CALCULUS STUDY MATERIAL. B.Sc. MATHEMATICS III SEMESTER UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION

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1 CALCULUS STUDY MATERIAL BS MATHEMATICS III SEMESTER UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY PO MALAPPURAM, KERALA, INDIA

2 UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION STUDY MATERIAL BS MATHEMATICS III SEMESTER CALCULUS Lessons prepred y: SriNndkumr M, Assistnt Professor Dept of Mthemtis, NAM College, Kllikkndy Ly out & Settings Computer Setion, SDE Copyright Reserved Clulus Pge

3 CALCULUS CONTENTS Module I Foundtions of Clulus 5 The Rolle s nd Men Vlue Theorems 6 Module II Asymptotes Optimiztions 5 Integrtion Prt One 5 Module III 6 Integrtion Prt Two 65 Module IV 7 Integrtion Prt Three 77 8 Moments nd Work 85 Clulus Pge

4 Clulus Pge

5 MODULE - I CHAPTER FOUNDATIONS OF CALCULUS Quik Review of Funtions In this hpter we first disuss funtion, one of the most importnt onepts in mthemtis Definition Let A nd B e two non-empty sets A funtion (mp, mpping or trnsformtion) from A to B is rule whih ssigns to eh element of A unique element of B The set A is lled the domin of the funtion, nd the set B is lled the trget set Nottions/Definitions/Remrks Funtions re ordinrily denoted y symols Let f denote funtion from A into B Then we write f : Afi B whih is red: f is funtion from A into B, or f tkes A into B, or f mps A into B Suppose f : Afi B nd A Then f ( ) [whih is red f of ] will denote the unique element of B whih f ssigns to This element f ( ) in B is lled the imge of under f or the vlue of f t We lso sy tht f sends or mps into f ( ) The set of ll suh imge vlues is lled the rnge or imge of f, nd it is denoted y Rn (f ), Im (f) or f ( A ) Tht is Im( f ) is suset of the trget set B Im( f ) { B: there eists A for whih f ( ) } Frequently, funtion n e epressed y mens of mthemtil formul: Consider the funtion whih sends eh rel numer into its squre We my desrie this funtion y writing f ( ) or or y In the first nottion, is lled vrile nd the letter f denotes the funtion In the seond nottion, the rred rrow is red goes into In the lst nottion, is lled the independent vrile nd y is lled the dependent vrile sine the vlue of y will depend on the vlue of Furthermore, suppose funtion is given y formul in terms of vrile Then we ssume, unless otherwise stted, tht the domin of the funtion is or the lrgest suset of for whih the formul hs mening nd tht the trget set is Suppose f : Afi B If A ' is suset of A, then f ( A ') denotes the set of imges of - elements in A ' nd if B ' is suset of B, then f ( B') denotes the set of elements of A eh whose imge elongs to B ' Tht is, We ll f ( A ') the imge of Emple f A f A f B A f B - ( ') { ( ): '} nd ( ') { : ( ) '} Consider the funtion A ', nd ll f ( ) - f ( B') the inverse imge or preimge of B ', ie, f ssigns to eh rel numer its ue Then the imge of is 8, nd so we my write () 8 f Similrly, ( ) 7 f - -, nd () f Identity Funtion Clulus Pge 5

6 Consider ny set A Then there is funtion from A into A whih sends eh element into itself It is lled the identity funtion on A nd it is usully denoted y A or I A or simply In other words, the identity funtion A : A fi A is defined y for every element A Composition of Funtions () A Consider funtions f : Afi B nd g : Bfi C, tht is, where the trget set B of f is the domin of g This reltionship n e pitured y the following digrm: f g A fi B fi C Let A; then its imge f ( ) under f is in B whih is the domin of g Aordingly, we n find the imge of ( ) g f( ) Thus we hve f under the funtion g, tht is, we n find ( ) rule whih ssigns to eh element in A n element g ( f( )) in C or, in other words, f nd g gives rise to well defined funtion from A to C The new funtion is lled the omposition of f nd g, nd it is denoted y g o f More riefly, if f : Afi B nd g : Bfi C, then we define new funtion g o f : Afi C y ( g o f)( ) g( f( )) Here is used to men equl y definition Note tht we n now dd the funtion g o f to the ove digrm of f nd g s follows: ONE TO ONE, ONTO AND INVERTIBLE FUNCTIONS Definition A funtion f : Afi B is sid to e one-to-one (written -) if f ( ) f( ') implies ' Tht is, f : Afi B is - if different elements in the domin A hve distint imges Definition A funtion f : Afi B is sid to e n onto funtion if " B, $ A f ( ) Tht is, f : Afi B is onto if every element of B is the imge of some elements in A or, in other words, if the imge of f is the entire trget set B In suh se we sy tht f is funtion of A onto B or tht f mps A onto B Remrks The term injetive is used for for one-to-one funtion, surjetive for n onto funtion, nd ijetive for one-to-one orrespondene If f : Afi B is oth one-to-one nd onto, then f is lled one-to-one orrespondene etween A nd B This terminology omes from the ft tht eh element of A will orrespond to unique element of B nd vie vers LIMITS OF FUNCTIONS Nottion lim f ( ) L denotes f ( ) pprohes the limit L s pprohes A f B go f g C Clulus Pge 6

7 Limits of Powers nd Algeri Comintions Theorem : Properties of Limits The following rules hold if lim f ( ) L fi nd lim g ( ) M fi (L nd M rel numers) Sum Rule: lim[ f ( ) + g( )] L+ M ie, the limit of the sum of two funtions is the sum of their limits Differene Rule: lim[ f ( ) - g( )] L- M ie, the limit of the differene of two funtions is the differene of their limits Produt Rule: lim f ( ) g( ) ] L M ie, the limit of the produt of two funtions is the produt of their limits Constnt Multiple Rule: lim kf ( ) kl (ny numer k) ie, the limit of onstnt times funtion is tht onstnt times the limit of the funtion f( ) L 5 Quotient Rule: lim, M fi g ( ) M ie, the limit of the quotient of two funtions is the quotient of their limits, provided the limit of the denomintor is not zero 6 Power Rule: If m nd n re integers, then lim[ ( )] m m m n n n f L, provided L is rel numer ie, the limit of ny rtionl power of funtion is tht power of the limit of the funtion, provided the ltter is rel numer + - Emple Find lim + 5 We note tht lim fi nd lim fi k k, where k is onstnt (we will prove this in Emple of Chpter ) Now we use vrious prts of Theorem for the evlution of the required limit lim( + -) + - fi lim fi, using quotient rule + 5 lim( + 5) lim + lim 5 lim + lim( ) + lim( -), using sum rule lim + lim 5 lim + lim + lim( -), using rule + -, sine + 5 lim ( lim )( lim ) nd ( ) lim lim, using produt rule, using power rule Clulus Pge 7

8 Emple Evlute lim - lim ( ) / ( ) lim - lim - lim ( ) lim lim ( -) / - lim -, using power rule with n fi ( -) Theorem : Limits of Polynomils y sustitution n n- If P( ) + + L+, then n n- n- n- L lim P( ) P( ) n n Theorem : Limits of Rtionl Funtions y sustitution (if the limit of the denomintor is not zero) If P( ) nd Q ( ) re polynomils nd Q (), then P( ) P( ) lim fi Q ( ) Q ( ) Emple + - () + () - lim () This is the limit in Emple with -, now done in one step Identifying ommon ftors It n e shown tht if ( ) Q is polynomil nd Q (), then ( -) is ftor of Q ( ) Thus, if the numertor nd denomintor of rtionl funtion of re oth zero t, then ( -) is ommon ftor Eliminting Zero Denomintors Algerilly Theorem pplies only when the denomintor of the rtionl funtion is not zero t the limit point If the denomintor is zero, neling ommon ftors in the numertor nd denomintor will sometimes redue the frtion to one whose denomintor is no longer zero t When this hppens, we n find the limit y sustitution in the simplified frtion + - Emple Evlute lim - We nnot just sustitute, euse it mkes the denomintor zero However, we n ftor the numertor nd denomintor nd nel the ommon ftor to otin + - ( - )( + ) +, if - ( -) Thus, y pplying Theorem, we otin Clulus Pge 8

9 lim lim - + h - Emple Find lim hfi h We nnot find the limit y sustituting h, nd the numertor nd denomintor do not hve ovious ftors However, we n rete ommon ftor in the numertor y multiplying it (nd the denomintor) y the so-lled onjugte epression + h +, otined y hnging the sign etween the squre roots: + h - + h - + h + h h + h + + h - h( + h + ) h, with ommon ftor of h h( + h + ), nelling the ftor h + h + Therefore, we otin + h - lim lim (The denomintor is no longer t h, so we n hfi h hfi + h + use Theorem nd n sustitute h ) + + Theorem : The sndwih Theorem Suppose tht g( ) f( ) h( ) for ll in some open intervl ontining, eept possily t itself Suppose lso tht lim g( ) lim h( ) L Then lim f ( ) L Emple Given tht - u ( ) + for ll Find lim u ( ) Sine lim - - fi implies tht lim u ( ) nd Emple Show tht if lim f( ), then lim f( ) lim + + fi, the Sndwih Theorem By ssumption lim f( ), so tht - f ( ) nd f ( ) oth hve limit s pprohes This omining with the ft - f ( ) f( ) f( ) gives lim f( ) y the Sndwih Theorem Eerises Find the limits in Eerises -8 lim( - ) lim ( ) - Clulus Pge 9

10 lim s(s- ) / y + 5 lim 6 yfi y + 5y+ 6 7 lim(z 8) zfi / - 8 lim lim ( + ) 5 98 lim hfi 5 h + + lim y + 8y lim yfi 6 y - y lim Suppose lim f( ) nd lim g ( ) - Find ) lim( g( ) + ) ) lim f( ) g( ) ) lim( g( )) d) lim fi f ( ) - 7 Suppose tht lim p ( ), lim r ( ), nd lim s ( ) 7 Find Clulus Pge ) lim( p ( ) + r ( ) + s ( )) ) lim p ( ) r ( ) s ( ) - Find the limits in Eerises lim t + t+ lim tfi- t -t- v -8 lim vfi v lim (- p( ) + 5 r( )) ) lim - s ( ) Evlute the limit for the given vlue of nd funtion f 8 f ( ), - 9 f( ), - f ( ) +, If - g ( ) osfor ll, find lim g( ) -os The inequlities - < < hold for ll vlues of lose to - os zero Wht if nything, does this tell you out lim? fi Suppose tht g( ) f( ) h( ) for ll nd suppose tht lim g( ) lim h( ) - 5 Cn we onlude nything out the vlues of f, g, nd h t? Could f ()? Could lim f( )? Give resons for your nswers f( ) f ( ) If lim, find () lim f ( ) nd () lim f( ) f ( ) 5 If lim, find () lim f ( ) nd () lim fi

11 FORMAL DEFINITION OF LIMIT Definition (A forml Definition of Limit) Let f ( ) e defined on n open intervl out, eept possily t itself We sy tht lim f ( ) L, if, for every numer e >, there eists orresponding numer d > suh tht for ll with < - < d f( ) - L < e Emple Using the Definition, show tht lim (5 - ) Set, f ( ) 5-, nd L in the definition of limit For ny given e > we hve to find suitle d > so tht if nd is within distne d of, tht is, if < - < d, then f ( ) is within distne e of L, tht is f( )- < e We find d y working kwrds from the e -inequlity f( ) - L (5-) < e 5 - < e e - < 5 e e Thus we n tke d If < - < d, then 5 5 e (5-) < 5 e 5 This proves tht lim (5 - ) e Remrk In the ove Emple, the vlue of d is not the only vlue tht will mke 5 < - < d imply 5-5 < e Any smller positive d will do s well The definition does not sk for est positive d, just one tht will work Clulus Pge

12 Emple Verify the following two importnt limits: () lim () lim k k (k onstnt) ) Let e > e given We must find d > suh tht for ll < - < d implies - < e The implition will hold if d e or ny smller positive numer This proves tht lim fi ) Let e > e given We must find d > suh tht for ll, < - < d implies k - k < e Sine k - k, we n use ny positive numer for d nd the implition will hold This proves tht lim - k k Finding Delts Algerilly for Given Epsilons In the ove emples, the intervl of vlues out for whih f ( ) - L ws less thn e ws symmetri out nd we ould tke d to e hlf the length of the intervl When suh symmetry is sent, s it usully is, we n tke d to e the distne from to the intervl s nerer endpoint Clulus Pge

13 Emple For the limit lim 5 -, find d > tht works for e We hve to find d > suh tht for ll < - 5 < d -- < We orgnize the serh into two steps First we solve the inequlity -- < to find n intervl (, ) out 5 on whih the inequlity holds for ll Then we find vlue of d > tht ples the intervl 5- d < < 5+ d (entered t 5) inside the intervl (, ) Step : We solve the inequlity -- < to find n intervl out 5 on whih the inequlity holds for ll -- < - < -- < < - < < - < 9 < < The inequlity holds for ll in the open intervl (, ), so it holds for ll 5 in this intervl s well Step : We find vlue of d > tht ples the entered intervl 5- d < < 5+ dinside the intervl (, ) The distne from 5 to the nerer endpoint of (, ) is If we tke d or ny smller positive numer, then the inequlity < - 5 < d will utomtilly ple etween nd to mke -- < < - 5 < -- < Emple Prove tht lim f( ) if, f( ), Our tsk is to show tht given e > there eists d > suh tht for ll < - < d f( )- < e Step : We solve the inequlity f( )- < e to find on open intervl out on whih the inequlity holds for ll For, we hve f ( ), nd the inequlity to solve is - < e : - < e - e< - < e - e< < + e - e< < + e here we ssume tht e < - e< < + e this is n open intervl out tht solves the inequlity Clulus Pge

14 The inequlity f( )- < e holds for ll in the open intervl ( - e, + e) Step : We find vlue of d > tht ples the entered intervl ( - d, + d) inside the intervl ( - e, + e) : to the nerer endpoint of ( e, e) Tke d to e the distne from words, tke d min - - e, + e-, { } - + In other the minimum (the smller) of the two numers - - e nd + e - If d hs this or ny smller positive vlue, the inequlity < - < d will utomtilly ple etween - e nd + e to mke f( )- < e For ll, < - < d f( )- < e This ompletes the proof We n ssume e < due to the following reson: In finding d suh tht for ll, < - < d implied f( )- < e <, we found d tht would work for ny lrger e s well Finlly, notie the freedom we gined in letting d min{ - - e, + e- } We did not hve to spend time deiding whih, if either, numer ws the smller of the two We just let d represent the smller nd went on to finish the rgument Emple Given tht lim f ( ) L fi nd lim g ( ) M fi, prove tht lim( f ( ) + g( )) L+ M Let e > e given We wnt to find positive numer d suh tht for ll < - < d f( ) + g( ) -( L+ M ) < e Regrouping terms, we get f ( ) + g( ) -( L+ M) ( f( ) - L) + ( g( ) - M) f ( ) - L + g( ) - M, using the tringle inequlity + + Sine lim f ( ) L, there eists numer d > suh tht for ll e < - < d f( ) - L < Similrly, sine lim g ( ) M fi, there eists numer d > suh tht for ll e < - < d g ( )- M < Let d min{ d, d}, the smller of d nd d If < - < d then - < d, so e e f( ) - L <, nd - < d, so g ( )- M < Therefore f( ) + g( ) -( L+ M) < f( ) - L + g( ) - M < e + e e lim f ( ) + g( ) L+ M This shows tht ( ) Clulus Pge

15 Eerises Eh of Eerises -8 gives funtion f ( ) nd numers L,, nd e > In eh se, find n open intervl out on whih the inequlity f( ) - L < e holds Then give vlue for d > suh tht for ll stisfying < - < d the inequlity f( ) - L < e holds f( ) -, L - 6, -, e f ( ), L,, e f( ) - 7, L,, e f ( ), L,, e 5 f( ), L -, -, e 6 f( ), L 5,, e 7 f( ) m, m> L m,, e > 8 f( ) m+, m> L m+,, e 5 Eh of Eerises 9- gives funtion f ( ), point, nd positive numer e Find L lim f( ) Then find numer d > suh tht for ll < - < d f( ) - L < e 9 f( ) --, -, e f( ), - 5, e f( ),, e Prove the limit sttements in Eerises -8 lim( - 7) lim -, - lim f( ) if f( ) -, lim 6 lim lim f ( ) if lim sin, < Let h ( ),, > Show tht ) lim h ( ) ), < f( ), lim h ( ) ) lim h ( ) Clulus Pge 5

16 EXTENSIONS OF LIMIT CONCEPTS One-Sided Limits Definition (Informl Definition of Right-hnd nd Left-hnd Limits) Let f ( ) e defined on n intervl (, ) where < If f ( ) pprohes ritrrily lose to L s pprohes from within tht intervl, then we sy tht f hs right-hnd limit L t, nd we write lim f ( ) L + Let f ( ) e defined on n intervl (, ) where < If f ( ) pprohes ritrrily lose to M s pprohes, from within the intervl (, ), then we sy tht f hs left-hnd limit M t, nd we write lim f ( ) M - Emple For the f( ) in Fig6, we hve lim f( ) nd + lim f( ) - - funtion A funtion nnot hve n ordinry limit t n endpoint of its domin, ut it n hve one-sided limit (This is illustrted in the following Emple) y - Fig 7 Clulus Pge 6

17 Emple The domin of hve lim + - nd f ( ) lim is [, ] - ; its grph is the semiirle in Fig 7 We The funtion does not hve left-hnd limit t - or right-hnd limit t It does not hve ordinry two-sided limits t either - or One-sided limits hve ll the limit properties listed in Theorem, Chpter The righthnd limit of the sum of two funtions is the sum of their right-hnd limits, nd so on The theorems for limits of polynomils nd rtionl funtions hold with one-sided limits, s does the Sndwih Theorem The onnetion etween one-sided nd two-sided limits is stted in the following theorem Theorem 5 (One-sided vs Two-sided Limits) A funtion f ( ) hs limit s pprohes if nd only if it hs left-hnd nd right-hnd limits there, nd these one-sided limits re equl: lim f ( ) L lim f ( ) L nd lim f ( ) L - + Infinite Limits + As fi, the vlues of f grow without ound, eventully rehing nd surpssing every positive rel numer Tht is, given ny positive rel numer B, however lrge, the vlues + of f eome lrger still (Fig) Thus f hs no limit s fi It is nevertheless + onvenient to desrie the ehvior of f y sying tht f ( ) pprohes s fi We write lim f( ) lim + + In writing this, we re not sying tht the limit eists Nor re we sying tht there is rel numer, for there is no suh numer Rther, we re sying tht lim + fi does not eist euse eomes ritrrily lrge nd positive s + fi Clulus Pge 7

18 - As fi, the vlues of f( ) eome ritrrily lrge nd negtive Given ny negtive rel numer - B, the vlues of f eventully lie elow - B (See Fig ) We write lim f( ) lim Agin, we re not sying tht the limit eists nd equls the numer - There is no rel - numer - We re desriing the ehvior of funtion whose limit s fi does not eist euse its vlues eome ritrrily lrge nd negtive Emple (One-sides infinite limits) Find lim+ - nd lim- fi - Geometri : The grph of y - is the grph of y shifted unit to the right Therefore, y - ehves ner etly the wy y ehves ner : lim nd lim Anlyti : Think out the numer - nd its reiprol As fi, we hve ( -) fi nd fi As fi, we hve ( -) fi nd fi Emple (Two-sided infinite limits) Disuss the ehvior of ) f( ) ner, ) g ( ) ner - ( + ) ) As pprohes zero from either side, the vlues of re positive nd eome ritrrily lrge (Fig) Hene lim f( ) lim ) The grph of g ( ) is the grph of f( ) shifted units of the left ( + ) Therefore, g ehves ner - etly the wy f ehves ner Hene lim g ( ) lim - - ( ) + The funtion y shows no onsistent ehvior s fi We hve + fi if fi, ut - fi - if fi All we n sy out lim fi is tht it does not eist The funtion Clulus Pge 8

19 y is different Its vlues pproh infinity s pprohes zero from either side, so we n sy tht lim fi Emple (Rtionl funtions n ehve in vrious wys ner zeros of their denomintors) ) ( -) ( -) - lim lim lim - ( - )( + ) + ) - - lim lim lim - ( - )( + ) + ) - - lim lim -, s the vlues re negtive for >, ner - ( - )( + ) d) - - lim lim, s the vlues re positive for <, ner - ( - )( + ) e) - - lim lim does not eist, s, y () nd (d), the left hnd nd right hnd - ( - )( + ) limits t re not equl f) - -( -) - lim lim lim ( -) ( -) ( -) - In prts () nd () the effet of the zero in the denomintor t is neled euse the numertor is zero there lso Thus finite limit eists This is not true in prt (f), where nelltion still leves zero in the denomintor Preise Definitions of One-sides Limits The forml definition of two-sided limit in Chpter is redily modified for one-sided limits Definition (Right-hnd Limit) We sy tht f ( ) hs right-hnd limit L t, nd write lim f ( ) L + if for every numer e > there eists orresponding numer d > suh tht for ll < < + d f( ) - L < e () Definition (Left-hnd Limit) We sy tht f hs left-hnd limit L t, nd write lim f ( ) L if for every numer e > there eists orresponding numerd > suh tht for ll - d < < f( ) - L < e () Definitions (Infinite Limits) We sy tht f ( ) pprohes infinity s pprohes, nd write lim f( ), if for every positive rel numer B there eists orresponding d > suh tht for ll < - < d f ( ) > B Clulus Pge 9

20 We sy tht f ( ) pprohes minus infinity s pprohes, nd write lim f( ) -, if for every negtive rel numer - B there eists orresponding d > suh tht for ll < - < d f ( ) <- B Eerises Whih of the following sttements out the funtion y f( ) grphed here re true, nd whih re flse? ) lim f( ) ) ) d) e) g) lim f ( ) does not eist lim f( ) lim f( ) - lim f( ) f) + lim f ( ) lim f( ) + - lim f ( ) does not eist h) lim f ( ) eists t every in the open intervl (-,) i) lim f ( ) eists t every in the open intervl (, ) j) lim f( ) k) lim f ( ) does not eist Find the limits in Eerises -6 - lim lim- hfi h ( -) 5 ) lim + fi - lim t- t 6 5h h lim ) ( -) lim - fi - 6 ) ( ) ) lim ( t- t ) + tfi - tfi Find the limits in Eerises lim 8 lim 9 lim lim - fi ( + ) ) lim ) lim + lim fi Find the limits in Eerises - lim se lim( - ot q) (- p ) + qfi Find the limits in Eerises lim - s ) fi ) fi ) fi - d) fi - y - o y f() Clulus Pge -

21 6 lim - + s ) fi - ) fi - ) fi d) fi lim s ) fi ) fi - ) fi d) fi e) Wht, if nything, n e sid out the limit s fi? Find the limits in Eerises lim + 7 s ) 5 t t + - fi ) t fi 9 lim - s ( -) ) fi ) fi ) fi d) fi Given e >, find n intervl I ( - d, ), d >, suh tht if lies in I, then - < e Wht limit is eing verified nd wht is its vlue? - Use the definitions of right-hnd nd left-hnd limits to prove tht lim + - sin, Let f( ) < Find () lim + f ( ) nd () lim - f ( ) ; then use limit, > definitions to verify your findings () Bsed on your onlusions in () nd (), n nything e sid out lim f ( )? Give resons for your nswer Use forml definitions to prove the limit sttements in Eerises - - lim fi - lim -5 ( + 5) Use forml definitions from Eerise 5 in Set A to prove the limit sttements in the following Eerises lim 6 lim - 7 lim CONTINUITY Continuity t point In prtie, most funtions of rel vrile hve domins tht re intervls or unions of seprte intervls, nd it is nturl to restrit our study of ontinuity to funtions with these domins This leves us with only three kinds of points to onsider: interior points (points Clulus Pge

22 tht lie in n open intervl in the domin), left endpoints, nd right endpoints Definition A funtion f is ontinuous t n interior point of its domin if lim f ( ) f ( ) The funtion in Fig is ontinuous t The funtion in Fig would e ontinuous if it hd f () The funtion in Fig would e ontinuous if f () were insted of The disontinuities in Fig nd Fig re removle Eh funtion hs limit s fi, nd we n remove the disontinuity y setting f () equl to this limit The funtion f( ) hs n infinite disontinuity Jumps nd infinite disontinuities re the ones most frequently enountered, ut there re others Definition A funtion f is ontinuous t left endpoint of its domin if lim f ( ) f( ) nd ontinuous t right endpoint of its domin if lim f ( ) f( ) + In generl, funtion f is right-ontinuous (ontinuous from the right) t point in its domin if lim + f ( ) f( ) It is left-ontinuous (ontinuous from the left) t if lim - f ( ) f () Thus, funtion is ontinuous t left endpoint of its domin if it is left-ontinuous t A funtion is ontinuous t n interior point of its domin if nd only if it is oth right-ontinuous nd left-ontinuous t Emple The funtion f ( ) - is ontinuous t every point of its domin [-, ] This inludes -, where f is right-ontinuous, nd, where f is left-ontinuous Continuity Test A funtion ( ) - f is ontinuous t if nd only if it meets the following three onditions f () eists ( lies in the domin of f ) lim f ( fi ) eists ( f hs limit s fi ) lim ( ) ( fi ) (the limit equls the funtion vlue) Theorem 6 : Continuity of Algeri Comintions If funtions f nd g re ontinuous t, then the following funtions re ontinuous t : f + g nd f - g fg kf, where k is ny numer f g (provided g () ) 5 ( f ( )) m n (provided ( ( )) m n f is defined on n intervl ontining, nd m nd n re integers) As onsequene, polynomils nd rtionl funtions re ontinuous t every point where they re defined Theorem 7 : Continuity of Polynomils nd Rtionl Funtions Clulus Pge

23 Every polynomil is ontinuous t every point of the rel line Every rtionl funtion is ontinuous t every point where its denomintor is different from zero Emple The following funtions re ontinuous everywhere on their respetive domins ) y Using Theorems 6 nd 7 (rtionl power of polynomil) ) y -- 5 Using Theorems 6 nd 7, (power of polynomil) Emple The funtions f( ) + nd g ( ) 5 ( - ) re ontinuous t every vlue of The funtion f( ) + r ( ) g ( ) 5 ( - ) is ontinuous t every vlue of eept nd, where the denomintor is Emple The funtion f ( ) is ontinuous t every vlue of If >, we hve f ( ), polynomil If <, we hve f ( ) -, nother polynomil Finlly, t the origin, lim f () Emple (Continuity of trigonometri funtions) The funtions sin nd os re ontinuous t every vlue of Aordingly, the quotients sin os tn ot os sin se s os sin re ontinuous t every point where they re defined Theorem 8 tells us tht ontinuity is preserved under the opertion of omposition Theorem 8 : Continuity of Composites If f is ontinuous t, nd g is ontinuous t f (), then g o f is ontinuous t Emple The following funtions re ontinuous every where on their respetive domins ) ) y Using Theorems 7 nd 8 (omposition with the squre root) os( ) y + - ) y - Using Theorems 6, 7, nd 8 (power, omposite, produt, polynomil nd quotient) Using Theorems 7 nd 8 (omposite of solute vlue nd rtionl funtion) Continuous Etension to Point As we sw in Chpter, rtionl funtion my hve limit even t point where its denomintor is zero If f () is not defined, ut lim f ( ) L fi eists, we n define new funtion F( ) y the rule f ( ) if isin the domin of f F( ) L if The funtion F is ontinuous t It is lled the ontinuous etension of f to For rtionl funtions f, ontinuous etensions re usully found y neling ommon ftors Clulus Pge

24 Emple Show tht + -6 f( ) - is not ontinuous t, ut hs ontinuous etension to, nd find tht etension Sine f () is not defined, f is not ontinuous t Although f () is not defined, if we hve ( - )( + ) + f( ) - ( - )( + ) + Here note tht lim f( ) The funtion F( ) + + is equl to f ( ) for, ut is lso ontinuous t F is the ontinuous etension of f to, nd lim lim ( ) f fi -, hving there the vlue of 5 Thus Continuity on Intervls A funtion is lled ontinuous if it is ontinuous everywhere in its domin A funtion tht is not ontinuous throughout its entire domin my still e ontinuous when restrited to prtiulr intervls within the domin A funtion f is sid to e ontinuous on n intervl I in its domin if lim f ( ) f ( fi ) t every interior point nd if the pproprite one-sided limits equl the funtion vlues t ny endpoints I my ontin A funtion ontinuous on n intervl I is utomtilly ontinuous on ny intervl ontined in I Polynomils re ontinuous on every intervl, nd rtionl funtions re ontinuous on every intervl on whih they re defined Emple The funtion - y - is ontinuous on the intervl [, ] The funtion y is ontinuous on the intervls (-, ) nd (, ) The funtion y os is ontinuous on the intervl (-, ) Theorem 9 : The Intermedite Vlue Theorem Suppose f ( ) is ontinuous on n intervl I, nd nd re ny two points of I Then if y is numer etween f ( ) nd f ( ), there eists numer etween nd suh tht f () y Eerises At wht points re the funtions in Eerises -8 ontinuous? Clulus Pge

25 y os y y - + sin y - + tn 5 y s 6 y 7 y + 8 y (- ) + Find the limits in Eerises 9-9 limsin( -sin ) limse( yse y-tn y- ) p yfi p limos tfi 9 - se t ( -9) Define g () in wy tht etends g ( ) ( -) to e ontinuous t ( s -) Define f () in wy tht etends f() s to e ontinuous t s ( s -) For wht vlue of is -, < f( ), ontinuous t every? Answers All eept All eept, All All eept np np 5 All eept, n ny integer 6 All eept, n n odd integer 7 All - > 8 All 9 g () 6 f () Clulus Pge 5

26 CHAPTER THE ROLLE S AND MEAN VALUE THEOREMS EXTREME VALUES OF FUNCTIONS Theorem If f is ontinuous t every point of losed intervl I, then f ssumes oth n solute mimum vlue M nd n solute minimum vlue m somewhere in I Tht is, there re numers nd in I with f ( ) m, f( ) M, nd m f( ) M for every other in I Emple On - p, p, the funtion f ( ) os tkes on mimum vlue of (one) nd minimum vlue of (twie) The funtion g( ) sin tkes on mimum vlue of (one) nd minimum vlue of (one) (Fig ) Remrk The requirements tht the intervl e losed nd the funtion ontinuous re key ingredients of Theorem On n open intervl, ontinuous funtion need not hve either mimum or minimum vlue For emple, the funtion f ( ) hs neither lrgest nor smllest vlue on (, ) Even single point of disontinuity n keep funtion from hving either mimum or minimum vlue on losed intervl Definition Let f e funtion with domin D Then f hs n solute mimum vlue on D t point if f ( ) f( ) for ll in D nd n solute minimum vlue on D t if f ( ) f( ) for ll in D Asolute mimum nd minimum vlues re lled solute etrem (or glol etrem) Definition A funtion f hs lol mimum vlue t n interior point of its domin if f ( ) f( ) for ll in some open intervl ontining A funtion f hs lol minimum vlue t n interior point of its domin if f ( ) f( ) for ll in some open intervl ontining Remrks We n etend the definitions of lol etrem to the endpoints of intervls y defining f to hve lol mimum or lol minimum vlue t n endpoint if the pproprite inequlity holds for ll in some hlf-open intervl in its domin ontining In Fig 5, the funtion f hs lol mim t nd d nd lol minim t e, nd An solute mimum is lso lol mimum Being the lrgest vlue overll, it is lso the lrgest vlue in its immedite neighorhood Hene, list of ll lol mim will utomtilly inlude the solute mimum if there is one Similrly, list of ll minim will inlude the solute minimum if there is one Clulus Pge 6

27 Theorem If f hs lol mimum or minimum vlue t n interior point of its domin, nd if f is defined t, then f () Definition An interior point of the domin of funtion f where f is zero or undefined is ritil point of f Remrk The only domin points where funtion n ssume etreme vlues re ritil points nd endpoints Emple Find the solute mimum nd minimum vlues of f ( ) on [-,] The funtion is differentile over its entire domin, so the only ritil point is where f '( ), nmely We need to hek the funtion s vlues t nd t the endpoints - nd : Critil point vlue: f () ; Endpoint vlues: f (- ) ; f () The funtion hs n solute mimum vlue of t - nd n solute minimum vlue of t Emple Find the solute etrem vlues of g() t 8t- t on [-,] The funtion is differentile on its entire domin, so the only ritil point our where g'( t ) Solving this eqution gives / 8- t or t or t, point not in the given domin The funtion s lol etrem therefore ours t the endpoints, where we find g( - ) - (As minimum); g () 7 (As mimum) / Emple Find the solute etrem of h ( ) on [-,] / The first derivtive - h ( ) / hs no zeros ut is undefined t The vlues of h t this one ritil point nd t the endpoints - nd re h () ; h( - ) (- ) nd h () () 9 The solute mimum vlue is 9, ssumed t ; the solute minimum is, ssumed t THE ROLLE S AND MEAN VALUE THEOREMS Theorem (Rolle s Theorem) Suppose tht y f( ) is ontinuous t every point of the losed intervl [, ] nd differentile t every point of its interior (, ) If f( ) f( ), then there is t lest one numer in (, ) t whih f () Remrk Insted of f( ) f( ), if f( ) f( ) l, then lso Rolle s theorem holds The hypothesis of Rolle s Theorem re essentil If they fil t even one point, the grph my not hve horizontl tngent Emple Verify Rolle s Theorem for the funtion f defined y Clulus Pge 7

28 ƒ() ( - ) m ( - ) n, where m nd n eing positive integers nd [, ] ƒ() ( - ) m ( - ) n is polynomil funtion in the vrile nd sine every polynomil funtion in the vrile is ontinuous, ƒ is ontinuous on [, ] Now differentiting ƒ with m- n m n- respet to, we get, f ( ) m( -) ( - ) + n( -) ( - ) ie, ƒ () ( - ) m - ( - ) n - [(m( - ) + n ( - )], () gin polynomil funtion, so tht ƒ () eists t every point in the open intervl (, ) Also ƒ() ƒ() Net to verify whether there eists point C(,ƒ()) t whih f ( ) From (), ƒ () implies ( - ) m - ( - ) n - [(m( - ) + n ( - )], whih implies m( - ) + n ( - ), sine (- ), (- ), s < <, m + n whih implies tht, point within the intervl (, ) dividing the intervl m + n in the rtio m : n The onlusion is tht there is point suh tht < <, where f ( ) Hene Rolle s theorem is verified Emple Emine whether Rolle`s theorem n e pplied to the funtion ƒ() tn for the intervl [, π] There is disontinuity t p to the funtion f ( ) tn Also f ( ) se whih does not eist t p HeneRolle`s theorem nnot e pplied to f ( ) in [, p ] Theorem (The Men Vlue Theorem) Suppose y ƒ() e funtion of sujet to the onditions () ƒ is ontinuous in the losed intervl [, ] () f is differentile in the open intervl (, ) Then there is point in the open intervl (, ) t whih f( ) - f( ) f () () - Proof We piture the grph of f s urve in the plne nd drw line through the point A(, f( )) nd B (, f( )) The line is the grph of the funtion f( ) - f( ) g( ) f( ) + ( -) - (point-slope eqution) The vertil differene etween the grphs of f nd g t is h ( ) f( ) - g ( ) Clulus Pge 8

29 f( ) - f( ) f ( ) - f( ) - ( -) () - The funtion h stisfies the hypotheses of Rolle s theorem on [, ]: It is ontinuous on [, ] nd differentile on (, ) euse oth f nd g re Also, h ( ) h ( ) euse the grphs of f nd g oth pss through A nd B Therefore, h ' t some point in (, ) This is the point we wnt for eqution () To verify eqution (), we differentite oth sides of eqution () with respet to, nd otin f ( ) - f( ) h ( ) f ( ) - - Setting, we otin f ( ) - f( ) h () f () - - f ( ) - f( ) implies f ( ) - - f ( ) - f( ) implies f (), - nd the proof is omplete Remrk If ƒ stisfies the two onditions s in Theorem, then ƒ( + h) ƒ() + hƒ ( + θh), where - h nd + θh, where < q < Emple 6 Verify Lgrnge s men vlue theorem for the funtion ƒ() e on [, ] Here ƒ() e Sine the eponentil funtion is ontinuous t every point on the rel line, in prtiulr ƒ is ontinuous in the losed intervl [,] Also ƒ () e eists in the open intervl (, ) Hene y Lgrnge s men vlue theorem, we hve f () - f () f ( ) - ie, e e - so tht log (e - ), whih lies in the open intervl (, ) Hene Lgrnge s men vlue theorem is verified for the given funtion Emple In the men vlue theorem ƒ( + h) ƒ() + hƒ ( + qh), show tht q if ƒ() is qudrti epression : Suppose f e the qudrti epression ƒ() + +, Then ƒ () + By the men vlue theorem, ƒ( + h) ƒ() + h ƒ ( + qh), or (+h) + ( + h) h [ (+ q h) + ] Clulus Pge 9

30 ie, h h q Hene q Emple If r elerting from zero tkes 8 se to go 5 ft, its verge veloity for the 8- se intervl is 5/8 ft/se At some point during the elertion, the Men Vlue Theorem sys, the speedometer must red etly mph ( ft/se) Corollry If f '( ) t eh point of n intervl I, then f ( ) C for ll in I, where C is onstnt Proof We wnt to show tht f hs onstnt vlue on I We do so y showing tht nd re ny two points in I, then f ( ) f( ) Suppose tht nd re two points in I, numered from left to right so tht < Then f stisfies the hypotheses of the Men Vlue Theorem on [, ]: It is differentile t every point of [, ], nd hene ontinuous t every point s well Therefore, f( ) - f( ) f () () - t some point etween nd Sine f throughout I, () eomes f( ) - f( ), or f( ) - f( ), - implies f ( ) f( ) Remrk We know tht if funtion f hs onstnt vlue on n intervl I, then f is differentile on I nd f '( ) for ll in I Corollry provides the onverse Corollry If f '( ) g'( ) t eh point of n intervl I, then there eists onstnt C suh tht f ( ) g( ) + C for ll in I Proof At eh point in I the derivtive of the differene funtion h f - g is h'( ) f '( ) - g'( ) Thus, h ( ) Con I (Corollry ) Tht is, f ( ) - g( ) C on I, so f ( ) g( ) + C Remrk Corollry sys tht funtions n hve identil derivtives on n intervl only if their vlues on the intervl hve onstnt differene We know, for instne, tht the derivtive of f ( ) on (-, ) is Any other funtion with derivtive on (-, ) must hve the formul + C for some vlue of C Emple Find the funtion f ( ) whose derivtive is sin nd whose grph psses through the point (, ) Sine f ( ) hs the sme derivtive s g( ) - os, we know tht f ( ) - os+ C for some onstnt C The vlue of C n e determined from the ondition tht f () (the grph of f psses through (,) ) f() - os() + C, or - + so C The formul for f is f( ) - os+ Definitions Let f e funtion defined on n intervl I nd let nd e ny two points in I f inreses on I if < f( ) < f( ) Clulus Pge

31 f dereses on I if < f ( ) < f( ) Corollry Suppose tht f is ontinuous on [, ] nd differentile on (, ) If f ' > t eh point of (, ) then f inreses on [, ] If f ' < t eh point of (, ), then f dereses on [, ] Proof Let nd e two points in [, ] with < The Men vlue Theorem pplied to f on [, ] sys tht f ( ) - f( ) f ( )( - ) (5) for some etween nd The sign of the right- hnd side of eqution (5) is the sme s the sign of f '( ) euse - is positive Therefore, f ( ) > f( ) if f is positive on (, ), nd f ( ) < f( ) if f is negtive on (, ) Emple The funtion f ( ) dereses on (-,), where f '( ) < It inreses on (, ), where f '( ) > Eerises Verify Rolle`s Theorem for the following funtions: (i) f (), [, ] (ii) f () + ( - ) /, [,] (iii) f () ( - ) on the intervl [, ] Verify Lgrnge s men vlue theorem for the funtion (i) f () - -, [,] (ii) f () - +, (, ) (iii) f () + + 9, [,5] In Eerise -, find the vlue or vlues of tht stisfy the eqution f( ) - f( ) f () - in the onlusion of the Men Vlue Theorem for the funtions nd intervls f( ) + -, [, ] f( ) +,, THE FIRST DERIVATIVE TEST FOR LOCAL EXTREME VALUES This hpter shows how to test funtion s ritil points for the presene of lol etreme vlues Theorem (The First Derivtive Test for Lol Etreme Vlues) The following test pplies to ontinuous funtion f ( ) At ritil point : If f ' hnges from positive to negtive t ( f ' > for < nd f ' < for > ), then f hs lol mimum vlue t If ' f hnges from negtive to positive t ( ' f < for < nd ' f > for > ), then f hs lol minimum vlue t Clulus Pge

32 If f ' does not hnge sign t ( f ' hs the sme sign on oth sides of ), then f hs no lol etreme vlue t At left endpoint : If f ' < ( f ' > ) for >, then f hs lol mimum (minimum) vlue t At right endpoint : If f ' < ( f ' > ) for <, then f hs lol minimum (mimum) vlue t Emple Find the ritil points of f ( ) ( - ) - Identify the intervls on whih f is inresing nd deresing Find the funtion s lol nd solute etreme vlues The funtion f is defined for ll rel numers nd is ontinuous The first derivtive '( ) d f - ) - d - ( -) ( - ) - is zero t nd undefined t There re no endpoints in f ' s domin, so the ritil points, nd, re the only ples where f might hve n etreme vlue of ny kind These ritil points divide the - is into intervls on whih f ' is either positive or negtive The sign pttern of f ' revels the ehvior of f oth etween nd t the ritil points We disply the informtion in piture like the following To mke the piture, we mrked the ritil points on the - is, noted the sign of eh ftors of f ' on the intervls etween the points, nd multiplied the signs of the ftors to find the sign of f ' We then pplied Corollry of the Men Vlue Theorem to determine tht f dereses ( ) on (,), - dereses on (,), nd inreses ( ) on (, ) Theorem tells us tht f hs no etreme t (s f ' does not hnge sign) nd tht the f hs lol minimum t f ' hnges from negtive to positive ) Clulus Pge

33 The vlue of the lol minimum is f () (- ) - This is lso n solute minimumm euse the funtion s vlues fll towrds it from the left nd rise wy from it on the right Emple Find the intervls on whih g( ) , - is inresing nd deresing Where does the funtion ssume etreme vlues nd wht re these vlues? The funtion f is ontinuous on its domin, [-,] The first derivtive g '( ) - + -( - ) - ( + )( - ), defined t ll points of [-,], is zero t - nd These ritil points divide the domin of g into intervls on whih g is either positive or negtive We nlyze the ehvior of g y pituring the sign pttern of g : We onlude tht g hs lol mim t - nd nd lol minim t - nd The orresponding vlues of g( ) re Lol mim: Lol minim: g( - ) -, g( - ) -, g() ) g () Sine g is defined on losed intervl, we lso know tht g(-) is the solute minimumm nd g() is the solute mimumm shows these vlues in reltion to the funtion s grph Clulus Pge

34 MODULE - II CHAPTER ASYMPTOTES Convity nd Conveity A urve is sid to e onve upwrds (or onve downwrds) t or ner P when t ll points ner P on it lies ove the tngent t P A urve is sid to e onve downwrds (or onve upwrds) t or ner P when t ll points ner P it lies elow the tngent t P (Fig nd Fig ) The Seond Derivtive Test for Convity Let y f( ) e twie differentile on n intervl I If y " > on I, the grph of f over I is onve up If y " < on I, the grph of f over I is onve down Emple The urve y is onve down on (-,) where y" 6< nd onve up on (, ) where y" 6 > Emple The prol y is onve up on every intervl euse y > Emple For wht vlues of is the urve y onve or onve to the foot of the ordinte Here y suessive differentition we otin d y - / d d y Hene is negtive for ll positive vlues (nd negtive vlues of re not dmissile) d so tht the urve in the neighourhood of ny speified point is onve downwrd (ie, onve to the foot of the ordinte of tht point) Points of Infletion A point of urve t whih the urve hnges its diretion from onve upwrds to onve downwrds s t P in Fig7 or from onve downwrds to onve upwrds s t P in Fig8, is lled point of infletion The tngent t the point is lled n infletionl tngent Remrk The urve rosses its tngent t point of infletion On the grph of twie-differentile funtion, y " t point of infletion Clulus Pge

35 Emple For the funtion s + os t, t, s" - ost nd s " when ost ie, when t p, p, Hene the grph of s + os t, t hnges onvity t t p, p, K ie, the points of infletion re t p, p, K Emple Find the points of infletion on the urve y nd show tht they lie on + stright line dy ( - ) Here d ( + ) d y - ( - ) d ( + ) d y At the point of infletion, we must hve d \ ( - ) or When, y ;when, y Hene the points of infletion re (,),, nd -, - These three points oviously lie on the stright line y Emple Prove tht for the urve y sin, every point t whih it meets the - is is point of infletion sin o or np or np dy d y d y Now os, - sin, - os d d d d y For points of infletion \ sin or d np d y Also when np, - os np d \ np gives the point of infletion These re sme s the points were the urve meets the -is Emple Show tht the points of infletion of the urve y ( - ) ( -) lie on the line + Given urve is y ( -) ( -) Hene y ( -) ( - ) nd dy ( - ) + ( -) ( - ) + ( - ) d ( -) - Clulus Pge 5

36 d y - - { ( ) ( ) } nd ( - - ) d - ( -) - (- - ) + ( - ) + - ( -) d y For points of infletion, ; hene d + - or + Hene the point of infletion lie on the line + Emple Prove tht the urve y hs three points of infletion nd they re olliner + Given urve is y dy - dy \ nd -6 + d + d + ( ) d y At the point of infletion ie, ( -) d \ or, when, y nd when, y ( ) - Hene the point of infletion re (, ),, nd -, Oviously, the ove points lie on the line y Hene the points of infletion re olliner Grphing with y'nd y"algorithm for Grphing y f() Step : Find y ' nd y " Step : Find the rise nd fll of the urve Step : Determine the onvity of the urve Step : Mke summry nd show the urve s generl shpe Step 5: Plot speifi points nd sketh the urve Emple Grph the funtion y - + Step Find y ' nd y " y - + y - ( - ) Critil points:, y" - ( - ) Possile infletion points, Step (Rise nd fll) Sketh the sign pttern for y ' nd use it to desrie the ehvior of y Clulus Pge 6

37 Clulus Pge 7

38 Step ( Convity) Sketh the sign pttern ends for y" nd use it to desrie the wy the grph Step (Summry nd generl shpe) Summrize the informtionn from steps nd Show the shpe over eh intervl Then omine the shpess to show the urve s generl form Step 5 (Speifi points nd urve) Plot the urve s interepts (if onvenient) nd the points where y ' nd y " re zero Indite ny lol etreme vlues nd infletion points Use the generl shpe in step s guide to sketh the urve (Plot dditionl points s needed ) 5 Emple 5 Grph y - 5 Step : 5 Find y ' nd y " y - 5 ( - 5) The - interepts re t nd 5 Clulus Pge 8

39 - - ' 5 y - 5 ( ) - Critil points: nd y " + ( ) Possile infletion point: nd Step (Rise nd fll) Sketh the sign pttern for y ' nd use it to desrie the ehvior of y - Step (Convity) Sketh the sign pttern for y" nd use it to desrie the wy the grph ends From the sign pttern for y", we see tht there is n infletion points t -, ut not However, knowing tht the funtion y y ' fi 5-5 is ontinuous, s fi - nd y ' fi - s fi + (See the formul for y ' in step ), nd the onvity does not hnge t (step) tells us tht the grph hs usp t Clulus Pge 9

40 ASYMPTOTES Definitions We sy tht f ( ) hs the limit L s pprohes infinity nd write lim f ( ) L if, for every numer e >, there eists orresponding numer M suh tht for ll > M f( ) - L < e We sy tht f ( ) hs the limit L s pprohes minus infinity nd write lim f ( ) L - if, for every numer e >, there eists orresponding numer N suh tht for ll < N f( ) - L < e Emple Show tht lim Let e > e given We must find numer M suh tht for ll > M - < e The implition will hold if M / e or ny lrger positive numer This proves lim Emple Show tht lim - (Ref Fig ) Let e > e given We must find numer N suh tht for ll < N - < e The implition will hold if N - / e or ny numer less thn - / e This proves lim - Theorem (Properties of Limit s ± ) The following rules hold if lim f ( ) L nd lim g( ) M ( L nd M rel numers) Sum rule lim [ f ( ) + g( )] L+ M Differene rule lim [ f ( ) - g( )] L- M Produt Rule lim f ( ) g( )] L M Constnt Multiple Rule lim kf ( ) kl (nynumer k) 5 Quotient Rule: f( ) lim L, if M g ( ) M 6 Power Rule: If mnd n re integers, then Clulus Pge

41 / lim [ f( )] m n m / L n provided m/ n L is rel numer Emple Evlute lim 5 + lim 5 lim 5 lim + +, using sum rule 5+ 5 Emple Evlute lim p - lim p lim fi- p - lim p lim lim, using produt rule p Limits of Rtionl Funtion s fi To determine the limit of rtionl funtion s fi, we n divide the numertor nd denomintor y the highest power of in the denomintor Wht hppens then depends on the degrees of the polynomils involved Emple Evlute lim (8/ ) (/ ) lim lim fi, y dividing + + (/ ) numertor nd denomintor y Emple Evlute + lim fi- - (/ ) ( / ) lim + + lim, y dividing numertor nd denomintor y (/ ) Emple Evlute lim fi- 7 + (/ ) lim - - lim, the numertor now pprohes - while the denomintor (/ ) pprohes 7, so the rtio pprohes - - Emple Evlute lim fi- - - Clulus Pge

42 (7/ ) lim lim (/ ) -(/ ) Asymptotes fi, numertor fi -, denomintor fi, so rtio An symptote of urve of infinite etent is line whose position is pprohed s limit y tngent to the urve s the point of ontt reedes indefinitely long the urve Definition If the distne etween the grph of funtion nd some fied line pprohes zero s the grph moves inresingly fr from the origin, we sy tht the grph pprohes the line symptotilly nd tht the line is n symptote of the grph y Emple Consider the hyperol - The eqution n e written s y - () Therefore, s tends to infinity, the eqution () tends to the equtions Hene the lines y nd y - re symptotes to the urve Emple Consider the eqution y ( ) y - The given eqution n e written s y nd from this it n e seen tht s fi, y fi, so tht the line is n - symptote to the given urve (Fig ) In Fig, dshed line denotes the symptote of the urve y ( - ) Definitions A line y is horizontl symptote of the grph of funtion y f( ) if either lim f ( ) or lim f ( ) - A line is vertil symptote of the grph if either lim f( ) or lim f( ) + - Emple The oordinte es re symptotes of the urve y The - is is symptote of the urve on the right euse lim nd on the left euse lim - The y-is is n symptote of the urve oth ove nd elow euse lim nd lim Notie tht the denomintor is zero t nd the funtion is undefined Emple Find the symptotes of the urve y + + Clulus Pge

43 We re interested in the ehvior s fi nd s fi -, where the denomintor is zero By tul division y + eomes + y + + Hene lim f ( ) nd lim f( ), lim f( ) nd hene, y the Definition, the symptotes re the lines y nd - Emple Verify tht the grph of Sine lim f( ) f( ) ( )( ) , - ( - )( + ) + hs vertil symptote t - ut not t nd hene the grph hs vertil symptote - But is not n symptote s lim f ( ), finite limit Olique Asymptotes If the degree of the numertor of rtionl funtion is one greter thn the degree of the denomintor, the grph hs n olique symptotes, tht is liner symptote tht is neither vertil nor horizontl Emple Find the symptotes of the grph of f( ) - - We re interested in the ehvior s fi nd lso s fi, where the denomintor is zero By tul division, we otin f( ) () - { - liner reminder Sine lim + f( ) nd lim - f( ) -, the line is two-sided symptote As fi, the reminder pprohes nd f( ) fi ( /) + The line y ( /) + is n symptote oth to the right nd to the left GRAPHING WITH ASYMPTOTES AND DOMINANT TERMS Emple Grph the funtion y + We investigte symmetry, dominnt terms, symptotes, rise, fll, etreme vlues, nd onvity Step Symmetry There is none Step Find ny dominnt terms nd symptotes We write the rtionl funtion s polynomil plus reminder: Clulus Pge

44 y + () For lrge, y» For ner zero, y» / Eqution () revels vertil symptote t, where the denomintor of the reminder is zero Step Find y ' nd nlyze the funtion s ritil points Where does the urve rise nd fll? The first derivtive y ' - - is undefined t nd zero when - - ie, when ie, when 8» Step Find y " nd determine the urve s onvity ie, when The seond derivtive y" + + is undefined t nd zero when + ie, when + ie, when - ie, when - Clulus Pge

45 CHAPTER OPTIMIZATION Strtegy for Solving Optimiztion Prolems We follow the following steps for solving optimiztion prolems (Mimize or minimize prolems): Red the prolem Red the prolem until you understnd it Wht is unknown? Wht is given? Wht is sought? Drw piture Lel ny prt tht my e importnt to the prolem Introdue vriles List every reltion in the piture nd in the prolem s n eqution or lgeri epression Identify the unknown Write n eqution for it If you n, epress the unknown s funtion of single vrile or in two equtions in two unknowns This my require onsiderle mnipultion 5 Test the ritil points nd endpoints Use wht you know out the shpe of the funtion s grph nd the physis of the prolem Use the first nd seond derivtives to identify nd lssify ritil points (where f or does not eist) Whenever you re mimizing or minimizing funtion of single vrile, we urge you to grph it over the domin tht is pproprite to the prolem you re solving The grph will provide insight efore you lulte nd will furnish visul ontet for understnding your nswer Emple Find two positive numers whose sum is nd whose produt is s lrge s possile If one numer is, then the other is ( - ) Their produt is f ( ) (- ) - We wnt the vlue or vlues of tht mke f ( ) s lrge s possile The domin of f is the losed intervl We evlute f t the ritil points nd endpoints The first derivtive, f ( ) -, is defined t every point of the intervl nd is f ( ) only t Listing the vlues of f t this one ritil point nd the endpoints gives Critil-point vlue : f () () -() Endpoint vlues : f(), f () We onlude tht the mimum vlue is f () The orresponding numers re nd ( - ) (Fig) Emple A retngle is to e insried in semiirle of rdius Wht is the lrgest re the retngle n hve, nd wht re its dimensions? Clulus Pge 5