# soluciones ejercicios calculo 1 en ingles, Ejercicios de Ingeniería de Telecomunicaciones. Universidad Carlos III de Madrid (UC3M)

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Bank of Problems (solutions)

CALCULUS I Academic Year 2017-2018 ‖ First Semester ‖ Updated to : September 5, 2017

A great discovery solves a great problem but there is a grain of discovery in the solution of any problem. Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discov- ery.

G. Polya (How to solve it)

Index

1 2 1.1 The real number line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Elementary functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Limits of functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 17 2.1 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2 Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.3 Graphical representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.4 Taylor polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 31 3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.2 Infinite series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.3 Power series and Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 42 4.1 Indefinite and definite integrals . . . . . . . . . . . . . . . . . . . . . . . . 42 4.2 Fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . 50 4.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2

1

Solutions to Chapter 1

1.1 The real number line

Problem 1.1.0.1. i) Straightforward computation; for instance,

a < √ ab⇐⇒ a2 < ab⇐⇒ a < b.

√ ab <

a+ b

2 ⇐⇒ 4ab < a2 + b2 + 2ab⇐⇒ 0 < a2 + b2− 2ab⇐⇒ 0 < (a− b)2 ⇐⇒ a 6= b.

a

b < a+ k

b+ k ⇐⇒ a(b+ k) < b(a+ k)⇐⇒ ak < bk ⇐⇒ a < b.

We have used the hypothesis about the signs of a, b and k; some of the above operations are not true with a different sign. ii)

|a+ b| = |a|+ |b| ⇐⇒ a2 + b2 + 2ab = a2 + b2 + 2|a| |b| ⇐⇒ ab = |ab| ⇐⇒ ab ≥ 0.

iii) For the triangle inequality, |a| = |a− b + b| ≤ |a− b| + |b|, so |a− b| ≥ |a| − |b|; for the same reason, |a− b| ≥ |b| − |a|.

iv) x+ y + |x− y|

2 =

 x+ y + x− y

2 si x ≥ y

x+ y − x+ y 2

si x ≤ y. v) f(x) = max{x, 0} =

x+ |x| 2

.

Problem 1.1.0.2. i) n2 − n = n(n− 1). One of the two factors is even. ii) n3 − n = n(n+ 1)(n− 1). At least one of the factors is even and one is a multiple of three. iii) n2 − 1 = (n + 1)(n − 1). Both factors are even, and moreover one is a multiple of four.

Problem 1.1.0.3. i) n+1∑ j=1

j = n+ 1 + n∑ j=1

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

2 =

(n+ 1)(n+ 2)

2 .

2

Solutions to Chapter 1. 3

ii) n+1∑ j=1

j2 = (n+ 1)2 + n∑ j=1

j2 = (n+ 1)2 + n(n+ 1)(2n+ 1)

6 =

(n+ 1)(n+ 2)(2n+ 3)

6 .

iii) n+1∑ j=0

rj = rn+1 + n∑ j=0

rj = rn+1 + rn+1 − 1 r − 1

= rn+2 − 1 r − 1

.

Problem 1.1.0.4. i) If 10n−1 = 9k,then 10n+1−1 = 10n+1−10n+10n−1 = 9(10n+k).

ii) If N∑ j=0

aj = 9m,then n = N∑ j=0

aj10 j = 9m +

N∑ j=1

(10j − 1)aj = 9(m + N∑ j=1

mjaj), since

10j − 1 = 9mj for the previous question.

Problem 1.1.0.5. Let n = z2r, where r does not contain any square fector. If there exist p, q ∈ ZZ, with g.c.d(p, q) = 1, such that p2/q2 = n, we have p2 = z2q2r. What implies p = kr for some k ∈ ZZ. Then, k2r2 = z2q2r, what implies q = mr for some m ∈ ZZ. Finally g.c.d(p, q) ≥ r.

Problem 1.1.0.6. i) A = {−8 ≤ x− 3 ≤ 8} = [−5, 11]. ii) B = (3/2, 2) ∪ (2, 5/2). iii) C = { (x− 2)(x− 3) ≥ 0 } = (−∞, 2] ∪ [3,∞). iv) D = (−∞,−3) ∪ (0, 5). v) E = { x+ 4

(x+ 1)(x+ 7) > 0 } = (−7,−4) ∪ (−1,∞).

vi) F = {x2 > 4, x > 0 } ∪ { x2 < 4, x < 0 } = (−2, 0) ∪ (2,∞). vii) G = [−1, 1/2). viii) H = (1−

√ 2, 1) ∪ (1, 1 +

√ 2). x) I = { 3, −4 }.

ix)

J = {x− 1 + x− 2 > 1, x ≥ 1, x ≥ 2} ∪ {x− 1 + 2− x > 1, x ≥ 1, x ≤ 2} ∪{1− x+ x− 2 > 1, x ≤ 1, x ≥ 2} ∪ {1− x+ 2− x > 1, x ≤ 1, x ≤ 2}

= {x > 2} ∪ ∅ ∪ ∅ ∪ {x < 1} = (−∞, 1) ∪ (2,∞).

Problem 1.1.0.7. i) A = { a, a+ b 2

, b}. ii) B = (a, b). iii) C = (−∞, a). iv) D = (b,∞).

Problem 1.1.0.8. i) supA = 3, inf A = minA = −1, there is no minimum. ii) supB = maxB = 3, inf B = minB = −1. iii) supC = maxC = 3, inf C = 2, there is no minimum. iv) inf D = minD = 2, there are no maximum neither supremum. v) supE = maxE = 3, inf E = minE = 1/3. vi) supF = d, inf F = a, there is no maximum, neither minimum. vii) supG = maxG = 7/10, inf G = 0, there is no minimum. viii) supH = maxH = 2, inf H = −1, there is no minimum.

Problem 1.1.0.9.

Solutions to Chapter 1. 4

i) The interior of a strip. ii) The region bounded between a

parabola and a straight line.

iii) Infinite parallel straight lines.

iv) A rhombus.

v) The interior of a circle. vi) Two straights that intersect.

Solutions to Chapter 1. 5

vii) The interior of two quarters of an ellipse, including the boundary. viii) The interior of a half ring,

including the bottom boundary and the minor semicircle.

Problem 1.1.0.10. The director vector are (1,m) and (1, n). They are orthogonal if their inner product is zero, 1 +mn = 0.

Problem 1.1.0.11. i) P1 = {x = 0} ∩ {y = a

c (x+ b)} = (0, ab

c ).

ii) P2 = {y = c − 2c

a− b x} ∩ {y = c

a+ 2b (x + b)} = (a− b

3 , c

3 ). iii) a = b, c = a

√ 3,

that is, the triangle is equilateral.

Problem 1.1.0.12. i) x2 + (x2 − 1/4)2 = (x2 − λ)2 ⇒ λ = −1/4. ii) (x − a)2 + (y − b)2 = (y − λ)2 ⇒ y = αx2 + βx + γ with α = 1

2(b− λ) , β =

a

λ− b ,

γ = a2 + b2 − λ2

2(b− λ) .

Problem 1.1.0.13. i) √

(x− c)2 + y2 + √

(x+ c)2 + y2 = 2a⇒ x 2

a2 +

y2

a2 − c2 = 1, an

ellipse. ii) √

(x− c)2 + y2 − √

(x− c)2 + y2 = 2a⇒ x 2

a2 − y

2

c2 − a2 = 1, an hyperbola.

1.2 Elementary functions.

Problem 1.2.0.1. i) IR− {2, 3}. ii) {−1, 1}. iii) [−1, 1/ √

2) ∪ (1/ √

2, 1]. iv) {

√ 3 ≤ |x| ≤ 2}. v) (0, e) ∪ (e,∞). vi) (0, 1). vii) (0, 1) ∪ (1, 5]. viii) [1/e, e].

Problem 1.2.0.2. i) f + g is odd, fg is even and f ◦ g is odd. For instace, for the composition,

f ◦ g(−x) = f(g(−x)) = f(−g(x)) = −f(g(x)) = −f ◦ g(x).

ii) f+g is not even neither odd, fg is odd and f ◦g is even. For instance, for the product f(−x)g(−x) = f(x)(−g(x)) = −f(x)g(x).

Solutions to Chapter 1. 6

Problem 1.2.0.3. iii) and iv) even; i) and vi) odd. For the last one note that

log ( 1√

x2 + 1− x

) = − log(

√ x2 + 1− x) = log

( 1√ x2 + 1− x

) = log(

√ x2 + 1 + x).

after multiplying and dividing by the conjugate.

Problem 1.2.0.4. a ax+b cx+d

+ b

c ax+b cx+d

+ d = x⇒ a = −d con a2+bc 6= 0 (otherwise the denominator

is zero, or in other words, the function is constant and the conclusion does not follow), or a = ±d, b = c = 0.

Problem 1.2.0.5. It is injective because x1 + 3

1 + 2x1 =

x2 + 3

1 + 2x2 implies that x1 + 2x1x2 +

3 + 6x2 = x2 + 2x1x2 + 3 + 6x1, that is, x1 = x2. The image is IR − {1/2} because f−1(y) =

y − 3 1− 2y

is defined for all y 6= 1/2.

Problem 1.2.0.6. i) a) Injective, because 7x1 − 4 = 7x2 − 4 ⇒ x1 = x2; the inverse is f−1(y) =

y + 4

7 . b) It is not injective, f(x) = f(x + 2π/7). c) Injective,

f−1(y) = (y − 2)1/3 − 1. d) Injective, f−1(y) = y − 2 1− y

. e) No injective, f(0) = f(3). f)

No injective, f(3) = f(1/3). g) Injective, f−1(y) = − log y. h) Injective, f−1(y) = ey−1. ii) f(x1) = f(x2) implies x1 + x2 = 3, what is impossible if x1, x2 > 3/2. iii) f(x1) = f(x2) implies x1x2 = 1, what is impossible if x1, x2 > 1. f

−1( √

2/3) = √

2. iii) a) One-to-one. b) No surjective: Im(f) = [−1, 1]. c) One-to-one. d) No surjective: Im(f) = IR − {1}. e) No surjective: Im(f) = [−1/4,∞). f) No surjective: Im(f) = [−1/2, 1/2]. g) No surjective: Im(f) = (0,∞). h) One-to-one.

Problem 1.2.0.7. i) tg x = 1 2

+ 1 3

1− 1 2 · 1 3

= 1, so x = π

4 +kπ. As both arctg

1

2 and arctg

1

3

lie in (0, π

4 ), we have x ∈ (0, π

2 ), then x =

π

4 .

ii) tg x = 2 + 3

1− 2 · 3 = −1 ⇒ x = −π

4 + kπ;

arctg 2, arctg 3 ∈ (π 4 , π

2 ) ⇒ x ∈ (π

2 , π)⇒ x = 3π

4 .

iii) tg x =

1 2

+ 1/5+1/8 1−1/40

1− 1 2 · 1/5+1/8

1−1/40

= 1 ⇒ x = π 4

+ kπ;

arctg 1

2 , arctg

1

5 , arctg

1

8 ∈ (0, π

4 )⇒ x ∈ (0, 3π

4 )⇒ x = π

4 .

Problem 1.2.0.8. You must be careful with the sign. i) arccosx ∈ [0, π]⇒ sen(arccosx) ≥ 0⇒ sen(arccosx) =

√ 1− x2.

ii) arcsenx ∈ [−π/2, π/2]⇒ cos(arcsenx) ≥ 0⇒ sen(2 arcsenx) = 2 sen(arcsenx) cos(arcsenx) = 2x

√ 1− x2.

Solutions to Chapter 1. 7

iii) For the first question, tg(arccosx) = sen(arccosx)

cos(arccosx) =

√ 1− x2 x

.

iv) sen(2 arctg x) = 2 tg(arctg x) cos2(arctg x) = 2x

x2 + 1 .

v) cos2(2 arctg x) = 1 − (

2x

x2 + 1

)2 =

( x2 − 1 1 + x2

)2 ; to know the sign when the square

root is taken we note that |x| ≤ 1 ⇒ 2 arctg x ∈ [−π/2, π/2] ⇒ cos(2 arctg x) ≥

0 ⇒ cos(2 arctg x) = 1− x 2

1 + x2 ; analogously cos(2 arctg x) ≤ 0 para |x| ≥ 1, that is,

cos(2 arctg x) = 1− x2

1 + x2 in all the cases. vi) e4 log x = x4.

Problem 1.2.0.9. x3x = (3x)x ⇒ 3x log x = x log(3x)⇒ 2 log x = log 3⇒ x = √

3, y = 3 √

3.

Problem 1.2.0.10.

i) Vertical traslation. ii) Horizontal traslation.

Solutions to Chapter 1. 8

iii) Horizontal con- traction or dilation.

iv) Inversion on the horizontal axis.

v) Replace the region {x < 0} by the symmetric one with respect to the ver- tical axis in the region {x > 0}.

vi) Symmetry with respect to the horizontal axis of the negative part of the function.

vii) Inverstion with re- spect to the vertical axis.

viii) Positive area.

Problem 1.2.0.11.

Solutions to Chapter 1. 9

Solutions to Chapter 1. 10

Solutions to Chapter 1. 11

Problem 1.2.0.12.

i) senhx is odd and coshx is even; for instance

senh(−x) = e −x − ex

2 = − senh(x).

ii) a) 1

4 (e2x + e−2x + 2)− 1

4 (e2x + e−2x − 2) = 1.

b) 2 · 1 2

(ex − e−x) · 1 2

(ex + e−x) = 1

2 (e2x − e−2x).

iii) x = senh(y) = 1

2 (ey − e−y); taken t = ey > 0 we have t − 1/t = 2x ⇒ t =

x± √ x2 + 1⇒ t = x+

√ x2 + 1⇒ senh−1 x = log(x+

√ x2 + 1).

Problem 1.2.0.13. i) Half circle. ii) A half-line.

iii) A circle. iv) A spiral.

Solutions to Chapter 1. 12

v) A spiral. From θ = −2π to 0:

From θ = 0 to 2π:

vi) A half-line.

vii) A cardioid.

viii) A lemniscate.

ix) A rose with four petals.

Solutions to Chapter 1. 13

x) A rose with three petals.

Problem 1.2.0.14.

i) The interior of a ring. ii) A circular sector.

iii) The interior of an arc of spiral. iv) A triangle.

1.3 Limits of functions.

Problem 1.3.0.1. i) If δ = min{1, ε/5} and 0 < |x− 2| < δ, then |x+ 2| < 5, so

|x2 − 4| = |x+ 2||x− 2| < 5δ ≤ ε.

ii) Let ε = 1. For any δ > 0, we take x = 3− δ/2 and we obtain f(x) < 14. So

|x− 3| = δ/2 < δ and |f(x)− 16| > 2 > ε.

Solutions to Chapter 1. 14

iii) It is enough to take δ = ε because | x 1 + sen2 x

| ≤ |x|. iv) If δ = min{9, 3ε} and 0 < |x − 9| < δ, then x > 0 (the square root is well-defined) and |

√ x+ 3| > 3, so

| √ x− 3| = |x− 9|

| √ x+ 3|

< δ/3 ≤ ε.

Problem 1.3.0.2. i) By Ruffini,

lim x→a

xn − an

x− a = lim

x→a

n∑ k=1

ak−1xn−k = nan−1,

or by the binomial theorem

lim x→a

xn − an

x− a = lim

h→0

(a+ h)n − an

h = lim

h→0

n∑ k=1

( n k

) an−khk−1 = nan−1.

The common factor is (x − a). ii) Multiplying by the conjugate ( √ x + √ a), or written

y = √ x we use the previous theorem, L =

1

2 √ a

. The common factor is ( √ x− √ a). iii)

Take z = x1/6 so that

L = lim z→2

z3 − 8 z2 − 4

= lim z→2

z2 + 2z + 4

z + 2 = 3.

The common factor is (x1/6 − 2). iv) Multiplying by the conjugate, L = 1/2. The common factor is x2. v) By the binomial theorem

L = lim h→0

h2 − 3h+ 3 (1− h)3

= 3.

The common factor is h. vi) L = lim x→1

√ x− 1 x− 1

= 1/2. The common factor is ( √ x− 1).

Problem 1.3.0.3. i) L = 4 (

lim x→0

sen 2x3

2x3

)2 = 4. ii) Multiplying by the conjugate,

L = lim x→0

sen2 x

x2(1 + cos x) = 1/2. iii) L = lim

x→0

senx2

x cosx2 + 2

1 + x = 2. iv) By the formula of the

sine of the sum, L = cos a. v) Take y = log(1 + x), L = lim y→0

y

ey − 1 = 1. vi) Using the

exponential form and the previous limit, L = e. vii) Multiplying and dividing by −2x, L = −2. viii) Writting in exponential form, multiplying and dividing the power by sen x, L = e2, or also L = exp(lim

x→0

2 senx

x ) = e2. ix) Extracting the common factor esenx and

taking y = x− senx, L = 1. x) L = 1/2. xi) L = exp(lim x→0

senx

senx− x ( x

senx − 1)) = e−1.

xii) L = exp(lim x→0

cosx− 1 x2

) = e−1/2. xiii) Taking y = x− π, L = 1/8.

xiv) L = lim x→0

ex log a − 1 + 1− ex log b

x = log(a/b).

Solutions to Chapter 1. 15

Problem 1.3.0.4. i) Dividing the numerator and the denominator by x3, L = − 1√ 2

.

ii) Dividing the numerator and the denominator by x, L = 1/5 since senx3

x → 0. iii)

Dividing the numerator and the denominator by √ x, L = 1. iv) Multiplying by the

conjugate, L = 2. v) L = 1. vi) L = 0. vii) L = 1/2. viii) Note that

√ x2 = |x| = −x if x < 0, L = −1/2.

Problem 1.3.0.5. i) L = 1 since (1 t

)[t] = 1 if 0 < t < 1. ii) L = 0 since(1

t

)[t] = t si −1 < t < 0. iii) L = ∞. iv) L = 0. v) L = lim

x→∞

1− x 1 + x

= −1 .

vi) L = lim x→0+

1− x 1 + x

= 1 . vii) L = exp (

lim x→∞

13 √

4x2 + x− 3 2x− 6

) = e13. viii) L =

exp (

lim x→−∞

13 √

4x2 + x− 3 2x− 6

) = e−13.

1.4 Continuity

Problem 1.4.0.1. i) Given ε > 0, there exists δ1 > 0 such that if |x− f(a)| < δ1, we have |g(x)− g(f(a))| < ε. At the same time there exists δ2 > 0 such that if |x− a| < δ2, we have |f(x)− f(a)| < δ1. Thus, if |x− a| < δ2, we have |g(f(x))− g(f(a))| < ε. ii) Applying the previous question to the function h(x) = |x|, which is continuous. The

inverse is not true, for instance consider the function f(x) = sign(x) =

{ 1 if x ≥ 0 −1 if x < 0 .

iii) If it is continuous and take two different values, it must take all the intermediate values. However, between two rational numbers there exist infinite irrational numbers. Thus the function must be constant.

Problem 1.4.0.2. If λ = 0, the function is trivially continuous because it is constant, b(x) = 1. If λ 6= 0 the roots of the denominator are 1±

√ 1− 1/λ. i) The denominator

does not vanish in IR if its roots are not real, that is, the required set is D = {0 ≤ λ < 1}. ii) The roots do not lie in the interval [0, 1] if λ < 0.

Problem 1.4.0.3. The first six functions are continuous on their domains, which are

i) D(f) = IR−{2, 6} , ii) D(g) = IR−{0} , iii) D(h) = IR−{3x+ 2 = π 2

+ kπ, k ∈ ZZ} , iv) D(j) = (−∞, 2] ∪ [3,∞), v) D(k) = [−1, 1] , vi) D(m) = (3/8,∞) . vii) Continuous on IR− ZZ, because in each a ∈ ZZ there is a unit jump. viii) Continuous on IR, because the limit in a = 0 is zero, and the remaining points do not give trouble.

ix) Continuous on IR − {π 2

+ kπ, k = 0, 1, · · ·}, because the one sided-limits in zero match and they are zero. It is only necessary to consider the positive discontinuities of the tangent function. x) Continuous only at x = 0: if x 6= 0, the function oscillates between the values −x and

Solutions to Chapter 1. 16

x; if x = 0, the limit is zero by the sandwich lemma, −|x| ≤ f(x) ≤ |x|. xi) Continous everywhere except at x = −1. xii) Continous everywhere except at x = −0. xiii) Continous everywhere except at x = −2. xiv) Continous everywhere except at x = 0 and x = 1.

Problem 1.4.0.4. i) Apply the Bolzano theorem to the function g(x) = f(x) − x in [0, 1]: it is continuous and the sign changes. ii) Apply the Bolzano theorem to the function h(x) = f(x)− g(x) in [a, b].

Problem 1.4.0.5. The behavior of a polynomial p(x) = anx n+an−1x

n−1 + . . .+a1x+a0 for large values of x is given by the term with the largest exponent. If the polynomial has degree odd, then

lim x→∞

p(x) = sgn(an)∞ and lim x→−∞

p(x) = −sgn(an)∞.

Bolzano theorem concludes the reasoning.

Solutions to Chapter 2

2.1 Differentiability

Problem 2.1.0.1. i) h′(x) = f(x)f ′(x) + g(x)g′(x)√

f 2(x) + g2(x) .

ii) h′(x) = f ′(x)g(x)− f(x)g′(x)

f 2(x) + g2(x) .

iii) h′(x) = [ f ′(g(x))g′(x) + f(g(x))f ′(x)

] ef(x).

iv) h′(x) = g′(x) sen(f(x)) + g(x)f ′(x) cos(f(x))

g(x) sen(f(x)) .

v) h′(x) = (f(x))g(x)[g′(x) log(f(x)) + g(x)f ′(x)

f(x) ].

vi) h′(x) = −[f ′(x) + 2g(x)g′(x)]

[log(f(x) + g2(x))]2 [f(x) + g2(x)] .

Problem 2.1.0.2. i) For instance, for 1 ≤ |x| ≤ 2, we can take f(x) = 2− |x|. ii) For instance, for 1 ≤ x ≤ 2, we can take f(x) = 2x3−9x2+12x−4, and the symmetric one on −2 ≤ x ≤ −1. Another easier one would be f(x) = sen2(πx/2) in 1 ≤ |x| ≤ 2.

Problem 2.1.0.3. It is straightforward. For instance,

(senhx)′ =

( ex − e−x

2

)′ =

ex + e−x

2 = coshx.

Problem 2.1.0.4. It is straightforward. For instance,

ii) xf ′ − f − f 2 − x2 = x (

tg x+ x(1 + tg2 x) ) − x tg x− (x tg x)2 − x2 = 0.

Problem 2.1.0.5. The differential of the three functions is zero in the domain. Thus they must be constant. For instance,

(arctg x+ arctg 1

x )′ =

1

1 + x2 + −1/x2

1 + (1/x)2 =

1

1 + x2 − 1 x2 + 1

= 0.

17

Solutions to Chapter 2. 18

i) In x = 1 we have arctg 1 + arctg 1 = π/2 (we could also take the limit for x→ 0+, or for x→∞). ii) In x = 0 we have arctg 1 + arctg 0 = π/4. iii) In x = 1 we have 2 arctg 1 + arcsen 1 = π.

Note for instance that the first function is even, so for x < 0 it takes the value −π 2

and

it is not continuous at x = 0.

Solutions to Chapter 2. 19

Problem 2.1.0.6.

The system

{ ax2 = log x 2ax = 1/x

,

has the solution x = √

e, a = 1/2e,

The tangent line is y = x√ e − 1

2 .

Problem 2.1.0.7.

The points where the sine is zero are x = kπ, k ∈ ZZ.

Problem 2.1.0.8.

lim h→0+

1

1 + e1/h = 0, lim

h→0−

1

1 + e1/h = 1.

The angle between the tangent lines are arctg 1 = π/4.

Problem 2.1.0.9. i) It is continuous and differentiable in all IR.

Solutions to Chapter 2. 20

ii) Yes, the mean value theorem can be applied in [0, 2]:

f(2)− f(0) 2− 0

= −1/2 = f ′(c)

=

{ −c if c < 1, −1/c2 if c > 1.

We have two values: c = 1/2 < 1 and c = √

2 > 1.

Problem 2.1.0.10. It is continuous in [−2,−1], because {x+2 ≥ 0} ⋂ {−1 ≤ x+2 ≤ 1},

and it is differentiable in (−2,−1) because f ′(x) = 1 2 √ x+ 2

− √ x+ 2√

1− (x+ 2)2 . Note that

the denominators vanish respectively in x = −2 and in x = −1.

Problem 2.1.0.11. f is differentiable if and only if the equation αx2−x+3 = 0 does not have two different roots. Using the discriminant condition 1− 12α ≤ 0, that is, α ≥ 1

12 .

Problem 2.1.0.12. f is not differentiable in x = 0.

Problem 2.1.0.13. If c ≤ 0 the function can be written as f(x) = |x|−1. This function is not continuous in x = 0. For c > 0, as f is symmetric it is enough to study x = c,

where we have a+ bc2 = 1

c , 2bc = − 1

c2 . Then a =

3

2c , b = − 1

2c3 .

Problem 2.1.0.14. Take f(x) = x2/3. By the mean value theorem in [26, 27] we have

272/3 − 262/3

27− 26 =

2

3 x−1/3, for some x ∈ (26, 27) ⊂ (8, 27).

Then 9− 262/3 ∈ (2 9 ,

1

3 ). Finally,

26

3 < 262/3 <

79

9 .

If we take now g(x) = log x and apply the mean value theorem in [1, 3/2], we have 1

3 < log(3/2) <

1

2 .

Problem 2.1.0.15. The limits of the exercise 1.3.2 are straightforward by L’Hôpital; for instance,

iii) L = lim x→64

1/(2x1/2)

1/(3x2/3) = 3.

In some limits of 1.3.3 it will be necessary to operate first to obtain a fraction. For instance, in vi), viii), xi) and xii) we must write in exponential form. If we have to apply l’Hôpital more than once, try to simplify as much as possible before applying l’hôpital again. For instance,

i) L = lim x→0

12x2 sen 2x3 cos 2x3

6x5 = 2 lim

x→0

sen 2x3

x3 = 2 lim

x→0

6x2 cos 2x3

3x2 = 4.

Solutions to Chapter 2. 21

Problem 2.1.0.16. i) Differentiate twice L = 1/2. ii) L = 1.

iii) L = lim x→1+

log(x− 1) 1/ log x

= lim x→1

1/(x− 1) −1/(x log2 x)

= lim x→1

log2 x

1− x = − lim

x→1

2 log x

x = 0.

iv) L = exp( lim x→∞

log x

x ) = 1. v) Take x+ 1 = t to simplify the derivatives.

L = lim t→1

tt − t2 + t− 1 (t− 1)3

= lim t→1

tt(log t+ 1)− 2t+ 1 3(t− 1)2

= lim t→1

tt(log t+ 1)2 + tt−1 − 2 6(t− 1)

= 1

6 lim t→1

( tt(log t+ 1)3 + 2tt−1(log t+ 1) + tt−1(log t+ (t− 1)/t

) = 1/2.

vi) Take t = 1/x, L = 1.

Problem 2.1.0.17. i) L = lim x→∞

1

x

( x x− 1

)x = 0. ii) L = 0. iii) L = 1/2.

iv) L = exp(lim x→0

3x2

2 arcsenx ) = 1. v) L = lim

x→1/2

2x2 + 3x− 2 cosπx

= − 5 π . vi) L = lim

x→0

x sen 2x

1− cosx =

4. vii) L = lim x→∞

( √ x2 + 1−

√ x2 + 4x) = lim

x→−∞

1− 4x√ x2 + 1 +

√ x2 + 4x

= 2. viii) L = e.

Problem 2.1.0.18. As lim x→0

h(x)

x2 = 1, we have h(0) = lim

x→0 h(x) = lim

x→0

h(x)

x = 0. So,

h′(0) = lim x→0

h(x)

x = 0. Finally h′′(0) = lim

x→0

h′(x)

x . Applying L’Hôpital, we have 1 =

lim x→0

h(x)

x2 = lim

x→0

h′(x)

2x . Hence, h′′(0) = 2.

Problem 2.1.0.19.

L = lim x→0

aeax − ex − 1 2x

= lim x→0

a2eax − ex

2 = a2 − 1

2 .

In order to apply L’Hôpital a second time we must have lim x→0

(aeax− ex−1) = 0 (otherwise the limit does not exist). Thus a = 2 and L = 3/2.

Problem 2.1.0.20.

i) L = lim t→0

(1 + t)1/t − e t

= lim t→0

(1 + t)1/t ( 1 t(t+ 1)

− log(1 + t) t2

) =

= e lim t→0

t− (1 + t) log(1 + t) t2

= e lim t→0

− log(1 + t) 2t

= −e 2 .

ii) L = lim x→∞

[(1 + 1/x)x e

]x = exp

[ lim x→∞

x ((1 + 1/x)x

e − 1 )]

=

= exp

[ 1

e lim x→∞

x (

(1 + 1/x)x − e )]

= exp

[ 1

e (−e

2 )

] = e−1/2,

Solutions to Chapter 2. 22

using the above limit.

iii) L = exp

[ lim x→∞

x (21/x + 181/x

2 − 1 )]

= exp

[ lim t→0

2t + 18t − 2 2t

] = exp

[ log 2 + log 18

2

] = 6 .

In fact, it is a particular case of the next one L = √

2 · 18 = 6.

iv) L = exp

[ lim x→∞

x (1 p

p∑ i=1

a 1/x i − 1

)] = exp

limt→0 p∑ i=1

ati − p

pt

 = exp

[ lim t→0

1

p

p∑ i=1

ati log ai

] = exp

[1 p

p∑ i=1

log ai

] = ( p∏ i=1

ai

)1/p ,

the geometric mean of ai.

Problem 2.1.0.21. i) f(0) = 0 so that the given limit exists. ii) f ′(0) = lim x→0

f(x)

x =

lim x→0

f(2x3)

2x3 =

5

2 .

iii) lim x→0

(f ◦ f)(2x) f−1(3x)

= lim x→0

(f ′(f(2x))

f ′(2x)2

3

f−1(3x) =

125

12 .

Problem 2.1.0.22. First note that if f is continuous, so is f ′ because f ′ = ef (2 + tg x). The inverse g is also continuous. We have

f(0) = 1, g(0) = f−1(1) = 0, f ′(0) = lim x→0

f ′(x) = 2ef(0) = 2e.

Moreover, as f ′(0) 6= 0, the following limit exists

lim x→0

g′(x) = lim x→0

1

f ′(g(x)) =

1

2e .

Now applying L’Hôpital:

L = lim x→0

ex − e− senx

g(x) = lim

x→0

ex + cosx e− senx

g′(x) = 4e.

If we do not want to apply L’Hôpital, we can decompose the limit as follows

L = lim x→0

ex − e− senx

g(x) = lim

x→0

ex − e− senx

x lim x→0

x

g(x) .

The first limit is 2 and the second one is

lim x→0

x

g(x) = lim

z→0

f(z)− 1 z

= f ′(0) = 2e.

Solutions to Chapter 2. 23

2.2 Extrema

Problem 2.2.0.1.

i) f is continuous in IR and differentiable in IR− {4}. ii) There is a local maximum in x = 3, local and absolute minimum in x = 0, x = 4. iii) f increases in [0, 1], with f(0) = −1, f(1) = 2.

Problem 2.2.0.2. The lateral area A(r, h) = 2πr2 + 2πrh must be minimized with the

constraint of fix volume V = πr2h. That is, we must minimize f(r) = 2π(r2 + V

πr ) for

r ∈ (0,∞). We have r = (V/2π)1/3, h = 2r = 3 √

4V

π .

Problem 2.2.0.3. Let x be the length of the square in the basis and let h be the height of the container. The volume is 32 litres, that is, 32 dm3. Thus, 32 = x2h if x and h are measured in dm. The function to be minimized is the area A(x, h) = x2 + 4hx since it is open on the top. Solution is x = 4 dm and h = 2 dm.

Problem 2.2.0.4. Let x and y be two positive numbers such that x+y = 20. We must maximize the value of the function f(x, y) = x2y3. The possible solutions are 0 and 20, or 8 and 12.

Problem 2.2.0.5.

Given a point in the first quarter P = (x, y), we must maximize the area A(x, y) = 4xy with the con- straint that the point lies in the ellipse (x/a)2+(y/b)2 =

1. That is, to maximize f(x) = 4bx

a

√ a2 − x2 for

x ∈ [0, a]. We then have x = a√ 2

, y = b√ 2

, and area

A = 2ab.

Problem 2.2.0.6. Let l be the height of the window and r be the length of the basis.

The function to be maximized if the area, that is, A(r, l) = lr + π r2

8 . The solution is

l = 4

4 + π and r =

8

4 + π .

Problem 2.2.0.7. Given a point in the first quarter P = (x0, y0) in the parabola y = 6 − x2, the tangent line to it at the point P is y = 6 + x20 − 2x0x. The area of

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