Figure 8
Figure ‘7
3-
y
y
= x
.3
=
-
2
.t
S
1
x
‘
x
Fixed-Point Algorithm
—
—
=
g(x,)
=
(2Vx,, + 1)1/2
=
(x + 1)’/
=
2 cos x using the Fixed-Point Algorithm.
I
SOLUTION Note first that solving this equation is equivalent to solving the pair
2 cos x. Thus, to get our initial value, we graph these
x and y
of equations y
two equations (Figure 8) and observe that the two curves cross at approximately
x
1. Taking x
11 = 2 cos x,, we obtain the
1 = 1 and applying the algorithm x,
results in the following table.
Solve x
4
5
6
• EXAMPLE 5
1.8350868
1.8350867
1.8350867
1.8350867
9
10
11
12
1.8392994
1.8357680
1,8351969
1.8351045
3
1.8350871
8
1.8612097
2
1.8350896
2.0
7
n
1
n
Figure 7 suggests that the point of intersection of the curves y
x and
1/4
‘+
occurs
between
1
and
2,
probably
(x
closer
to
2,
so
we
i)
take x
1 = 2
y
‘V
as our starting point. This leads to the following table. The solution is approxi
mately 1.8350867.
1
x÷
2 = 2Vx -1- 1, which leads to x
± (2VJ)h/2.
SOLUTION We write x
Since we know the solution is positive, we take the positive square root and write
the iteration as
• EXAMPLE 4 Use the Fixed-Point Algorithm to approximate the solution of
2ViJ = 0.
f(x)
1.
—
1 be an initial approximation to the
Let g(x) be a continuous function, and let x
root r of x
g(x). Let E denote a bound for the error
x,I.
1, 2,... until Ix,
Repeat the following step for n
i x,j < E:
1
Al orithm
—
The Fixed-Point Algorithm The Fixed-Point Algorithm is simple and
straightforward, but it often works.
Suppose that an equation can be written in the form x
g(x). To solve this
equation is to find a number r that is unchanged by the function g. We call such a
number a fixed point of g. To find this number, we propose the following algo
. Then let x
1
2 = g(x
rithin. Make a first guess x
), x
1
3 = g(x
), and so on. If we are
2
Do.
lucky, x, will converge to the root r as n
1 is
wrong (see also Problem 22). For the function in Figure 6, the difficulty is that x
not close enough to r to get a convergent process started. Other difficulties arise if
f’(x) is zero or undefined at or near r. When Newton’s Method fails to produce
approximations that converge to the solution, then you can retry Newton’s
Method with a different starting point, or use a different method such as the
Bisection Method.
n— OC
Newton’s Method creates a sequence of successive approximations to the root.
(We mentioned sequences briefly in Section 1.5.) It is often the case that Newton’s
Method produces a sequence {x} that converges to the root of f(x)
0, that is,
urn .x, = r. This is not always the case, however. Figure 6 illustrates what can go
Chapter 3 Applications of the Derivative
Figure 6
194
r
—
3. 2 cos x sin x = 0; [1,2]
4. x—2+2cosx=0;[1,2]
8
9
10
1.4394614
0.2619155
1.9317916
—0.7064109
1.5213931
=
2
x,, + 2 cos x,,
7
8
9
10
11
12
1
1.0403023
1.0261107
1.0312046
1.0293881
1.0300374
1
2
3
4
5
6
1.0298054
1.0298883
1.0298588
1.0298693
1.0298655
1.0298668
x
13
14
15
16
n
1.0298665
1.0298666
1.0298665
1.0298666
X,,
=
x is called a
of g.
—
+
5
+
—
0
+
2 cos x
=
+ 9x
2
6x
6. The real root of 7x
3 + x
7. The largest root of x 2
5. Thelargestrootofx
3
=
0 (see Problem 4)
1=0
1 In Problems 5—14, use Nesvtonr Method to approximate the
indicated root of the given equation accurate to five decimal
places. Begin by sketching a graph.
4. A point x satisfying g(x)
...,,.__;
3. The Bisection Method, Newton’s Method, and the FixedPoint Algorithm are examples of
that is, they provide a fi
nite sequence of steps that, if followed, will produce a root of an
equation to desired accuracy.
Now we raise an obvious question. Why did the second algorithm yield a con
vergent sequence, whereas the first one failed to do so? Whether or not the FixedPoint Algorithm works depends on two factors. One is the formulation of the
equation x
g(x). Example 5 demonstrates that an equation such as x
2 cos x
can be rewritten in a form that yields a different sequence of approximations. In
Example 5 the reformulation was x = (x + 2 cos x)/2. In general, there may be
many ways to write the equation and the trick is to find one that works. Another
factor that affects whether the Fixed-Point Algorithm converges is the closeness of
the starting point x
1 to the root r. As we suggested for Newton’s Method, if the
Fixed-Point Algorithm fails with one starting point, you can try a different one.
n
xn
ii
In Problems 1—4, use the Bisection Method to approximate the
real root of the givelz equation on the given interval. Each answer
should be accurate to two decimal places.
1. x
3 +2x—6=0;[1,2]
2. 3
-1-5x 1 = 0;[—1,0]
4
x
+
Problem Set 3.7
.
6
7
x
This process produces a convergent sequence, shown in the following table. (The
oscillation in the last digit is probably due to round-off errors.)
1. The virtues of the Bisection Method are its simplicity and
reliability; its vice is its
2. If f is continuous on [a, bi, and f(a) and f(b) have oppoSite signs, then there is a
of f(x) = 0 between a and b. This
follows from the
Theorem.
Concepts Review
1
1.0806046
0.9415902
1.1770062
0.7673820
n
195
Quite clearly the process is unstable, even though our initial guess is very close to
the actual root.
Let’s take a different tack. Rewrite the equation x = 2 cos x as x =
(x -i- 2 cos x)/2 and use the algorithm
5
1
2
3
4
n
Section 3.7 Solving Equations Numerically
-I-
—
1
—
24x
=
0
=
2 ± 24.v
2.v
sinx
+
—
14. The smallest positive root of 2 cot x
—
3
6x
2
3 H- 22x
8x
=
2
13. The positive rootof 2x
4
12. Alireal roots ofx
11. All real roots of
4
—
8
-f 8
=
sin x
0
0
=
0 (see
to five decimal
=
=
=
f(s)
f(s)
18.
19.
20. f(x)
H-
+
3
x
+-
2
x
2 sin; [0,4ir]
x
-;[1T,31rj
4
x
•
[—1,1]
1
(l+i)k
H- 1)24
—
(1 + i)
24 -1- 1
=
0
L
500i
—
4
23
E20i H- 19i —1 -4- (1 ± i,j—
(c) Find i accurate to five decimal places starting with
0.012, and then give the annual rate r as a percent
(r
1200i).
1,,
24. In applying Newton’s Method to solve f(x) = 0, one can
,x
1
,x
2
,
3
usually tell by simply looking at the numbers x
whether the sequence is converging. But even if it converges, say
to T, can we be sure that ? is a solution? Show that the answer is
0.
yesprovidedfandf’ are continuousatT andf’()
E1
=
(b) Show that Newton’s Method for this equation reduces to
20i(1
where i is the interest rate per month. Tom bought a used car for
$2000 and agreed to pay for it with $100 payments at the end of
each of the next 24 months.
(a) Show that i satisfies the equation
i
23. In installment buying. one would like to figure out the
real interest rate (effective rate), but unfortunately this involves
solving a complicated equation. If one buys an item worth $P
today and agrees to pay for it with payments of $R at the end of
each month for k months, then
173 Obviously, its only x
x
22. Sketch the graph of y = .
intercept is zero. Convince yourself that Newton’s Method fails
to converge. Explain this failure,
21. Kepler’s equation x = in -I- E sin x is important in as
tronomy. Use the Fixed-Point Algorithm to solve this equation
forx when in = 0.8 and E = 0.2.
L1
=
17. f(x)
In Problems 17—2 approximate the values of x that give
maximnuni and niininiumn values of the function on the indicated
intervals.
E
i. Use Newton’s Method to calculate
places.
F1
15. Use Newton’s Method to calculate “/ to five decimal
places. Flint: Solve x
3 6
0.
E1
2x
10. Therootof2x —sin x
9. The root of cos x
—
Chapter 3 Applications of the Derivative
S. ‘The smallest positive root of 2 cos x
Problem 3)
196
=
=
1
\/.2 + x; x
\/2.7 + x; x
1
=
=
47
1
—
=
5(x
—
+ 1 +
33. Consider x
=
+
1 +
+
+
=
1 +
1
S
—.
1
x
1
x
=
=
=
1 to
0 to
0 to
)
2
x
—
1+
1+
1
1
1+•
1
=
1
2x,
—
ax
using several different values of a.
(a) Make a conjecture about what this algorithm computes.
(b) Prove your conjecture.
x+
35. Experiment with the algorithm
34. Consider the equation x = x — f(x)/f’(x) and sup
pose that f’(x)
0 in an interval [a, b].
(a) Show that if r is in [a, b] then r is a root of the equation
x = x
f(x)/f’(x) if and only if f(r) = 0.
(b) Show that Newton’s Method is a special case of the FixedPoint Algorithm, in which g’(r) = 0.
1+
(c) Evaluate the following expression. (An expression like this
is called a continued fraction.)
(b) Algebraically solve for x in x
(a) Apply the Fixed-Point Algorithm starting with
, x
3
, and x
4
.
5
find x
,x
2
E
(c) Evaluate
32. Consider x =
(a) Apply the Fixed-Point Algorithm starting with
find 2, x
, x
3
, and x
4
.
5
(b) Algebraically solve for x in x =
1
(c) Evaluate 1
(a) Apply the Fixed-Point Algorithm starting with x
1
find x
.
5
,x
2
,x
3
, and x
4
(b) Algebraically_solve for x in x =
1 31. Consider x
=
30. Follow the directions of Problem 29 for x
g(x).
E
) = g(x).
2
x
29. Consider the equation x = 2(x
(a) Sketch the graph of y = x and y g(x) using the same co
ordinate system, and thereby approximately locate the posi
tive root of x = g(x).
(b) Try solving the equation by the Fixed-Point Algorithm start
ing with x
1
0.7.
(c) Solve the equation algebraically.
11
28. x
27. x
2
x=2—sinx;x
=
26. 1
25. x=cosx;x
=1
1
In Problems 25—28, use the Fixed-Point Algorithm with x
1 as
indicated to solve the equations to five decimal places.
L1
Figure 1
Figure 9
3.8
Figure 10
F’(x)
=
3
4x
x
Antiderivatives
1050
Answers to Concepts Review: 1. slowness of convergence
2. root: Intermediate Value 3. algorithms 4. fixed point
(a) Find the position of the object when it is closest to the
observer.
(b) Find the position of the object when it is farthest from the
observer.
stands 3 feet from the bottom of the cliff.
—--
39. An object thrown from the edge of a 42-foot cliff follows
2
2x
+ x + 42 (Figure 10). An observer
the path given by y =
197
=
f(x) on I, that is, if
I
Find
an antiderivative of the function f(x)
=
3
4x
on
A moment’s thought will suggest other solutions to Example 1. The function
F(x) = x
4 + 6 also satisfies F’(x) = 4x
; it too is an antiderivative of
3
+
(x)
.
3
4x
where
In
fact,
F(x)
=
4
x
C is any constant, is an antiderivative
C,
f
3 on (—, no) (see Figure 1).
of 4x
SOLUTION We seek a function F satisfying F’(x) = 4x
3 for all real x. From
our experience with differentiation, we know that F(x) = x
4 is one such
function.
(-oo, Do).
• EXAMPLE 1
We said an antiderivative rather than the antiderivative in our definition. You
will soon see why.
We callFan antiderivative off on the interval Ill DF(x)
F’(x)
f(x)forallxinl.
Definition
Most of the mathematical operations that we work with come in inverse pairs:
addition and subtraction, multiplication and division, and exponentiation and root
taking. In each case, the second operation undoes the first, and vice versa. One rea
son for our interest in inverse operations is their usefulness in solving equations.
For example, solving x
3
8 involves taking roots. We have been studying differen
tiation in this chapter and the previous one. If we want to solve equations involving
derivatives we will need its inverse, called antidifferentiation or integration.
E1 After differentiating and setting the result equal to zero, many
practical max—mm problems lead to an equation that cannot be
solved exactly. For the following problems, use a numerical
method to approximate the solution to the problem.
36. A rectangle has two corners on the x-axis and the other
. What are the
2
two on the curve y = cos x, with —n’/2 < x < iT/
dimensions of the rectangle of this type with maximum area?
(See Figure 24 of Section 3.4.)
37. Two hallways meet in a right angle as shown in Figure 6 of
Section 3.4, except the widths of the hallways are 8.6 feet and 6.2
feet. What is the length of the longest thin rod that can be carried
around the corner?
38. An 8-foot-wide hallway makes a turn as shown in Fig
ure 9. What is the length of the longest thin rod that can be car
ried around the corner?
Section 3.8 Antiderivatives