Solution Manual to Elementary Analysis, 2 Ed., by Kenneth ...

Solution Manual to Elementary Analysis, 2nd Ed., by Kenneth A. Ross

David Buch December 18, 2018

Contents

1 Basic Properties of the Derivative

1

2 The Mean Value Theorem

7

iii

1 Basic Properties of the Derivative

Note: In this section, we make routine use of the fact that limxa is evaluated on sets

J

=

I

\ {a}

so

that,

for

example,

limx0

x2 x

=

limx0 x

is

allowed,

despite

the

cancella-

tion not being valid at x = 0.

28.1 a) {0} b) {0} c) {n|n Z} d) {0, 1} e) {-1, 1} f) {2}

28.2

a)(x3 - 8) = (x - 2)(x2 + 4x + 4),

so,

limx2

x3-8 x-2

=

limx2(x2

+

2x

+

4)

=

(by

20.4)

4

+

4

+

4

=

12

b)limxa

(x+2)-(a+2) x-a

=

limxa 1

=

1

c)limx0

x2 cos(x)-0 x-0

=

limx0

x cos(x)

=

0

1

=

0

d)limx1

3x+4 2x-1

-7

x-1

=

limx1

(3x+4)-(14x-7) (2x-1)(x-1)

= limx1

(-11)(x-1) (2x-1)(x-1)

= -11

28.3

a)(x - a) =( x + a)( x - a),

so, limxa

x- a x-a

= limxa

1 x+ a

=

1 2a

for

a

>

0.

Note: For limxa f (x) to exist, we require that the limit converge for all (xn) in a set J = I \ {a} for some open interval I containing a. However, x is not defined on an

open interval around 0, so the limit does not exist there.

b) Similarly

to

part

(a),

limxa

x1/3 -a1/3 x-a

=

limxa

1 x2/3 +x1/3 a1/3 +a2/3

=

1 3a2/3

for a = 0.

c) f (x) = x1/3 is not differentiable at x = 0.

limx0

x1/3 -01/3 x-0

=

limx0

1 x2/3

and this

limit does not exist, because the limits approaching from the left and right are distinct

(- and ).

1

1 Basic Properties of the Derivative

28.4

a) First, notice sin(x) is differentiable on R and therefore on R \ {0}.

Since

1

and

x

are

differentiable

on

R

and

x

=

0

on

R \ {0},

1 x

,

by

theorem

28.3,

is

differentiable on R \ {0}.

Clearly x2 is differentiable on R \ {0}.

Therefore,

by

28.3,

f (x)

=

x2

sin(

1 x

)

is

differentiable

on

R

\

{0}

and

f

(x)

=

2x

sin(

1 x

)

-

cos(

1 x

).

b)

By

def

28.1

f

(0)

=

limx0

x2 sin(1/x)-0 x-0

=

limx0

x

sin(

1 x

).

By the squeeze theorem,

this limit is 0. c) Since our choice of sequence (xn) 0 can influence lim n f (xn),

we conclude lim n 0f (x) does not exist, so f (x) is discontinuous there.

Note: Recall theorem 28.2 showed that, if f is differentiable at a, then it is continuous

there. This exercise has shown that while any f differentiable at a must be continuous

at a, the corresponding derivative function, f , need not be.

28.5 a) f is differentiable on R by exercise 28.4. g is alos differentiable on R, obviously. b) 0. c) We can find x arbitrarily close to 0 for which the argument of the limit is not defined. Hence, there is no open interval I around 0 on which the argument of the limit is defined, so the limit cannot be evaluated.

28.6

a) f is continuous, since limx0 f (x) = 0 by the squeeze theorem, and f (0) = 0.

b) f is not differentiable at x = 0.

limx0

x sin(1/x)-f (0) x-0

=

limx0

x sin(1/x) x

=

limx0 sin(1/x)

which

does

not

exist.

28.7

a) [Graph Not Shown]

b)

limx0

f (x)-f (0) x-0

is

not

obvious,

but

limx0-

f (x)-f (0) x-0

=

0

and

limx0+

f (x)-f (0) x-0

=

limx0+

x

=

0,

so,

by

theorem,

limx0

f (x)-f (0) x-0

= 0.

Therefore, f

is differentiable at

x = 0.

c) By section 28, example 3, we have f (x) = 2x for x > 0. In part (b), we showed

f (0) = 0, and clearly f (x) = 0 for x < 0.

d) Continuous: Yes. Differentiable: No.

2

28.8 a) f (0) = 0. Let

> 0.

Let = .

|f (x) - 0| = ||x|2 - 0| or |0 - 0|.

Clearly,

||x|2 - 0| = |x2 - 0|. If |x - 0| < then |x2 - 0| < |2| = . |0 - 0| < trivially. Hence,

|f (x) - 0| < .

b) Since the rationals and irrationals are dense, there are rational numbers arbitrarily

close to each irrational number, and there are irrational numbers arbitrarily close to each

rational number. Let x0 = 0, we know x2 > 0, so there is an open interval I around x2 such that x2 > for some > 0. Since x2 is continuous, there is a corresponding

open interval G around x0 that is the inverse image of I. Select rational xr from G. For

arbitrarily close irrational numbers xi G, f (xr) - f (xi) > - 0 so f is discontinuous

for all x = 0.

c)

limx0

f (x)-f (0) x-0

=

limx0 f (x)x

limx0Q

f (x) x

= limx0Q

x2 x

=

limx0Q x = 0

limx0R\Q

f (x) x

=

0

So,

by

theorem,

limx0

f (x)-f (0) x-0

=

0

=

f

(0).

28.9 a)h (x) = 7(x4 + 13x)6(4x3 + 13) b)h(x) = g f (x) where g(x) = x7 and f (x) = x4 + 13x

28.10 a)h (x) = 12(cos(x) + ex)11(- sin(x) + ex) b)h(x) = g f (x) where g(x) = x12 and f (x) = cos(x) + ex

28.11 (h g f ) (a) : Extended Chain Rule By theorem 28.4, since f is differentiable at x, and g is differentiable at f (a), g f is differentiable at a. This, along with the fact that h is differentiable at g f (a), shows that, by theorem 28.4, h(g f ) = h circg f is differentiable at a, and (h g f ) (a) = h (g f (a))(g f ) (a). As stated before, by 28.4 we have g f differentiable at a and (g f ) (a) = g (f (a))f (a). Hence, (h g f ) (a) = h (g f (a))g (f (a))f (a).

28.12 a) - sin(ex5-3x)ex5-3x(5x4 - 3) b) cos(ex5-3x) = h g f (x) where h(x) = cos(x), g(x) = ex, and f (x) = x5 - 3x.

28.13 Let I be an open interval containing f (a) on which g is defined. Naturally, we can find some > 0 such that (f (a) - , f (a) + ) I. Since f is continuous at a, there exists such that |x - a| < implies |f (x) - f (a)| < . Since f is defined on some open interval G containing a, it is defined on G (a - , a + ). Notice that x G (a - , a + ) implies f (x) (f (a) - , f (a) + ), so g is defined at f (x). Therefore, we conclude g f is defined on x G (a - , a + ).

3

1 Basic Properties of the Derivative

28.14

a) For every sequence (hn) in R that converges to 0, there exists (xn) in R such that

xn = hn + a. Clearly, (xn) 0 + a = a. Since f is differentiable at a, we know

limn

f (xn)-f (a) xn-a

exists

and

is

finite.

So,

for

all

(hn) 0,

lim f (a + hn) - f (a) = lim f (a + hn) + f (a)

n

hn

n a + hn - a

= lim f (xn) - f (a) n xn - a

f (x) - f (a) = lim

xa x - a

= f (a).

Therefore,

limh0

f (a+h)-f (a) h

=

f

(a).

b) From part (a),

limh0

f (a+h)-f (a) h

=

f

(a).

So,

by renaming h as

-h,

lim-h0

f (a-h)-f (a) -h

=

f (a).

Further,

-h

0

as

h

-0,

so

limh-0

f (a-h)-f (a) -h

=

limh0

f (a)-f (a-h) h

=

f (a). Thus,

f (a + h) - f (a - h)

f (a + h) - f (a) f (a) - f (a - h)

lim

= lim

+

h0

2h

h0

2h

2h

f (a + h) - f (a)

f (a) - f (a - h)

= lim

+ lim

h0

2h

h0

2h

= f (a)/2 + f (a)/2

= f (a).

28.15

Assume f and g have n derivatives at a. By the product rule (theorem 28.3 iii) (f g) (a) =

f (a)g (a) + f (a)g(a). Suppose now that (f g)(n-1)(a) =

n-1 k=0

n-1 k

f (k)(a)g(n-1-k)(a).

Then,

(f g)(n)(a) = [(f g)(n-1)] (a)

n-1

=

k=0

n - 1 f (k)(a)g(n-k)(a) + k

n-1

= f (a)g(n)(a) +

k=1

n-1 +

k

n

=

n f (k)(a)g(n-k)(a)

k

k=0

n-1 k

n-1 k-1

f (k+1)(a)g(n-k-1)(a) f (k)(a)g(n-k)(a) + f (n)(a)g(a)

So the theorem holds true by mathematical induction.

4

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