EE 518 Homework 5 Due on Monday 2 00pm October 3 2016 Problem 1 Suppose that f 0 1 R is uniformly continuous on 0 1 If an is a Cauchy sequence in 0 1 and bn f an show that bn is a Cauchy sequence in R Let 0 be an arbitrary real number Since f is uniformly continuous on 0 1 then there exists a 0 such that if x y and x y 0 1 f x f y Since an is a Cauchy sequence in 0 1 then for every 0 there is an integer N such that an am if n m N Then from the uniform continuity of f it follows that f an f am if n m N which implies that f an is a Cauchy sequence in R Note The above result is not true if f is only continuous For example f x x1 is continuous on 0 1 and let an n1 then bn f an n is not a Cauchy sequence Problem 2 True or false Justify your answer Give a counterexample if it is false a If limx a f x 0 and limx a g x 0 then limx a f x g x does not exist b If neither limx a f x nor limx a g x exists then limx a f x g x does not exist c If limx a f x exists but limx a g x does not exist then limx a f x g x does not exist d If limx a f x g x exists then the limit must be equal to f a g a e If limx a f x and limx a g x then limx a f x g x 0 a False Counterexample limx 1 x x 1 x 1 or limx 1 b False Counterexample Let f x x1 and g x but limx 0 f x g x limx 0 0 0 exists sin x 1 x 1 1 x then limx 0 f x and limx 0 g x do not exist c True By contradiction assume limx a f x g x exists then using the properties of limit we can write limx a g x limx a f x g x f x limx a f x g x limx a f x Since limx a f x g x and limx a f x both exist then limx a g x exists This contradiction shows that our first assumption that limx a f x g x exists is wrong d False Counterexample Let f x x 1 and g x f 1 0 and g 1 is undefined e False Counterexample Let f x x12 and g x 2 limx 0 f x g x limx 0 x x 1 4 1 x4 1 x 1 then limx 1 f x g x 1 exists but then limx 0 f x limx 0 g x but Problem 3 Find any local minimum or local maximum of following functions if exist i f x 1 2 x x e 2 f x x 1 x2 ii Page 1 of 4 EE 518 Homework 5 Due on Monday 2 00pm October 3 2016 i Since f x is well defined and differentiable on R we only consider when f 0 x 0 Let 1 f 0 x xe x x2 e x 0 2 we have 2 solutions x1 0 x2 2 Since 1 f 00 x 1 2x x2 e x 2 f 00 0 1 0 f 00 2 e 2 0 Therefore x1 0 is a local minimum and x2 2 is a local maximum ii Let f 0 x 1 x2 0 1 x2 2 We have x1 1 x2 1 since f 1 1 2 f 1 1 2 therefore x1 1 is a local minimum and x2 1 is a local maximum This only works for continuous functions and only two local optimum exists Problem 4 a A critical point of a function f X R is a point c X such that either f 0 c 0 or f 0 c does not exist Consider the function f x ax3 bx2 cx d where a 6 0 Show that f can have two one or no critical points give examples and sketches to illustrate the three possibilities How many local extreme values can f have b Show that the equation ex 2 cos x has at least one root in 0 1 c Consider the function f x x4 4x3 4x2 c where c is a constant Show that f has at most one zero in the interval 1 2 For what interval of values of c does the equation f x 0 have exactly one root in 1 2 a f 0 x 3ax2 2bx c is a 2nd order polynomial so it has either two one or zero real roots Thus f can have either two one or zero critical points For example i f x x3 3x has two critical points ii f x x3 has one critical point and iii f x x3 3x has no critical point It can have at most two local extreme values case i For cases ii and iii there is no local extreme value See Fig 1 b The function f x ex 2 cos x is continuous on 0 1 f 0 2 and f 1 0 178 Since f 0 0 f 1 then from the Intermediate Value Theorem there exists a c 0 1 such that f c 0 Then ex 2 cos x has at least one root in 0 1 c f 0 x 4x3 12x2 8x 4x x2 3x 2 4x x 1 x 2 0 on 1 2 Since f is strictly decreasing on 1 2 then if there is any root it is unique So f has at most one zero in the interval 1 2 Since f is continuous and strictly decreasing on 1 2 and f 1 1 c and f 2 c then from the Intermediate Value Theorem f x 0 has exactly one root if and only if f 1 and f 2 have opposite signs i e c 0 c 1 or 1 c 0 Page 2 of 4 EE 518 Homework 5 Due on Monday 2 00pm October 3 2016 Figure 1 Sketch for Problem 4a Problem 5 Suppose function f has n derivatives on the interval 0 a and limx 0 x2n f n x L exists in R where f n denotes the n th derivative of f Find limx 0 xn f x in terms of L Apply L Hospital s rule n times in succession to f x x n we obtain lim xn f x lim x 0 x 0 1 n f n x 1 n lim x2n f n x n n 1 2n 1 x 2n n n 1 2n 1 x 0 1 n 1 n n 1 L L n n 1 2n 1 2n 1 Problem 6 Given values of variables x and y x 0 …