The First Half of Calculus in 10 (or 13) pages

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1 Limits n Continuity Rtes of chnge n its: The First Hlf of Clculus in 10 (or 13) pges Limit of function f t point = the vlue the function shoul tke t the point = the vlue tht the points ner tell you f shoul hve t f(x) = L mens f(x) is close to L when x is close to (but not equl to) x Ie: slopes of tngent lines secnt line y=f(x) (,f()) tngent line (x,f(x)) The closer x is to, the better the slope of the secnt line will pproximte the slope of the tngent line. The slope of the tngent line = it of slopes of the secnt lines ( through (,f()) ) f(x) = L oes not cre wht f() is; it ignores it x f(x) nee not exist! (function cn t mke up it s min?) x Rules for fining its: If two functions f(x) n g(x) gree (re equl) for every x ner (but mybe not t ), then f(x) = g(x) x x Ex.: x 2 x 2 3x+2 x 2 4 (x 1)(x 2) = x 2 (x+2)(x 2) = x 1 x 2 x+2 = 1 4 If f(x) L n g(x) M s x (n c is constnt), then f(x)+g(x) L+M ; f(x) g(x) L M ; cf(x) cl ; f(x)g(x) LM ; n f(x)/g(x) L/M provie M 0 If f(x) is polynomil, then f(x)= f(x 0 ) x x 0 Bsic principle: to solve, plug in x = x 0! x x 0 If (n when) you get 0/0, try something else! (Fctor (x ) out of top n bottom...) If function hs something like x in it, try multiplying (top n bottom) with x+ (ie: u = x,v =, then x = u 2 v 2 = (u v)(u+v).) 1

2 Snwich Theorem: If f(x) g(x) h(x), for ll x ner (but not t ), n f(x) = h(x) = L, then g(x) = L. x x x One-sie its: Motivtion: the Hevisie function 1 y=h(x) The Hevisie function hs no it t 0; it cn't mke up its min whether to be 0 or 1. But if we just look to either sie of 0, everything is fine; on the left, H(0) `wnts' to be 0, while on the right, H(0) `wnts' to be 1. It's becuse these numbers re ifferent tht the it s we pproch 0 oes not exist; but the `one-sie' its DO exist. Limit from the right: x + f(x) = L mens f(x) is close to L when x is close to, n bigger thn, Limit from the left: x f(x) = M mens f(x) is close to M when x is close to, n smller thn, f(x) = L then mens f(x) = f(x) = L x x + x (i.e., both one-sie its exist, n re equl) Limits t infinity / infinite its: represents something bigger thn ny number we cn think of. f(x) = L mens f(x) is close of L when x is relly lrge. x f(x) = M mens f(x) is close of M when x is relly lrge n negtive. x 1 x = x 1 x = 0 Bsic fct: x More complicte functions: ivie by the highest power of x in the enomentor. f(x),g(x) polynomils, egree of f = n, egree of g = m f(x) x ± g(x) = 0 if n < m f(x) = (coeff of highest power in f)/(coeff of highest power in g) if n = m x ± g(x) f(x) = ± if n > m x ± g(x) f(x) = mens f(x) gets relly lrge s x gets close to x Also hve f(x) = ; f(x) = ; f(x) = ; etc... x x + x Typiclly, n infinite it occurs where the enomentor of f(x) is zero (lthough not lwys) 2

3 Asymptotes: The line y = is horizontl symptote for function f if f(x) or f(x) is equl to. x x I.e., the grph of f gets relly close to y = s x or The line x = b is verticl symptote for f if f ± s x b from the right or left. If numertor n enomentor of rtionl function hve no common roots, then verticl symptotes = roots of enom. Continuity: A function f is continuous (cts) t if x f(x) = f() This mens: (1) x f(x) exists ; (2) f() exists ; n (3) they re equl. Limit theorems sy (sum, ifference, prouct, quotient) of cts functions re cts. Polynomils re continuous t every point; rtionl functions re continuous except where enom=0. Points where function is not continuous re clle iscontinuities Four flvors: removble: both one-sie its re the sme jump: one-sie ts exist, not the sme infinite: one or both one-sie its is or oscillting: one or both one-sie its DNE Intermeite Vlue Theorem: If f(x) is cts t every point in n intervl [,b], n M is between f() n f(b), then there is (t lest one) c between n b so tht f(c) = M. Appliction: fining roots of polynomils Tngent lines: Slope of tngent line = it of slopes of secnt lines; t (,f() : f(x) f() x x Nottion: cll this it f () = erivtive of f t Different formultion: h = x, x = +h f f(+h) f() () = = it of ifference quotient h 0 h If y = f(x) = position t time x, then ifference quotient = verge velocity; it = instntneous velocity. Derivtives The erivtive of function: erivtive = it of ifference quotient (two flvors: h 0, x ) If f () exists, we sy f is ifferentible t Fct: f ifferentible (iff ble) t, then f cts t Using h 0 nottion: replce with x (= vrible), get f (x) = new function 3

4 Or: f f(z) f(x) (x) = z x z x f (x) = the erivtive of f = function whose vlues re the slopes of the tngent lines to the grph of y=f(x). Domin = every point where the it exists Nottion: f (x) = y x = f (f(x)) = x x = y = D x f = Df = (f(x)) Differentition rules: x (constnt) = 0 x (x) = 1 (f(x)+g(x)) = (f(x)) + (g(x)) (f(x)-g(x)) = (f(x)) - (g(x)) (cf(x)) = c(f(x)) (f(x)g(x)) = (f(x)) g(x)+ f(x)(g(x)) ( f(x) g(x) ) = f (x)g(x) f(x)g (x) g 2 (x) (x n ) = nx n 1, for n = nturl number - integer rtionl number e x ) = e x ( x ) = x ln [see below!] [[ (1/g(x)) = -g (x)/(g(x)) 2 ]] Higher erivtives: f (x) is just function, so we cn tke its erivtive! (f (x)) = f (x) (= y = 2 y x 2 = 2 f x 2) = secon erivtive of f Keep going! f (x) = 3r erivtive, f (n) (x) = nth erivtive Rtes of chnge Physicl interprettion: f(t)= position t time t f (t)= rte of chnge of position = velocity f (t)= rte of chnge of velocity = ccelertion f (t) = spee Bsic principle: for object to chnge irection (velocity chnges sign), f (t)= 0 somewhere (IVT!) Exmples: Free-fll: object flling ner erth; s(t) = s 0 +v 0 t g 2 t2 s 0 = s(0) = initil position; v 0 = initil velocity; g= ccelertion ue to grvity Economics: C(x) = cost of mking x objects; R(x) = revenue from selling x objects; P = R C = profit C (x) = mrginl cost = cost of mking one more object R (x) = mrginl revenue ; profit is mximize when P (x) = 0 ; i.e., R (x) = C (x) 4

5 Derivtives of trigonometric functions sinx 1 cosh Bsic it: = 1 ; everything else comes from this! = 0 x 0 x h 0 h Note: this uses rin mesure! sin(bx) = b sin(bx) = b sin(u) = b x 0 x x 0 bx u 0 u Then we get: (sinx) = cosx (cosx) = sinx (tnx) = sec 2 x (cotx) = csc 2 x (secx) = secxtnx (cscx) = cscxcotx The Chin Rule Composition (g f)(x 0 ) = g(f(x 0 )) ; (note: we on t know wht g(x 0 ) is.) (g f) ought to hve something to o with g (x) n f (x) in prticulr, (g f) (x 0 ) shoul epen on f (x 0 ) n g (f(x 0 )) Chin Rule: (g f) (x 0 ) = g (f(x 0 ))f (x 0 ) = ((outsie) evl t insie fcn) ((insie)) Ex: ((x 3 +x 1) 4 ) = (4(x ) 3 )(3x 2 +1) Different nottion: y = g(f(x)) = g(u), where u = f(x), then y x = y u ux Prmetric equtions: generl curve neen t be the grph of function. But we cn imgine ourselves trvelling long curve, n then x = x(t) n y = y(t) re functions of t=time. We still my hve resonble tngent line to the grph, n its slope shoul still be (chnge in y/chnge in x = t t0 y(t) y(t 0 ) x(t) x(t 0 ) = t t 0 (y(t) y(t 0 ))/(t t 0 ) (x(t) x(t 0 ))/(t t 0 ) = t t 0 (y(t) y(t 0 ))/(t t 0 ) t t0 (x(t) x(t 0 ))/(t t 0 ) = y (t 0 ) x (t 0 ) Implicit ifferentition We cn ifferentite functions; wht bout equtions? (e.g., x 2 +y 2 = 1) grph looks like it hs tngent lines tngent line? (,b) 5

6 Ie: Preten eqution efines y s function of x : x 2 +(f(x)) 2 = 1 n ifferentite! 2x+2f(x)f (x) = 0 ; so f (x) = x f(x) = x y Different nottion: x 2 +xy 2 y 3 = 6 ; then 2x+(y 2 +x(2y y x ) 3y2y x = 0 y x = 2x y2 2xy 3y 2 Appliction: exten the power rule x (xr ) = rx r 1 works for ny rtionl number r (y = x p/q mens y q = x p ; ifferentite!) Inverse functions n their erivtives Bsic ie: run function bckwrs y=f(x) ; ssign the vlue x to the input y ; x=g(y) nee g function; so nee f is one-to-one f is one-to-one: if f(x)=f(y) then x=y ; if x y then f(x) f(y) g = f 1, then g(f(x)) = x n f(g(x)) = x (i.e., g f=i n f g=i) fining inverses: rewrite y=f(x) s x=some expression in y grphs: if (,b) on grph of f, then (b,) on grph of f 1 grph of f 1 is grph of f, reflecte cross line y=x horizontl lines go to verticl lines; horizontl line test for inverse erivtive of the inverse: f (f 1 (x)) (f 1 ) (x) = 1 if f() = b, then (f 1 ) (b) = 1/f () Logrithms f(x)= x is either lwys incresing ( > 1) or lwys ecresing ( < 1) inverse is g(x) = log x = lnx ln lnx is the inverse of e x. lnx is log; it turns proucts into sums: ln(b) = ln()+ln(b) ln( b ) = bln() ; ln(/b) = ln() ln(b) e lnx = x n (e x ) = e x, so 1 = (e lnx ) = (e lnx )(lnx) = x(lnx), so (lnx) = 1/x. x (lnx) = 1/x ; x (ln(f(x))) = f (x) This gives us: f(x) Logrithmic ifferentition: f (x) = f(x) x (ln(f(x))) useful for tking the erivtive of proucts, powers, n quotients ln( b ) shoul be bln, so b = e bln b+c = b c ; bc = ( b ) c x = e xln ; x (x ) = x ln x r = e rlnx (mkes sense for ny rel number r) ; 6 x (xr ) = e rlnx (r)( 1 x ) = rxr 1

7 Inverse trigonometric functions Trig functions (sinx, cosx, tnx, etc.) ren t one-to-one; mke them! sinx, π/2 x π/2 is one-to-one; inverse is Arcsin x sin(arcsin x)=x, ll x; Arcsin(sinx)=x IF x in rnge bove tnx, π/2 < x < π/2 is one-to-one; inverse is Arctn x tn(arctn x)=x, ll x; Arctn(tnx)=x IF x in rnge bove secx, 0 x < π/2 n π/2 < x π, is one-to-one; inverse is Arcsec x sec(arcsecx)=x, ll x; Arcsec(sec x)=x IF x in rnge bove Computing cos(arcsin x), tn(arcsec x), etc.; use right tringles The other inverse trig functions ren t very useful, they re essentilly the negtives of the functions bove. Derivtives of inverse trig functions They re the erivtives of inverse functions! Use right tringles to simplify. x (rcsinx) = 1 cos(rcsin(x)) = 1 1 x 2 x (rctnx) = 1 sec 2 (rctnx) = 1 x 2 +1 x (rcsec x) = 1 sec(rcsec x)tn(rcsec x) = 1 x x 2 1 Relte Rtes Ie: If two (or more) quntities re relte ( chnge in one vlue mens chnge in others), then their rtes of chnge re relte, too. xyz = 3 ; preten ech is function of t, n ifferentite (implicitly). Generl proceure: Drw picture, escribing the sitution; lbel things with vribles. Which vribles, rtes of chnge o you know, or wnt to know? Fin n eqution relting the vribles whose rtes of chnge you know or wnt to know. Differentite! Plug in the vlues tht you know. Liner pproximtion n ifferentils Ie: Thetngentlinetogrphoffunctionmkesgoopproximtiontothefunction, ner the point of tngency. Tngent line to y = f(x) t (x 0,f(x 0 ) : L(x) = f(x 0 )+f (x 0 )(x x 0 ) f(x) L(x) for x ner x 0 Ex.: (27 25), using f(x) = x (1+x) k 1+kx, using x 0 =0 f = f(x 0 + x) f(x 0 ), then f(x 0 + x) L(x 0 + x) trnsltes to f f (x 0 ) x ifferentil nottion: f = f (x 0 )x So f f, when δx = x is smll 7

8 In fct, f f = (iffrnce quot f (x 0 )) x = (smll) (smll) = relly smll, goes like ( x) 2 Applictions of Derivtives Extreme Vlues c is n (bsolute) mximum for function f(x) if f(c) f(x) for every other x is n (bsolute) minimum for function f(x) if f() f(x) for every other x mx or min = extremum Extreme Vlue Theorem: If f is continuous function efine on close intervl [,b], then f ctully hs mx n min. Gol: figure out where they re! c is reltive mx (or min) if f(c) is f(x) (or f(x)) for every x ner c. Rel mx or min = rel extremum. An bsolute extremum is either rel extremum or n enpoint of the intervl. c is criticl point if f (c) = 0 or oes not exist. A rel extremum is criticl point. So bsolute extrem occur either t criticl points or t the enpoints. So to fin the bs mx or min of function f on n intervl [,b] : (1) Tke erivtive, fin the criticl points. (2) Evlute f t ech criticl point n enpoint. (3) Biggest vlue is mximum vlue, smllest is minimum vlue. The Men Vlue Theorem You cn (lmost) recrete function by knowing its erivtive Men Vlue Theorem: if f is continuous on [,b] n ifferentible on (,b), then there is t lest one c in (,b) so tht f (c) = f(b) f() b Consequences: Rolle s Theorem: f() = f(b) = 0; between two roots there is criticl point. So: If function hs no criticl points, it hs t most one root! A function with f (x)=0 is constnt. Functions with the sme erivtive (on n intervl) iffer by constnt. The First Derivtive Test f is incresing on n intervl if x > y implies f(x) > f(y) f is ecresing on n intervl if x > y implies f(x) < f(y) If f (x) > 0 on n intervl, then f is incresing If f (x) < 0 on n intervl, then f is ecresing Locl mx s / min s occur t criticl points; how o you tell them prt? Ner locl mx, f is incresing, then ecresing; f (x) > 0 to the left of the criticl point, n f (x) < 0 to the right. 8

9 Ner locl min, the opposite is true; f (x) < 0 to the left of the criticl point, n f (x) > 0 to the right. If the erivtive oes not chnge sign s you cross criticl point, then the criticl point is not rel extremum. Bsic use: plot where function is incresing/ecresing: plot criticl points; in between them, sign of erivtive oes not chnge. The secon erivtive test n grphing When we look t grph, we see where function is incresing/ecresing. We lso see: f is concve up on n intervl if f (x) > 0 on the intervl Mens: f is incresing; f is bening up. f is concve own on n intervl if f (x) < 0 on the intervl Mens: f is ecresing; f is bening own. A point where the concvity chnges is clle point of inflection Grphing: Fin where f (x) n f (x) re 0 or DNE Plot on the sme line. In between points, erivtive n secon erivtive on t chnge sign, so grph looks like one of: f ' f '' ecresing, concve own ecresing, concve up incresing, concve own incresing, concve up Then string together the pieces! Use informtion bout verticl n horizontl symptotes to finish sketching the grph. Secon erivtive test: If c is criticl point n f (c) > 0, then c is rel min (smiling!) f (c) < 0, then c is rel mx (frowning!) Newton s metho: A relly fst wy to pproximte roots of function. Ie: tngent line to the grph of function points towrs root of the function. But: roots of (tngent) lines re computtionlly strighforwr to fin! 9

10 L(x) = f(x 0 )+f (x 0 )(x x 0 ) ; root is x 1 = x 0 f(x 0) f (x 0 ) Now use x 1 s strting point for new tngent line; keep repeting! x n+1 = x n f(x n) f (x n ) Bsic fct: if x n pproximtes root to k eciml plces, then x n+1 tens to pproximte it to 2k eciml plces! BUT: Newton s metho might fin the wrong root: Int Vlue Thm might fin one, but N.M. fins ifferent one! Newton s metho might crsh: if f (x n ) = 0, then we cn t fin x n+1 (horizontl lines on t hve roots!) Newton s metho might wner off to infinity, if f hs horizontl symptote; n initil guess too fr out the line will generte numbers even frther out. Newton s metho cn t fin wht oesn t exist! If f hs no roots, Newton s metho will try to fin the function s closest pproch to the x-xis; but everytime it gets close, nerly horizontl tngent line sens it zooming off gin... Optimiztion This is relly just fining the mx or min of function on n intervl, with the e compliction tht you nee to figure out which function, n which intervl! Solution strtegy is similr to relte rtes problem: Drw picture; lbel things. Wht o you nee to mximize/minimize? Write own formul for the quntity. Use other informtion to einte vribles, so your quntity epens on only one vrible. Determine the lrgest/smllest tht the vrible cn resonbly be (i.e., fin your intervl) Turn on the mx/min mchine! L Hôpitl s Rule sinx ineterminte forms: its which evlute to 0/0 ; e.g. x 0 x LR# 1: If f() = g() = 0, f n g both ifferentible ner, then f(x) x g(x) = f (x) x g (x) Note: we cn repetely pply L Hôpitl s rule to compute it, so long s the conition tht top n bottom both ten to 0 hols for the new it. Once this oesn t hol, L Hôpitl s rule cn no longer be pplie! Other ineterminte forms: LR#2: if f,g s x, then, 0,, 00, 1, 0 f(x) x g(x) = f (x) x g (x) Other cses: try to turn them into 0/0 or /. In the 0 cse, we cn o this by throwing one fctor or the other into the enomentor (whichever is more trctble. In the lst three cses, o this by tking logs, first. 10

11 Integrtion Antierivtives. Integrl clculus is ll bout fining res of things, e.g. the re between the grph of function f n the x-xis. This will, in the en, involve fining function F whose erivtive is f. F is n ntierivtive (or (inefinite) integrl) of f if F (x) =f(x). Nottion: F(x) = f(x) x ; it mens F (x)=f(x) ; the integrl of f of x ee x Every ifferentition formul we hve encountere cn be turne into n ntiifferentition formul; if g is the erivtive of f, then f is n ntierivtive of g. Two functions with the sme erivtive (on n intervl) iffer by constnt, so ll ntierivtives of function cn be foun by fining one of them, n then ing n rbitrry constnt C. Bsic list: x n x = xn+1 n+1 + C (provie n 1) 1/x x = ln x +C cos(kx sin(kx) sin(kx) x = + C cos(kx) x = + C k k sec 2 x x = tnx + C csc 2 x x = cotx + C secxtnx x = secx + C cscxcotx x = cscx + C e x x = e x +C Most ifferentition rules cn be turne into integrtion rules (lthough some re hrer thn others; some we will wit while to iscover). Bsic integrtion rules: sum n constnt multiple rules re strighforwr to reverse: for k=constnt, k f(x) x = k f(x) x (f(x)±g(x) x = f(x) x ± g(x) x Sums n Sigm Nottion. Ie: lot of things cn estimte by ing up lot of tiny pieces. Sigm nottion: i = 1 + n ; just the numbers up Forml properties: k i = k i n ( i ±b i ) = Some things worth ing up: length of curve: pproximte curve by collection of stright line segments length of curve (length of line segments) istnce trvelle = (verge velocity)(time of trvel) over short perios of time, vg. vel. instntneous vel. so istnce trvelle (inst. vel.)(short time intervls) Averge vlue of function: i ± Averge of n numbers: the numbers, ivie by n. For function, up lots of vlues of f, ivie by number of vlues. 11 b i

12 vg. vlue of f 1 n f(c i ) Are n Definite Integrls. Probbly the most importnt thing to pproximte by sums: re uner curve. Ie: pproximte region b/w curve n x-xis by things whose res we cn esily clculte: rectngles! y=f(x) b Are between grph n x-xis (res of the rectngles) = f(c i ) x i where c i is chosen insie of the i-th intervl tht we cut [,b] up into. This is Riemnn sum for the function f on the intervl [,b].) We efine the re to be the it of these sums s the lengths of the subintervls gets smll (so the number of rectngles goes to, n cll this the efinite integrl of f from to b: f(x) x = f(c i ) x i n More precisely, we cn t ll Riemnn sums, n look t wht hppens when the length x i of the lrgest subintervl (cll it ) gets smll. If the Riemnn sums ll pproximte some number I when is smll enough, then we cll I the efinite integrl of f from to b. But when o such its exist? Theorem If f is continuous on the intervl [,b], then (i.e., the re uner the grph is pproximte by rectngles.) f(x) x exists. But this isn t how we wnt to compute these integrls! Limits of sums is very cumbersome. Inste, we try to be more systemtic. Properties of efinite integrls: First note: the sum use to efine efinite integrl oesn t nee to hve f(x) 0; the it still mkes sense. When f is bigger thn 0, we interpret the integrl s re uner the grph. 12

13 Bsic properties of efinite integrls: b b f(x) x =0 kf(x) x = k f(x) x + c b f(x) x f(x) x = c f(x) x If m f(x) M for ll x in [,b], then m(b ) More generlly, if f(x) g(x) for ll x in [,b], then f(x)±g(x) x = b b f(x) x = f(x) x ± f(x) x M(b ) f(x) x g(x) x b f(x) x g(x) x Averge vlue of f : formlize our ol ie! vg(f) = 1 f(x) x b Men Vlue Theorem for integrls: If f is continuous in [,b], then there is c in [,b] so tht f(c) = 1 f(x) x b The funmentl theorems of clculus. Formlly, x f(x) x epens on n b. Mke this explicit: f(t) t = F(x) is function of x. F(x) = the re uner the grph of f, from to x. Fun. Thm. of Clc (# 1): If f is continuous, then F (x) = f(x) ntierivtive of f!) Since ny two ntierivtives iffer by constnt, n F(b) = f(t) t, we get (F is n Fun. Thm. of Clc (# 2): If f is continuous, n F is n ntierivtive of f, then Ex: f(x) x = F(b) F() = F(x) b π 0 sinx x = ( cosπ) ( cos0) =2 FTC # 2 mkes fining ntierivtives very importnt! FTC # 1 gives metho for builing ntierivtives: x F(x)= sint t is n ntierivtive of f(x) = sinx x 3 G(x) = 1+t2 t = F(x 3 ) F(x 2 ), where x 2 F (x) = 1+x 2, so G (x) = F (x 3 )(3x 2 ) F (x 2 )(2x)... 13