Transcription of MATH2060A Solution to Assignment 1
1 MATH2060A Solution to Assignment 1 Section Letf:R Rbe defined byf(x) =x2for rationalxandf(x) = 0 for irrationalx. Showthatfis differentiable atx= 0 and findf (0).We claim thatfis differentiable at 0 withf (0) = 0. Consider the difference quotientf(x) f(0)x 0x6= rational, it is equal toxand, whenxis irrational, it is equal to 0. Therefore, f(x) f(0)x 0 0 |x|.For every >0, we take = , then f(x) f(0)x 0 0 |x|< , x6= 0,|x|< .We conclude thatf (0) = (x) g(c)x c=|f(x)| |f(c)|x c= sgn(x c) f(x) f(c)x c ,sincef(c) = +(c) = limx c+sgn(x c) f(x) f(c)x c =|f (c)|.g (c) = limx c sgn(x c) f(x) f(c)x c = |f (c)|.Hencegis differentiable atciffg +(c) =g (c) |f (c)|= |f (c)| f (c) = (a)f(x) =|x|+|x+ 1|= 2x+ 1,forx 01,for 1 x <0 2x 1,forx < 1 Clearly,f (x) = 2,forx >01,for 1< x <0 2,forx < 1 Forx >0,f(x) f(0)x 0=(2x+ 1) 1x 0= 2 f +(0) = 2 Forx <0,f(x) f(0)x 0=1 1x 0= 0 f (0) = 06= 2 =f +(0).
2 Similar procedures proceed forx < 1,x > differentiable except 0, 1.(b)g(x) = 2x+|x|={3x,forx 0x,forx <0 Clearly,g (x) ={3,forx >01,forx <0 Forx >0,g(x) g(0)x 0=3x 0x 0= 3 g +(0) = 3 Forx <0,g(x) g(0)x 0=x 1x 0= 1 g (0) = differentiable except 2019 MATH2060A2(c)h(x) =x|x|={x2,forx 0 x2,forx <0 Clearly,h (x) ={2x,forx >0 2x,forx <0 Forx >0,h(x) h(0)x 0=x2 0x 0=x h +(0) = 0 Forx <0,h(x) h(0)x 0= x2 0x 0= x h (0) = differentiable on the wholeR.(d)k(x) =|sinx|={sinx,for sinx 0 2n x (2n+ 1) sinx,for sinx <0 (2n 1) < x <2n , n ,k (x) ={cosx,for 2n < x <(2n+ 1) cosx,for (2n 1) < x <2n , n Zandx >2n ,k(x) k(2n )x 2n =sinxx 2n =sin(x 2n )x 2n k +(2n ) = 1 Forn Zandx <2n ,k(x) k(2n )x 2n = sinxx 2n = sin(x 2n )x 2n k (2n ) = 1 Similar procedures proceed forx <(2n+ 1) ,x >(2n+ 1) ,n ,kis differentiable exceptn forn ( x) =f( x+h) f( x)h= limh 0f(x h) f(x) h= limh 0f(x+h ) f(x)h = f (x).}}}}}}
3 Hencef is an odd ( x) =g( x+h) g( x)h= limh 0[ g(x h)] [ g(x)] ( h)= limh 0g(x+h ) g(x)h =g (x).Henceg is an even Denoteg(h) :=f(c+h) f(c) limh 0g(h) = limh 0f(c+h) f(c)h=f (c) sequential criterion for limits (Theorem page 101), denotehn:= 1/n6= 0 for alln, and limhn= lim1n= 0, we have limg(hn) = limh 0g(h) =f (c), whereg(hn) =f(c+ 1/n) f(c)1/n=n{f(c+ 1/n) f(c)}.Hencef (c) = lim (n{f(c+ 1/n) f(c)}).Takef(x) :={sin /x, x >00,x 0,n{f(1/n) f(0)}=n(0 0) = 0 , lim (n{f(c+ 1/n) f(c)}) = ,f (c) doesn t exist becausefis not continuous , we may takef:= Q= Dirichlet function. Fixc {f(c+ 1/n) f(c)}={n(1 1), c Qn(0 0), c6 Q= 0 Dirichlet function Qis not rational andyis irrational, why isx+yirrational?Spring 2019 MATH2060A314. Nowh (x) = 3x2+ 2>0 x , by Theorem ,h 1is differentiable and(h 1) (y) =1h (x)=13x2+ 2 x R,whereyis related toxbyy=h(x).}}
4 Forx= 0, we havey=h(0) = 1, and (h 1) (1) =13(0) + 2=12 Forx= 1, we havey=h(1) = 4, and (h 1) (4) =13(1) + 2=15 Forx= 1, we havey=h( 1) = 2, and (h 1) ( 1) =13(1) + 2= Exercises1. Consider the functionfdefined on [0, )f(x) =x sin1x, >0,andf(0) = 0. Determine the range of in which(a)fis continuous on [0, ),(b)fis differentiable on [0, ), and(c)f exists and is differentiable on [0, ). function is smooth, that is, infinitely many times differentiable on (0, ).It suffices to consider the case atx= 0.(a) As|x sin1x| x ,by Sandwich rulelimx 0+x sin1x= 0,sofis continuous atx= 0 hence we conclude that it is continuous on [0, ).(b) By definition,f (0) = limx 0+x sin1x 0x 0= limx 0+x 1sin1x= 0,when >1. This limit does not exist when (0,1]. Sofis differentiable on [0, )if and only if (1, ).(c) The derivative offisf (x) = x 1sin1x x 2cos1x, x (0, ),andf (0) = 0.]]]]]
5 Atx= 0, using the definition of the derivative, we have, for >1,f (0) = limx 0+ x 1sin1x x 2cos1x 0x 0= limx 0+ x 2sin1x x 3cos1x= 0,when (3, ).The limit does not exist when (0,3]. We conclude thatf isdifferentiable on [0, ) if and only if (3, ).Spring 2019 MATH2060A42. Find (a) the maximal domain on which the function is well-defined, (b) the domain onwhich it is continuous and (c) the domain on which it is differentiable in each of thefollowing cases. Justify your answer in (c).(a)f(x) =|x2 5x+ 6|.(b)h(x) = log(16 x2).(c)j(x) = cos|x|. Solution .(a) The function is the composition of two functionsf(x) =g(h(x)) whereh(x) =x2 5x+ 6 andg(y) =|y|. Bothgandhare continuous onR. As continuity if preservedunder composition,fis continuous on ( , ).Next, writef(x) =|x2 5x+ 6|=|x 2||x 3|. It is known thatx7 |x 2|is notdifferentiable at 2 andx7 |x 3|is non-zero and differentiable at 2.
6 It follows thatfis not differentiable at 2. (See the proposition on next page.) By the same reasonfisnot differentiable at 3. We conclude thatfis differentiable on ( ,2) (2,3) (3, ).(b) The functionh= log(16 x2) = log(k(x)) wherek(x) = 16 x2is differentiableeverywhere. Using the fact that the log function is defined and smooth only forpositive number,his defined, continuous and differentiable as long as 16 x2>0,that is, on ( 4,4).(c)jis defined and continuous everywhere. The functionx7 |x|is differentiable exceptatx= 0 andy7 cosyis differentiable everywhere. Sojis differentiable at allnon-zero x. However, as the derivative of cosyis equal to 0 aty= 0. We mustexamine the differentiability ofjat 0 using definiton. Indeed, using the fact thecosine function is even,limh 0cos|h| cos 0h 0= limh 0cosh 1h= 0,from which we conclude thatjis also differentiable atx= 0.
7 Hencejis shortcut is to realize that the cosine is an even function, soj(x) = cosxis dif-ferentiable everywhere. In this approach we do not viewjas the composite of Find a function which is not differentiable exactly at the following points on ( , ) ineach of the following cases:(a)n-many distinct points{a1,a2, ,an},(b) The set of integersZ, and(c){0,1,12, ,1n, ,}. forgot to require these functions to be continuous. In the following functionsare continuous.(a)f(x) =n k=1|x ak|.Spring 2019 MATH2060A5(b)g(x) = k= (x k),where is a function which makes a corner at 0 but otherwise smooth and vanishesoutside [ 1,1].(c) You may try thish(x) = xsin x .Of course, seth(0) = A functionf: (a,b) Rhas a symmetric derivative atc (a,b) iff s(c) = limh 0f(c+h) f(c h)2hexists. Show thatf s(c) =f (c) if the latter exists.
8 Butf s(c) may exist even thoughfisnot differentiable atc. Can you give an example? (c+h) f(c h)2h=f(c+h) f(c) +f(c) f(c h)2h=12f(c+h) f(c)h+12f(c h) f(c) we havef s(c) = limh 0f(c+h) f(c h)2h=12limh 0f(c+h) f(c)h+12limh 0f(c h) f(c) h=12limh 0f(c+h) f(c)h+12limh 0f(c+h ) f(c)h =12f (c) +12f (c) =f (c) set-up forf s(c) = limh 0f(c+h) f(c h)2hdoesn t involve the valuef(c), a simple idea to construct a counter example is by changing the valuef(c) from adifferentiable functionf, so that the new function is not differentiable (x) ={1,forx=c0,forx6=c. Thenf s(c) = limh 0f(c+h) f(c h)2h= (c) doesn t exist sincefis not continuous atx= Letf:R Rsatisfyf(x+y) =f(x)f(y) for allx,y R. Supposefis differentiable at0 withf (0) = 1. Show thatfis differentiable onRandf (x) =f(x) for allx 0, thenf (0) = 06= 1, contradiction arises.}
9 Hence x0 (x0)6= 2019 MATH2060A6 Thenf(x0) =f(x0+ 0) =f(x0)f(0) f(0) = ,fis differentiable at 0, hence limh 0f(h) 1h= limh 0f(0 +h) f(0)h=f (0) = For allh6= 0,f(x+h) f(x)h=f(x)f(h) f(x)h=f(x)f(h) 1h f (x) = limh 0f(x+h) f(x)h=f(x) limh 0f(h) 1h=f(x).Hence,fis differentiable 2019 MATH2060A7 The following observation was discussed in class. I formulate it as a proposition defined on (a,b) such thatfis not differentiable atc (a,b)butgis differentiable atcandg(c)6= 0. Thenfgis not differentiable on the contrary thath(x) =f(x)g(x) is differentiable atc. Thenf(x) =h(x)g(x)is differentiable atcby the quotient rule, contradiction holds.
