Continuity and Differentiability

Transcription

1 Continuity and Dierentiability Tis capter requires a good understanding o its. Te concepts o continuity and dierentiability are more or less obvious etensions o te concept o its. Section - INTRODUCTION TO CONTINUITY We start wit a very intuitive introduction to continuity. Consider te two graps given in te igure below: Our purpose is to analyse te beaviour o tese unctions around te region =. Te obvious visual dierence between te two graps around = is tat wereas te irst grap passes uninterrupted (witout a break) troug =, te second unction suers a break at = (tere is a jump). Tis visual dierence, put into matematical language, gives us te concept and deinition o continuity. Matematically, we say tat te unction is continuous at = wile is discontinuous at = For, LHL RHL at and at and () = Mats / Continuity and Dierentiability

2 LHL = RHL = () For, LHL at and RHL at () = and LHL RHL = From te discussion above, try to see tat or a unction to be continuous at = a, all te tree quantities, namely, LHL, RHL and (a) sould be equal. In any oter scenario, te unction becomes discontinuous. Discontinuities tereore arise in te ollowing cases: (a) (i) One or more tan one o te tree quantities, LHL, RHL and (a) is not deined. Lets consider some eamples: around =. LHL=, RHL =, is not deined. Tereore, is obvious rom te grap: is discontinuous at = wic (ii) or around = Mats / Continuity and Dierentiability

3 LHL = RHL = but () is not deined. Tereore, tis unction s grap as a ole at = ; it is discontinuous at = : 3 (b) All te tree quantities are deined, but any pair o tem is unequal (or all tree are unequal). Lets go over some eamples again: (i) () = [] around any integer I LHL = I, RHL = I, (I) = I LHL RHL = I so tis unction is discontinuous at all integers as we already know. (ii) () = {}around any integer I LHL =, RHL =, (I) = LHL RHL= I so tis unction is also discontinuous at all integers. (iii),, around any integer I From te igure, we notice tat at any integer I, LHL =, RHL =, (I) = LHL=RHL I so tat tis unction is again discontinuous. Mats / Continuity and Dierentiability

4 4 (iv), around At =, we see tat LHL =, RHL =, () = LHL RHL and tis unction is discontinuous. To summarize, i we intend to evaluate te continuity o a unction at = a, wic means tat we want to determine weter () will be continuous at = a or not, we ave to evaluate all te tree quantities, LHL, RHL and (a). I tese tree quantities are inite and equal, () is continuous at = a. In all oter cases, it is discontinuous at = a Eample Find te value o () i te unction m m m, is continuous at = Solution: We are going to encounter a lot o questions similar to te one above over te net ew pages; we are given a deinition o () and are asked to determine (a) so tat () becomes continuous at = a. We simply evaluate a and i it eists, we let (a) equal to tis it. Tis ensures tat te necessary and suicient condition or continuity at = a, i.e is satisied. a a Mats / Continuity and Dierentiability

5 5 For tis question, m m m m m m m m m... m m m... m m... Tereore m m m m Eample Find te value o is continuous at Solution: i te unction 4 cos, cot 4 4 Tis question is very similar to te previous one: cos 4 4 cot Mats / Continuity and Dierentiability

6 6 4 cos sin cos sin = cos sin cos 4 cos sin cos sin 4 cos sin cos sin.. Tereore, 4 Eample 3 Find te values o a and b i a cos bsin 3,, is continuous at = Solution: We obviously require a Mats / Continuity and Dierentiability

7 (Note tat in all te tree eamples above, we ave not ound te LHL and RHL separately; we only determine te it assuming it eists, or in oter words assuming tat LHL = RHL. We will soon encounter cases were LHL and RHL need to be separately determined) We use te epansion series or sin and cos to get: 7 Must be a b a b! 3! 4! 5! a b... a Now notice tat ( + a b) must be necessarily oterwise 3 Hence, Also, + a b =...(i) a b! 3! 6a b + = Solving (i) and (ii), we get a 5 b 3...(ii) b will become ininite, Eample 4 Find te condition on () and g () wic makes te unction Solution: F n n g n continuous everywere. Note rom te deinition o F() tat we ave a variable n present. Lets irst try to make F() independent o n. Te obvious way is to consider tree separate cases or :,, : F g g Mats / Continuity and Dierentiability

8 8 : F n n g n n n g n = g () n because as n or : F n n g n = () n because as n or Tereore, we can now rewrite F() independently o n in te ollowing manner wen or g F wen or, g wen or, F() could be discontinuous at only points, = or = To ensure continuity at tese points LHL at RHL at F g g g LHL at RHL at F g g g Tereore, or continuity o F (), () = g () ( ) = g( ) Mats / Continuity and Dierentiability

9 9 Eample 5 I a sin sin, 6 b, e tan tan3, 6 is continuous at =, ind te values o a and b. Solution: For continuity at =, LHL (at = ) = () = RHL (at = ) LHL at sin a sin y y. a y were sin y; as, y a e RHL at e e / 3 e tan tan 3 tan 3. 3 tan 3 a /3 e b e a and b e 3 /3 Eample 6 Let : be a unction satisying y y or all, y. I te unction () is continuous at = sow tat it is continuous or all Solution: In questions tat ave suc unctional equations, we sould try to substitute certain trial values or te variables to gain useul inormation; in tis particular question, since we are given some condition about () at =, we sould try to ind () Mats / Continuity and Dierentiability

10 Putting = y =, we get () = () () = Now, since () is continuous at =, or equivalently LHL = RHL = () For continuity at an arbitrary value o, te LHL, RHL and () sould be equal. Similarly, LHL at RHL (at ) = () Tereore, () is continuous or all values o Eample 7 Let sin sin a. Find te values o a and b so tat () is continuous at =. b b 3/ Solution: LHL at sin sin = 3.sin sin Mats / Continuity and Dierentiability

11 () = a RHL at = b b 3/ b b b b 3/ b 3/ b b b b Since LHL at = RHL, no suc values o a and b eist tat could make te unction continuous Eample 8 Let F, and G, e 3, k, were () is some unction o. I F() is continuous at =, ind te value o k so tat G () is also continuous at =. Solution: Since F() is continuous at =, F F e 3 e 3 Mats / Continuity and Dierentiability

12 e. 3 e 3 For G() to be continuous at =, k sould be equal to G G e. e Tereore, e k = e Eample 9 Sow tat te unction e e / / wen wen is discontinuous at = Solution: LHL at e e / / Now, as, Tereore, LHL = so e / Mats / Continuity and Dierentiability

13 3 Let Similarly, RHL Mats / Continuity and Dierentiability e e = Eample / / e e / / Since LHL RHL, is discontinuous at = a Solution: n a. Evaluate te continuity o () at =. Te LHL and RHL migt dier due to te discontinuous nature o [] and {}. Lets determine teir value: LHL at a For >, it is obvious tat and. a Tereore, LHL Similarly, RHL a a = a a a = ln a Since, LHL RHL; is discontinuous at =.

14 4 Note: (i) Some autors talk about removable and irremovable discontinuities. Let us discuss wat tis means: Consider, Te LHL and RHL at = eist and bot are equal to. Tere is a ole in te grap at = and tereore te unction is discontinuous at = We can, i we want to, ill tis ole by redeining te unction in te ollowing manner: Te additional deinition at = ills te ole and removes te discontinuity. Hence, suc a discontinuity would be called a removable discontinuity. By now, you sould ave realised tat a discontinuity is removable only i te LHL and RHL are equal, since only ten we can redeine (a) to make all te tree quantities equal. I te LHL and RHL are temselves non-equal, no redeinition o (a) could possibly make te unction continuous and ence, suc a discontinuity would be called irremovable. For eample, () = [] and () = {} suer rom irremovable discontinuities at all integers. (ii) I and g are two continuous unctions at a point = a (wic is common to teir domains), ten g and g will also be continuous at = a. Furtermore, i g a, ten g will also be continuous at = a. (iii) I g is continuous at = a and is continuous at = g(a), ten (g()) will be continuous at = a. (iv) Any polynomial unction is continuous or all values o. (v) Te unctions sin, cos and e (or a ) are continuous or all values o. ln (or log ) is continuous or a all >. Mats / Continuity and Dierentiability

15 5 TRY YOURSELF - I Q. Find te values o a and () i () is continuous at =, were sin a sin, 3 Q. Find te value o a i () is continuous at =, were cos 4 a 6 4 Q. 3 I te unction () deined by log a log k b is continuous at =, ind k. Q. 4 Let (y) = () (y) or every, y. I () is continuous at any one point = a, ten prove tat () is continuous or all ~ {}. Q. 5 Find te values o a and b so tat () is continuous at =, were a b 3 3 /. Q. 6 Discuss te continuity o () = [] + [ ] at integer points. Q. 7 Discuss te continuity o () in [, ] were cos 3 Q. 8 I, g, g ( ())., and, ten discuss te continuity o (g()), and Mats / Continuity and Dierentiability

16 6 Section - DIFFERENTIABILITY Having seen te concept and te pysical (grapical) signiicance o continuity, we now turn our attention to te concept o dierentiability. We will again start wit an intuitive, grapical introduction. Consider te two graps given in te igure below: Wat is te striking dierence between te two graps at te origin (apart rom one being linear and te oter, nonlinear)? Te irst as a sarp, sudden turn at te y-ais wile te second passes smootly troug te y-ais. Lets make tis idea more concrete. Imagine a person called Teta walking on te grap o () =, towards te y-ais, once rom te let and once rom te rigt: Wile walking rom te let, Teta will be moving in a nort-east direction as e approaces te y-ais; wile walking rom te rigt, e will be walking in a nort-west direction. Now consider Teta walking on () =, once rom te let and once rom te rigt, towards =. Mats / Continuity and Dierentiability

17 As Teta approaces te y-ais, we see tat e moves almost orizontally near te y-ais, in bot te cases. Te line o travel becomes almost te same rom eiter side near =. At =, te line o travel becomes precisely orizontal (or an instant), weter Teta is walking rom te let or te rigt. (Tis unique line o travel is obviously te tangent drawn at = ) 7 In matematical language, since te lines o travel rom bot sides tend to become te same as te y-ais is approaced or more precisely, te tangent drawn to te immediate let o = and te one to te immediate rigt, become precisely te same at = (a unique tangent), we say tat te unction () = is dierentiable at =. Tis means tat te grap is smootly varying around = or tere is no sarp turn at =. In te case o () =, te lines o travel rom te let and and te rigt and sides are dierent. Te line o travel (or tangent) to te immediate let o = is inclined at 45º to te -ais wile te one to te immediate rigt is inclined at 35º. Precisely at =, tere is no unique tangent tat can be drawn to (). We tereore say tat () = is non-dierentiable at =. Tis means tat te grap as a sarp point (or turn) at =, as is evident rom Fig Section - 3 LEFT AND RIGHT DERIVATIVES Let us now put te discussions we ave done above in concrete matematical orm. Consider te curve () = +. Teta is walking on tis curve towards te y-ais rom te rigt. Wen e is ininitesmally close to te y-ais, is direction o travel will be along te tangent drawn to te rigt segment o te grap, at an -coordinate in te Mats / Continuity and Dierentiability

18 immediate rigt neigbourood o te origin ; or equivalently, at a point on te rigt segment o te grap wic is ininitesmally close to te point (, ). 8 How do we ind out tis direction o travel near te point (, )? In oter wards, ow do we ind out te slope o a tangent drawn to te rigt part o te grap, at a point etremely (ininitesmally) near to (, )? To evaluate tis slope, we irst drawn a secant on tis grap, passing troug (, ), as sown in te igure below: Let te -coordinate o te point Q be. Te slope o te secant PQ is QR tan PR Notice tat QR is () () and PR is. Tereore, tan...(i) Mats / Continuity and Dierentiability

19 Now we make tis secant closer to a tangent by reducing : look at te igure below; as is reduced or as, te secant PQ tends to become a tangent drawn at P (or more accurately, a tangent at a point ininitesmally close to te point P): 9 We see tat as te point Q P or as, te secant PQ tends to become a tangent to te curve; to ind te slope o tis tangent, we ind tan were tan is given by (i) a Slope o tangent Tis it gives us te slope o te tangent at te point P. (By at we mean just near ). Lets evaluate tis it or tis particular unction: so and Mats / Continuity and Dierentiability Slope o tangent = = = Tis means tat te tangent drawn at P (on te rigt part o te grap) is inclined at 45º to te -ais. In matematical jargon, te it we ave just evaluated is called te Rigt Hand Derivative (RHD) o () at =. Tis quantity, as we ave seen, gives us te beaviour o te curve (its slope) in te immediate rigt side vicinity o =.

20 Obviously, tere will eist a Let Hand Derivative (LHD) also tat will give us te beaviour o te curve in te immediate let side vicinity o =. In oter words, te LHD will give us te direction o travel o Teta as e is just about to reac te point (, ) travelling rom te let towards te y-ais. To evaluate te LHD, we ollow a procedure similar to te one we used to evaluate te RHD ; only tis time we will drawn te secant PQ on te let side o te grap. As te point Q P or as, te secant PQ again tends to become a tangent. As or te previous case, te slope o tis tangent will be given by; For tis particular case: LHD = Mats / Continuity and Dierentiability

21 Te tangent drawn to te let part o te grap just near P will be inclined at 35º to te -ais. Notice tat because tere is a sarp point at =, or in oter words Teta s direction o travel will cange wen crossing te y-ais, te LHD and RHD ave dierent values. We would say tat tis unction is non-dierentiable at =. No tangent can be drawn to () precisely at =. On te oter and, or a smoot unction, te LHD and RHD at tat point will be equal and suc a unction would be dierentiable at tat point. Tis means tat a unique tangent can be drawn at tat point. Beore summarizing, let us write down te general epressions or te LHD and RHD LHDat or y a tan a a a a LHD at a a a RHDat a or y Mats / Continuity and Dierentiability

22 tan a a RHD at a a a To summarize: Consider a continuous unction y = () * I at a particular point = a, te LHD and RHD ave equal numerical values, we say tat () is dierentiable at = a. In grapical terms, tis means tat te grap crosses = a smootly, witout any sarp turn. Tis also means tat a unique tangent can be drawn to te curve y = () at = a. Te derivative at = a implies te slope o te tangent at = a ; i.e. Derivative (at = a) = LHD = RHD {Obviously, te derivative eists only i () is dierentiable at = a} Te derivative at = a is denoted by '(a) * I at a particular point = a, te LHD and RHD ave non-equal values or one (or bot) o tem does not eist, we say tat () is non dierentiable at = a. Grapically, tis means tat te grap does not pass troug = a smootly, tere is a sarp turn at = a. For a discontinuous unction () at = a we can deine LHD and RHD separately. (We ll ave to sligtly modiy our tecnique to evaluate LHD and RHD; can you see wat modiication is required?). But or any discontinuous unction at = a, () would always be non dierentiable at = a since no unique tangent could be drawn to () at = a. Tereore, or dierentiability at = a te necessary and suicient conditions tat () as to satisy are: (i) (i) () must be continuous at = a. (ii) LHD = RHD at = a. We will consider some eamples to make tis discussion more clear. Mats / Continuity and Dierentiability

23 Te grap o () is smoot. Lets veriy tis by evaluating te LHD and RHD at any general -coordinate. LHD RHD We see tat LHD = RHD or every value. Hence () is everywere dierentiable (smoot). Also te slope at any coordinate as a numerical value (equal to te LHD and RHD). We say tat te derivate o is. 3 Mats / Continuity and Dierentiability

24 4 (ii) Te grap as a sarp turn (corner) at =. Tis means tat at =, sould be non-dierentiable. Lets veriy tis again : LHD we want toevaluate LHD at = Similarly, = RHD = Tis result is visually obvious rom Fig - ; te let segment is y rigt segment is y, wic as slope. wic as a slope, wile te (iii) and Mats / Continuity and Dierentiability

25 5 Te analysis o tese two unctions is straigtorward grapically. For, te LHD and RHD at any integer are bot, but () is discontinuous too. Hence, altoug LHD = RHD, () is non-dierentiable at integral points. At all oter values o, it is dierentiable wit te derivative s value being. Similarly, or, te LHD and RHD at any integer are bot, but due to discontinuity at all integers, () is non-dierentiable at all integers. At all oter values o, () is dierentiable wit te derivative s value being. (iv) sin Tis unction seems smoot everywere. We will now veriy tis: LHD sin sin cos = cos sin RHD sin sin cos = cos sin Hence, LHD = RHD or all values o ; sin is everywere dierentiable. Te derivative at (slope o tangent at ) as te numerical value cos. Mats / Continuity and Dierentiability

26 6 Te derivative o sin is cos. By now you sould be pretty clear about te meaning and grapical signiicance o dierentiability and evaluating te derivative. Evaluating te derivative using te process we ave described above is called dierentiation using irst principles. Tis discussion is etremely important to watever we ll study aterwards. You must ensure tat you ve ully understand tis. Eample I 3, 4, 4, discuss its continuity and dierentiability. Solution: In questions involving evaluation o continuity and dierentiability, we can o course proceed analytically by writing down te epressions or LHD and RHD (at tose points were we eel te unction migt not be continuous or dierentiable); but we will always try to ollow a grapical approac also, werever possible, since it gives muc useul inormation about te beaviour o te unction. Note tat in piecewise deined unctions suc as te one above we ave to use dierent deinitions o () in dierent intervals. So or eample, to evaluate (a ) we will ave to use te unction deinition or < a, wereas or (a + ) we ave to use te deinition or > a. In tis question, tere is only one point, namely =, were tis unction could be possibly discontinuous and /or non-dierentiable. Te two unctions 3 and 4 are oterwise continuous and dierentiable in teir separate intervals. Hence, we need to analyse te continuity and dierentiability only at = {suc points are sometimes reerred to as critical points} Analytical approac: LHL at 3 = RHL at Mats / Continuity and Dierentiability

27 7 4 = = Tis unction is discontinuous at = and tereore also non-dierentiable. LHD at Tis sould be te epression to evaluate LHD at =. But notice tat or (), we cannot substitute since = is part o te oter segment i.e. = does not lie in te domain o te let and unction. So wat sould we use or ()? Tink about tis careully and you will realise tat we would ave to use or LHL at = in place o (). (Tis is te modiication we mentioned a little wile back). Tis LHL value is. Tereore, LHD at 3 Mats / Continuity and Dierentiability

28 8 4 4 RHD at Eample A unction () is deined as:, 5 4, 4 3, 3 5,. Discuss te continuity and dierentiability o (). Solution: Te critical points or tis unction are =,,. Lets analyse () or eac o tese critical points separately. (i) LHL LHD RHL RHD RHL at Tereore, tis unction is non-continuous (and non-dierentiable) at =. (ii) LHL 5 4 Mats / Continuity and Dierentiability

29 9 LHD 5 4 = 5 RHL 4 3 RHD RHL at 4 3 = 5 Since LHL = RHL and LHD = RHD, () is continuous and dierentiable at = (iii) LHL 4 3 LHD = RHL 3 5 = Mats / Continuity and Dierentiability

30 3 We see tat () is non-continuous (and non-dierentiable) at = and =. Te grap is plotted below: Eample 3 Te elaborate analysis done or tis question will not be done or subsequent questions were we will try ocus more on te grapical approac. Let, and, g. Evaluate te continuity and dierentiability o g in te interval [, ] by drawing te grap. Mats / Continuity and Dierentiability

31 3 Solution: From te grap o, we can easily derive te graps o and, and add tem point by point to get te grap o g. Veriy or yoursel te result o te addition o te two graps. It is obvious rom te resultant grap tat g is continuous every were but non-dierentiable at = and = Eample 4 I 6, 4, 4, 3, evaluate te continuity and dierentiability o () Mats / Continuity and Dierentiability

32 3 Solution: Tis grap will ave 3 dierent segments wic can easily be plotted as sown in below (you are urged to carry out te plotting on your own): We irst draw all te tree unctions graps on te ais using dotted curves and ten darken te relevant portions as speciied in te piecewise domain. For eample, we selected only tat segment o 6 or wic 4, and so on. It is clear rom te grap tat () is continuous in its entire domain [ 4, 3] but is non-dierentiable at 3 points =,,. Te evaluation o let and rigt derivatives on eac o tese tree points is let to you as an eercise. Te results are: LHD 4 RHD LHD RHD LHD RHD Eample 5 A unction : satisies 3 y y or all, y. Prove tat is a constant unction. Solution: For a constant unction () = k, te tangent drawn at any point would be te same line y = k itsel. Hence, te slope o tangent at any point (te derivative at any point) as te value. Tereore, our aim in tis question sould be to someow sow tat () as derivative at all points. Let us evaluate te RHD at any : Mats / Continuity and Dierentiability

33 33 RHD at Tereore, RHD at 3 Using te given relation RHD Since RHD cannot be negative, it as to be. Similarly, LHD = or all RHD or all () is a constant unction. Eample 6 Let e,,, Evaluate te continuity and dierentiability o () Solution: We sould irst o all write () separately or > and < ; using,, e Mats / Continuity and Dierentiability

34 At tis juncture, we do not ave suicient knowledge to accurately plot te grap or > ; we will ence ollow te analytical approac. Te critical point is only = LHL at / RHL at e Veriy 34 LHD at = RHD at = e Tereore, tis unction is continuous at = (and everywere else) but not dierentiable at = / Mats / Continuity and Dierentiability

35 35 TRY YOURSELF - II Q. I min,,, draw te grap o and discuss its continuity and dierentiability. Q. I ( ) e, (3e 4), discuss te continuity and dierentiability o ( ) at =. Q. 3 Let ( ) be deined in te interval [, ] suc tat ( ) [ ],, and g( ) ( ) ( ). Discuss te dierentiability o g( ) in [, ]. Q. 4 I [ ], ( ), test its dierentiability at =. Q. 5 Given ( ) ( y) y, and (), ind ( ). Q. 6 I ( ) k l l l discuss te continuity and dierentiability o g( ) ( ) ( ) Q. 7 I : satisies ( ) is continuous or all. y ( ) ( y) or all, y, and '(), prove tat Q. 8 Find te values o a, b and c i te unction ( ) a sin be c is dierentiable at =. Mats / Continuity and Dierentiability

36 36 SOLVED EXAMPLES Eample I ( ) sin 3 a sin bsin 5, c, is continuous at =, ind a, b and c. Solution: For continuity, we require: ( ) c LHL and RHL will be te same so we do not need to evaluate tem separately ( ) sin 3 a sin b sin 5 = As in Section -, eample-3, note tat 3 + a + b =... (i) 7 + 8a + b =... (ii) I tis does not old, (3 a b) (7 8 a b) and 5 5 will become ininite 3 Solving (i) and (ii), we get a = 4 b = 5 Mats / Continuity and Dierentiability

37 37 Te it now reduces to 43 3a b 43 3a b = Hence, c = iger order terms Eample Let : be a unction deined by y y or all, y. I is continuous at =, sow tat is continuous or all. Solution: Since is continuous at,... (i) To evaluate, we substitute y in te given unctional relation to get. or (i) I, ten i.e or all values o so tat is continuous everywere. (ii) We now assume. Now, LHL at any (From (i)) Mats / Continuity and Dierentiability

38 38 Similarly, RHL at any Hence, is continuous or all Eample 3 Let sin, and, g( ) sin,., Evaluate te continuity and dierentiability o ( ) and g( ). Solution: We will irst try to grapically understand te beaviour o tese two unctions and ten veriy our results analytically. Notice tat no matter wat te argument o te sin unction is, its magnitude will always remain between and. Tereore, and sin sin Tis means tat te grap o sin will always lie between te lines y = ± and te grap o sin will always lie between te two curves y. Also, notice tat as increases, decreases in a progressively slower manner wile wen is close to, te increase in is very ast (as decreases visualise te grap o y ). Tis means tat near te origin, te variation in te grap o sin will be etremely rapid because te successive zeroes o te grap will become closer and closer. As we keep on increasing, te variation will become slower and slower and te grap will spread out. For eample, or > tere will be no Mats / Continuity and Dierentiability

39 39 inite root o te unction. Only wen will sin again approac. Mats / Continuity and Dierentiability

40 4 Figures 3 and 3 sow te approimation variation we sould epect or tese two unctions. Notice ow te lines y = ± envelope te grap o te unction in te irst case and te curves y envelope te grap o te unction in te second case. Te envelopes srink to zero vertical widt at te origin in bot cases. Tereore, we must ave: sin and sin (Tis is also analytically obvious; sin is a inite number between and ; wen it gets multiplied by (were ), te wole product gets ininitesmally small). Now lets try to get a eel on wat will appen to te derivatives o tese two unctions at te origin. For ( ) sin, te slope o te envelope is constant (±). Tus, te sinusoidal unction inside te envelope will keep on oscillating as we approac te origin, wile srinking in widt due to te srinking envelope. Te slope o te curve also keeps on canging and does not approac a ied value. However, or g( ) sin, te slope o te envelope is itsel decreasing as we approac te origin, apart rom srinking in widt. Tis envelope will compress or ammer out or latten te sin oscillations near te origin. Wat sould tereore appen to te derivative? It sould become at te origin! Let us zoom in on te graps o bot te unctions around te origin, to see wat is appening: Mats / Continuity and Dierentiability

41 Mats / Continuity and Dierentiability 4

42 Tese graps are not very accurate and are only o an approimate nature; but tey do give us some eeling on te beaviour o tese two unctions near te origin. Te sin grap keeps continuing in te same manner no matter ow muc we zoom in; owever, in te sin grap, te decreasing slope o te envelope itsel tends to latten out te curve and make its slope tend to. Hence, te derivative o sin at te origin will not ave any deinite value, wile te derivative o sin will be at te origin. Lets veriy tis analytically: 4 (i) sin at LHD = RHD sin sin Tis it, as we know, does not eist; ence, te derivative or () does not eist at = (ii) sin at LHD=RHD sin sin = Tis veriies our earlier assertion Mats / Continuity and Dierentiability

43 43 Eample 4 I y y or all, y and g G,were prove tat () is continuous or all. g and G eists, Solution: We ave g g and G G k some inite number Now, LHL o () (at any ) g G Similarly, RHL (at any ) o () = () is continuous or all Eample 5 I, 3,, discuss te continuity and dierentiability o (). Solution: Note tat () can equivalently be rewritten as: 3/ 3 5 3/ Mats / Continuity and Dierentiability

44 44 We will now simply draw te grap: We see tat () is non-continuous at = and non-dierentiable at 4 points, =,, 3, Eample 6 Let a unction : be given by ( y) ( ) ( y ) or all, y and ( ) or any. I te unction ( ) is dierentiable at =, sow tat ( ) '() ( ) or all. Solution: Substituting = y = in te given relation, we get () () () {since () or any } It is given tat ( ) is dierentiable at =, i.e., '() ( ) () ( ) Now we write down te epression or '( ) : '( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). '() eists. ( ) '() {rom (i)} '( ) '() ( ) or all...(i) Mats / Continuity and Dierentiability

45 45 Eample 7 I ( y) ( ) ( y ) or all, y > and ( ) is dierentiable at =, ten prove tat ( ) is dierentiable or all > Solution: Substituting y, we get () Since ( ) is dierentiable at, '() eists '() ( ) () ( ) ( ) ( ) Now, '( ) To evaluate '( ), we ave to someow manipulate its epression so tat we are able to use te epression or '() we evaluated in (i) We do tis as ollows: '( ) ( ) ( ) ( )...(i) ( ). Introduction o in te denominator ( ) '() Using (i) Tereore, '( ) as a inite value or all > ( ) is dierentiable everywere. Mats / Continuity and Dierentiability

46 46 Eample 8 I ( ) and g( ) its continuity and dierentiability. Solution: Min{ ( t) : t, } Ma{ ( t) : t, 3} Te grap o () is sketced below (in te relevant domain): draw te grap o g( ) and discuss Now we must understand wat te deinition o g() means. Consider te upper deinition o g(): g( ) Min{ ( t) : t, } To evaluate g( ) at any, we scan te entire interval rom to, (tis is wat te variable t is or), select tat value o (t) wic is minimum in tis interval; tis minimum value o (t) becomes te value o te unction at. Te igure below illustrates tis grapically or our dierent values o (we are considering te interval, te upper deinition o g()): Mats / Continuity and Dierentiability

47 Notice tat were crosses (wen < < ), te minimum value o te unction in te interval [, ] becomes ied at t =. Wen < <, te minimum value o te unction in te interval [, ] is at t =. So, ow do we draw te grap o g ()? In [, ], te grap o g () will te same as () (because te minimum value o () is at t = itsel, as described above). In [, ], te minimum value becomes ied at t =, equal to, so tat in tis interval te grap o g() is constant; g () = or,. Te grap o g () or, is sketced below; 47 For 3, te deinition is g Ma t ; t, 3. Te igure below illustrates ow to obtain g() in tis case or our dierent values o : Mats / Continuity and Dierentiability

48 Notice ten untill lies in te interval [, ], te maimum value o () in te interval [, ] is at t =, equal to. As soon as becomes greater tan, te maimum value o () in te interval [, ] is now at t =. Te grap o g() or te interval [, 3] is sketced below: 48 Te overall grap or g () is tereore: We see tat g () is discontinuous at = and non-dierentiable at =,. Try to draw te ollowing graps on your own: (i) ma ; t, (ii) min ; t, Eample 9 a,, Let and g, b, were a and b are non-negative real numbers. Determine te composite unction g( ()). I g( ()) is continuous or all real, determine te values o a and b. For tese values o a and b, will g( ()) be dierentiable at =? Mats / Continuity and Dierentiability

49 49 Solution: Evaluation o te composition o piecewise deined unctions can be tricky; ence ollow te solution to tis problem careully. g b Te conditions () < and ave to be written in terms o. Notice rom te deinition o () tat () < only wen + a < i.e. a. so, wen a. But also notice tat te deinition o () canges at =. Hence, g( ()) can be rewritten as. g a b a a a a b a b a a a b a b b Tis is te simpliied deinition o g( ()). Reread te wole discussion above careully till you ully understand it. Now our task is easy. We just need to equate LHL and RHL at eac o te critical points = a,, to ind out a and b. Te remaining part o tis question is let as an eercise to you. Te answers are: a = b = For tese values, () is non dierentiable at Mats / Continuity and Dierentiability

50 5 Eample Let () be te unction (deined in Eample -4, Section-3) 6, 4, 4, 3. Plot [ ()] and discuss its continuity. Solution: We drew te grap o tis unction in Fig. 9. We are replicating it in more detail ere: To draw te required grap, notice tat te value o [ ()] will cange every time (te value o ) () crosses an integer. For eample, notice rom te grap te ollowing acts: Wen 4,, Wen,, and so on. Wen or, and can be evaluated by solving , Mats / Continuity and Dierentiability

51 5 For te sake o completeness, te complete deinition o [ ()] is given below : Te grap o [ ()] is: Eample Tis unction is discontinuous at te ollowing points: 4,,,,,, Let. Prove tat a unction : is dierentiable at i and only i tere eists a unction g : wic is continuous at and satisies g or all. Mats / Continuity and Dierentiability

52 5 Solution: Te proo tat we seek is two-way tat is, wen we say tat (P is true) i and only i (Q is true) we mean tat (P is true) implies (Q is true) and (Q is true)implies (P is true). For tis question, we irst assume te eistence o a unction g () were g () satisies Due to tis continuity, g and g() is continuous at. g eists But, g Eample Hence, eists or () is dierentiable at = Te oter way proo is let as an eercise to te reader Evaluate te dierentiability o Solution: We drew te grap o tis unction in te unit on Functions. Tat grap is replicated ere: () is obviously continuous every were but non-dierentiable at all integer points. Let us evaluate te LHD and RHD at eac integer. For tat,we analyse te segment o te curve between any two adjacent integers. Lets pick up te segment between and ; tis segment is part o te segment : Mats / Continuity and Dierentiability

53 53 Te lines L and L correspond to te RHD at = and LHD at = respectively. g g RHD(at = ) (Tis tangent is vertical). LHD (at = ) g g (By rationalization) Tereore LHD at eac integer is and RHD at eac integer is Mats / Continuity and Dierentiability

54 54 EXERCISE Q. Find te value o a and b i is continuous at /, were 6 5 tan6 tan5 b tan a / a / cot Q. Evaluate te continuity o te unction. Q. 3 Find te value o () so tat e ln, is continuous at =. Q. 4 A unction () is deined as. Discuss te continuity o () at =. Q. 5 Discuss te continuity o te unction ln n n sin n at =. Q. 6 Discuss te continuity o n sin in te interval [, ]. n Q. 7 Let m sin,,. Find te set o values o m or wic (i) () is continuous at = (ii) () is dierentiable at = (iii) () is continuous but not dierentiable at =. Mats / Continuity and Dierentiability

55 55 Q. 8 Let 4,,. Discuss te continuity and dierentiability o g () in (, ). Q. 9 Let 3 and ma t : t ; g 3, Discuss te continuity and dierentiability o g () in te interval (, ). Q. Let. Draw te grap o () and analyze its continuity and dierentiability. min t : t, Q. I 3 3 and g 5, 3, 3, draw te grap o g () and discuss its continuity and dierentiability. Q. Let g () be a polynomial o degree one and () be deined as: g, /,. Find g () suc tat () is continuous and '() = ( ). Q. 3 Eamine te continuous unction, 3 a b, or dierentiability. Q. 4 Consider te unction min,,, Plot te grap or and discuss its continuity and dierentiability. Q. 5 I a unction : a, a is an odd unction suc tat a or a,a and te let and derivative at = a is, ten ind te let and derivative at = a. Mats / Continuity and Dierentiability