Unformatted text preview: _________’_’_’——————7 148 CHAPTERZ PHOBLEMSPLUS
(c)ﬂx+y]]2=1 => ﬂw+y]]::l:1 (d)FornSa:<n+1.[[:c]]:n.Thenﬂa:]]+ﬂy]]:1 =>
:' 1£m+y<20r [[yﬂ:1—n => 1—n§y<2mn.Choosinginteger
—1Sx+y<0 values for n produces the graph. 7. f is continuous on (—00. a) and (a, 00). To make f continuous on R. we must have continuity at (1. Thus, limf(a:): 1imf(m) => lim 3:22 lim(:c+1) => a2=a+1 ::> a2—a’1:0 => $Aa+ mad :v——>a+ zra‘ [by the quadratic formula] (1 = (1 i \/S )/2 m 1.618 or —0.618. 8. (a) Here are a few possibilities: (b) The “obstacle” is the line m : 3; (see diagram). Any intersection of the graph of f with the line y : :3
constitutes a ﬁxed point, and if the graph of the function does not cross the line somewhere in (0, 1), then it must
either start at (O. 0) (in which case 0 is a ﬁxed point) or ﬁnish at (1, 1) (in which case 1 is a ﬁxed point). (c) Consider the function F (x) : f (ac) — as. where f is any continuous function with domain [0, 1] and range in
[0. 1]. We shall prove that f has a ﬁxed point.
Now if f (0) : 0 then we are done: f has a ﬁxed point (the number 0), which is what we are trying to prove.
So assume f(0) 3£ 0. For the same reason we can assume that f(1) 7E 1. Then F(0) = f(0) > O and
F (1) = f (1) 7 1 < 0. So by the Intermediate Value Theoremt there exists some number 0 in the interval (07 1)
such that F (c) = f(c)  c : 0. So f(c) : c. and therefore f has a ﬁxed point. 9. 53; me) = gym (a W) + g($)l + a W) — 9mm
: % 53 W) + gm] + % 3113; um — mu 1 _3
_22+§~1—§.and hm gm) 2 hm (We) + 9th  f(:v)) : hm W) + gm] — hm fa) : 2 — g : 24m z—va (z—>a, a;—)a So lim [f(a:)g(m)] : [11m foo] [hm g(:1:)] : g . g : (E—NI (C—‘Va [E—Hl NIH H3190 ...
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 Calculus

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