337
Answers
501–600
Answers and Explanations
517.
1.69562
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). In general, a bit
of trial and error is involved, but in this case, you’re given a closed interval to work
with, which considerably narrows down the choices!
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = x
3
– x
2
– 2 and f
'(x) = 3x
2
– 2x. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. You can choose any
value in the interval [1, 2] for x 1 , but x 1 = 1 makes the computations a bit easier for the
first step, so use that value:
x
x
2
3
2
2
3
3
2
2
1
1
2
3 1
2 1
3
3
3
3
2
3 3
2 3
47
2
= −
− (1) −
−
=
= −
−
−
−
=
( )
( )
( )
( ) ( )
( )
( )
1 1
2 23810
2 23810
2 23810
2 23810
2
3 2 23810
2
4
3
2
2
≈ .
.
( .
)
.
( .
)
(
x =
−
− (
) −
− 2 2 23810
1 83987
1 83987
1 83987
1 83987
2
3 1 839
5
3
2
.
)
.
.
( .
)
.
( .
≈
=
−
− (
) −
x
8 87
2 1 83987
1 70968
1 70968
1 70968
1 70968
2
2
6
3
2
)
( .
)
.
.
( .
)
.
−
≈
=
−
− (
) −
x
3 3 1 70968
2 1 70968
1 69577
1 69577
1 69577
1 69
2
7
3
( .
)
( .
)
.
.
( .
)
.
−
≈
=
−
− (
x
5 577
2
3 1 69577
2 1 69577
1 69562
1 69562
1 69562
2
2
8
) −
−
≈
=
−
( .
)
( .
)
.
.
( .
)
x
3 3
2
2
1 69562
2
3 1 69562
2 1 69562
1 69562
− (
) −
−
≈
.
( .
)
( .
)
.
The approximation is no longer changing, so the solution is 1.69562.
518.
1.22074
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). There’s certainly
a bit of trial and error involved; in this case, you could graph both y
x
=
+1 and y = x
2
and look for a point of intersection to get a rough idea of what the root may be.
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f x
x
x
( ) = + −
1
2 and ′ ( )
f x
x
x
=
+
−
1
2
1
2 . Note
that the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. Notice that
f ( )
1
1 1 1
2 1
2
= + − =
− , which is close to the desired value of 0, so you can start
with x 1 = 1:
Answers
501–600
Answers and Explanations
517.
1.69562
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). In general, a bit
of trial and error is involved, but in this case, you’re given a closed interval to work
with, which considerably narrows down the choices!
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = x
3
– x
2
– 2 and f
'(x) = 3x
2
– 2x. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. You can choose any
value in the interval [1, 2] for x 1 , but x 1 = 1 makes the computations a bit easier for the
first step, so use that value:
x
x
2
3
2
2
3
3
2
2
1
1
2
3 1
2 1
3
3
3
3
2
3 3
2 3
47
2
= −
− (1) −
−
=
= −
−
−
−
=
( )
( )
( )
( ) ( )
( )
( )
1 1
2 23810
2 23810
2 23810
2 23810
2
3 2 23810
2
4
3
2
2
≈ .
.
( .
)
.
( .
)
(
x =
−
− (
) −
− 2 2 23810
1 83987
1 83987
1 83987
1 83987
2
3 1 839
5
3
2
.
)
.
.
( .
)
.
( .
≈
=
−
− (
) −
x
8 87
2 1 83987
1 70968
1 70968
1 70968
1 70968
2
2
6
3
2
)
( .
)
.
.
( .
)
.
−
≈
=
−
− (
) −
x
3 3 1 70968
2 1 70968
1 69577
1 69577
1 69577
1 69
2
7
3
( .
)
( .
)
.
.
( .
)
.
−
≈
=
−
− (
x
5 577
2
3 1 69577
2 1 69577
1 69562
1 69562
1 69562
2
2
8
) −
−
≈
=
−
( .
)
( .
)
.
.
( .
)
x
3 3
2
2
1 69562
2
3 1 69562
2 1 69562
1 69562
− (
) −
−
≈
.
( .
)
( .
)
.
The approximation is no longer changing, so the solution is 1.69562.
518.
1.22074
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). There’s certainly
a bit of trial and error involved; in this case, you could graph both y
x
=
+1 and y = x
2
and look for a point of intersection to get a rough idea of what the root may be.
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f x
x
x
( ) = + −
1
2 and ′ ( )
f x
x
x
=
+
−
1
2
1
2 . Note
that the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. Notice that
f ( )
1
1 1 1
2 1
2
= + − =
− , which is close to the desired value of 0, so you can start
with x 1 = 1: