MUST-KNOW MATERIAL FOR CALCULUS

MUST-KNOW MATERIAL FOR CALCULUS MISCELLANEOUS: intervl nottion: (, b), [, b], (, b], (, ), etc. Rewrite ricls s frctionl exponents: 3 x = x 1/3, x3 = x 3/2 etc. An impliction If A then B is equivlent to its contrpositive If (not B) then (not A) To go from grph of y = f(x) to grph of x = f(y) : tke the (fmilir) grph of y = f(x), rotte 90 egrees clockwise, then flip bout the horizontl xis. TEST POINT METHOD: for solving f(x) > 0 : there re only two types of plces where function cn chnge from positive to negtive (or vice vers): where it equls zero, or t brek. Locte ll such plces, n check the resulting subintervls. GEOMETRY: Circle with rius r : AREA = πr 2, CIRCUMFERENCE = 2πr, DIAMETER = 2r Sphere with rius r : VOLUME = 4 3 πr3, SURFACE AREA = 4πr 2 Are of tringle: bse b n height h, AREA = 1 2 bh sies n b with inclue ngle θ : AREA = 1 2b sin θ Tringles: ngles sum to 180 ; longest sie is opposite biggest ngle, etc. Consier n rbitrry tringle with ngles A, B, C n opposite sies, b, c : sin A Lw of Sines: = sin B = sin C b c Lw of Cosines: 2 = b 2 + c 2 2bc cos A Similr tringles: hve the sme ngles; scling fctor in going from one to the other Right tringles: hve 90 ngle; longest sie is clle the hypotenuse; the Pythgoren Theorem: 2 + b 2 = c 2 Trpezoi with bses b 1 n b 2 n height h : AREA = (b 1+b 2 ) 2 h (verge the bses n multiply by the height) cyliner (2 prllel congruent plne figures of re A, perpeniculr istnce between plnes is h): VOLUME = Ah right circulr cyliner: VOLUME = πr 2 h cone ( plne figure of re A, point, ll lines connecting; h is perpeniculr istnce from point to plne): VOLUME = 1 3 Ah right circulr cone: VOLUME = 1 3 πr2 h http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 1

TRIGONOMETRY: RADIAN MEASURE: the rin mesure of n ngle is the length of the rc on the unit circle: positive is counterclockwise. Right tringle efinitions: SOHCAHTOA Unit circle efinitions; ly off ngle x sin x is the y-vlue of the point cos x is the x-vlue of the point tn x = sin x cos x cot x = 1 tn x = cos x sin x sec x = 1 cos x csc x = 1 sin x sin 1 x = rcsin x is the ngle between π 2 n π 2 whose sine is x. cos 1 x = rccos x is the ngle between 0 n π whose cosine is x. tn 1 x = rctn x is the ngle between π 2 n π 2 Note: sin 2 x mens ( sin x ) 2 etc. whose tngent is x. Double-ngle formuls: sin 2x = 2 sin x cos x ; cos 2x = cos 2 x sin 2 x the Pythgoren Ientity: sin 2 x + cos 2 x = 1 Two specil tringles: 30-60 -90 n 45-45 -90 FUNCTIONS: f(x) is the output from the function f when the input is x. Functions hve the property tht ech input hs exctly one corresponing output. ZERO of function: n input whose output is zero DOMAIN of function: set of llowble inputs RANGE of function: its output set GRAPH of function: the picture of its (input,output) pirs GRAPHS of BASIC MODELS: constnt (y = k) y = x 2 n higher powers y = x 3 n higher powers y = 1 x y = x y = x http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 2

y = e x y = ln x y = sin x y = cos x y = tn x y = sec x y = [[x]], the gretest integer function, [[x]] is the gretest integer less thn or equl to x COMPOSITIONS OF FUNCTIONS: f(g(x)) mens g cts first, f cts lst EVEN function: f( x) = f(x) ; when inputs re opposites, outputs re the sme ODD function: f( x) = f(x) ; when inputs re opposites, outputs re opposites ONE-TO-ONE FUNCTION: Ech output hs exctly one corresponing input; grph psses both horizontl n verticl line test; the inputs n outputs cn be tie together with strings INVERSE FUNCTIONS: If f is 1-1, then its inverse f 1 unoes wht f i: f(f 1 (x)) = x n f 1 (f(x)) = x the omins n rnges of f n f 1 re switche the grphs of f n f 1 re reflections bout the line y = x if (, b) is on the grph of f, then (b, ) is on the grph of f 1 TRANSFORMATIONS of functions: strt with y = f(x) working with y is intuitive: y = f(x) + 3 moves up 3 y = 3f(x) multiplies ll y-vlues by 3 (verticl stretch) y = f(x) multiplies the y-vlues by 1 ; reflects bout the x-xis working with x is counter-intuitive: y = f(x 3) ; replce every x with x 3 ; moves to the RIGHT 3 y = f(3x) ; replce every x with 3x ; (, b) ( 3, b) ; horizontl compression y = f( x) ; replce every x with x ; reflects bout the y-xis LINES: liner functions: y = mx + b or x + by + c = 0 ; chnges in y equl chnges in x give rise to equl slope: m = y 2 y 1 x 2 x 1 ; if m = 3, then the y-vlues re chnging 3 times s fst s the x-vlues point-slope form: y y 1 = m(x x 1 ) prllel lines hve the sme slope; perpeniculr lines hve slopes tht re opposite reciprocls horizontl lines: y = c ; hve zero slope verticl lines: x = c ; hve no slope http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 3

QUADRATIC FUNCTIONS: f(x) = x 2 + bx + c ; grph s prbols; > 0 hols wter, < 0 shes wter vertex: t x = b 2 POLYNOMIALS: Let P (x) = 0 + 1 x + 2 x 2 + + n x n. egree of P : highest exponent As x ±, polynomil looks like its highest power term. The following re equivlent: c is zero of P x c is fctor of P (x) P (c) = 0 the point (c, 0) is on the grph of P the grph of P crosses the x-xis t c x c goes into P (x) evenly (reminer = 0) EXPONENTIAL FUNCTIONS: Allowble bses: b > 0, b 1 y = b t Incresing when b > 1 ; ecresing when 0 < b < 1 Common form: A(t) = A 0 e kt ; A 0 is the mount t time 0 Every exponentil function cn be written with ANY llowble bse, so use whtever bse is convenient. For equl chnges in x, y gets MULTIPLIED by constnt (tht epens both on the bse of the exponentil function, n the chnge in x) Doubling time: for n incresing exponentil function, it lwys tkes the sme mount of time for quntity to ouble hlf-life: for ecresing exponentil function, it lwys tkes the sme mount of time for quntity to be cut in hlf LOGARITHMS: y = log b x Allowble bses: b > 0, b 1 Incresing when b > 1 ; ecresing when 0 < b < 1 Lws work for ll llowble bses: ln x = log e x is the nturl logrithm ln xy = ln x + ln y (the log of prouct is the sum of the logs) ln x y = ln x ln y (the log of quotient is the ifference of the logs) ln x y = y ln x (you cn bring powers own) chnge-of-bse formul: log b x = log x log b http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 4

y = e x n y = ln x re inverse functions; use this ie to solve exponentil n logrithmic equtions A log is n exponent! log 3 5 is the POWER tht 3 must be rise to, to get 5 EXPONENTIAL FUNCTIONS grow fster thn POWER FUNCTIONS grow fster thn LOGARITHMIC FUNCTIONS ABSOLUTE { VALUE: x if x 0 x = x if x < 0 For c > 0, x < c c < x < c x > c x > c or x < c x = c x = ±c 0 < x c < δ x (c δ, c) (c, c + δ) (puncture neighborhoo bout c) LIMITS: Consier the limit sttement: lim f(x) = l x c low-level unerstning: when x is close to c, f(x) is close to l higher level: we cn mke the vlues of f(x) s close to l s we like, by tking x to be sufficiently close to c, but not equl to c Precisely: ɛ > 0 δ > 0 s.t. if 0 < x c < δ then f(x) l < ɛ When we evlute limit s x c, we never let x equl c x c + mens x pproches c from the right-hn sie x c mens x pproches c from the left-hn sie LIMIT LAWS: Work nicely! Proviing the iniviul limits exist, the limit of sum is the sum of the limits (sme for ifference, proucts, quotients, etc.) If you hve continuous function (see below) then evluting limit is s esy s DIRECT SUBSTITUTION. BE CAREFUL! If you re working with iscontinuous function (e.g., gretest integer function, some piecewise-efine functions), then irect substitution MAY NOT WORK. Try l Hospitl s rule, renming, grphing, etc. sin x An importnt limit: lim x 0 x = 1 http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 5

CONTINUITY: Low-level unerstning: no breks in the grph higher level: when inputs re close, outputs re close The following re equivlent: f is continuous t c lim x c f(x) = f(c) (when f is continuous t c, then evluting the limit is s esy s irect substitution) lim h 0 f(c + h) = f(c) As x c, f(x) f(c) INTERMEDIATE VALUE THEOREM: Suppose f in continuous on [, b], n N is number between f() n f(b). Then there exists number c between n b for which f(c) = N. (If grph hs no breks, n you trvel long the grph from one point to nother, you must pss through ALL the y-vlues in between; i.e., ll the intermeite vlues.) EXTREME VALUE (MAX/MIN) THEOREM: Let f be continuous on close intervl [, b]. Then f ttins both n bsolute mximum vlue f(c) n bsolute minimum vlue f() for some c n in [, b]. (This theorem GUARANTEES the existence of highest n lowest point on grph uner pproprite conitions.) MEAN VALUE THEOREM: Let f be ifferentible on [, b]. Then there exists number c between n b for which f f(b) f() (c) =. b (This theorem gurntees plce where the instntneous rte of chnge is the sme s the verge rte of chnge uner pproprite conitions.) http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 6

DERIVATIVES: The following re equivlent: f (c) = m f(c + h) f(c) lim = m h 0 h The slope of the tngent line to the grph of f t the point (c, f(c)) is m f is ifferentible t c (n the vlue of the erivtive is m) f(x) f(c) lim = m x c x c The instntneous rte of chnge of f t (c, f(c)) is m When x = c, the function vlues re chnging m times s fst s the inputs Suppose f (2) = 5. Roughly: when x chnges by 1 (from 2 to 3), we expect y to go up by ABOUT 5. Or, when x chnges by 1 (from 2 to 1), we expect y to go own by ABOUT 5. The UNITS of f (c) re the units of f(x) (the outputs from f) ivie by the units of x (the inputs to f) If function is ifferentible, then its grph is SMOOTH: it hs non-verticl tngent lines everywhere. A function is NON-DIFFERENTIABLE t: verticl tngent lines; kinks; iscontinuities Theorem: If f is ifferentible t x, then f is continuous t x Contrpositive: If f is not continuous t x, then f is not ifferentible t x Leibnitz nottion versus prime nottion: y = y x, y = 2 y x etc. 2 Liner Approximtion (lineriztion): function is best pproximte t point by its tngent line: t (c, f(c)) we hve: L(x) = f(c) + f (c)(x c) f(x) If f (x) > 0 then f is incresing If f (x) < 0 then f is ecresing If the SLOPES re INCREASING (f incresing; f > 0) then f is concve up If the SLOPES re DECREASING (f ecresing; f < 0) then f is concve own Remember: function cn increse in bsiclly three ifferent wys: linerly, concve up; concve own Inflection point: where the concvity chnges (from concve up to own, or own to up): cnites re where f (c) = 0 or f (c) oes not exist. LOCAL MAX/MIN: A locl mx/min for function cn only occur t three types of plces (clle the criticl points ): where f (c) = 0 where f (c) oes not exist t ENDPOINTS of omin of f http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 7

So, to fin mx/min, locte ll cnites, n check them. Creful: criticl point is not necessrily mx or min! FIRST DERIVATIVE TEST: Check signs of first erivtive to the left/right of cnite (where the function is continuous) to ecie if it is mx or min. Why is continuity neee? See the sketch below the test woul tell us tht there s locl mx t c! SECOND DERIVATIVE TEST: If concve up t cnite; it s min. If concve own t cnite, it s mx. AVERAGE RATE OF CHANGE: The verge rte of chnge of f on the intervl f(b) f() [, b] is ; this is the slope of the line between (, f()) n (b, f(b)) b DIFFERENTIATION FORMULAS: Be ble to GENERALIZE ll these formuls: replce x by f(x), n multiply by f (x) ) x xn = nx n 1 n ( ) n 1 generlize: x( f(x) = n f(x) f (x) x cf(x) = c xf(x) (you cn slie constnts out) the erivtive of sum/ifference is the sum/ifference of the erivtives x ex = e x (the y-vlue of the point tells you how fst the function is chnging t tht point) PRODUCT RULE: x f(x)g(x) = f(x)g (x) + g(x)f (x) (the erivtive of prouct is NOT!! NOT!! NOT!! the prouct of the erivtives) f(x) QUOTIENT RULE: x g(x) = g(x)f (x) f(x)g (x) (g(x)) 2 NOT!! NOT!! NOT!! the quotient of the erivtives) (the erivtive of quotient is x MUST BE MEASURED IN RADIANS FOR THESE FORMULAS TO BE CORRECT: x x sin x = cos x cos x = sin x x tn x = sec2 x x cot x = csc2 x x x sec x = sec x tn x csc x = csc x cot x Chin Rule: x x = x ln x f(g(x)) = f (g(x) g (x) ; how to ifferentite composite functions x rcsin x = x sin 1 x = 1 1 x 2 http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 8

x rccos x = 1 x cos 1 x = 1 x 2 x rctn x = x tn 1 x = 1 1 + x 2 x ln x = 1 x x log x = 1 x ln erivtive of n inverse function: (f 1 ) (x) = 1 f (f 1 (x)) So if (, b) is on the grph of f with slope of tngent line m, then (b, ) is on the grph of f 1 with slope of tngent line 1 m! IMPLICIT DIFFERENTIATION: Whenever you see y, tret it s function of x n ifferentite ccoringly. For exmple: For exmple: x y2 = 2y y x y xxy = x x + y LOGARITHMIC DIFFERENTIATION: Use this to ifferentite complicte proucts or quotients; lso to ifferentite vrible stuff rise to vrible power. First tke logs, then ifferentite! PARTICLE MOVING ON A NUMBER LINE: Let s(t) enote the position t time t. Then, s (t) = v(t) is the velocity; positive is moving to the right; negtive to the left. s (t) = v (t) = (t) is the ccelertion. Speeing up mens moving to the right fster n fster (v(t) > 0 n (t) > 0) or moving to the left fster n fster (v(t) < 0 n (t) < 0). Thus, the prticle spees up when velocity n ccelertion hve the sme sign (both positive, or both negtive). Note: spee = v(t) Suppose you re given the velocity of prticle trveling long number line, v(t). Then, totl istnce trvele from t 1 to t 2 is given by t 2 t 1 v(t) t ; i.e., integrte the spee. However the totl DISPLACEMENT is t 2 t 1 v(t) t = s(t 2 ) s(t 1 ). Notice tht if you strt t 0, move to the right 5 n then to the left 5, your totl isplcement is 0 but the totl istnce trvele is 10. RELATED RATE PROBLEMS: Ask: Wht is chnging with time? Rtes re erivtives! Write own SOMETHING THAT IS TRUE tht involves wht you re intereste in. (Look for: similr figures, right tringles, etc.) Remember: if x is chnging with time, then the erivtive of x 2 is 2x x t. http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 9

OPTIMIZATION PROBLEMS: Fin the CANDIDATES for locl mx/min: enpoints, plces where the erivtive is zero, plces where the erivtive oesn t exist. Use the First Derivtive Test or Secon Derivtive Test to check whether they re mx or min. If you wnt n ABSOLUTE mx/min, fin ALL the locl mx/min, n choose the highest/lowest from these. Remember, you CAN T USE YOUR CALCULATOR to locte mx/mins; this is NOT n llowble opertion! ANTIDERIVATIVES: A function F (x) is n ANTIDERIVATIVE of f(x) if n only if F (x) = f(x) ; i.e., F is function whose erivtive is f. Antierivtives uno erivtives. An ntierivtive hs specifie erivtive, n this erivtive etermines the shpe, but not the verticl trnsltion. So, if you hve ONE ntierivtive, then you hve n infinite number they ll iffer by constnt. The symbol f(x) x enotes ll the ntierivtives of f(x). x n x = 1 n+1 xn+1 + C for n 1 1 x = ln x + C sin x x = cos x + C cos x x = sin x + C e kx x = 1 k ekx + C 1 1+x 2 x = tn 1 x + C 1 1 x 2 x = sin 1 x + C Any CONTINUOUS function f hs n ntierivtive: the function x f(t) t is n ntierivtive of f(x). Tht is, the function tht fins AREA uner the grph of f is n ANTIDERIVATIVE of f! DEFINITE INTEGRALS: the efinite integrl of f from to b is enote by b f(x) x n is efine s follows: Divie [, b] into n equl subintervls, ech of length x = b n. Choose x i from the ith subintervl. Then, b f(x) x = lim n n f(x i ) x You cn pproximte efinite integrls with rectngles (left-hn; right-hn; mipoint), with trpezois, even with prbols. The efinite integrl gives informtion bout the (signe) re trppe between the grph of f n the x-xis: re bove is trete s positive; re below is negtive. i=1 http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 10

Cution: if b f(x) x = 0, this only mens tht there is the sme mount of re ABOVE the x-xis s BELOW on the intervl [, b]. When you hve efinite integrl problem tht cn be solve with simple geometry formuls (tringles, trpezois, circles) then USE GEOMETRY to fin the efinite integrl it s much more efficient! EVALUATION THEOREM: If F is ny ntierivtive of f, then b f(x) x = F (b) F (). TOTAL CHANGE THEOREM: When you integrte rte of chnge, you get totl chnge: Rewrite this s b f (x) x = f(b) f() f(b) = f() + b f (x) x n think of it like this: If you wnt to know the vlue of f t b, first fin the vlue of f t someplce you know (), n then see how much f hs CHANGED BY in going from to b ( b f (x) x). SUBSTITUTION METHOD for ntiifferentition/integrtion: Choose u to be something whose erivtive is in the integrn, perhps off by constnt. Often, u is something in prentheses, the rgument of function, something in n exponent, etc. Be sure to chnge the limits if you hve efinite integrl. APPLICATIONS OF INTEGRATION: AREAS BETWEEN CURVES: Fin the intersection points, write the re of typicl slice n sum ppropritely. verticl slices: AREA = b( ) f(x) g(x) x horizontl slices: AREA = ( ) c f(y) g(y) y (Will nee to solve for x in terms of y) VOLUMES OF REVOLUTION: DISK METHOD: revolve y = f(x) bout the x-xis on [, b] ; volume of the resulting soli is b π( f(x) ) 2 x. SHELL METHOD: revolve y = f(x) on [, b] bout the y-xis ; volume of the resulting soli is b 2πxf(x) x. 1 AVERAGE VALUE OF A FUNCTION: the verge vlue of f(x) on [, b] is b f(x) x ; you re summing up the outputs from to b (the integrl), n then iviing by how mny you hve (the length of the intervl). If you smush the re into rectngulr shpe, the verge vlue gives the height of the rectngle. Cution: Don t mix up verge rte of chnge n verge vlue! CONNECTION BETWEEN verge vlue n the verge rte of chnge: b 1 f f(b) f() (x) x = : verging the vlues of f (x) on [, b] gives the verge b b rte of chnge of f on [, b] b http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 11

SEPARABLE DIFFERENTIAL EQUATIONS: Get ll the y s on one sie, n ll the x s on the other sie. Integrte. Don t forget the constnt of integrtion. Use given conition to solve for this constnt. SLOPE FIELDS: Slope fiels help us to visulize the solutions to first-orer ifferentil equtions. Get formul for the erivtive in terms of x n y; fin the slope t mny ifferent points. The resulting fiel of slopes helps us to see the shpes of the solution curves. http://www.onemthemticlct.org/getprereqsanobjectivesheets.htm pge 12