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 asequence of intervals {I,}, I, = [a, bn}, as follows. Let a, = aand b, = b, and set c, = (a, + b1)/2. If f(c,) = k, we are done; iff(c,) < k, set a2 = C1 and b2 = b1, and if f(c,) > k, set az = a, andb2= cq. Proceeding in this manner, we either find a point c, forwhich f(c,) = k, or else construct a nested sequence {In} ofintervals. Show that, in the latter case,(a)(b) (c) = k, where c = =1n%Dconsists of a single point, and%D

Suppose that f is continuous on a,b and that fa<k<fb Let us construct a sequence of intervals…

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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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

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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