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Unformatted text preview: MAA 4200: Exam 2 Solutions Summer 2009 [1] (34 points) [3.] Give the exact statement of the Extreme Value Theorem. If f : E —> R is continuous and E is compact, then there exist 331,132 6 [(1, (3]
such that ﬁrm) 5 f(m) S f(m2) for all as 6 [a,b]. [b] Give a speciﬁc example of a continuous function f : (0,1) —> R whose range is
unbounded. Why does this not violate the Extreme Value Theorem? f(w) = 1/11; does not violate EVT since (0,1) is not compact (not closed). [0] Give a speciﬁc example of a continuous function g : R —> R Whose range is unbounded.
Why does this not violate the Extreme Value Theorem? 9(33) = a: does not violate EVT since IR is not compact (not bounded). [d] Give a speciﬁc example of a function g : [0, 1] —> R whose range is unbounded. Why
does this not violate the Extreme Value Theorem? g(:c) = 1/11: for $750 and 9(0) =0 does not violate EVT since 9 is not continuous. [e] Give a speciﬁc example of a function f : R —> R which is continuous at exactly one
point in R, i.e. f is continuous at 3:0 but not continuous at any other point. You do not
need to prove your answer works. Hm) =33 if a: is rational and f(:1:)=0 if a: is irrational. [f] Explain the relationship between the continuity of f : D —> R at 3:0 and limxnxo f (:L'),
i.e. compare and contrast the two concepts. (1) To be continuous 930 E D, but 230 need not be in D for limmnmo f(:1:) to exist.
(2) For limanD f(a:) to exist 320 must be an accumulation point of D, but if :30 E
D is not an accumulation point, f is continuous there. (3) Finally, at an accumulation point $0 6 D, the function f is continuous
at 330 if and only if limmnxo f(m) = f(:c0). [2] (33 points) Let f : (0,00) —> R be deﬁned by [a] Show that f satisﬁes a Lipschitz condition at 320 = 1. Give the value of the Lipschitz
constant C and the neighborhood Q on which it is valid. _WH i_1_1—\/§ 1+\/§ <a3—1
V5 _ $5 1+VE m+¢E— m For Q: (1/2,oo) if a; E Q, than 1/3: >2 so that 0:2. [b] Sketch the graph of f and the cone which the Lipschitz condition represents. [0] Give an 66 proof that f is continuous at 330 = 1. Let E > 0.
Choose 6 = min{1/2,e/2}.
Then for m—1< 6 we have 32> 1/2 so that 1
m——dgzm—u<2. ﬂ 5
2 =6. [3] (33 points) [a] Suplaose f : R —> R is continuous and f (7') = 7’2 for each rational number 7‘. Determine
f (ﬂ) and justify (Le. prove) your answer. There exists a sequence of rationals Tn —) ﬂ. Then since f is continuous everywhere
f(\/§) = limnnoo f(7"n) = limnqm r: = (J22 = 2 by the Sequential Limit Theorem
(or Theorem 3.1). [b] Prove the following generalization of [a]: If f,g : R —> R are continuous and ﬂ?) = g0“) for all 7‘ E Q, then f(3:) = 9(m) for all
a: E R. Let cc 6 R. There exists a sequence of rationals r” —> x. Then since f,g are
continuous everywhere f (3:) = lirn,H00 ﬁr”) =1imnnoog(7‘n) = 9(513) by the Sequential
Limit Theorem (or Theorem 3.1). [4] (33 points) [a] Use the deﬁnition of continuity to prove that:
If f : D —> R and g : E —> IR with range( f) C E where f is continuous at 5120 and g is
continuous at f (11:0), then 9 o f is continuous at 330. Let E > 0.
Set 190 = “330) Since g is continuous at yo there exists ,o> 0 (I used ,0 in place of 6 which
will occur later) such that g(y) —g(y0) < 6 whenever y E E with y—y0[ < ,0. Since f is continuous at 11:0, there exists 6 > 0 such that f(a:) — f(330) < p
whenever :1: E D with cc — 330] < 6. Putting these two statements together: when :1: E D with x — m0 < 6 we have
that ﬁx) E E with f(m) — f(m0) < p which implies g(f(a:)) — g(f(a:0)) < e. [b] Show that if f satisﬁes a Lipschitz condition at 2:0 and 9 satisﬁes a Lipschitz condition
at f($0) then 9 o f satisﬁes a Lipschitz condition at 11:0. Let yo = f(930)
Suppose g(y) — g(yg)l S Ogly  yo for all y 6 Q9, a neighborhood of yo.
Suppose f(33) — f(:t0) 3 CH3: — $0 for all a: E Qf, a neighborhood of 320. By the continuity of f, there exists a neighborhood Q C Qf of 3:0 such that
f(a:) 6 Q3 for all :6 E Q. Then for a: E Q, 9(f($)) — 9(f(~'vo)) S Cglﬂﬂ?) — f($o) S 09 ' Cflm — fEol so that 0:0901: is a Lipschitz constant for gof on Q. ...
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 Spring '07
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