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t2_ans_scan - MAA 4200 Exam 2 Solutions Summer 2009[1(34...

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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. _W-H 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 6-6 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 0:09-01: is a Lipschitz constant for gof on Q. ...
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