Calculus I
Fall 2005
MW 8:5510:45, Silver 710
Section 3.9
6. Approximation using the derivative of
f
(
x
) :=
x
2
/
3
at the point
x
= 27:
(26)
2
3
≈
(27)
2
3

2
3(27)
1
3
=
79
9
= 8
.
7777
...
On the other hand, the calculator gives (26)
2
/
3
= 8
.
7763
...
.
12. Approximation using the derivative:
sin
43
π
180
≈
sin
45
π
180

2
π
180
cos
45
π
180
=
√
2
2

√
2
π
180
= 0
.
682424
...
The calculator gives sin(43
π/
180) = 0
.
681998
...
.
26.
(a) Approximating using the derivative we obtain
(
x
+
h
)
n

x
n
≈
nhx
n

1
.
Hence, if the error of the left hand side is required to be less than 1% of
x
n
we get the
approximate inequality
n

hx
n

1


x
n

100
.
From this we conclude that the error
h
when estimating
x
has to be (approximately) at
most

x

100
n
, ie. approximately within
1
n
% of the correct value
x
.
(b) Proceeding in the same way as above we obtain for the approximation of
x
1
/n
that the
error in estimating
x
should be (approximately) less than
n
% of the correct value
x
.
30. The NewtonRaphson method reads as
x
n
+1
=
x
n

x
5
n

30
5
x
4
n
.
Starting with
x
1
= 2 we therefore obtain:
x
2
=
79
40
= 1
.
975,
x
3
= 1
.
9743509
...
,
x
4
≈
1
.
97435
.
32. The NewtonRaphson method reads as
x
n
+1
=
x
n

sin
x
n

x
2
n
cos
x
n

2
x
n
.
Thus, starting with
x
1
= 1 with obtain
x
2
= 0
.
891395995
...
,
x
3
= 0
.
876984844
...
,
x
4
≈
0
.
876726.
34.
(a) Since
f
is continuous everywhere and since
f
(1) =

2 and
f
(2) = 3, by the intermediate
value theorem,
f
attains all values between

2 and 3 on the interval [1
,
2]. In particular,
it attains 0.
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(b) The NewtonRaphson method gives
x
n
+1
=
x
n

2
x
3
n

3
x
2
n

1
6
x
2
n

6
x
n
.
Since
f
(1) = 0 we cannot start the NewtonRaphson method with
x
1
= 1.
(c) However, starting with
x
1
= 2 we obtain
x
2
= 1
.
75,
x
3
= 1
.
682539
...
,
x
4
≈
1
.
67768.
Evaluating
f
(
x
4
) = 0
.
000199874 we see that
x
4
is a fairly good approximation for the
root of
f
.
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