Math 161 (33) - Final exam

Transcription

1 Name: Id #: Mat 161 (33) - Final exam Fall Quarter 2015 Wednesday December 9, :30am to 12:30am Instructions: Prob. Points Score possible TOTAL 75 (BEST 3) Read eac problem carefully. Write legibly. Sow all your work on tese seets. Feel free to use te opposite side. Tis exam as 12 pages, and 4 problems. Please make sure tat all pages are included. You may not use books, notes, calculators, etc. Cite teorems from class or from te texts as appropriate. Proofs sould be presented clearly (in te style used in lectures) and explained using complete Englis sentences. Your final score will be te sum of te best 3 scores for te questions. Good luck!

2 Mat 161 (33) - Final exam Fall Quarter 2015 Page 2 of 12 Question 1. (Total of 25 points) a) (8 points) By using only te axioms (P1)-(P12) and indicating were eac axiom is applied, prove tat if a < b and c > d, ten a c < b d. Solution. Since a < b and c > d, it follows by definition tat b a, c d P. By (P11), it follows tat (b a) + (c d) P. By repeated application of (P1) one deduces (b a) + (c d) = b + ( a + (c d)) = b + (( a + c) d) wilst by (P4) and furter application of (P1), b + (( a + c) d) = b + ( d + ( a + c)) = (b d) + ( a + c). We claim ( a + c) = (a c). To see tis note tat by repeated application of (P1) and (P4) one deduces tat ( a + c) + (a c) = (a + ( a)) + (c + ( c)). Applying first (P3) and ten (P2), and so (a + ( a)) + (c + ( c)) = = 0 ( a + c) + (a c) = 0. Terefore, again applying (P2) and (P3), ( a + c) = ( a + c) + 0 = ( a + c) + ((a c) (a c)) but, by (P1) and te above observations, ( a+c)+((a c) (a c)) = (( a+c)+(a c)) (a c) = 0 (a c) = (a c), as claimed. Hence, (b d) (a c) P so, by definition, b d > a c.

3 Mat 161 (33) - Final exam Fall Quarter 2015 Page 3 of 12 For te remainder of te exam you may use axioms (P1)- (P12) witout comment. b) (5 points) Let a, b R. Use te triangle inequality to prove tat a b a b. Solution. Let a, b R. By te triangle inequality, a = a b + b a b + b and so a b a b Similarly, by te triangle inequality, b = b a + a a b + a. and so Combining tese estimates b a a b. a b a b, as required. c) (4 points) For tis problem you may use te following fact: If n N is not of te form n = k 2 irrational. Prove is irrational. for some k N, ten n is Solution. Suppose is rational. Ten ( 5 + 7) 2 = = is rational. Rearranging, one deduces 35 = ( 5 + 7) is rational. However, since 35 is not a perfect square, 35 is irrational by te above fact, a contradiction. Tus, must be irrational.

4 Mat 161 (33) - Final exam Fall Quarter 2015 Page 4 of 12 d) (8 points) Using (strong) induction, sow tat for every n N tere exists a, b N suc tat 5 n = a 2 + b 2. Solution. For eac n N let P (n) be te proposition tat tere exists a, b N suc tat 5 n = a 2 + b 2. Note tat 5 = and 25 = and so P (1) and P (2) old. Now suppose tat P (1),..., P (n) are true for some n N wit n 2. In particular, since P (n 1) olds tere exist u, v N suc tat 5 n 1 = u 2 + v 2. Note tat 5 n+1 = n 1 = 5 2 (u 2 + v 2 ) = (5u) 2 + (5v) 2. Hence, taking a = 5u and b = 5v it follows tat 5 n+1 = a 2 + b 2 and so P (n + 1) olds. By te principle of strong induction, P (n) olds for all n N, as required.

5 Mat 161 (33) - Final exam Fall Quarter 2015 Page 5 of 12 Question 2. (Total of 25 points) a) Let f : R R be a function and a R. i) (6 points) State te definition of lim x a f(x) =. Solution. For every R > 0 tere exists some δ > 0 suc tat if x R and 0 < x a < δ, ten f(x) > R. ii) (6 points) Suppose lim f(x) = and g : R R satisfies lim g(x) = l for x a x a some l > 0. Sow tat lim f.g(x) =. x a Solution. Let R > 0 be given. By definition one may coose δ 1 > 0 suc tat if x R and 0 < x a < δ 1, ten f(x) > 2R/l. One may also coose δ 2 > 0 suc tat if x R and 0 < x a < δ 2, ten g(x) l < l/2. Consequently, if x R and 0 < x a < δ := min{δ 1, δ 2 }, ten and so f(x) > 2R l > 0 and g(x) l g(x) l > l 2 > 0 Tus, by definition, lim x a f.g(x) =. f(x)g(x) > 2R l. l 2 = R. b) Let f : R R be a function and a R. i) (3 points) State wat it means for f to be continuous at a. Solution. lim x a f(x) = f(a). ii) (6 points) For te remainder of te question let f be given by { 1 if x = 1/n for some n Z \ {0} f(x) =. 0 oterwise Using te ε-δ definition, determine weter lim x 0 f(x) exists. Solution. Te limit does not exist. Indeed, let l R and define ε = 1/2. Let δ > 0 be given, (by te Arcimedean property) coose n N suc tat 0 < 1/n < δ and define { 1/n if l < 1/2 x := 1/ 2n if l 1/2.

6 Mat 161 (33) - Final exam Fall Quarter 2015 Page 6 of 12 Observe tat 0 < x < δ. If l < 1/2, ten f(x) = 1 and so f(x) l > 1/2. On te oter and, if l 1/2, ten f(x) = 0, since x is clearly irrational, and so f(x) l 1/2. In eiter case, f(x) l ε and so, by definition, te limit fails to exist. iii) (2 points) State witout proof te set of a R for wic lim x a f(x) exists. For eac a R belonging to tis set state (again witout proof) te value of te corresponding limit. Solution. Te limit lim x a f(x) exists for all a R \ {0} and equals 0. iv) (2 points) At wic points a R is f continuous? Solution. Te function is continuous at all points belonging to te set {x R : x 1/n for some n Z and x 0}.

7 Mat 161 (33) - Final exam Fall Quarter 2015 Page 7 of 12 Question 3. (Total of 25 points) a) (12 points) State and prove te Intermediate value teorem. In your proof, you may use te following result: Suppose f : [a, b] R is continuous at z [a, b] and f(z) > 0 (respectively, f(z) < 0). Tere exists δ > 0 suc tat if x [a, b] (z δ, z+δ), ten f(x) > 0 (respectively, f(x) < 0). Solution. Teorem (Intermediate value teorem). Let f : [a, b] R be continuous were a < b and suppose min{f(a), f(b)} < y < max{f(a), f(b)}. Ten tere exists some a < x < b suc tat f(x) = y. Proof. Suppose f(a) < y < f(b) and let A := {z [a, b] : f(x) < y for all x [a, z]}. Clearly a A so A and A is bounded above by b so, by te LUB axiom, α := sup A exists and a α b. Suppose f(α) < y. In tis case α < b and by te lemma tere exists some δ > 0 suc tat f(x) y < 0 for all x [a, b] (α δ, α + δ) and α + δ < b. Since α is te LUB for A, tere exists some z 0 (α δ, α] suc tat z 0 A. Tus, f(x) < y for all x [a, z 0 ] and so f(x) < y for all x [a, α + δ/2] = [a, z 0 ] [z 0, α + δ/2]. But in tis case α + δ/2 A, contradicting α is an upperbound for A and so f(α) y. Now suppose f(α) > y. In tis case a < α and by te lemma tere exists some δ > 0 suc tat f(x) y > 0 for all x [a, b] (α δ, α + δ) and a < α δ. If f(x) < y for all x [a, z], it follows tat z α δ. Consequently, α δ is an upperbound for A, contradicting α is te least upperbound for A. Hence f(α) y. Consequently, f(α) = y, as required. For te case f(b) < y < f(a), apply te previous argument to te function f : [a, b] R.

8 Mat 161 (33) - Final exam Fall Quarter 2015 Page 8 of 12 b) (8 points) Let f, g : R R be continuous functions, M > 0 and suppose te following old: g(x) M for all x R; Bot l 1 := l 1 < M and l 2 > M. lim f(x) and l 2 := lim f(x) exist; x x + Sow tat tere exists some x 0 R suc tat f(x 0 ) = g(x 0 ). Solution. Let ε 1 := ( l 1 M)/2, ε 2 := (l 2 M)/2 and note tat ε 1, ε 2 > 0 by ypotesis. By definition, tere exist R 1 > 0 suc tat if x < R 1, ten f(x) l 1 < ε 1 and so f(x) = f(x) l 1 + l 1 f(x) l 1 + l 1 < ε 1 + l 1 = (l 1 M)/2 < M. Hence f(x) g(x) < M + g(x) 0 for all x < R 1. Similarly, tere exist R 2 > 0 suc tat if x > R 2, ten f(x) l 2 < ε 2 and so f(x) = l 2 (l 2 f(x)) l 2 f(x) l 2 > l 2 ε 2 = (l 2 + M)/2 > M. Hence f(x) g(x) > M g(x) 0 for all x > R 2. Combining tese observations, := f g is continuous on [ 2R 1, 2R 2 ] and ( 2R 1 ) < 0 wilst (2R 2 ) > 0. Terefore, by te intermediate value teorem, tere exists some point x 0 R suc tat (x 0 ) = 0; tat is, f(x 0 ) g(x 0 ) = 0, as required.

9 Mat 161 (33) - Final exam Fall Quarter 2015 Page 9 of 12 c) (5 points) Let R > 0. By using te ε-δ definition, sow tat te function f : (0, R) R given by f(x) := x 3 + 3x is uniformly continuous. Solution. Let ε > 0 be given and coose δ := 1 6(R 2 + 1). Suppose x, y (0, R) wit x y < δ. Ten f(x) f(y) = x 3 y 3 + 3(x y) = x 2 + xy + y x y Hence, by te triangle inequality, f(x) f(y) ( x 2 + x y + y 2 + 3) x y 3(R 2 + 1) x y < ϵ. Tus, by definition, te function f is uniformly continuous.

10 Mat 161 (33) - Final exam Fall Quarter 2015 Page 10 of 12 Question 4. (Total 25 points) a) (6 points) Let f : R R be given by f(x) := { sin x x if x R \ {0} 1 if x = 0. Using te definition, sow f is differentiable at 0 and determine f (0). Solution. Let R \ {0} and consider f() f(0) = sin 1 = sin 2. Now p(x) := sin x x and q(x) := x 2 are infinitely differentiable and p (x) = cos x 1, p (x) = sin x and q (x) = 2x, q (x) = 2. on R. Tus, However, so tat By l Hôpital s rule, lim p() = lim q() = 0 and lim p () = lim q () = 0. 0 lim 0 p () = 0 and lim q () = 2 0 p () lim 0 q () = 0. p () lim 0 q () = lim p () 0 q () = 0. and so, by a second application of l Hôpital s rule, p() lim 0 q() = lim p () 0 q () = 0. Tus, f() f(0) lim 0 and so, by definition, f (0) = 0, as required. = 0,

11 Mat 161 (33) - Final exam Fall Quarter 2015 Page 11 of 12 b) (5 points) State te mean value teorem. Solution. Suppose f : [a, b] R is continuous on [a, b] and differentiable on (a, b). Ten tere exists some ξ (a, b) suc tat f (ξ) = f(b) f(a). b a c) (4 points) Suppose I is an open interval, f : I R is differentiable and f : I R is a bounded function. Sow tere exists some M > 0 suc tat f(x) f(y) M x y for all x, y I. Solution. Since f is bounded tere exists some M > 0 suc tat f (x) M for all x I. Given x, y I wit x < y, since f is differentiable on I it must be continuous on [x, y] and it follows from te mean value teorem tat tere exists some ξ (x, y) suc tat f(y) f(x) = f (ξ)(y x). Consequently, as required. f(x) f(y) f (ξ) y x M x y, d) (4 points) Now suppose I is an open interval and f : I R satisfies f(x) f(y) M x y for all x, y I. for some M > 0. Does it follow tat f must be differentiable? Using te definition of a derivative, eiter give a proof or sow tere exists a counterexample. (Hint: part of Question 1 migt be useful ere). Solution. Consider te function f : R R given by f(x) := x. question 1 b), f(x) f(y) x y Ten, by for all x, y R. In particular, f satisfies te desired property wit M = 1. However, we claim f is not differentiable at 0. Indeed, if > 0, ten f() f(0) = = 1

12 Mat 161 (33) - Final exam Fall Quarter 2015 Page 12 of 12 and so, taking te limit, lim 0 + f() = 1. However, if < 0, ten f() f(0) = = 1 and so, taking te limit, lim 0 f() = 1. Since te left and rigt-and limits do not agree, lim 0 f() does not exist, and so f is not differentiable at 0. e) (6 points) Suppose I is an open interval, f : I R and for some M > 0 and α > 1 te function f satisfies f(x) f(y) M x y α for all x, y I. Sow tat f is differentiable and compute its derivative. Wat kind of function is f? Solution. Observe tat if x I and R \ {0} satisfies x + I, ten f(x + ) f(x) = f(x + ) f(x) x + x α α 1 M α 1. Consequently, given ε > 0, if one cooses 0 < δ to be given by δ := (ε/m) 1/(α 1), ten wenever 0 < < δ, it follows tat Tus, by definition, f(x + ) f(x) < ϵ. f f(x + ) f(x) (x) = lim = 0 0 and terefore f is differentiable on I wit everywere zero derivative. It follows tat f is a constant function.