Clulus Finl Exm Stuy Guie This stuy guie is in no wy exhustive. As stte in lss, ny type of question from lss, quizzes, exms, n homeworks re fir gme. Every problem n hve vrition or require multiple tehniques.. Some Algebr Review () Logrithm Properties i. y = log b x is equivlent to x = b y. ii. log b b = iii. log b = 0, ln = 0 iv. log b (b r ) = r, ln(e r ) = r v. log b (x r ) = r log b (x), ln(x r ) = r ln(x) vi. log b (MN) = log b (M) + log b (N) ( ) M vii. log b = log N b (M) log b (N) viii. The omin of log b (u) is u > 0 ix. All the rules bove hol for ln x. Remember tht ln x = log e x x. Chnge of Bse log (M) = ln M ln = log b M log b This is useful if you re ske to fin the erivtive of y = log 5 (x + ) (b) Exponentil Properties i. n m = n+m ii. ( n ) m = nm iii. (b) n = n b n n iv. m = n m = m n v. n = n vi. = n n vii. m/n = ( m ) /n = ( /n ) m () Properties of Rils i. n = /n ii. n m = m/n. Limits () Nottion i. Generl Limit Nottion: x f(x) = L ii. Left Hn Limit: f(x) = L x iii. Right Hn Limit: f(x) = L x + iv. If x f(x) = f(x) = L then f(x) = L + x x (b) Limits t ±. Assume ll polynomils re in esening orer.
Clulus Finl Exm Stuy Guie i. x x r = 0, r > 0 ii. If n > m, then x x m +... bx n +... = 0 iii. x n +... x bx n +... = b iv. If m > n, then x x m +... bx n +... = ± () Evlution Tehniques i. If f(x) is ontinuous, then x f(x) = f() ii. Ftor n Cnel x + 4 = 3 + 4 = 3 x 3 If you evlute rtionl funtion n get 0, then try to ftor. 0 x 9 x 3 x 3 = (x 3)(x + 3) (x + 3) = = 6 x 3 x 3 x 3 iii. Rtionlizing Numertors / Denomintors Try this tehnique if you hve rils. Multiply top n bottom by the onjugte. x 4 x 4 x + 4 x 4 x 6 = x 4 x 6 x 6 = x + 4 x 4 (x 6)( x + 4) = x 4 x + 4 iv. Combine by Using Common Denomintors Try this when you nee to ombine frtions within frtions = x x x+4 +4 x = x x+4 +4 x x (x )(x + 4)( + 4) = x (x + 4)( + 4) (x + 4)( + 4) = ( + 4) (x + 4) x (x )(x + 4)( + 4) (x ) (x )(x + 4)( + 4) = x (x + 4)( + 4) = ( + 4)( + 4) () Pieewise Funtions - Know how to grph n evlute Pieewise Funtions. These re goo ones to test your unerstning of left-hn, right-hn, n generl its. They re lso use to test your unerstning of ontinuity. (e) Definition of Continuity A funtion f is ontinuous t x = if the following three onitions re stisfie: i. f() must exist ii. x f(x) must exist iii. x f(x) = f()
Clulus Finl Exm Stuy Guie 3. Derivtives () Limit Definition of Derivtive (b) Tngent Lines f (x) = h 0 f(x + h) f(x) h or f () = x f(x) f() x When ske to fin the eqution of tngent line on f t x =, you nee two things: A point n slope. i. You re usully given the x vlue. If they on t tell you the y vlue, you must plug the x vlue into f(x) to get the y vlue. Now you hve point (, f()) ii. To fin the slope m, you fin m = f (). iii. The eqution of the tngent line is y f() = f ()(x ). You my nee to write this in slope-interept form. () Derivtive Formuls i. x () = 0 ii. x (f ± g) = f ± g iii. x (x) = iv. x (kx) = k v. Power Rule: x (xn ) = nx n vi. x ([f(x)]n ) = n[f(x)] n f (x) vii. Prout Rule: (f g) = f g + fg ( ) f viii. Quotient Rule: = f g fg g g ix. Chin Rule: (f(g(x))) = f (g(x)) f (x) x. x (x ) = x ln xi. x (ex ) = e x xii. x (ln x) = x xiii. x (log x) = x ln xiv. ( ) e f(x) = f (x) e f(x) x xv. xvi. xvii. xviii. xix. xx. xxi. xxii. xxiii. xxiv. xxv. xxvi. xxvii. x (ln(f(x))) = f(x) f (x) (sin x) = os x x (os x) = sin x x x (tn x) = se x (se x) = se x tn x x (s x) = s x ot x x x (ot x) = s x ( sin x ) = x x ( os x ) = x x ( tn x ) = x + x ( ot x ) = x + x ( se x ) = x x x ( s x ) = x x x REMEMBER: ll of these erivtives hve version for the hin rule. For exmple: ( tn (e x ) ) = x + (e x ) ex 3
Clulus Finl Exm Stuy Guie () Critil Points x = is vlue of f(x) if f () = 0 or f () oes not exist. (e) Inresing / Deresing i. If f (x) > 0 on n intervl I, then f(x) is inresing. ii. If f (x) < 0 on n intervl I, then f(x) is eresing. (f) Conve Up / Conve Down i. If f (x) > 0 on n intervl I, then f(x) is onve up. ii. If f (x) < 0 on n intervl I, then f(x) is onve own. (g) Infletion Points x = is n infletion point of f if i. The point t x = must exist. ii. Convity hnges t x = (h) First Derivtive Test to fin Lol (Reltive) Extrem i. Fin ll ritil vlue of f(x) where f () = 0 ii. Lol Mx t x = if f (x) hnges from (+) to ( ) t x =. iii. Lol Min t x = if f (x) hnges from ( ) to (+) t x =. iv. If f (x) oes not hnge signs, it s still n importnt point to plot. It my be ple where the slope is 0, orner, n symptote, vertil tngent line, et. v. Mke sure you write your reltive mx n mins s points (, f()) (i) Seon Derivtive Test to fin Lol (Reltive) Extrem i. Fin ll ritil vlue of f(x) where f () = 0 ii. Lol Mx t x = if f () < 0 iii. Lol Min t x = if f (x) > 0 iv. Mke sure you write your reltive mx n mins s points (, f()) (j) Absolute Extrem i. (, f()) is n bsolute mximum of f(x) if f() f(x) for ll x in the omin. ii. (, f()) is n bsolute minimum of f(x) if f() f(x) for ll x in the omin. (k) Fining Absolute Extrem of ontinuous f(x) over [, b] i. Fin ll ritil vlues of f(x) on [, b]. ii. Evlute f(x) t ll ritil vlues. iii. Evlute f(x) t the enpoints, f() n f(b). 4
Clulus Finl Exm Stuy Guie iv. The bsolute mx is the lrgest funtion vlue n the bsolute min is the smllest funtion vlue. (l) L Hospitls Rule i. If x f(x) = 0 n x g(x) = 0, then f(x) x g(x) = f (x) x g (x), g (x) 0 ii. If x f(x) = ± n x g(x) = ±, then iii. Ineterminte Form of Type 0 f(x) x g(x) = f (x) x g (x) Given f(x)g(x), you o the following to put it in the form 0 x 0 or f(x) A. Rewrite f(x)g(x) s /g(x) g(x) B. Rewrite f(x)g(x) s /f(x) iv. Ineterminte Form of 0 0, 0, or Given x f(x) g(x), rewrite the it s then tke the it of the exponent ln(f(x)) eg(x) x g(x) ln(f(x)) x This shoul put the it in the Ineterminte Form of Type 0 v. Ineterminte Form of The first thing to try to ombine the funtions into One Big Frtion. L Hospitls Rule. Then try (m) Inverse Derivtive Theorem Let f(x) be one-to-one funtion. ( f (b) ) = f (f (b)) 5
Clulus Finl Exm Stuy Guie 4. Summry of Curve Skething Alwys strt by noting the omin of f(x) () x n y interepts i. x-interepts our when f(x) = 0 ii. y-interept ours when x = 0 (b) Fin ny vertil, horizontl symptotes, or slnt symptotes. i. Vertil Asymptote: Fin ll x-vlues where x f(x) = ±. Usully when the enomintor is 0 n the numertor is not 0. Rtionl funtion MUST be reue. ii. Horizontl Aymptotes: Fin f(x) n f(x). There re shortuts bse on the x x egree of the numertor n enomintor. () Fin f (x) iii. Slnt Asymptotes: Ours when the egree of the numertor is one lrger thn the enomintor. You must o long ivision to etermine the symptotes. i. Fin the ritil vlues, ll x-vlues where f (x) = 0 or when f (x) oes not exist. ii. Plot the ritil vlues on number line. iii. Fin inresing / eresing intervls using number line iv. Use The First Derivtive Test to fin lol mximums / minimums (if ny exist). () Fin f (x) Remember to write them s points. A. Lol Mx t x = if f (x) hnges from (+) to ( ) t x =. B. Lol Min t x = if f (x) hnges from ( ) to (+) t x =. C. Note: If f (x) oes not hnge signs, it s still n importnt point. It my be ple where the slope is 0, orner, n symptote, vertil tngent line, et. i. Fin ll x-vlues where f (x) = 0 or when f (x) oes not exist. ii. Plot these x-vlues on number line. iii. Fin intervls of onvity using the number line iv. Fin points of infletion (e) Sketh A. Must be ple where onvity hnges B. The point must exist (i.e, n t be n symptote, isontinuity) i. Drw every symptote ii. Plot ll interepts iii. Plot ll ritil points (even if they re not reltive extrem). They were ritil points for reson. iv. Plot ll infletion points. v. Connet points on the grph by using informtion bout inresing/eresing n its onvity. 6
Clulus Finl Exm Stuy Guie 5. Integrls () Definitions i. Antieritive: An ntierivtive of f(x) is funtion F (x), where F (x) = f(x). ii. Generl Antierivtive: The generl ntierivtive of f(x) is F (x) + C, where F (x) = f(x). Also known s the Inefinite Integrl f(x) x = F (x) + C iii. Definite Integrl: b (b) Approximtion Integrtion Tehniques b f(x) x = F (b) F () Given the integrl f(x) x n n (for Simpson s Rule n must be even), with x = b n n x i = + i x, then Left-hn b Right-hn Mipoint Trpezoi Simpson s b b b b () Common Integrls f(x) x = x [f(x 0 ) + f(x ) + f(x ) + f(x 3 ) +... + f(x n )] f(x) x = x [f(x ) + f(x ) + f(x 3 ) + f(x 4 ) +... + f(x n )] f(x) x = x [f(x ) + f(x ) + f(x 3) + f(x 4) +... + f(x n)], where x i is mipoint in [x i, x i ] f(x) x = x [f(x 0) + f(x ) + f(x ) + f(x 3 ) +... + f(x n ) + f(x n ) + f(x n )] f(x) x = x 3 [f(x 0) + 4f(x ) + f(x ) + 4f(x 3 ) +... + f(x n ) + 4f(x n ) + f(x n )] i. ii. iii. iv. v. vi. vii. viii. ix. x. k x = kx + C x n x = xn+ + C, n n + x = ln x + C x kx + b x = ln kx + b + C k e x x = e x + C e kx x = k ekx + C e kx+b x = k ekx+b + C x x = x ln + C kx x = k ln kx + C kx+b x = k ln kx+b + C xi. xii. xiii. xiv. xv. xvi. xvii. xviii. xix. xx. ln x x = x ln x x + C os x x = sin x + C sin x x = os x + C se x x = tn x + C s x x = ot x + C se x tn x x = se x + C s x ot x x = s x + C tn x x = ln se x + C ot x x = ln sin x + C s x = ln s x ot x + C 7
Clulus Finl Exm Stuy Guie xxi. xxii. xxiii. se x x = ln se x + tn x + C ( x x = x sin + x x = ( x tn ) + C ) + C xxiv. xxv. xxvi. + x x = ( x ) ot + C x x x = se (x) + C x x x = s (x) + C () Every integrl must be written into the proper form in orer to use the formuls. For exmple, x x = x / x (e) Fining Are uner or between Curves 3 3 5x 4 x = 5 x 4 x 8 5 x = 8x 9/5 x x 9 6. Solis of Revolution () Disk Metho - in terms of x () Disk Metho - in terms of y V = b πr(x) x (b) Wsher Metho - in terms of x V = πr(y) y () Wsher Metho - in terms of y V = b πr(x) πr(x) x 8 V = πr(y) πr(y) y
Clulus Finl Exm Stuy Guie You my hve to rotte roun n xis other thn the x or y xis. If you o, you nee to just the rius. For exmple, V = π( + R(y)) + π( + r(y)) y () Shell - in terms of x (b) Shell - in terms of y V = b πxh(x) x V = πyh(y) x 7. Ar Length () In terms of x (b) In terms of y L = b + (f (x)) x L = + (g (x)) y 8. Are of Surfe of Revolution () Rotte roun the x-xis 9
Clulus Finl Exm Stuy Guie SA = b πy + ( ) y x SA = πy + x ( ) x y y The first integrl is in terms of x. Tht mens the y in front of the squre root shoul be in terms of x. For exmple, y = x +. You use x + inste of y. (b) Rotte roun the y-xis. The formul oesn t hnge muh. SA = b πx + ( ) y x SA = πx + x ( ) x y y The seon integrl is in terms of y. Tht mens the x in front of the squre root shoul be in terms of y. For exmple, if y = x, solve for x (x = y) n use y inste of x. 9. Integrtion Tehniques () u Substitution Given b f(g(x))g (x) x, i. Let u = g(x) ii. Then u = g (x) x iii. If there re bouns, you must hnge them using u = g(b) n u = g() b f(g(x))g (x) x = g(b) g() f(u) u (b) Integrtion By Prts u v = uv v u Exmple: x e x x u = x u = x x x e x x = x e x v = e x x v = e x xe x x You my hve to o integrtion by prts more thn one. When trying to figure out wht to hoose for u, you n follow this guie: LIATE 0
Clulus Finl Exm Stuy Guie L Logs I Inverse Trig Funtions A Algebri (rils, rtionl funtions, polynomils) T Trig Funtions (sin x, os x) E Exponentil Funtions () Prouts of Trig Funtions i. sin n x os m x x ii. tn n x se m x x A. m is o (power of os x is o). Ftor out one os x n ple it in front of x. Rewrite ll remining os x s sin x by using os x = sin x. Then let u = sin x n u = os x x B. n is o (power of sin x is o). Ftor out one sin x n ple it in front of x. Rewrite ll remining sin x s os x by using sin x = os x. Then let u = os x n u = sin x x C. If n n m re both o, you n hoose either of the previous methos. D. If n n m re even, use the following trig ientities A. m is even (power of se x is even). Ftor out one se x n ple it in front of x. Rewrite ll remining se x s tn x by using se x = + tn x. Then let u = tn x n u = se x x B. n is o (power of tn x is o). Ftor out one se x tn x n ple it in front of x. Rewrite ll remining tn x s se x by using tn x = se x. Then let u = se x n u = se x tn x x C. If n o n m is even, you n use either of the previous methos. sin x = ( os(x)) os x = ( + os(x)) D. If n is even n m is o, the previous methos will not work. You n try to simplify or rewrite the integrls. You sin(x) = os x sin x my try other methos. () Prtil Frtions p(x) Use this metho when you re integrting x, the egree of p(x) must be smller thn q(x) the egree of q(x). Ftor the enomintor q(x) into prout of liner n qurti ftors. There re four senrios. Unique Liner Ftors: If your enomintor hs unique liner ftors x (x )(x 3) = A x + B x 3 To solve for A n B, multiply through by the ommon enomintor to get x = A(x 3) + B(x ) You n fin A by plugging in x =. Fin B by plugging in x = 3.
Clulus Finl Exm Stuy Guie Repete Liner Ftors: Every power of the liner ftor gets its own frtion (up to the highest power). Unique Qurti Ftor: x (x 3)(3x + 4) 3 = A x 3 + B 3x + 4 + 3x (x + 4)(x + 9) = A x + 4 + Bx + C x + 9 C (3x + 4) + D (3x + 4) 3 To solve or A, B, n C, multiply through by the ommon enomintor to get 3x = A(x + 9) + (Bx + C)(x + 4) Repete Qurti Ftor: Every power of the qurti ftor gets its own frtion (up to the highest power). (e) Trig Substitution x (3x + 4)(x + 9) 3 = A 3x + 4 + Bx + C x + 9 + Dx + E (x + 9) + F x + G (x + 9) 3 If you see Substitute Uses the following Ientity x x = sin(θ) sin (θ) = os (θ) + x x = tn(θ) + tn (θ) = se (θ) x x = se(θ) se (θ) = tn (θ) Exmple: x 3 6 x x Let x = 4 sin θ, x = 4 os θ θ. So x3 = 64 sin 3 θ x 3 x = 6 x 64 sin 3 θ 64 sin 3 6 6 sin θ 4 os θ θ = θ 4 os θθ = 4 os θ sin 3 θ θ Now you hve n integrl ontining powers of trig funtions. You n refer to tht metho to solve the rest of this integrl. sin 3 θ = os3 θ 3 os θ + C To get bk to x, we nee to use right tringle with the originl substitution x = 4 sin x. 6 x If os θ =, then 4 x 3 6 x x = ( 6 x ) 3 4 3 6 x + C 4
Clulus Finl Exm Stuy Guie 0. Improper Integrls () Infinity s boun i. ii. b f(x) x = f(x) x = t t b t t f(x) x f(x) x iii. f(x) x = f(x) x + f(x) x (b) Disontinuity t boun i. Disont. t : ii. Disont. t b: b b iii. Disont. t, < < b:. Sequenes n Series f(x) x = f(x) x = b t + t t t b b f(x) x = f(x) x f(x) x f(x) x + b f(x) x () Sequenes: A list of numbers in efinite orer, { 0,,, 3, 4,...} i. Sequeeze Theorem If n < b n < n, for n N n b n = L n ii. Useful Theorem n = n = L, then n n If n n = 0, then n n = 0 iii. Geometri Sequene n rn = 0, if < r <, if r = Diverge, elsewhere iv. A typil heirrhy of sequenes C < ln n < n p < n < n! < n n (b) Series i. Definition iii. Geometri Series: If < r <, then n = + + 3 + 4 +... n= ii. Prtil Sums N S N = n = + + 3 +... + N n= r n = r n=0 r n = r n= r n k = r n=k 3
Clulus Finl Exm Stuy Guie. Summry of Convergene Tests Test Theorem Comments Divergene Test P-Test Integrl Test Diret Comprison Test Limit Comprison Test Alternting Series Test Rtio Test Root Test Absolute Convergene Conitionl Convergene Given the series n, if n 0, the series n iverges. The series then n= n= onverges if p >. If p, np n p iverges. Given the series n, if you n write n = f(n) n If f(n) n onverges, then so oes n. f(n) n iverges, so oes n Let n n b n be series suh tht 0 n b n. If b n onverges, so oes n. If n iverges, so oes b n. Let n n b n be series where n 0 n n b n 0. If = L, where L is positive n b n finite number, then either both series onverge or they both iverge. Given the series ( ) n b n, if b n > 0, eresing, n b n 0, then the series onverges. Let n be series. Let L = n If L <, the series onverges bsolutely. If L >, the series iverges. n+ n If L =, the test is inonlusive. Let n n be series. Let L = n. n If L <, the series onverges bsolutely. If L >, the series iverges. If L =, the test is inonlusive. If the series n onverges bsolutely, then it onverges. If the series n onverges but n iverges,. then the series n onverges onitionlly. Cn only be use to test for ivergene. It n never be use to test for onvergene. You re normlly using this series s omprison for Diret or Limit Comprison Tests The funtion f shoul be esily integrte using one of our integrtion tehniques from previous hpters. This test shoul not be use when n hs ftorils. You nee to be ble to ntiipte the onvergene. This llows you to hoose n pproprite omprison. You shoul know the onvergene of one of the series. Cn be esier thn Diret Comprison Test sine getting the require inequlities to work n be time onsuming. You hve to hek for bsolute or onitionl onvergene by testing ( ) n b n = b n. Use this test when you hve exponentil funtions or ftorils. Use this test when n = (b n ) n. If n onverges, when n is bsolutely onvergent. You re usully heking if n onverges beuse n onverge by AST 4
Clulus Finl Exm Stuy Guie 3. Estimting Series () Reminer Estimte for Integrl Test: If n n= is onvergent, f(n) is ontinuous, positive, n eresing, then R N N f(n) n If you re ske to estimte n= n to 4 eimls (or hve n error less thn 0.000), then solve integrte n solve for N. N f(n) n < 0.000 (b) Alternting Series Estimtion Theorem: Given onvergent lternting series n = n= n=( ) n b n, then R N b N+ 4. Power Series () A power series entere t 0 the rius of onvergene is R = 0. n x n = 0 + x + x + 3 x 3 +... This hppens when the Rtio Test gives n=0 L >. (b) A power series entere t ii. The series onverges for ll x. The intervl of onvergene is (, ) n (x ) n = 0 + (x )+ (x ) + 3 (x ) 3 +... n n=0 () Rius n Intervl of Convergene: Perform the Rtio Test or Root Test (usully Rtio Test) For given power series n (x ) n, there re only three possibil- n= ities: i. The series onverges only when x =. The intervl of onvergene is {} n the rius of onvergene is R =. This hppens when the Rtio Test gives L = 0. iii. The series onverges on n intervl ( R, +R). This hppens when the Rtio Test gives K x, where you nee to solve K x <. n=0 5. Writing Funtions s Power Series Given f(x) = n (x ) n = 0 + (x ) + (x ) + 3 (x ) 3 +... with rius of onvergene R > 0, then f (x) = f(x) x = n n (x ) n = + (x ) + 3 3 (x ) + 4 4 (x ) 3 n= n (x ) n+ = 0 (x ) + n + (x ) + 3 (x ) 3 + 4 3(x ) 4 +... n=0 You use this tehnique if you re ske to fin power series of funtion like, ln( + x), ( + x) tn x, et. Your gol is to ifferentite or integrte your funtion f(x) until it s in the proper form A u 5
Clulus Finl Exm Stuy Guie A n u n be nything. One it s in the proper form, you n write it s power series. A u = n=0 A(u) n () If you h to integrte to get the proper form, then ifferentite the power series to get bk to the originl f(x). (b) If you ifferentite to get the proper form, then integrte the power series to get bk to the originl f(x). 6. Tylor Series / Mlurin Series () Tylor Series is entere t (b) Mlurin Series is entere t 0 f(x) = n=0 n (x ) n = 0 + (x )+ (x ) +... f(x) = n=0 n (x) n = 0 + x + x +... with n = f (n) () n! with n = f (n) () n! It my helfpul to etermine the oeffiients 0,,,... by using the tble n f n (x) f n () n = f n () n! 0 f(x) f() 0 = f() 0! f (x) f () = f ()! f (x) f () = f ()! 3 f (x) f () 3 = f () 3! () If you nee to fin the Tylor / Mlurin Series, fin pttern for n. Then you write the following series with the pproprite n f(x) = n=0 n (x ) n (b) If you nee to fin the n-th egree Tylor / Mlurin Polynomil, then just fin the oeffiients you nee. For exmple, 3r egree Tylor Polynomil requires the oeffiients up to 3. f(x) = 0 + (x ) + (x ) + 3 (x ) 3 6