link solutions manual for additional calculus topics 11th edition by raymond barnett karl byleen michael ziegler

Karl
Byleen
1

1.f(x) =

= x

-1

x

-2

f’(x) = -x

f”(x) = 2x

(3)

f
2.

(Using Power Rule)

-3

-4

(x) = -6x

6
= -x 4

f(x) = ln(1 + x)
1
-1
f'(x) = 1  x = (1 + x)
f"(x) = (-1)(1 + x)
(Using Power Rule)

(3)

f

(x) = (-1)(-2)(1 + x)

=
3.

2

(1  x)
-x

-3

(Using Power Rule)

3

f(x) = e

f’(x) = -e
f”(x) = e -x
(3)
f
4.

(x) = -e
f(x) = ln(1 + 3x)
f’(x) =

3

= 3(1 + 3x)

1  3x

-2

f”(x) = 3(-1)(3)(1 + 3x)

Rule) f

(4)

f

(3)

(x) = (-9)(-2)(3)(1 + 3x)

(x) = (54)(-3)(3)(1 + 3x)

-4

5x

5. f(x) = e

f’(x) = 5e

5x
5x

f”(x) = 5(5)e

(3)

f

2

3

(Using Power

-3

5x

(x) = 5 (5)e

5x

3 5x

= 5 e

4 5x

= 5 e

5x

= 625e

TAYLOR POLYNOMIALS AND INFINITE SERIES

-3

= 54(1 + 3x)

= -486(1 + 3x)

2 5x

= 5 e

(x) = 5 (5)e

(4)

f
96

-2

= -9(1 + 3x)

-4

486

= -

(1  3x)

4


1

-1

6. f(x) = 2  x = (2 + x)
f’(x) = (-1)(2 + x)
f”(x) = (-1)(-2)(2 + x)
f
(x) = (2)(-3)(2 + x)

(4)

f

-5

-5
-0

(x) = (-6)(-4)(2 + x)

= 24(2 + x)

-x

7.

f(x) = e
f'(x) = -e

f"(x) =

f

= 2(2 + x)
= -6(2 + x)

(3)

f'(0) = -e

-x
e
-x

(x) = -e

(4)

f(0) = e

-x

f
(x) = e
Using 2,

(3)

(4)

-0

f

(0) = -e

f

(0) = e

p (x) = f(0) + f'(0)x + f"(0) x
2!
4
Thus,

f(x) = e
f'(x) = 4e
f"(x) = 16e

(3)

f

9.

p (x) = 1 - x + 1 x
4
2!

- 1 x
3!

4x

8.

(x) = 64e

= -1

-0
e
= 1
-0

f"(0) =

-x

= 1

-0

= -1

+

+ 1x
4!

= 1
f

(3)
(0) x

+ f

3!
= 1 - x +

(4)
(0) x

4!

1 x
2

-1

x

6

+ 1x
24

f(0) = 1
f'(0) = 4
f"(0) = 16

4x

(3)

f

(0) =

64

(3)

3
f"(0) 2
f (0)
+
Thus, p 3 (x) = f(0) + f'(0)x +
2! x
3! x
2
= 1 + 4x + 16 x2 + 64 x3 = 1 + 4x + 8x +
3!
3 2!
f(x) = (x + 1) ,
f(0) = 1

32

3 x

3

2

f'(x) = 3(x + 1) ,
f'(0) = 3

f"(x) = 6(x + 1),
f"(0) = 6
f
(x) = 6,
f
(0) = 6
f
(x) = 0
f
(0) = 0
2
2
3
p (x) = 1 + 3x + 6 x + 6 x3 = 1 + 3x + 3x + x
4
2!
3!
EXERCISE 2-1

97


4

f(x) = (1 - x) ,
f'(x) = -4(1 - x) ,
f"(x) = 12(1 - x) ,

10.

f(0) = 1
f'(0) = -4
f"(0) = 12

(3)

f
Thus,

p 3 (x) = 1 - 4x +

11.

(3)

(x) = -24(1 - x),

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

12
2!

x2

f

(0) = -24

24

- 3!

x3

2

2

2

f'(x) = 1  2x

(2) =1  2x
4
-2
f"(x) = -2(1 + 2x) (2) = (1  2x)
(3)
-3
16
f
(x) = 8(1 + 2x)
(2) = (1  2x)

12.

f(x) =

2!

3x  1

= (x + 1)

1/3

+

(3)

f

(0)

3
x

= 1

2

f"(x) = 9(x  1)
10

(x) = 27(x  1)

x

f(0) = 31
1
f'(0) = 3

2

f

(3)
(0) = 16

3!
2
3
2
(x) = 0 + 2x - 4 x + 16 x = 2x - 2x
+ 8 x3
3!
2!
3

1
f'(x) = 3(x  1)

(3)

f
2

3

3

1  2  0
= 2 f"(0) = -4
f'(0) =

f"(0)

Using 2, p (x) = f(0) + f'(0)x +

Thus, p

3

= 1 - 4x + 6x - 4x
f(0) = ln(1) = 0

8/3

f"(0) = -9
f

(3)

10

(0) =

27

(3)
f"(0)

f (0)
p 3(x) = f(0) + f'(0)x +
x
2
+
2!
3! x3
1
1
1
Thus, p (x) = 1 + x 2 x
+
x
10 x = 1 + x 9  2!
27  3!
3
3
9
3

+

5 x

81


1/4

f(x) = 4 x  16 = (x + 16)

f(0) = 2
1
1
f'(x) = 4(x + 16)
f'(0) = 32
f"(x) = - 3 (x + 16)
f"(0) = - 3
16
2048
3 x
p (x) = 2 + 1 x 32
2
4096
14. (A)
f(x) = x
- 1
f(0) = -1
f'(x) = 4x
f'(0) = 0
f"(x) = 12x
f"(0) = 0
13.

-3/4

(3)

f

(3)

(x) = 24x

f

(0) = 0

Using 2, p (x) = f(0) + f'(0)x + f"(0)
3
2!
Thus, p3(x) = -1

|p 3(x) - f(x)| = |-1 - (x

Now

4

4

+

(3)

f

(0)

3!

x

4

- 1)| = |x | = |x|

< 0.1 implies |x| < (0.1)
and |x|
Therefore, -0.562 < x < 0.562
(B) From part (A),

(4)

2

1/4

3
x

4

≈ 0.562

(4)

(4)
f
(x) = 24
f

(0) = 24
(3)
f
(0)
Using 2, p (x) = f(0) + f'(0)x + f"(0) x + f (0) x +
x4
2!
3!
4
4!
24 4
4
4
Thus, p 4(x) = -1 + 4! x = -1 + x
= x - 1 = f(x)

|p4(x) - f(x)| = 0 < 0.1 for
f(0) =
f(x) = x
f'(x) = 5x
f'(0) =
f"(x) = 20x
f"(0) =
f
(x) = 60x
f
(0) =

all x.
0

0
0
0

and
15. (A)

(4)

f

(4)

(x) = 120x

f

p4(x) = 0

(0) = 0

5

5

|p4(x) - f(x)| = |0 - x | = |x|

1/5

< 0.01 or |x| < (0.01)

= 0.398
Therefore, -0.398 < x < 0.398

(3)

(B) f(0) = f'(0) = f"(0) = f
= 0 f

(5)

5!

x

5

(5)

5

|p5(x) - f(x)| = |x

5

(4)

(0) = f

(x) = 120 and hence f

120. p5(x) =

=

(0)

(0) =

= x

5

- x | = 0 < 0.01

Thus for all x, |p5(x) - f(x)| < 0.01.
EXERCISE 2-1

99


3
2

f(x) = x

16.

f

f'(x) = 3x

f"(x) = 6x
(3)
(x) = 6

f(1) = 1
f'(1) = 3
f"(1) = 6
(3)
f
(1) = 6
2

p (x) = 1 + 3(x - 1) + 6
2!
3

(x - 1)

2

= 1 + 3(x - 1) + 3(x - 1)
17.

+

6 (x - 1)
3!

2

f(x) = x - 6x + 10
f'(x) = 2x - 6
f"(x) = 2

p (x) = 1 +
2

2

(x - 3)

2!

3

+ (x - 1)

f(3) = 1
f'(3) = 0
f"(4) = 2
= 1 + (x - 3)

3


18.

f(x)
f'(x)
f"(x)

f
(x)

(4)

f

=
=
=
=

ln(2 - x)
-(2 - x)
-(2 - x)
-2(2 - x)

f(1) = 0
f'(1) = -1
f"(1) = -1

-4

(x) = -6(2 - x)

p (x) = -(x - 1) - 1 (x - 1)

2

19.

f(x) =

2

(x - 1)

-2x
e
-2x
-2x

f"(x) = 4e

f

- 2x

(x) = -8e

f

3

Thus, p (x) = 1 - 2x + 4
3

Now, e

= e

4

6 (x - 1)

4!
1

3

4

- 3 (x - 1) - 4(x - 1)

(3)

Using 2, p (x) = f(0) + f'(0)x +

- 0.5

(1) = -6

f(0) = 1
f'(0) = -2
f"(0) = 4

f'(x) = -2e
(3)

f

3!
1

2

1

= -(x - 1) -

(1) = -2

(4)

3
- 2 (x - 1) -

2!

4

f

-2(0.25)

2
x -

2!

(0)

= -8
(3)

f"(0) x + f (0) x
2!
3!
3
2 - 4
8 x = 1 - 2x + 2x

3!

3

3
x .

= f(0.25) ≈ p 3 (0.25)

= 1 - 2(0.25) + 2(0.25)

2

4
3
(0.25)
= 0.60416667.
-3

20.

f(x) =
1

f'(x) =

= x1/2

x
2

x-1/2
x-3/2

f"(x) = - 1
4
3

(3)

f

(4)

f(1) = 1
1
f'(1) = 2

(x) = 8 x-5/2
15

f
(x) = - 16 x
Thus,

-7/2

f"(1) = - 1
4
3

(3)

f

(4)

f

(1) = 8
15

(1) = - 16
(3)

(4)

f"(1)
f
(1)
f (1)
2
3
4
p4(x) = f(1) + f'(1)(x - 1) +
+
+
2! (x - 1)
3! (x - 1)
4! (x - 1)
2
3
4
1
= 1 +
(x - 1) (x - 1) +
15 (x - 1)
3 (x - 1) 1
16  4!
8  3!
4 2!
2
1
1
5
1
= 1 +

2

(x - 1) - 8 (x - 1)

Now,
1.2

= f(1.2) ≈ p4(1.2)

2

4

3

+

16 (x - 1)

2

- 128 (x - 1) .

(1.2 - 1)
+ 1 (1.2 - 1)
= 1
+ 1 (1.2 - 1) - 1
2
8

16
2
3
4
= 1 + 1 (0.2) - 1 (0.2) + 1 (0.2) 5 (0.2)
8
16
128
2
= 1.0954375.

3

4

- 5 (1.2 - 1)
128


-1

21.

f(x) =

f

= (4 - x)
1
4  x

-2

-2

(x) = 2(-3)(4 - x)

(4)

f

(n)

f

(-1) = 1·2·3(1 - x)

(x) = 2·3(-4)(4 - x)

-5

-4

(-1) = 1·2·3·4(1 - x)

-5

-(n+1)

(x) = n!(4 - x)

4
= 4(1 +x) -x
1  x
f'(x) = -4(1 + x)
f"(x) = (-1) (4)(2)(1 + x)

22.

-3

f'(x) = -1(4 - x)-3(-1) = (4 - x)
f"(x) = -2(4 - x) (-1) = 1·2(4 - x)
(3)
-4

f(x) =

f

(3)

= 4(2!)(-1) (1 + x)

3

-4

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

M

(n)

f

n

3

-(n+1)

(x) = 4(n!)(-1) (1 + x)

3x

23.

f(x) = e
3x
3x
f'(x)
= e (3) = 3e
3x
3x
2 3x
= 3 e
f"(x) = 3e (3) = 9e

(3)

3x

f

(x) = 9e

f

(x) = 27e

(4)

(n)

f
24.

(3) = 27e

3x

3x

(3) = 81e

3x

3 3x

= 3 e

4 3x

= 3 e

n 3x

(x) = 3 e

f(x) = ln(2x + 1)
-1

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

2

f"(x) = -(2) (2x + 1)

-2 (3)

1)

f

(n)

f
25.

4

(x) = (-1)

(x) = (-1)

2
3

= (-1) (2x + 1)

-2

= (-1)

-1

2

3

2 (2x +

2 (2!)(2x + 1)

-3

n+1 n

2 ((n - 1)!)(2x + 1)

-n

f(x) = ln(6 - x)
= -(6 - x)
f'(x) =
1 (-1) = - 1
6 x
6 x

-2 -2

f"(x) = (6 - x) (-1)

= -(6 - x)

(4)

f

-4

(x) = 1·2·3(6 - x)

-4 (n)

x)

f

-4

= 4(3!)(-1) (1 + x)

(-1) = -1·2·3(6 -

-n

(x) = -(n - 1)!(6 - x)


f(x)

26.

x/2

= e
1

f'(x) =

x/2

2

e

 1

1

2


f"(x) =

ex/2 =

2

(3)

2 ex/2

2

(x) = 13 ex/2
2
(n )
1
f

f

(x) =

n

ex/2

27.

From Problem 31,
1
1
2!
(3)
3!
(n)
n!
f(0) =
, f'(0) =
, f"(0) =
, f
(0) =
, … , f
(0) =
4
4
4
4
4
Thus,
1
1
1
x .
+ 1 x + 1
p (x) =
x +

x + … +
2

4

n

4

2

3

4

3

4

n 1

4

4

4n 1

-1

28. f(x) = 1  x = 4(1 + x)

From problem 32, f (x) = 4(n!)(-1) (1 + x) and
(n)
4
hence f
(0) = 4(n!)(-1)n.

n

The coefficient of x

2

(3)

3 , f
Thus,

= 1, f'(0) = 3e

3 0

0

2

3

x

2

+

3

3

2!

n

(n )
x

3

(0) = 3
n

+ … + 3

3!

n

= 4(-1) .

n!
2 0

= 3 , …, f

p (x) = 1 + 3x +



n!

n

4(n!)(1)

= 3, f"(0) = 3 e

3

(0) = 3 e

(0)

is

29.From Problem 33,
0
f(0) = e

(n)

f

=

n 0
e

n

= 3

n
x .

n!

30.f(x) = ln(2x +
1) From problem 34,
(n)
n
n+1 n
-n
n+1 2
(n )
f
(0)
n!
n
f
(x) = (-1)
2 ((n - 1)!)(2x + 1)
and

= (-1)
2
3
n+1
2n x n
Thus, p (x) = 2x - 2 x2 + 23 x - 24 x4 + … + (-1)
n
n
2
3
4
31. From Problem 35,

(3)
(n)
(n  1)!
1
1
2!
f(0) = ln 6, f'(0) = - , f"(0) = - 2 , f
(0) = - 3 , …, f
(0) = n
6
6
6
6
Thus,
1 x 1 x3 - … 1 x .
p (x) = ln 6 - 1 x n

6

2

2



6

3

3
6

n

n 6

EXERCISE 2-1

x/2

32. f(x) = e

From problem 36, f

(n)

(x) =

1 e x/2

and

(n)

f

(0)

=

1

. Thus,

105


p (x) = 1 +
n

1
2

1

x +

2!2

2

2x

f"(x) = (-1)

(3)

f

(x) = (-1)

f

(x) = (-1)

(n)

2

3

n

3!2

3x

2(2!)x

f"(x) = - x

(3)

f

(x) = 23
x

n!2

nx

n!2

n

+x

-4

2(n!)x

-(n+1)
n

= 2(-1)

and

2

3

- 2(x - 1)

n

+ … + 2(-1) (x - 1)

Step 2.
f(1) =
0

Step 3.
a
0 = f(1) = 0

f'(1) = 1

a1 = f'(1) = 1

1

1
2

2

-3

pn(x) = 2 - 2(x - 1) + 2(x - 1)

f'(x) = x

+ … +

1

f(1) = 2

n!

34. Step 1.
f(x) = ln x

3

2(3!)x

f(n) (1)

Therefore

1

+

-1

f(x) = 2 = 2x
x
f'(x) = -2x

33.

2

f"(1)

f"(1) = -1

(3)

f

(1) = 2

2

=
=

3

1

2! = -

2!

= - 2

(1)

2

(3)

f
a

1

3!

=

3!

1
=

3

n

(4)

(x) = - x 4

f
f

2  3

(n)

f

(x) =

n 1

(1)

f

(n  1)!

(4) (1)
(n)

= -3!

a

(1) =

n+1

(-1)

(4)

a

f

4

=

n

=

f

(1)

n!

xn

1

4! = -4! = - 4

(n)

(n - 1)!

3!

(1)

 1)!

n 1

= (1)

Step 4. The nth degree Taylor polynomial is:
2
3
p (x) = (x - 1) - 1 (x - 1) + 1 (x - 1) - 1 (x n
2
34

(n

n!

n 1

= (1)

(1)n 1
n
(x - 1) .
1)4 + … +
n

x

f(x) = e

35.
f

(n)

(n)

x
(x) = e

Thus, p

(x) =
n

f

and
1

e2

(2)

n!
+

1

e2

1
= n!e

(x + 2) +

1

2 (x

2!e

+ 2)

2

n

+ … +

1
n!e

2 (x

n

+ 2)


5

36.f(x) = x

+ 2x

3

+ 8x

2

+ 1

(A) Fourth-degree Taylor polynomial p4(x) for f at 0 is:

2
3
(3)

(4)
p (x) = f(0) + F’(0) x + F’’(0) x + f (0) x + f (0) x4
1!
2!
3!
4!
4
f(0) = 1
f’(x) = 5x
+ 6x + 16x ; f’(0) = 0
f”(x) = 20x
+ 12x + 16 ; f”(0) = 16
f
(x) = 60x + 12
;f
(0) = 12

(4)

f
(x) = 120x
Thus,
p 4(x) = 1 + 8x

3

; f

2
+

2

(4)

(0)

=0

3

2x

= 2x + 8x + 1
(B) The degree of the polynomial is 3.

6

37.f(x) = x
f(0) = 1
f’(x) =
f”(x) =
f
(x)
f
(x)
f
(x)
f
(x)

(n)

+ 2x

3

+ 1

6x
+ 6x
30x
+ 12x
= 120x + 12
= 360x
= 720x
= 720

;
;
;
;
;
;

f’(0) =
f”(0) =
f
(0)
f
(0)

f
(0)
f
(0)

0
0
=
=
=
=

12
0
0
720

f
(x) = 0
for
n ≥ 7.
Thus, for n = 0, 3 and 6, the nth-degree Taylor polynomial for f at 0
has degree n.

4

38.f(x) = x
f(0) = -1
f’(x) =
f”(x) =

f
(x)
f
(x)
f

(n)

– 1

4x
12x
= 24x
= 24

(x) = 0

;
;
;
;

f’(0) =
f”(0) =
f
(0)
f
(0)

0

0
= 0
= 24

for n ≥ 5.

Thus, for n = 0 and n = 4; the nth degree Taylor polynomial for f at
0 has degree n.
EXERCISE 2-1

10
9


39.

f

f(x) = ln(1 + x)
1
f'(x) = 1  x
1
f"(x) = (1  x)
(3)
2
(x) =
(1  x)

f(0) = 0
f'(0) = 1

f"(0) = -1
f

3

(3)

(0) = 2

2

Thus,

2

x
p1(x) = x, p
x
-0.2
-0.1
0
0.1
0.2

2(x)

= x -

p1(x)

p2(x)

-0.2
-0.1
0
0.1
0.2

-0.22
-0.105
0
0.095
0.18

x

p1(x) - f(x)

-0.2
-0.1
0
0.1
0.2

0.023144
0.005361
0
0.00469
0.017678

2

, p3(x) = x p3(x)

-0.222667
-0.105333
0
0.095333
0.182667

2

x

+

3 .

f(x)
-0.223144
-0.105361
0
0.09531
0.182322

p2(x) - f(x)
0.003144
0.000361
0
0.00031

0.002322

3

x

p3(x) - f(x)
0.000477
0.000028
0
0.000023
0.000345


40.

(3)

f(x) = ln(1 + x)
f'(x) = (1 + x)
f"(x) = -(1 + x)

-3

(3)

f(0) = 0
f'(0) = 1
f"(0) = -1

f
(x) = 2(1 + x)
f
(0) = 2
Thus,
p (x) = x - 1 x 2 + 1 x 3 .
3
2
3
Using a graphing utility, we find that
|p3(x) - ln(1 + x)| < 0.1 for

-0.654 < x < 0.910.


x

f(x) = e
f'(x) = e
f"(x) = e

41.

(3)

f

(x) = e

M

(n)

f
(x) = e
Thus,

f(a) = e
f'(a) = e
f"(a) = e

x

f

(3)

(a) = e

M

x

f

(n)

(a) = e

a

a
a
(3)

p (x) = f(a) + f'(a)(x - a) +
n

(n)
f (a)
+

+

n!

(x - a)2

2!

(x - a)

(a)

f

+

(x - a)

3!

n

a
a
2
3
a
a
= e + e (x - a) + e (x - a) + e (x - a) +
a 

= e

= e
42.

1 

(x  a) 

2!
1 (x  a)
2!
k


a n

1



(x - a) k 0

k !

3!
1 (x  a)
3!

f'(x) = x

(4)

f

f

4

-3

(3)

(x) = (-1) (2!)x

5

f

(x) = (-1)

Thus,
p (x) = ln a +

f

n+1

((n - 1)!)x

1

-n

2

2a

k  1

= ln a

f

(n)

1 (x - a)

(x - a) -

a

n

n (1)
+ 
k
k 1 ka

a

a

(x - a)

n!

n 

(x 

a)



(x - a)

2

4

a

3

5
(a) = (-1) (3!) 14
a

(a) = (-1)

2

1

(a) = (-1) (2!)

(4)

-4

(x) = (-1) (3!)x

( n)

e

f(a) = ln a
1
f'(a) = a
f"(a) = - 1

-1

(3)

 1
n!

+



f(x) = ln x

f

3

1

+

n+1

((n - 1)!)

(x - a)

1
n

a

3

3a

- … + (-1)

n+1 1
n

na

(x - a)

n

n


43.

f(x) = (x + c)

n

n

f(0) = c

f'(x) = n(x + c)

f'(0) = nc

n- 2

f"(x) = n(n - 1)(x + c)

n- 2

f"(0) = n(n - 1)c
M

f

(x) = n!(x + c)

f

(n)

f
(x) = n!
Thus,
p (x) = c + nc

n

= c
n

=

(n)

f

x + n(n  1) c

x

+ … +

2!
n!
n!
+
x +
cn-2x2
(n  1)!
2!(n  2)!
n!

cn-kxk.



k 0

(0) = n!c

k!(n  k)!

(0) = n!
n!
(n  1)!

+ … + x

cx

n!
n

+x

n!


44.Let f(x) be a polynomial of degree k, k ≥ 0. Then
f(x) = a 0 + a 1x + a 2

x

3

2

+ a 3x

f'(x) = a 1 + 2a 2x + 3a 3x2

(3)

xk

f(0) = a 0

+ … + ka kxk-1 f'(0) = a 1

+ 6a3x + … + k(k - 1)a kxk-2

f"(x) = 2a 2

f

+ … + a k

f"(0) = 2a 2 + 2!a 2

(3)

xk-3

(x) = 6a 3 + … + k(k - 1)(k - 2)a k

In general,
 m!am for m  0,1,2,K k
f (m)
(0) =
for m  k
 0
Since p (x) = f(0) + f'(0)x + f"(0) 2+ f
2!
n

f

(0) = 6a 3 = 3!a 3



(3)

(0)

3

x

x

3!

it follows that pn(x) = f(x) for all n ≥ k.
45.f(x) = e

f(0) = 1

x

x

f’(x) = e

f”(x) = e

(k)

x

f
(x) = e
Therefore,
p

x

(x) = 1 +

;
;

f’(0) = 1
f”(0) = 1

(k)

; f

1 x2 + … + 1 x10
1!
2!
10!
p (x) = 1 + 1 x + 1 x + … + 1 x +
1x11
1!
2!
10!
11!
11
10

1 x +

(0) = 1

+ … +

(n)

f

(0) n
x

n!

For x > 0, e
x
e

x

> p 11(x)

(x) > p (x) – p
10
11
1/11
Take x = 2(11!)
, then

e

– p

x

– p

(x) >
10

and hence
1x

11!

(x) =
10

1 (2(11!)
11!

1/11 11
)

= 2

11

11

= 2048.

So, there exist values of x for which |p

x

x

10(x) – e | = |e

1

46.f(x) =

= x

f(1) = 1
f’(x) = -x

– p 10 (x)| ≥ 100.

-1

-2

; f’(1) = -1 = -1! or

f (1)

= -1

1!
f”(x) = (-1)(-2)x
M

(k)

f

-3

k

f (1)

; f”(x) = 2 = 2! or
f

(1) = (-1) k!

or

(k)

(1)

k!

= (-1)

Therefore,
p 12(x) = 1 – (x – 1) + (x – 1)
and
|p12(x) – f(x)| =

2

-

1  (x  1)  (x  1)

=

2!

k

- (x – 1)

2

= 1

11

 L  (x  1)

1
2
x |x – x(x – 1) + x(x – 1) -

+ (x – 1)

12

…

1
x

12

,

12

+ x(x – 1)

– 1|

1
If we take x = 0.001, then x = 1000 and every term involving x on
the right-hand side of the above equation is positive
and smaller than x.
Thus,
|p12(x) – f(x)| ≥ 1000(1 – 13x) = 1000(1 – 0.013) =
987. So there exist values of x ≠ 0 for
which |p12(x) – f(x)| ≥ 100.

EXERCISE 2-1

115


47.ln 1.1
Let f(x) = ln(1 + x)
1
-1
=
(1
+
x)
f’(x) = 1  x

f”(x) = -(1 + x)

(3)

f

(x) = 2(1 + x)

(n)

f

Rn(x) =

-3

(x) = (n – 1)!(-1)

f(n 1)(t)xn 1

Note that f

(n  1)!
(n+1)

n-1

-n

(1 + x)

for some t between 0 and x.
n
-(n+1)

(t) = n!(-1) (1 + t)

(n+1)

n

-(n+1)

|f
(t)| = |n!(-1) (1 + t)
0. Therefore,
|R (x)| = f(n 1)(t)xn 1
(n  1)!
n
and

n! x n 1
< (n  1)!

and hence

| = n!(1 + t)

-(n+1)

< n! for t >

x n 1
=n  1

n 1

R (0.1) < (0.1)
n
(n  1)
5
(0.1)
For n = 4, |R (0.1)| <
5
polynomial with the lowest degree is p
4

which has degree 4.
ln(1.1) ≈ p (0.1) = 0.1 - 1 (0.1)
2
4
≈ 0.095308

(x) = x - 1 x
2
4

+ 1 (0.1)
3

2

3

+1x
3

= 0.000 002 < 0.000 005, and hence the

- 1

x4
4

- 1 (0.1)
4
CHAPTER 2 REVIEW

209