316 Functions, graphs, and numbers
If f(a) = M or f(b) = M, we set x2 = a or x2 = b. Otherwise, assuming
that f(a) < M and f(b) < M, we determine the required number x2
by means of a Dedekind partition. Put a number x' in A if x' S a and
also if a < x' _< b and for some e > 0 the interval a < x 5 x' contains
no point for which f(x) > M - e. Let B contain all other numbers, and
observe that b is in B. Let x2 be the partition number of this partition.
Clearly, f(x2) < M. If we assume that f(x2) < M, say f(x2) = M - eo,
where eo > 0, we can choose a positive number S such that a < xo - S <
x2 + S < b and f(x) < M - eo/2 whenever xo - S < x < x2 + S. The
fact that x2 - S is in .4 will then enable us to draw the erroneous con-
clusion that x2 + S is in A. Therefore, f(x2) = M. This completes the
proof of the part of Theorem 5.47 involving M. To prove the part of the
theorem involving m, we can use an analogous argument. We can,
alternatively and more easily, use the
fact that -f(x) must have a maximum
- m attained when x = x1 and hence that
f(x) must have a minimum m attained
when x = xi.
The following theorem is known as
the intermediate-value theorem; Figure
Figure 5.481 5.481 provides experimental evidence.
Theorem 5.48 If f is continuous over an interval a < x < b and if k
is a constant for which f(a) < k < f(b) or f(a) > k > f(b), then there exists
at least one number s; for which a < t < b and f(i;) = k.
Taking first the case in which f(a) < k < f(b), we prove the result with
the aid of a Dedekind partition. Let x1 be put in A if xl 5 a and also if
a S xl < b and f(x) < k whenever a <_ x < xi. Let x2 be put in B if
x2? b and also if a < X2 < b and the interval a 5 x < x2 contains a
number x for which f(x) >--_ k. Let be the partition number. Since
f(a) < k and f has right-hand continuity at a, we can let e = k - f(a)
and choose a positive number S such that If(x) - f(a) I < e and hencef (x) < k when a< x S a -{- S. Hence a+ S belongs to d and l
a + S > a. A similar argument shows that E < b. Therefore a < i <
b. If we suppose that f(s) > k, then we can choose a positive number S
such that a<i;-6<i;+8<band f(x)>kwhen g-SSx<i+S. This contradicts the fact that f(x) < k when a < x < i - S and
f(x) > k for some x in the interval a < x < E + S. If we suppose that
f() k, a similar argument leads to a contradiction. Therefore
f Q) = k. A very similar proof covers the case in which f (a) > k > f (b)
and Theorem 5.48 is proved.
In ordinary circumstances we try to be too efficient to clutter our books
and our memories with obvious corollaries and applications of our
theorems, but one corollary of the intermediate-value theorem is so
important that we relax to look at it. If a function f is negative at x1