Calculus: Analytic Geometry and Calculus, with Vectors

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(^312) Functions, graphs, and numbers
where h(x) = xf'(x) - f(x). We will know that g is decreasing over the interval
0 < x < a if we can show that h(x) < 0 when 0 < x < a. Since
h'(x) = x f "(x) + f'(x) fi(x) = x
our hypothesis that f"(x) < 0 when 0 < x < a implies that h'(x) < 0 when
0 < x < a. Thus h(x) is decreasing over the interval 0 < x < a. Since h is
continuous and h(0)
-f(0) 5 0, it follows that h(x) < 0 when 0 < x < a.
This gives the conclusion of the theorem.
19 Ideas of this section and the preceding one can be used to obtain informa-
tion about the graphs of the Bernoulli functions Bo(x), Bj(x), B2(x), that
appeared in Section 4.3, Problem 10. We recall that Bo(x) = 1, that
(1) B,',(x) =
(2) fat dx = 0
when n = 1, 2, 3, except that (1) fails to hold when n is 1 or 2 and x is
an integer, and that all of the functions except Bi(x) are continuous. To keep
our task within reasonable bounds, we suppose we know the fundamental fact
that B.(0) = 0, B (--) = 0, and B (1) = 0 when n is odd, that is, when n = 1,
3, 5, 7,. We want to show, without making tedious calculations, that the
miniature graphs of Bj(x), B2(x), , Bs(x) over the interval 0 5 x 5 1
appearing in Figure 5.394 give correct information about the trends and the
B1(x)
Bs(x)
Figure 5.394
B2(x)
v
B3(x) B4(x)
B6(x) B7(x) Bs(x)
zeros of these functions. When (12.384) and related formulas have been studied,
we will be able to see that scales on the vertical axes have been adjusted to make
the graphs visible; it can be shown that 4/(27r)n when n > 1 and hence
that numerical values cannot be estimated from the graphs in Figure 5.394.
Supposing that 0 < x < 1, show that the formulas B'(x) = 1 and foI Bi(x) dx =
0 imply that Bi(x) = x - 'ff and hence that the graph of Bi(x) is correct. Show
that the formula B'2(x) = Bi(x) implies that B2(x) is decreasing over the interval
0 < x < I and is increasing over the interval I < x < 1. Show that this fact
and the formula
o
i
B2(x) dx = 0 imply that B2(0) > 0 and B2(Y) < 0 and hence
that B2(x) has exactly two zeros between 0 and 1. Show that the formula
B'3 (x) = Bi(x) implies that the graph of B3(x) is bending downwardover the
interval 0 < x < - and is bending upward over the interval " < x < 1. Show
that the formula B3(x) = B2(x) implies that Ba(x) is increasing over the first
part of the interval 0 < x < 1, is decreasing over a central part, and is increasing
over the remaining part. Tell why B4(x) is increasing over the interval 0 < x <

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