Solutons to Homework 7, Mathematcs 1 Problem 1: a Prove that arccos 1 1 for 1, 1. b* Startng from the defnton of the dervatve, prove that arccos + 1, arccos 1. Hnt: For arccos arccos π + 1, the defnton leads to computng lm 1+. Use substtuton arccos π and pa attenton to from what sde approaches 0 ou can use the +1 correspondng substtuton theorem, even f we dd not formulate t. Soluton From Dr. Katarna Bellová : a Let arccos, 0, π. B the theorem for the dervatve of the nverse, snce cos sn 0 for 0, π, we get arccos 1 cos 1 sn 1 1 cos 1 1. b As n the hnt, wth arccos π 0 as 1 +, arccos + 1 arccos arccos 1 lm 1 + 1 arccos π lm 1 + + 1 lm 0 lm 0 lm 0. In a smlar wa, wth arccos 0 + as 1, cos + π + 1 cos + 1 1 cos 1 arccos arccos arccos1 1 lm 1 1 arccos lm 1 1 lm 0 + cos 1 lm 0 + 1 cos 1.
Problem : Fnd the defnton doman of the followng functons and compute ther dervatve at the ponts where t ests: a b [ ponts] f lncos, f cosln, c f f, d [3 ponts] f 1 + sn. Hnt: Epress an f g as e lnf g. Soluton: a The functon ln s defned for > 0. Therefore, we need cos > 0 for the functon to be defned, whch happens on the unon of ntervals kπ π, kπ + π, k Z whch s the mamal doman of the functon. Its dervatve can then be computed usng the chan rule lncos 1 cos cos sn tan, cos whch s defned for ever n the doman of f. b The functon s defned for ever > 0. Its dervatve s gven b cosln snln ln snln. cfrom Dr. Katarna Bellová Defnton doman: The general eponental functon α, f α / Z, s defned for > 0. Hence makes sense for ether Z, 0 optonall, 0 0 s sometmes defned as well, as 0 0 1, or R +. In order to dfferentate, the functon must be defned n a neghborhood around the pont. Hence, t onl makes snce to speak of f onl for > 0. Then f e ln e ln ln 1 ln + 1 ln + 1. dths functon s equvalent to n ths homework too I would lke to use the notaton e ep ep ln1 + sn. Snce s alwas postve, there s no ssue here wth the doman of the logarthmc functon, and therefore the functon s defned for all R. Let us dfferentate then. Just applng the chan rule ever da and a bt of the product rule, so please learn t well gus! ep ln1 + sn ep ln1 + sn ln1 + sn and t s also defned for all R. ep ln1 + sn ln1 + sn + ln1 + sn ep ln1 + sn 1 + 1 1 + sn + ln1 + cos 1 + sn 1 + sn + ln1 + cos, Problem 3: Let f arctan +1 1. a [3 ponts] Fnd the defnton doman of f and the dervatve f at ponts where t ests. Smplf.
b [ bonus ponts] What other functon has the same dervatve as f? Usng Corollar..14, fnd a smplfed epresson for f. Be carefull that Corollar..14 can be appled onl on ntervals! Soluton: a: We know that f s defned everwhere, ecept for 1. We can even see that the lmts from the rght and left are dfferent + 1 lm arctan π + 1 and lm 1+ 1 arctan π 1 1, though ths s not our focus rght now. Frst we use the result about dervatves of nverse functons to see that arctan 1 1 +. Then we agan appl a bunch of dfferentaton rules n order to derve f: arctan + 1 1 1 + 1 1 + +1 1 1 1 + 1 + + + 1 1 1 +. 1 + 1 1 b: Now, werdl, the functon above has the same dervatve as the functon arctan. So, t should be as a consequence of the mean value theorem equal to arctan + C for some constant C > 0, snce f arctan f + 1 1 + 0 and n an nterval [a, b] where both f and arctan are defned we have fb + arctan b fa + arctan a fc arctan c a b 0. But ths seems a lttle werd rght? In fact, sketchng a bt of the graph of the above functon we see that t s ndeed so, but we have to take nto account the fact that f s not defned n an nterval. In fact the doman of f conssts on the unon of two ntervals:, 1 1,. In each of these ntervals, f s ndeed equal to arctan plus some constant: Seeng that and that π 4 we see that so that we have f arctan + 1 1 f0 arctan { arctan + C 1 for < 1, arctan + C for > 1. 0 + 1 arctan 1 π 0 1 4, + 1 arctan 1 lm arctan lm + 1 arctan + C π + + C, f arctan C 1 π 4 and C 3π 4, { + 1 arctan π 4 1 arctan + 3π 4 for < 1, for > 1. 3
4 Problem 4 : Compute the frst and the second dervatves of the followng functon for each R. Does the thrd dervatve of f est at 0? Justf our answer. { 4 sn 1 f 0, f 0 f 0. Soluton B Dr. Katarna Bellová: 0 f 4 3 sn 1 + 4 cos 1 f 0 lm 0 4 sn 1 0 0 lm 0 3 sn 1 0 1 4 3 sn 1 cos 1, b the sandwch theorem as 3 3 sn 1 3. Then, n a smlar wa, 0 f 1 sn 1 + 43 cos 1 1 sn 1 6 cos 1 sn 1, f 0 lm 0 f f 0 0 1 cos 1 + sn 1 lm 0 4 sn 1 cos 1 0 1 agan b the sandwch theorem, as 4 4 sn 1 cos 1 4 +. Notce that f s not contnuous at 0, so f s not three tmes dfferentable at 0. Problem 5: a [3 ponts] Compute e 100. b : Fnd general formula for the n-th dervatve of the functon f ln, wherever t ests. Epress wth no dervatves and smplf as much as possble. Soluton: a: Ths problem can be solved b drect applcaton of the Lebnz formula and usng that n 0 for an n 3, e n 1 n e the latter easl followng b nducton. Wthout usng Lebnz formula, one can also prove a specfc verson of t, as n the soluton below: Assume we have some functon of that can be wrtten as fe for some f : R R that can be dfferentated as much as we want to, lke on the statement of ths eercse wth f. We have 1 fe f e fe e f f. We can now appl the above formula substtutng f bf f! We obtan f fe e f f f f e f f + f. See where ths s gong et? If no, we can now appl the same formula n 1 substtutng f b f f + f to obtan f f + fe e f f + f f f + f e f 3f + 3f f. There s a pattern here! We wll prove b nducton that n fe n e 1 n f. 0
We proved not onl the case for n 1, but for and 3 also. So let us assume that t s the case for n and prove that t also works for n + 1. n fe n+1 e 1 n f 0 n e 1 n f 0 0 n e 1 n +1 f + e 1 n f +1. Now we reorganze the sums above and use the facts that n + 1 n n n n + 1 +, 1 + 1 + 1 n n + 1 to see that fe n+1 0 0 n 0 n + 1 n e 1 n +1 f + e 1 n f +1 n n e 1 n+1 f + e 1 n+1 +1 f +1 0 n n+1 n e 1 n+1 f + e 1 n+1 k f k 0 n e 1 n+1 f 0 + 0 1 n + e 1 n+1 n+1 f n+1 n n+1 n + 1 0 k1 n e 1 n+1 f. k 1 + n 1 0, e 1 n+1 f Ths s a specfc eample of a more general wa to calculate the n-th dervatve of a product of two functons. Mabe ou have seen t, but anwa, ou can tr to guess how the general formula works and prove t n the same manner as above. The general formula has the name of one of the most famous ctzens of Lepzg, and a gu who faled to get a doctorate from ths same Unverst of Lepzg n the ear 1666 because the consdered hm too oung! One of the fathers of calculus and nventor of some trul nce notaton, Gottfred Wlhelm Lebnz. Ths whole calculaton above was mabe too nvolved and unnecessar, one could just dfferentate the functon man tmes and see a smple pattern emerge, but I thought mabe t could be nce to go about ths eercse wth a more general perspectve. Anwa, we see now that 100 100 e 100 e 1 100 0 e +, snce the dervatves of order hgher than 3 vansh for the functon. and bb Dr. Katarna Bellová: B nducton, 1 n 1 n n! n 1 for an n 0 ln n 1 n 1 n 1! n for an n 1. 5
6 Hence, b Lebnz formula n ln n k k0 n 0 ln k 1 n k ln 1 n n! n 1 + k1 1 n n! n 1 ln + 1n+1 n! n+1 n 1 k 1 k 1! k 1 n k n k! n+k 1 k 1 k. k1 Problem 6: Identf the ntervals on whch the followng functons are monotone ncreasng: a f +, b f sn, c [3 ponts] f. 1 As alwas, start wth the defnton doman of f! Soluton: The functons n tems a and b are defned for all real numbers, and the functon n tem c s defned onl for > 1. That beng sad, let us fnd some ntervals where these functons are monotone ncreasng. Snce the are all dfferentable n ther domans, ths s equvalent to fnd ntervals where ther respectve dervatve s nonnegatve. a f s monotone ncreasng when + + 1 0, therefore, f s monotone ncreasng f [ 1/,. There s another wa to see ths f ou know that the graph of f descrbes a parabola, a famous formula n hgh school... but ths s the easer, cleanest wa to see that the above statement s true n m opnon. b We must have sn cos 0, for f to be monotone ncreasng, and therefore [ kπ π, kπ + π ]. k Z c Agan, lookng for the nterval where the dervatve of f s nonnegatve, s equvalent of searchng for such that 1 1 1 1 1 1 3 1 3 0. Wthn the doman 1,, the above nequalt s satsfed for, and therefore f s monotone ncreasng on the nterval [,.
Problem 7 [ 3 ponts]: Fnd all local mama and mnma and decde whch of the two each s of the followng functons: Soluton a f 7 7 5, Hnt: t mght be computatonall easer to avod computng the second dervatve n the crtcal pont. b f cos. a We have that f s defned for ever R. Furthermore f 7 6 35 4 and f 4 5 140 3. Snce f s a polnomal wth odd degree, ts global mamum and mnmum are never reached b the functon or ou can nterpret that the occur on + and respectvel, but for now don t thnk lke that. Anwa, the queston s about local mama and mnma. Those can occur when f 0, whch means 7 6 35 4 7 4 5 0. We have as crtcal ponts for f the numbers 0, 5 and 5. We could compute f at these ponts to see the concavt of the graph of f and therefore f the have a mamum or mnmum at these ponts. But there are easer was of dong that. For eample, we can see that near 0 the functon f s alwas negatve, whch means that f s decreasng on a neghborhood of 0 and 0 s nether a mnmum nor a mamum pont. On the other hand, as goes through 5, f goes from postve to negatve and as goes through 5, f goes from negatve to postve. Ths means that f was ncreasng before 5 and decreasng afterwards, and that t was decreasng before 5 and ncreasng afterwards. Ths means that 5 s a local mamum and that 5 s a local mnmum. The above analss s made b lookng at the graph of the functon 7 4 5 and askng where s t negatve, where s t postve and where s t 0. Ths factored form s useful for ths purpose because we know that 7 4 s alwas nonnegatve and that 5 s postve before 5, negatve between 5 and 5, and postve after 5. b We have cos cos sn sn, and cos sn cos. We have that cos sn 0 f kπ/, for k Z. On the other hand, we know that cos kπ {, for k even, coskπ, for k odd. We therefore know that kπ/ s a local mamum for even k and a local mnmum for odd k. The are also global mama and mnma, because of the perodct of the cosne functon! Problem 8: Fnd the global mamum and mnmum of the followng functons on the specfed ntervals: a f 3 3, [ 1, 4]; b [3 ponts] f cos + sn, [ π, π ]. Soluton We are talkng about dfferentable, n partcular contnuous functons defned on a closed and bounded nterval, so the mnma and mama est. We have to check the crtcal ponts and the boundares of the domans - mnma and mama ponts must be among them, 7
8 so the ponts wth largest values wll be ponts of mama and ponts of smallest values wll be ponts of mnma. a f 3 3, [ 1, 4], f 3 6 3, We have 0 and as crtcal ponts. Furthermore, we have f0 0, f 4, f 1 4, f4 16. Therefore the mnmum value 4 s attaned at 1 and and the mamum value 16 s attaned at 4. b f cos + sn, [ π, π ], f sn + sn cos sn cos 1. We have as crtcal ponts wth π 3, 0, π 3, f0 1 f π π f 1 3 3 + 3 4 5 4, f π π f 1. Therefore we have the global mnmum 1 attaned at the ponts π, 0, π and the global mamum attaned at the ponts ±π/3. 5 4 Problem 9: Let a R and 1, 0. Use Lagrange s Mean Value Theorem to prove the followng nequaltes: a f a < 0 or a > 1, then 1 + a > 1 + a; b f 0 < a < 1, then 1 + a < 1 + a. Hnt: Appl Lagrange s Mean Value Theorem to the functon fz 1 + z a on the nterval [0, ] f > 0 or [, 0] f < 0. Soluton: Let us dfferentate and appl ths ver neat theorem, that most mathematcans I know appl on a regular bass I mean, actuall we use the Talor polnomal epanson usuall, where what we do here corresponds to the frst term, that s, we are appromatng the functon fz 1 + z a usng a lnear functon, whch n a sense s the whole essence of dfferentaton n all the contets I know. So, we have fz 1 + z a f z a1 + z a 1.
Therefore, on the nterval [0, ] or n the nterval [, 0] f < 0 there must est some 0 such that f f0 f 0 0 1 + a 1 a1 + 0 a 1 1 + a 1 + a1 + 0 a 1 And everthng follows b studng the sgn of for the possble dfferent values of a and. 1 + a 1 + a + a1 + 0 a 1 1 a1 + 0 a 1 1 9