line at the midpt.

Exercise 2. The contrapositive of a theorem is always true. So the contrapositive of Rolles Theorem is

If @ at least one c (a, b) s.t. f (c) = 0,then f(a) 6= f(b).

g = 3x2 3 = 3(x2 1) = g(1) = 0Suppose g(B) = 0, B (1, 1)

then x (1, 1), x 6= B, g(x) 6= g(B), so g(x) 6= 0 for x 6= Bso only at most one B (1, 1) s.t. g(B) = 0

Exercise 3. f(x) = 3x2

2 if x 1, f(x) = 1x if x 1.(1) See sketch.(2)

f(x) =

{3x2

2 if x 11/x if x 1 f(1) = 1 = f(1) = 1/1

f (x) =

{x; f (1) = 1 for x 11/x2; f (1) = 1 for x > 1

Then f(x) is cont. and diff. on [0, 2].

For 0 a < b 13b2

2 (

3a22

)b a =

(a+ b)2

= cNote that 1 f 0 for 0 x 1

For 1 a < b 21b 1ab a =

1ab

=1c2

= c =ab

Note that 1 f 1/4For 0 a 1 , 1 b 2

1b

(3a2

2

)b a =

2 (3 a2)b2b(b a) = c or

1c2

depending upon if 0 c 1 or 1 c 2, respectively54

For instance, for a = 0, b = 2, then f(b)f(a)ba = 1/2, so c = 1/2 or c =

2

Exercise 4.

f(1) = 1 12/3 = 0 = f(1) = 1 ((1)2)3 = 0f =

23x1/3 6= 0 for |x| 1

This is possible since f is not differentiable at x = 0.

Exercise 5. x2 = x sinx+ cosx. g = xS + C x2. g = S + xC S 2x = xC 2x = x(C 2). Since |C| 1 then(C 2) is negative for all x. Then for x 0, g 0. Since g(0) = 1 and for x , g , then we could concludethat g must become zero between 0 and and and 0.Exercise 6.

f(b) f(a)b a = f

(c)

b = x+ h

b a = ha = x

x < x+ h < x+ h

= f(x+ h) f(x) = hf (x+ h)

(1) f(x) = x2, f = 2x.

(x+ h)2 x2 = 2xh+ h2 = h(2(x+ h))2x+ h

2 x = h = = 1

2so then lim

h0 =

1

2

(2) f(x) = x3, f = 3x2.

(x+ h)3 x3 = 3x2h+ 3xh2 + h3 = h3(x+ h)2 =(

3x2 + 3xh+ h2

3 x)/h =

=

3x2 + 3xh+ h2

3h2 xh

=

x2 + xh+ h

2

3 xh

x2 + xh+ h

2

3 + xx2 + xh+ h

2

3 + x

==

x+ h3

x+x2 + hx+ h

2

3

= limh0

=1

2

Notice the trick of multiplying by the conjugate on top and bottom to get a way to evaluate the limit.

Exercise 7. f(x) = (x a1)(x a2) . . . (x ar)g(x).

(1) a1 < a2 < < ar.Since f(a1) = f(a2) = 0. f (c) = 0 for c1 (a1, a2).Consider that f(a2) = f(a3) = 0 as well as f (c2) = 0 for c (a2, a3).Indeed, since f(aj) = f(aj+1) = 0, f (c) = 0 for c (aj , aj+1).Thus, r 1 zeros.f (k) has r k zeros in [a, b].f (k) = (x a1)(x a2) . . . (x ark)gk(x)Since f(a1) = f(a2) = 0, f (k+1)(c1) = 0 for c1 (a1, a2).

f (k)(aj) = f(k)(aj+1) = 0, f

(k+1)(cj) = 0 for cj (aj , aj+1)= f (k)(x) has at least r k zeros in [a, b]

We had shown the above by induction.(2) We can conclude that theres at most r + k zeros for f (since f (k) has exactly r zeros, the intervals containing the r

zeros are definite).55

Exercise 8. Using the mean value theorem

(1)sinx sin y

x y = cos c = sinx sin yx y

= | cos c| 1= | sinx sin y| |x y|

(2) x y > 0.f(z) = zn is monotonically increasing for n Z.

By mean-value theorem,xn ynx y = nc

n1 for y < c < x

Since 0 < y < c < x; nyn1 xnynxy nxn1.

Exercise 9. Let g(x) =(f(b)f(a)

ba)x+

(bf(a)af(b)

ba)

.

f g = h h(a) = h(c)h(c) = h(b)

so c1 (a, c), c2 (c, b) s. t. h(c1) = h(c2) = 0 by Rolles Thm.Let h = H

since H(c1) = H(c2) = 0 and H is cont. diff. on (c1, c2). then

c3 (c1, c2) s.t. H (c3) = h(c3) = 0Now h = (f g) = f so f (c3) = 0

Weve shown one exists; thats enough.

Exercise 10. Assume f has a derivative everywhere on an open interval I .

g(x) =f(x) f(a)

x a if x 6= a; g(a) = f(a)

(1) g =(

1xa

)f 1xaf(a). f is cont. on (a, b] since f x (a, b).

1xa is cont. on (a, b]. Then g is cont. on (a, b] (remember, you can add, subtract, multiply, and divide cont. functionsto get cont. functions because the rules for taking limits allow so).g is cont. at a since limxa g = limxa

f(x)f(a)xa = f

(a).

By mean value theorem,(f(x) f(a)

x a)

= f (c) = g(x) c (a, x) x (a, b]

Then c [a, b], f (c) ranges from f (a) to g(b) since f (c) = g(x) so whatever g(x) ranges from and to, so doesf (c).

(2) Let h(x) = f(x)f(b)xb if x 6= b; h(b) = f (b).h is cont. on [a, b) since 1xb is cont., f(x) is cont.

limxb h = limxbf(x)f(b)

xb = f(b) so h is cont. at b.

h is cont. on [a, b] h takes all values from h(a) to f (b) on [a, b] (by intermediate value theorem).

By mean value theorem,

h(x) =f(b) f(x)

b x = f(c2) for c2 (x, b) x [a, b]

So then f ranges from h(a) to f (b) just like h.

h(a) = g(b). So then f must range from f (a) to f (b)56

4.19 Exercises - Applications of the mean-value theorem to geometric properties of functions, Second-derivative testfor extrema, Curve sketching.

Exercise 1. f(x) = x2 3x+ 2(1) f (x) = 2x 3 x0 = 32 .(2) f (x) 0 for x 32(3) f = 2 > 0 for x R(4) See sketch.

Exercise 2. f(x) = x3 4x(1) f = 3x2 4 xc = 23(2) f 0 when |x| 2

3

(3) f = 6x f 0 when x 0(4) See sketch.

Exercise 3. f(x) = (x 1)2(x+ 2)(1) f = 3(x 1)(x+ 1) f (x) = 0 when x = 1(2) f when |x| 1(3) f = 3(2x) = 6x f 0 when x 0(4) See sketch.

Exercise 4. f(x) = x3 6x2 + 9x+ 5(1) f = 3x2 12x+ 9 = 3(x 3)(x 1) f (x) = 0 when x = 3, 1(2)

f (x) > 0 when x < 1, x > 3

f (x) < 0 when 1 < x < 3(3) f = 6x 12 = 6(x 2) f 0 when x 2(4) See sketch.

Exercise 5. f(x) = 2 + (x 1)4

(1) f (x) = 4(x 1)3. f (0) = 0 when x = 1(2) f (x) 0 when |x| 1(3) f (x) = 12(x 1)2 > 0 x 6= 1(4) See sketch.

Exercise 6. f(x) = 1/x2

(1) f = 2x3 f(x) = 0 for no x

(2) f 0 when x 0(3) f = 6x4 > 0 x 6= 0(4) See sketch.

Exercise 7. f(x) = x+ 1/x2

(1) f = 1 + 2x3 f(x) = 0 = 1 2x3 = xc = 21/3

(2)f (x) > 0 when x < 0, 0 < x < 21/3

f (x) < 0 when x > 21/3

(3) f = 6x4 > 0 x 6= 0(4) See sketch.

Exercise 8. f(x) = 1(x1)(x3)

(1) f = 1(x1)2(x3)2 ((x 3) + x 1) = (2)(x2)(x1)2(x3)2f (x) = 0 when x = 2

(2) f 0 when x 257

(3)

f = (2)(

(x 1)2(x 3)2 (x 2)(2(x 1)(x 3)2 + 2(x 3)(x 1)2)(x 1)4(x 3)4

)=

= (6)

(x2 4x+ 133

(x 1)3(x 3)3)

x2 4x+ 133> 0 since 144 4(3)(13) = 144 + 12(13) < 0 so

f > 0 if x > 3, x < 1

f < 0 if 1 < x < 3(4) See sketch.

Exercise 9. f(x) = x/(1 + x2)

(1)

f =(1 + x2) x(2x)

(1 + x2)2=

1 x2(1 + x2)2

f (x) = 0 when x = 1(2) f 0 when |x| 1(3)

f =2x(1 + x2)2 2(1 + x2)(2x)(1 x2)

(1 + x2)4=

2x(x2 3)(1 + x2)3

f > 0 when x >

3

f < 0 when 0 < x 0 when

3 < x < 0

f < 0 when x <

3

(4) See sketch.

Exercise 10. f(x) = (x2 4)/(x2 9)(1)

f =2x(x2 9) (x2 4)(2x)

(x2 9)2 =10x

(x2 9)2f (0) = 0

(2) f 0 when x 0, x 6= 3(3)

f = (10)(

(x2 9)2 2(x2 9)(2x)x(x2 9)2

)= (30)

(x2 + 3)

(x2 9)3f 0 when |x| 3

(4) See the sketch.

Exercise 11. f(x) = sin2 x

(1) f = sin 2x So then f = 0 when x = pi2n(2)

f > 0 when0 < x 0 if x 6= pi2 + 2pin(3)

f = cosxf > 0 when

pi2

+ 2pin < x 0 for x > 0 so x =

A minimizes P

Exercise 2.A = xy L = 2x+ y

A = x(L 2x) = Lx 2x2 = dAdx

= L 4x = 0 when x = L4

y =L

2

A = 4 so x = L4

maximizes A

Exercise 3.

A = xy L = 2x+ y = 2x+A

x

dL

dx= 2 +

Ax2

= 0 when x =

A2

y =

2A

L =2A

x3> 0 for x =

A

2so x minimizes L

Exercise 4. f = x2 + y2 = x2 + (S x)2f = 2x+ 2(S x)(1) = 2S + 4x = x = S2f = 4 > 0 so x = S2 minimizes f

Exercise 5. x2 + y2 = R > 059

f = x+ y

f = 1 + y = 0 = 1 +xy

= 0 = y = x

f = y =1 y2

yfor y > 0, f < 0 so that f is max. when y = x

Note that

2x+ 2yy = 0

y =xy

x+ yy = 0

1 + y2 + yy = 0 = yy = 1 y2

Exercise 6.

l2 = (L x)2 + x2 = L2 2Lx+ 2x2 = AdA

dx= 2L+ 4x = 0 = x = L

2d2A

dx2= 4 > 0 = A minimized

l(x =L

2) =

L

2

2

Exercise 7.

(x+L2 x2)2 = A

A = 2(x+L2 x2)(1 + x

L2 x2 ) = 0 when L2 x2 = x2 or x = L

2

so then the side of the circumscribing and area-maximized square isL2

+

L2 L

2

2=

2L2

Exercise 8.

A = (2x)(2R2 x2) = 4x

R2 x2

A = 4(R2 x2 + x

2

R2 x2 ) = 4

(R2 2x2R2 x2

)= x = R

2

since A 0 when x R2, so A is maximized at x =

R2

2x = 2R2

; 2R2 x2 = 2R

2so then the rectangle that has maximum size is a square.

Exercise 9. Prove that among all rectangles of a given area, the square has the smallest circumscribed circle.

A0 = (2x)(2r2 x2) = 4xr2 x2 (fix the area to be A0)(

A04x

)2= r2 x2 = x4 x2r2 + A2016 = 0

= 0 = 2xr2 + x22r drdx 4x3

dr

dx= 0 (for extrema) = x = r

2and

r2 x2 = r

2

We could argue that we had found a minimum because at the infinity boundaries, the circumscribing circle would beinfinitely large.

Exercise 10. Given a sphere of radius R, find the radius r and altitude h of the right circular cylinder with the largest lateral60

surface area 2pirh that can be inscribed in the sphere.

R2 =

(h

2

)2+ r2

A = 2pirh = 2pir

4(R2 r2) = 4pirR2 r2

dA

dr= 4pi

(R2 r2 + r

2

R2 r2

)= 4pi

(R2 2r2R2 r2

)= r = R

2

= h =

2R

Exercise 11. Among all right circular cylinders of given lateral surface area, prove that the smallest circumscribed sphere has

radius

2 times that of the cylinder.

A0 = 2piRH (A0 is the total lateral area of the cylinder)

r2 = R2 +

(H

2

)2= R2 +

(A0

4piR

)2= R2 +

A2016pi2R2

2rdr

dR= 2R+

A2016pi2

(2R3

)= dr

dR= 0 = R =

A0

2pi

= H2

= R

r2 = R2 +R2 = 2R2 = r =

2R

Exercise 12. Given a right circular cone with radiusR and altitudeH . Find the radius and altitude of the right circular cylinder

of largest lateral surface area that can be inscribed in the cone.h

Rr =HR = is the constraint (look, directly at the side, at the similar triangles formed)

A = 2pirh = 2pir(R r) = 2pi(rR r2)dA

dr= 2pi(R 2r) = 0 = r = R

2; h =

H

2

A = 2pi(R 2r)

sincedA

dr 0 when r R

2, r =

R

2maximizes lateral surface area

Exercise 13. Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a right circular

cone of radius R and altitude H .Constraint: hRr =

HR =

V = pir2h = pir2(R r) = pir2(R r) = pi(Rr2 r3)dV

dr= pi(2Rr 3r2) = rpi(2R 3r) r = 2R

3

sincedV

dr 0 when r 2R

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