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364 Answers to Exercises and Solutions to Problems

as required.

5.4. Let c, m, n be fixed, and define g(x) = (x+c)“. Then f(m) = A”g(O),
which yields the result immediately.

5.5. We have that

g(x) = d-1)x(x - 1)

2 + dW - x2> +

dl)x(x + 1)

2

so that

9’64 = M-1) - 29(O) + !7Wlx + w2)M1) - d-l>1

= g(l)(x + l/2) +9(-1)(x - l/2) - 29(0)x.

Suppose that l/2 5 ]u] 5 1. Then u + l/2 and u - l/2 have the same
sign so that ]u + l/2] + ]u - l/2] = 2]u]. Thus

I!ml I I!dl>l b + WI + ISWI Iu - WI + %@>l I4

5 Iu + l/2] + ]u - l/2] + 2]U] = 4]U].

Equality occurs if g(x) = 2x2 - 1, so that K,, = 41~1.
Suppose 0 5 ]u] 5 l/2. S ince the bound for u is the same as that for
-u, with no loss of generality, we can take 0 5 u 5 l/2. There are several
types of quadratic functions to consider:

(1) g(x) is increasing on the interval [-l,l];

(2) g(x) is decreasing on the interval [-l,l];

(3) g(x) has a minimum at a point c in [-l,O];

(4) g(x) has a minimum at a point c in [0, 11;

(5) g(x) has a maximum at a point c in [-l,l].

Cases (2) and (5) need not be treated directly since -g(x) falls under one
of the other cases.
If g(x) satisfies (l), then

htx) = (^2) (
g(x) - d-1)
g(1) - g(-1) >
]
is a quadratic such that Ih( 5 1 on [-1, l] and jg’(u)l 5 IV(u)] for all U.
(The graph of h( x ) is obtained from that of g(x) by expanding the vertical
scale.) Since h(1) = 1 and h(-1) = -1, the graphs of two such functions
h(x) cross only when x = fl. Hence h(0) is as small as possible when
h/(-l) = 0 (i.e. when h(x) = -1 + (x + 1)2/2) and as large as possible
when h’(1) = 0 ( i.e. when h(x) = 1 -(x - 1)2/2). Thus -l/2 < h(0) 5 l/2.
Since
h’(u) = (u + l/2) - (U - l/2) - 2h(O)u = 1 - 2h(O)u,