Differentiation of Polynomials

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C H A P T E R 9 Differentition of Polnomils Ojetives To unerstn te onept of it. To unerstn te efinition of ifferentition. To unerstn n use te nottion for te erivtive of polnomil funtion. To e le to fin te grient of urve of polnomil funtion lulting its erivtive. To ppl te rules for ifferentiting polnomils to solving prolems. In te previous pter te rte of nge of one quntit wit respet to seon quntit s een onsiere. In tis pter tenique will e evelope for lulting te rte of nge for polnomil funtions. To illustrte tis n introutor emple will e onsiere. On plnet X,nojet flls istne of metres in t seons, were =.8t. (Note: On Ert te ommonl use moel is = 4.9t.) Cn generl epression for te spee of su n ojet fter t seons e foun? In te previous pter it ws foun tt te grient of te urve t given point P n e pproimte =.8t fining te grient of or PQ, were Q is point on te urve s lose Q s possile to P. Te grient of or PQ pproimtes te spee of te ojet t P. Te sorter tis or is me te etter te pproimtion. P 5 (5 + ) t 53

54 Essentil Mtemtil Metos &CAS Let P e te point on te urve were t = 5. Let Q e te point on te urve orresponing to seons fter t = 5, i.e. Q is te point on te urve were t = 5 +. Te grient of or PQ =.8 ( (5 + ) 5 ) (5 + ) 5 =.8 ( (5 + ) 5 ) =.8( + ) Te tle gives te grient for ifferent vlues of. Use our lultor to ek tese. If vlues of of smller n smller mgnitue re tken, it is foun tt te grient of or PQ gets loser n loser to 8. At te point were t = 5 te grient is 8. Tus te spee of te ojet t te moment t = 5is8metres per seon. Te spee of te ojet t te moment t = 5iste iting vlue of te grients of PQ s Q pproes P. Grient of PQ.7 8.56.6 8.48.5 8.4.4 8.3.3 8.4. 8.6. 8.8 A formul for te spee of te ojet t n time t is require. Let P e te point wit oorintes (t,.8t )onteurve n Q e te point wit oorintes (t +,.8(t + ) ). Te grient of or PQ =.8 ( (t + ) t ) (t + ) t =.8(t + ) From tis n epression for te spee n e foun. Consier te it s pproes, tt is, te vlue of.8(t + )s eomes ritrril smll. Te spee t time t is.6t metres per seon. (Te grient of te urve t te point orresponing to time t is.6t.) Tis tenique n e use to investigte te grient of similr funtions. 9. Te grient of urve t point, n te grient funtion Consier te funtion f : R R, f () =. Te grient of te or PQ in te f () = jent grp = ( + ) + = + + + = + ( +, ( + ) ) P(, ) n te grient t P n e seen to e. Te it of ( + ) spproes is. It n e seen tt tere is noting speil out.soifis rel numer similr formul ols. Q

Cpter 9 Differentition of Polnomils 55 Te grient of te funtion = t n point is equl to. It is si tt is te erivtive of wit respet to or, more riefl, te erivtive of is. Tis itself is lerl te rule for funtion of n we refer to tis funtion s te grient (or erive) funtion. Te strigt line tt psses troug P n s grient is lle te tngent to te urve t P. From te isussion t te eginning of te pter it n e seen tt te erivtive of.8t is.6t. Emple B first onsiering te grient of te or QP, fin te grient of = t te point Q wit oorintes (3, 3). Consier or PQ: Grient of PQ = (3 + ) (3 + ) 3 3 + 3 = 9 + 6 + 6 3 3 + 3 4 + = = 4 + = (3 +, (3 + ) (3 + )) P Q(3, 3) Consier s it pproes zero. Te grient t te point (3, 3) is 4. Emple Fin te grient of or PQ n ene te erivtive of +. = + P( +, ( + ) + ( + )) Q(, + )

56 Essentil Mtemtil Metos &CAS Te grient of or PQ = ( + ) + ( + ) ( + ) + = + + + + = + + = + + From tis it is seen tt te erivtive of + is +. Te nottion for it of + + s pproes is + +. Te erivtive of funtion wit rule f () m e foun first fining n epression for te grient of te or from Q(, f ()) to P( +, f ( + )) n ten fining te it of tis epression s pproes. Emple 3 B first onsiering te grient of te or from Q(, f ()) to P( +, f ( + )) for te urve f () = 3, fin te erivtive of 3. f () = 3 f ( + ) = ( + ) 3 Te grient of or PQ = f ( + ) f () ( + ) = ( + )3 3 ( + ) Te erivtive of f () = ( + ) 3 3 ( + ) = 3 + 3 + 3 + 3 3 = 3 + 3 + 3 = 3 + 3 + = 3 We ve foun tt te erivtive of 3 is 3. Te following emple provies prtie in etermining its.

Cpter 9 Differentition of Polnomils 57 Emple 4 Fin: 3 + + + 3 4 + + = 3 + 3 = 3 4 = 4 Using CAS lultor For Emple 4, selet 3:it( from te Cl, F3, menu n omplete s sown. = 3 + = 3 Eerise 9A A spe veile moves so tt te istne trvelle over its first minute of motion is given = 4t 4,were is te istne trvelle in metres n t te time in seons. B fining te grient of te or etween te points were t = 4 n t = 5, estimte te spee of te spe veile wen t = 5. A popultion of insets grows so tt te popultion, P, ttime t (s) is given P = + t + t,were t >. B fining te grient of te or etween te points were t = 3 n t = 3 +, fin n estimte for te rte of growt of te inset popultion t time t = 3. Emple 4 3 Fin: 3 + + 3 + 3 + + e 5

58 Essentil Mtemtil Metos &CAS Emple 3 Emple 4 Fin: ( + ) + ( + ) ( + ) (5 + ) + 3(5 + ) 4 ( + ) 3 + ( + ) ( 3 + ) 5 Forurve wit eqution = 3 : i.e. te erivtive of = + i.e. te grient of = + 3, were = 5 i.e. te erivtive of = 3 + fin te grient of or PQ, were P is te point (, ) n Q is te point (( + ), 3( + ) ( + )) fin te grient of PQ wen =. fin te grient of te urve t P. 6 Forurve wit eqution = : fin ( te grient) of or AB were A is te point (, ) n B is te point ( + ), + fin te grient of AB wen =. fin te grient of te urve t A. 7 Forurve wit eqution = + 3: fin te grient of or PQ, were P is te point (, 5) n Q is te point (( + ), ( + ) + ( + ) 3) fin te grient of PQ wen =. fin te grient of te urve t P. 9. Te erive funtion In Setion 9. we sw ow te grient funtion of funtion wit rule f () oul e erive onsiering te grient of or PQ on te urve of = f (). Consier te grp = f ()ofte funtion f : R R. Te grient of te or PQ = = f ( + ) f () + f ( + ) f () Terefore te grient of te grp t P is given f ( + ) f () P (, f ()) = f () Q ( +, f ( + ))

Cpter 9 Differentition of Polnomils 59 Te reer is referre to Setion 9.4 for furter isussion of its. Te grient or erive funtion is enote f, were f : R R n f f ( + ) f () () = In tis pter onl polnomil funtions re onsiere. For polnomil funtion te erive funtion lws eists n is efine for ever numer in te omin of f. Determining te erivtive of n epression or te erive funtion evluting te it is lle ifferentition first priniples. Emple 5 For f () = + fin f ()first priniples. f () = f ( + ) f () = ( + ) + ( + ) ( + ) = + + + + = + + = + + = + f () = + Emple 6 For f () = 3 fin f ()first priniples. f () = f ( + ) f () = ( + ) 3 ( 3 ) = ( 3 + 3 + 3 + 3 ) ( 3 ) = 3 3 3 = 3 3 = 3

5 Essentil Mtemtil Metos &CAS Using CAS lultor Define f () = 3. Ten, from te Cl menu selet A:nDeriv( f ()), were f ()isefine s 3, n tke te it s. Te following results ve een otine: For f () =, f () =. For f () =, f () =. For f () = 3, f () = 3. For f () = 4, f () = 4 3. For f () =, f () =. Tis suggests te following generl result: For f () = n, f () = n n, n =,, 3,... n for f () =, f () = From te previous setion it n e seen tt for k, onstnt: If f () = k n, te erivtive funtion f s rule f () = kn n. It is wort mking speil note of te results: If g() = kf(), were k is onstnt, ten g () = kf (). Tt is, te erivtive of numer multiple is te multiple of te erivtive. Foremple, for g() = 5, te erive funtion g () = 5() =. Anoter importnt rule for ifferentition is: If f () = g() + (), ten f () = g () + (). Tt is, te erivtive of te sum is te sum of te erivtives. Foremple, for f () = +, te erive funtion f () = +. Te proess of fining te erivtive funtion is lle ifferentition.

Cpter 9 Differentition of Polnomils 5 Emple 7 Fin te erivtive of 5 3 +, i.e. ifferentite 5 3 + wit respet to. f () = 5 3 + ten f () = 5 4 (3 ) + () = 5 4 6 Using CAS lultor Selet :( from te Cl menu ( F3 ) n omplete s sown. Emple 8 Fin te erivtive of f () = 3 3 6 + n f (). f () = 3 3 6 + ten f () = 3(3 ) 6() + () = 9 f () = 9 = 3 Using CAS lultor Define f () = 3 3 6 +. Fin te erivtive n store it s f(). Ten fin f().

5 Essentil Mtemtil Metos &CAS Emple 9 Fin te grient of te urve etermine te rule f () = 3 3 6 + tte point (, ). Now f () = 9 n f () = 9 = 3. Te grient of te urve is 3 tte point (, ). An lterntive nottion for te erivtive is te following: If = 3, ten te erivtive n e enote, so tt = 3. In generl, if is funtion of, te erivtive of wit respet to is enote n wit te use of ifferent smols z, were z is funtion of t. Te erivtive of z wit respet to t is z t. In tis nottion is not ftor n nnot e nelle. P Tis me out euse in te eigteent entur te δ stnr igrm for fining te iting grient ws lelle s in te figure. ( is te lower se Greek letter Q δ for, n is pronoune elt.) mens ifferene in. mens ifferene in. Emple If = t, fin t. If = t 3 + t, fin t. If z = 3 3 +, fin z. = t ten t = t = t 3 + t ten = 3t + t z = 3 3 + ten z = + Emple For = ( + 3), fin. For z = (t ) (t + ), fin z t. For = + 3, fin. Differentite = 3 wit respet to.

Cpter 9 Differentition of Polnomils 53 It is first neessr to write = ( + 3) in epne form. = + 6 + 9 n = + 6 First ivie : = + 3 = Epning: n z = (4t 4t + )(t + ) = 4t 3 4t + t + 8t 8t + = 4t 3 + 4t 7t + z t = t + 8t 7 = 3 = 6 Opertor nottion Fin te erivtive of 4 wit respet to n lso e written s generl, ( f ()) = f (). Emple ( 4), n, in Fin: (5 4 3 ) z (5z 4z) z (6z3 4z ) (5 4 3 ) = 5 z (5z 4z) = z 4 z (6z3 4z ) = 8z 8z Emple 3 Fin te oorintes of te points on urves etermine e of te following equtions t wi te grient s te given vlue: = 3 ;grient = 8 = 4 + ; grient = = 4 3 ;grient = 6

54 Essentil Mtemtil Metos &CAS = 3 implies = 3 3 = 8 8 =± 3 = ± 6 3 oorintes re ( 6 3, 6 ) ( 6 n 9 ) 6, 6 6 3 9 = 4 3 implies = 3 3 = 6 = =±. oorintes re = 4 + implies = 4 4 = = oorintes re (, ) ( ) ( ), 4 3 n, 4 + 3 Using CAS lultor Define f () = 4 3.Tke te erivtive n store s f(). Now solve te eqution f() = 6. Sustitute in f () tofin te -oorintes. Eerise 9B Emples 5, 6 Emple 7 For e of te following, fin f f ( + ) f () () fining : f () = 3 f () = 4 f () = 3 f () = 3 + 4 + 3 e f () = 3 4 f f () = 4 5 g f () = 3 + Fin te erivtive of e of te following wit respet to. + 4 + 3 3 + 4 e 5 3 + 3 f 3 +

Cpter 9 Differentition of Polnomils 55 3 For e of te following fin f (): f () = f () = 3 7 f () = 5 f () = 5 + 3 e f () = 3 f f () = 5 3 g f () = 5 + 3 4 f () = 4 3 3 4 + Emple 4 For e of te following, fin : = = = 4 3 3 + = 3 ( 3 3 + 6) e = ( + )( + ) f = (3 4) Emple 9 g = 5 + 3 4, 5 For te urve wit eqution = 3 + fin te grient t points: i (, ) ii (, 3 + ) Fin te erivtive of 3 + wit respet to. 6 Given tt = 3 3 + 3, fin. Hene sow tt interpret tis in terms of te grp of = 3 3 + 3. Given tt = +, for, fin. Differentite = (3 + ) wit respet to. for ll, n 7 At te points on te following urves orresponing to te given vlues of, fin te -oorinte n te grient. = +, = = + +, = =, = = ( + )( 4), = 3 e = 3 3, = f = (4 5), = 8 For e of te following, fin f () n f (), if = f () ten fin te {(, ): f () = } i.e. te oorintes of te points were te grient is. i = ii = + + 3 iii = 3 + iv = 4 3 Wt is te interprettion of {(, ): f () = } in terms of te grps? Emple 9 Fin: t (3t 4t) (3 3 ) e (4 + 3 ) ( 3 4 ) f z (5 z z 4 ) t (9.8t t)

56 Essentil Mtemtil Metos &CAS Emple 3 Fin te oorintes of te points on te urves given te following equtions t wi te grient s te given vlues: = ;grient = 8 = 3 ;grient = = ( ); grient = = 3 + ; grient = e = 3 6 + 4; grient = f = 3 ;grient = 9.3 Grps of te erive or grient funtion Consier te grient for ifferent intervls of te grp of = g() sown opposite. At point (, g()) of te grp = g() te grient is g (). Some of te fetures of te grp re: < te grient is positive, i.e. g () > = te grient is zero, i.e. g () = < < te grient is negtive, i.e. g () < = te grient is zero, i.e. g () = > te grient is positive, i.e. g () > R(, g ()) S(, g()) = g() Emple 4 For te grp of f : R R, fin: {: f () > } {: f () < } {: f () = } (5, 6) {: f () > } = {: < < 5} = (, 5) {: f () < } = {: < } { : > 5} = (, ) (5, ) = f () {: f () = } = {, 5} (, 7) Emple 5 Sket te grp of = f () for e of te following. (It is impossile to etermine ll fetures.) = f () = f () (.5, 4) = f () 4 3 4 (3, ) (, 4)

Cpter 9 Differentition of Polnomils 57 f () > for > 3 f () < for < 3 f () = for = 3 f () > for > f () < for.5 < < f () > for <.5 f (.5) = n f () = 3 = f'().5 f () = for ll = f () = f () An ngle ssoite wit te grient of urve t point Te grient of urve t point is te grient of te tngent t tt point. A strigt line, te tngent, is ssoite wit e point on te urve. If is te ngle strigt line mkes wit te positive iretion of te -is, ten te grient, m,ofte strigt line is equl to tn, i.e., tn = m. If = 45 ten tn = n te grient is. If = ten te grient of te strigt line is tn. If = 35 ten tn = n te grient is. Emple 6 Fin te oorintes of te points on te urve wit eqution = 7 + 8twi te tngent: mkes n ngle of 45 wit te positive iretion of te -is is prllel to te line = + 6. = 7 7 = (tn 45 = ) n = 8 = 4 = 4 7 4 + 8 = 4 oorintes re (4, 4) Note: = + 6 s grient 7 = n = 5 = 5 ( 5 oorintes re, 3 ) 4

58 Essentil Mtemtil Metos &CAS Emple 7 Te plnne pt for fling suer leving plnet is efine te eqution = 4 4 + 3 3 for >. Te units re kilometres. (Te -is is orizontl n te -is vertil.) Fin te iretion of motion wen te -vlue is: i ii 3 Fin point on te fling suer s pt were te pt is inline t 45 to te positive -is. (i.e. were te grient of te pt is ). Are tere n oter points on te pt wi stisf te sitution esrie in prt? = 3 + i Wen =, = 8 + 8 = 6 tn 6 = 86.4 (to te -is) ii Wen = 3, = 7 + 8 = 45 tn 45 = 88.73 (to te -is), Wen te fling suer is fling t 45 to te iretion of te -is, te grient of te urve of its pt is given tn 45. Tus to fin te point t wi tis ppens we onsier eqution = tn 45. 3 + = 3 + = ( + )( + ) = = or = ± 5 Te onl eptle solution is = + 5 (.6) s te oter two possiilities give negtive vlues for n we re onl onsiering positive vlues for. Eerise 9C On wi of te following urves is positive for ll vlues of?

Cpter 9 Differentition of Polnomils 59 e On wi of te following urves is negtive for ll vlues of? e f 3 For te funtion f () = ( ) fin te vlues of for wi: f () = f () = f () > f () < e f () = Emple 5 4 For te grp of = () sown ere fin: {: () > } {: () < } {: () = } 3, 5 (4, 6), 3

5 Essentil Mtemtil Metos &CAS 5 Wi of te grps lelle A F orrespon to e of te grps lelle f? A D e B E f C F Emple 5 6 For te grp of = f () fin: {: f () > } {: f () < } {: f () = } (.5, 3) (, ) Emple 6 7 Sket te grp of = f () for e of te following (3, 4) (3, 4) = f () 5 3 = f () (, 3) Emple 7 8 Fin te oorintes of te points on te urve = 5 + 6twi te tngent: mkes n ngle of 45 wit te positive iretion of te -is, i.e. were te grient is is prllel to te line = 3 + 4.

Cpter 9 Differentition of Polnomils 5 Emple 7 9 Fin te oorintes of te points on te prol = 6twi: te grient is zero te tngent is prllel to te line + = 6. Use lultor to plot te grp of = f () = were: f () = sin f () = os f () = A r moves w from set of trffi ligts so tt te istne S(t) metres overe fter t seons is moelle S(t) = (.)t 3. Fin its spee fter t seons. Wt will its spee e wen t =, 3, 5? A roket is lune from Cpe York Peninsul so tt fter t seons its eigt (t) metres is given (t) = t, t 5. After minutes tis moel is no longer pproprite. Fin te eigt n te spee of te roket wen t = 5. After ow long will its spee e m/s? 3 Te urve wit eqution = + s grient of 3 t te point (, ). Fin te vlues of n. Fin te oorintes of te point were te grient is. 9.4 Limits n ontinuit Limits We onsier te it of funtion f ()toete vlue tt f () pproes s pproes given vlue. f () = p mens tt, s pproes, f () pproes p. Itisimportnt to unerstn tt it is possile to get s lose s esire to p s pproes. Note tt f () m or m not e efine t =. For mn funtions f ()isefine, so to evlute te it we simpl sustitute te vlue into te rule for te funtion. Emple 8 If f () = 3 fin 3. Sine f () = 3 is efine t = 3 = 3() = If te funtion is not efine t te vlue for wi te it is to e foun, ifferent proeure is use.

5 Essentil Mtemtil Metos &CAS Emple 9 For f () = 5 +,, fin f (). Oserve tt f ()isefine for R \ {}. Emine te eviour of f () for vlues of ner. < > f (.7) =.4 f (.3) = 3.6 f (.8) =.6 f (.) = 3.4 f (.9) =.8 f (.) = 3. f (.99) =.98 f (.) = 3. f (.999) =.998 f (.) = 3. 3 f From te tle it is pprent tt s tkes vlues loser n loser to, regrless of weter pproes from te left or from te rigt, te vlues of f () eome loser n loser to 3. i.e. f () = 3 Tis m lso e seen oserving tt: ( )( ) f () =, n =,, Te grp of f : R \ {} R, f () = issown. Te following importnt results re useful for te evlution of its: ( f () + g()) = f () + g() i.e. te it of te sum is te sum of te its. kf() = k f (), k eing given numer ( f ()g()) = f () g() i.e. te it of te prout is te prout of te its. f () f () g() = g(), provie g() i.e. te it of te quotient is te quotient of te its.

Cpter 9 Differentition of Polnomils 53 Emple Fin: ( 3 + ) 3 3 7 + ( + )(3 ) e 3 3 5 e Emple ( + ) = + = + = 3 3 3 = 3 ( 3) 3 = 3 = 3 ( )( + ) = = ( + ) = 3 ( + )(3 ) = ( + ) (3 ) = 7 7 = 49 3 3 3 7 + ( )( 5) ( ) = 3 5 3 ( + 5)( 5) = 3 = ( + 5) 8 3 Fin: 5 + (3 + 4) 4( + ) 3 (3 + 4) = (3) + (4) = + 4 = 4 4( + ) = (4) ( + ) = 8 4 = 3 5 + 3 = (5 + ) ( ) 3 3 = 7 = 7 Te nottion of its is use to esrie te eviour of grps, n similr nottion s een use previousl in te ook. Consier f : R \{} R, f () =. Oserve tt s, ot from te left n from te rigt, f () inreses witout oun. Te it nottion for tis is f () =.

54 Essentil Mtemtil Metos &CAS For g: R \{} R, g() =, te eviour of g() spproes from te left is ifferent from te eviour s pproes from te rigt. Wit it nottion tis is written s: g() = n f () = + Now emine tis funtion s te mgnitue of eomes ver lrge. It n e seen tt s inreses witout oun troug positive vlues, te orresponing vlues of g() ppro zero. Likewise s ereses witout oun troug negtive vlues, te orresponing vlues of g() lso ppro zero. Smolill tis is written s: g() = g() = n g() = Mn funtions ppro iting vlue or it s pproes ±. Left n rigt its An ie wi is useful in te following isussion is te eistene of its from te left n from te rigt. If te vlue of f () pproes te numer p s pproes from te rigt-n sie, ten it is written s + f () = p. If te vlue of f () pproes te numer p s pproes from te left-n sie, ten it is written s f () = p. Te it s pproes eists onl if te its from te left n te rigt ot eist n re equl. Ten f () = p. Te following is n emple of wen te it oes not eist for prtiulr vlue. 3 for < Let f () = 5 for = 6 for < It is ler from te grp of f tt te f () oes not eist. However, if is llowe to ppro from te left, ten f () pproes. On te oter n if is llowe to ppro from te rigt, ten f () pproes 6. Also note tt f () = 5. 6 5 4 3

Cpter 9 Differentition of Polnomils 55 Continuit t point: informl efinition A funtion wit rule f ()issi to e ontinuous wen = if te grp of = f () n e rwn troug te point wit oorintes (, f ()) witout rek. Oterwise tere is si to e isontinuit t =. Most of te funtions onsiere in tis ourse re ontinuous for teir omins. A more forml efinition of ontinuit follows. A funtion f is ontinuous t point if f (), re equl. Or equivlentl: + f () n f () ll eist n A funtion f is ontinuous t te point = if te following tree onitions re met: f ()isefine t = f () eists 3 f () = f () Te funtion is si to e isontinuous t point if it is not ontinuous t tt point. A funtion is si to e ontinuous everwere if it is ontinuous for ll rel numers. Te polnomil funtions re ll ontinuous for R. Te funtion wit rule f () = s isontinuit t =, s f () is not efine. It is ontinuous elsewere in its omin. Hri funtions, s introue in Cpter 6, provie emples of funtions wi ve points of isontinuit were te funtion is efine. Emple Stte te vlues for for wi te funtions wose grps re sown elow ve isontinuit. 3 3 3 Tere is isontinuit t =, s f () = 3ut + Tere is isontinuit t =, s f ( ) = n + ut f () = f () =. f () = ut f ()= n isontinuit t = s f () = n f () = f () = 3. + Tere is isontinuit t =, s f () = n f () =. + f () = ut

56 Essentil Mtemtil Metos &CAS Emple 3 For e of te following funtions stte te vlues of for wi tere is isontinuit n use te efinition of ontinuit in terms of f (), f () n f ()toeplin w e + is isontinuous: { { if if f () = f () = + if < + if < if { f () = + if if < < f () = + if < + if { if e f () = if < f () = ut f () =, terefore tere is isontinuit t = f () = ut f () =, terefore tere is isontinuit t = f ( ) = ut f () =, terefore tere is isontinuit t = + f () = ut f () =, terefore tere is isontinuit t = No isontinuit e No isontinuit Eerise 9D Emples 8 Emple Fin te following its: g j 5 ( 5) 3 6 (t ) t 3 (t + 5) t t t 3 8 e k t + t + t t + f (3 5) ( + ) 4 9 + 3 i 3 + 5 4 l 3 + 6 + 5 For e of te following grps give te vlues of t wi isontinuit ours. Give resons. 6 3 4 5 7

Cpter 9 Differentition of Polnomils 57 Emple 3 3 For e of te following funtions stte te vlues of t wi tere is isontinuit n use te efinition of ontinuit in terms of f (), f () n f () toeplin + w e stte vlue of orrespons to isontinuit. { { 3 if + if f () = f () = + if < + if < if f () = if < < 3 + if 4 Te rule of prtiulr funtion is given elow. For wt vlues of is te grp of tis funtion isontinuous?, < = ( 4) 9, < 7 7, 7 9.5 Wen is funtion ifferentile? f ( + ) f () A funtion f is si to e ifferentile t if eists. Te polnomil funtions onsiere in tis pter re ifferentile for ll. However tis is not true for ll funtions. { if Let f : R R, f () = if < Tis funtion is lle te moulus funtion or solute vlue funtion n it is enote f () =, f is not ifferentile t = : f ( + ) f () > = = < { > = < f ( + ) f () So oes not eist. i.e. f is not ifferentile t =. Note: Te grient to te left of is n to te rigt of te grient is.

58 Essentil Mtemtil Metos &CAS Emple 4 { if Let f : R R, f () = (= ) if < Sket te grp of te erivtive for suitle omin. { f if > () = if < f ()isnot efine t = Emple 5 Drw sket grp of f were te grp of f is s illustrte. Inite were f is not efine. Te erivtive oes not eist t = ; i.e. te funtion is not ifferentile t =. It ws sown in te previous setion tt some ri funtions re ontinuous for R. Tere re ri funtions wi re ifferentile for R. Te smootness of te joins etermines if tis is te se. Emple 6 For te funtion wit following rule fin f () n sket te grp of = f (): { + + if f () = + if < f () = { + if if < In prtiulr f () is efine n is equl to. Also f () =. Te two setions of te grp of = f () join smootl t (, ).

Cpter 9 Differentition of Polnomils 59 Emple 7 For te funtion wit rule { + + if f () = + if < stte te set of vlues for wi te erivtive is efine, fin f () for tis set of vlues n sket te grp of = f (). f () = { + if > if < f () is not efine s te its from te left n rigt re not equl. Te funtion is ifferentile for R\{}. Eerise 9E Emples 4 6 In e of te figures elow funtion grp f is given. Sket te grp of f.oviousl our sket of f nnot e et, ut f () soul e t vlues of for wi te grient of f is zero; f () soul e < were te originl grp slopes ownwr, n so on. f f 4 4 f e f f f f Emple 6 For te funtion wit following rule fin f () n sket te grp of = f (): { + 3 + if f () = 3 + if <

53 Essentil Mtemtil Metos &CAS Emple 7 Emple 7 3 For te funtion wit following rule stte te set of vlues for wi te erivtive is efine, fin f () for tis set of vlues n sket te grp of = f (): { + + if f () = + 3 if < 4 For te funtion wit following rule stte te set of vlues for wi te erivtive is efine, fin f () for tis set of vlues n sket te grp of = f (): { 3 + if f () = + 3 if <

Cpter 9 Differentition of Polnomils 53 Cpter summr Te nottion for it s pproes is written s. Te following re importnt results for its: ( f () + g()) = f () + g() i.e. te it of te sum is te sum of te its. kf() = k f (), k eing given numer ( f ()g()) = f () g() i.e. te it of te prout is te prout of te its. f () g() = f () g(), provie g() i.e. te it of te quotient is te quotient of te its. A funtion f is efine s ontinuous t te point = if tree onitions re met: f ()isefine t = f ()eists 3 f () = f () Te funtion is si to e isontinuous t point if it is not ontinuous t tt point. We s tt funtion is ontinuous everwere if it is ontinuous for ll rel numers. For te grp of = f ()ofte funtion f : R R: f ( + ) f () Te grient of te or PQ =. Te grient of te grp t P is given f ( + ) f (). Tis it gives rule for te erive funtion enote f,were f : R R n f f ( + ) f () () = Te generl rule of te erive funtion of f () = n, n =,, 3,... For f () = n, f = n n, n =,, 3,... For f () =, f () = P (, f ()) Foremple: For f () =, f () =. For f () = 3, f () = 3. For f () = 4, f () = 4 3. For f () =, f () =. Te erivtive of numer multiple is te multiple of te erivtive: For g() = kf(), were k is onstnt g () = kf () Foremple: g() = 3, g () = 3() = 6 Q ( +, f (+)) Review

53 Essentil Mtemtil Metos &CAS Review Te erivtive of onstnt is lws zero: For g() =, g () = Foremple: f () = 7.3, f () = Te erivtive of te sum is te sum of te erivtives: If f () = g() + () ten f () = g () + (). Foremple: f () = + 3, f () = + 3 g() = 3 + 4 3, g () = 3() + 4(3 ) = 6 + At point (, g()) on te urve = g() te grient is g (). For < te grient is positive, i.e. g () > R(, g()) = te grient is zero, i.e. g () = = g() < < te grient is negtive, i.e. g () < = te grient is zero, i.e. g () = > te grient is positive, i.e. g () > S(, g()) Multiple-oie questions Te grient of te urve = 3 + 4 t te point were = is A B 4 C D 6 E 8 Te grient of te or of te urve = etween te points were = n = + is given A ( + ) B 4 + C 4 D 4 E 4 + 3 If = 4 5 3 +, ten equls A 8 3 5 + B 4 4 5 + C 4 4 D 8 3 5 + E 8 3 5 4 If f () = ( + ) ten f ( ) equls A B C D E 5 5 If f () = ( 3), ten f () equls A 3 B 6 C 6 D + 9 E 6 If = 4 + 9 3, ten equls A 4 3 + 6 B + 3 C + 3 D 8 3 + 8 E 8 3 + 8 3 7 Given tt = 6 + 9, te vlues of for wi re A 3 B > 3 C 3 D 3 E < 3 8 If = 4 36, te points t wi te tngent to te urve is prllel to te -is re A, n 3 B n 3 C 3 n 3 D n 3 E 3, n 3

Cpter 9 Differentition of Polnomils 533 9 Te oorintes of te point on te grp of = + 6 5twi te tngent is prllel to te line = 4 re A (, ) B (, ) C (, ) D (, 4) E (, ) If = 3 + 3 +, ten is equls A 6 + 6 B 6 + 6 C 6 + 3 D 6 + 6 E 6 6 Review Sort-nswer questions (tenolog-free) Fin wen: = 3 + 6 = 5 = ( ) = 4( )(5 + ) e = ( + )(3 ) f = ( + )( 3) Fin wen: ( + 3)( + ) = = = 4 = 3 e = 4 + 3 3 3 For e of te following funtions fin te -oorintes n te grient t te point on te urve for te given vlue of : = +, = =, = = ( + )( 4), = 3 = 3 3, = 4 Fin te oorintes of te points on te urves given te following equtions t wi te grient s te given vlue: = 3 + ; = = 3 6 + 4; = = 3 ; = = 3 + 7; = e = 4 3 + ; = f = ( 3) ; = 5 For te funtion wit rule f () = 3( ) fin te vlues of for wi: f () = f () = f () > f () < e f () > f f () = 3 6 Te urve wit eqution = + s grient of 3 t te point (, ). Fin: te vlues of n te oorintes of te points were te grient is. 7 Sket te grp of = f (). (All etils nnot e etermine ut te is interepts n spe of grp n e etermine.) 5

534 Essentil Mtemtil Metos &CAS Review 8 For te grp of = () fin: {: () > } {: () < } {: () = } Etene-response questions (, 4) (4, 3) = () Te igrm to te rigt sows prt of te grp of ginst. Sket possile spe of ginst over te sme intervl if: = wen = 5 = wen = = wen =. Te grp sown is tt of polnomil of te form P() = 3 + + +. Fin te vlues of,, n. Note: Q(, ) is not turning point. R(, 3) Q(, ) 3 A o moves in pt esrie te eqution = 5 5 + 4,. Units re in kilometres n n re te orizontl n vertil es respetivel. Wt will e te iretion of motion (give te nswer s ngle etween iretion of motion n te -is) wen te -vlue is i km? ii 3 km? Fin vlue of for wi te grient of te pt is 3. 4 A tril over mountin pss n e moelle te urve wit eqution = +.. 3,were n re, respetivel, te orizontl n vertil istnes mesure in kilometres, 3. Fin te grients t te eginning n te en of te tril. Clulte te point were te grient is zero, n lulte lso te eigt of te pss. 5 A tpole egins to swim vertill upwrs in pon n fter t seons it is 5.t 3 m elow te surfe. How long oes te tpole tke to re te surfe, n wt is its spee ten? Wt is te verge spee over tis time? 6 Sow tt te grients of te urve = ( ) t te points (, ) n (, ) onl iffer in sign. Wt is te geometril interprettion for tis? If te grient of te urve = ( )( 5) t te points (, ), (, ) n (5, ) re l, m n n respetivel, sow tt l + m + n =.

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