7. The IVT can be proved using the Nested Interval Theorem. Suppose that f is continuous on [a, b], and that f(a) < k < f(b). To determine a point c e (a, b) for which f(c) = k, construct a sequence of intervals {I,}, I, = [a, bn}, as follows. Let a, = a and b, = b, and set c, = (a, + b,)/2. If f(c,) = k, we are done; if f(c,) < k, set a2 = c, and b2 = b1, and if f(c,) > k, set a2 = a, and %3D %3D %3D %3D %3D %3D %3D %3D b2 = c1. Proceeding in this manner, we either find a point c, for %3D which f(c,) = k, or else construct a nested sequence {In} of intervals. Show that, in the latter case, (a) =1n consists of a single point, and (b) f(c) = k, where c = N-.
7. The IVT can be proved using the Nested Interval Theorem. Suppose that f is continuous on [a, b], and that f(a) < k < f(b). To determine a point c e (a, b) for which f(c) = k, construct a sequence of intervals {I,}, I, = [a, bn}, as follows. Let a, = a and b, = b, and set c, = (a, + b,)/2. If f(c,) = k, we are done; if f(c,) < k, set a2 = c, and b2 = b1, and if f(c,) > k, set a2 = a, and %3D %3D %3D %3D %3D %3D %3D %3D b2 = c1. Proceeding in this manner, we either find a point c, for %3D which f(c,) = k, or else construct a nested sequence {In} of intervals. Show that, in the latter case, (a) =1n consists of a single point, and (b) f(c) = k, where c = N-.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![27. The IVT can be proved using the Nested Interval Theorem.
Suppose that f is continuous on [a, b], and that f(a) < k < f(b).
To determine a point c e (a, b) for which f(c) = k, construct a
sequence of intervals {I,}, I, = [a, bn}, as follows. Let a, = a
and b, = b, and set c, = (a, + b1)/2. If f(c,) = k, we are done; if
f(c,) < k, set a2 = C1 and b2 = b1, and if f(c,) > k, set az = a, and
b2
= cq. Proceeding in this manner, we either find a point c, for
which f(c,) = k, or else construct a nested sequence {In} of
intervals. Show that, in the latter case,
(a)
(b) (c) = k, where c = =1n
%D
consists of a single point, and
%D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2ab050a-d13e-4a73-9fee-dc44d5d071e8%2Fb35a5bc1-bf9d-44d5-aac3-e463547128e6%2F4obxxco_processed.png&w=3840&q=75)
Transcribed Image Text:27. The IVT can be proved using the Nested Interval Theorem.
Suppose that f is continuous on [a, b], and that f(a) < k < f(b).
To determine a point c e (a, b) for which f(c) = k, construct a
sequence of intervals {I,}, I, = [a, bn}, as follows. Let a, = a
and b, = b, and set c, = (a, + b1)/2. If f(c,) = k, we are done; if
f(c,) < k, set a2 = C1 and b2 = b1, and if f(c,) > k, set az = a, and
b2
= cq. Proceeding in this manner, we either find a point c, for
which f(c,) = k, or else construct a nested sequence {In} of
intervals. Show that, in the latter case,
(a)
(b) (c) = k, where c = =1n
%D
consists of a single point, and
%D
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