0.1 Chapters 1: Limits and continuity

Transcription

1 1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem Sndwich Theorem(F 96 # 20, F 97 # 12) If f(x) g(x) h(x) (or f(x) g(x) h(x)) on n intervl contining c nd x c f(x) x c h(x) b, then x c g(x) b. Definition A function f(x) is defined on n intervl contining c is continuous t c if nd only if x c f(x) f(c). Theorem If f(x) nd g(x) re both continuous t c, then so re the functions i) f(x) ± g(x), ii) f(x) g(x) nd iii) f(x)/g(x), in cse iii), provided tht g(c) 0. If f(x) is continuous t b nd g(x) is continuous t c, with g(c) b, then f(g(x)) is continuous t c. Polynomils re continuous t every point, ll trigonometric functions re continuous t ll points where the definition does not involve division by zero, nd by the previous theorem, so ll sums products, differences, quotients (excluding division by 0), nd compositions of these. Once we hve differentition t our disposl, there is the following theorem bout continuity. Theorem If f(x) is differentible t c, then f is continuous t c. Asymptotes(S 97 # 15) Given polynomils, f(x) n x n nd g(x) b m x m + + b 0, n 0 nd b m 0. f(x) x g(x) i) n /b n if m n, ii)0 if n<m, iii) ± if n>m f(x) iv) x g(x) ( n x)c< if m n 1. b n 1 When the function f(x) is grphed, cses i) ndii) correspond to horizontl symptotes y g(x) n bn nd y 0, respectively, nd cse iv) corresponds to n oblique symptote y n b n 1 x + c. Theorem Intermedite Vlue Theorem If f(x) is continuous on the intervl [, b] nd d is some number between f() nd f(b), then there exist number c in the intervl such tht f(c) d. Theorem Extreme Vlue Theorem If f(x) is continuous on the intervl [, b], then there exist numbers c, d in the intervl such tht f(c) f(x) f(d) for ll x in the intervl.

2 2 0.2 Chpters 2-3 nd prts of Chpter 6:Differentition Rules for derivtives(f 96 #14, F 97 #3, S 98 #20) 1. Product rule: [f(x)g(x)] f (x)g(x)+f(x)g (x). 2. Quotient rule: [ ] f(x) g(x) f (x)g(x) f(x)g (x). (g(x)) 2 3. Chin rule: [f(g(x))] f (g(x))g (x). Used in relted rtes problems. 4. Some derivtives: (x ) x 1, (e x ) e x, for ll rel 0,nd( x ) ln() x, [ln(x)] 1, [log x (x)] 1 for >0. ln()x Tngent line nd liner pproximtion (F 98 # 5, F 98 #20, F 99 #6, S 2000 #16) If the function f(x) is differentible t x, the tngent line to the grph y f(x) t the point (, f()) is y f()+f () (x ). The function l(x) f()+f () (x ) is clled the liner pproximtion to f(x) tx. The expression df f () is clled the differentil. Substituting x, the differentil is the liner pproximtion to f f(x) f(). Implicit differentition(f 96 #6, S 97 #19, F 97 #13, S 98 #15, S 99 #4) The curve described by f(x, y) 0 hs tngent line t the point (, b) givenbythe eqution y b + y () (x ), where y () solves the eqution given by differentiting the originl eqution with respect to x, treting y s function of x nd using the stndrd rules, nd finlly substituting x, y b. For exmple, the curve x 2 + xy + y 3 7 hs tngent line t the point (2, 1) given by the eqution y 1 (x 2) 3 x, s we see from differentiting the eqution to get 2x + y + xy +3y 2 y 0, substituting x 2,y 1 nd solving for y (2), which gives y (2) 1. Theorem Men Vlue Theorem If f(x) is differentible on the open intervl (, b) nd continuous on the closed intervl [, b], then there is number c in the intervl such tht f (c) [f(b) f()]/(b ). Some useful consequences of the Men Vlue Theorem Let f(x) be differentible function defined on the intervl I 1. If f (x) 0onI, thenf(x) is constnt. 2. If f (x) > 0onI, thenf(x) is incresing. 3. If f (x) < 0onI, thenf(x) is decresing.( S 97 #14) 4. If f (x) > 0onI, then the grph of f(x) is concve up.(f 96 #18) 5. If f (x) < 0onI, then the grph of f(x) is concve down.(s 98 #7. F 99 #7)

3 0.2. CHAPTERS 2-3 AND PARTS OF CHAPTER 6:DIFFERENTIATION 3 6. If f (c) 0ndf (c) > 0, then c is locl minimum. (F 98 #14, F 96 #4, Remember to check endpoints when looking for mx-min on [,b]) 7. If f (c) 0ndf (c) < 0, then c is locl mximum. (S 98 #18) 8. If f (c) 0, nd f (x) is of opposite sign for x<cnd x>c,thenc is point of inflection.(s 98 #19) Theorem l Hôpitl s rule (Chpter 6) (S 98 #2, F 98 # 1,2,4, F 99 #16) If x f(x) x g(x) 0or both its equl, nd g(x) 0on n intervl contining, then f(x) x g(x) f (x) x g (x) Some exmples: 1. x 0 sin(x)/x x 0 cos(x)/ x 0 x ln(x) ln(x)/x [1/x]/[( )x 1 1 ] x 0 x 0 x 0 ( ) x 0forll>0. x x n differentiting n-times ex x n! 0 for ll positive integers n. ex 4. x 0 (1 + x) 1 x x (1 + x )x e, for ll. The third exmple is proved using the following useful fct If x ln(f(x)) L, then x f(x) e L. Exponentil growth nd decy: ( F 97 # 19, S 98 # 3 (growth), F 99 #14) A(t) A 0 e kt, where A 0 is the initil mount. For rdioctive decy, k ln(2) where T is the hlf-life. T Derivtive of the inverse function(f 96 #23, F 97 #16, S 97 #6, S 99 #15) If f(x) is one-to-one on the intervl I, then there is n inverse function on the imge of the intervl, f(i). If f(x) is differentible t the point in I nd f () 0, then the inverse function f 1 (x) is differentible nd [f 1 (x)] xf() 1/[f ()].

4 4 0.3 Chpters 4-7 : Integrtion Riemnn sums: (F 96 #11) If f(x) is continuous on the intervl [0, 1], then n n i1 1 n f( i 1 n ) 0 f(x). Initil vlue problems (S 98 #16) The initil vlue problem y f(x), y() b is solved by first finding the indefinite integrl (set of ntiderivtives) of f(x) nd then choosing the free constnt so tht the vlue of the ntiderivtive t x is b. Exmple: y 2x 3 + x, y(1) 2. The indefinite integrl of 2x 3 + x is 1 2 (x4 + x 2 )+c. Setting x 1,weseetht1+c 2,soc 1. The solution is y 1 2 (x4 + x 2 )+1. Integrtion by substitution or reding the chin rule bckwrds(mny exmples, S 97 #3, F 97 #10, S 98 #6, F 98 #19, S 99, #11, F 99 #10, 11) b u(b) f(u(x))u (x) f(u)du, or s definite integrl f(u(x))u (x) f(u)du. Integrtion by prts or reding the product rule bckwrds (F 96 #13, S 97 #1) udv uv vdu, s definite integrl b u() b u(x)v (x) u(b)v(b) u()v() v(x)u (x). Averge vlue of f(x) min <x<b f(x) 1 b f(x) mx <x<b f(x). b Theorem Fundmentl Theorem of Clculus (F 97 #6,S 97 #11) ( d x ) f(t)dt) f(x). Note: To differentite u(x) f(t)dt with respect to x, when the upper it is function u(x), use the chin rule. Are between curves(f 97#15, F 99 #9) If f(x) ndg(x) re continuous on [, b] ndf(x) g(x) on tht intervl, then the re of the region bounded by the grphs y f(x), y g(x) nd the verticl lines x, x b is given by the definite integrl b [f(x) g(x)]. If the region is between curves x h(y) n k(y) which re grphs over the x-xis with h(y) k(y) on the intervl c y d then the re of the region bounded by the grphs nd the horizontl lines y c nd y d is d c [k(y) h(y)]dy. A region my stisfy both conditions. For exmple the region between the curves y 2x nd y x 2 lies over the intervl 0 x 2, but it is lso the region between x y nd 2 x y over the intervl 0 y 4. The re is clculted either by 2 0 [2x x 2 ] or [ y y 2 ]dy

5 0.3. CHAPTERS 4-7 : INTEGRATION 5 Volumes by slicing( F 96 #16,19, F 97 B2, F 98 #9) The volume of known cross-sectionl re A(x) fromx to x b is given by the integrl b A(x). ThisgivestheWsher formul for the volume of solid of revolution given by rotting the region between the curves y f(x) ndy g(x) over the intervl [, b] round the x-xis, (where f(x) g(x) on[, b]) : V b π[f(x) 2 g(x) 2 ]. If the region is lso described s being between the curves x k(y) n h(y) over the intervl c y d, but we still rotte it round the x-xis, then the volume is given by the method of Cylindricl shells: V d c 2πy[k(y) h(y)]dy. Note tht the formul in the book on pge 389 uses f(y) insted of k(y) h(y). These both men the height of the cylindricl shell. In the wsher formul the generting segment is perpendiculr to the xis of rottion, wheres, in the cylindricl shell formul the generting segment is prllel to the xis. See pge 391 in the book. Prtil frctions (S 97 #4, F 97#9,10 F 98 #6) To integrte rtionl function f(x)/g(x), first mke sure the expression is in reduced form with deg(f) < deg(g) (dividing if necessry). Then fctor the denomintor g(x) into product of liner fctors x r corresponding to rel roots nd qudrtic fctors x 2 + bx + c corresponding to compex roots. Then expnd the expression in prtil frctions s sum of C terms,wherem runs from 1 to the highest power of x r in the fctoriztion of g(x) (x r) m Ax+B nd terms,wherek is the highest power of x 2 + bx + c in the fctoriztion of g(x). (x 2 +bx+c) k In most of the exmples, the highest power for both cses is 1. Remember tht when x r occurs just once s fctor in g(x), then the coefficient C in C is esily clculted by cncelling the term x r in g(x) nd substituting x r in the (x r) remining terms of f(x) ndg(x). For exmple: 1/[(x 1)(x 2)(x 3)] + b + c. So 1/[( 1)( 2)], (x 1) (x 2) (x 3) b 1/[(1)( 1)] nd c 1/[(2)(1)]. /[(x 1)(x 2)(x 3)] 1/2 (x 1) + 1 (x 2) + 1/2 (x 3) b 1 2 ln x 1 ln x ln x 3 2 ln( (x 1)(x 3)/ x 2 ). Improper integrls (S 97 #9, F 98 # 8, S 98 # 5, S 99 #24) If f(x) is continuous on the hlf open intervl (, b] nd x + f(x) ± then the f(x) is clled n improper integrl nd it is defined s it: b b f(x) f(x) c + c

6 6 Similrly, if f(x) is continuous on the hlf open intervl [, b) nd x b f(x) ± b c f(x) f(x) c b Integrls over n infinite intervl re lso improper integrls, nd re defined s its b f(x) f(x), b b f(x) b f(x). Exmples: b x k b { x k 1 b 1 k x1 k b b 1 1 k b1 k 1 1 k 1 k for k 1 b lnb ln for k 1. { 1 k /(k 1) for k>1 for k 1. b 0 b x k 0 + { x k k b1 k 1 1 k 1 k for k lnb ln for k 1. { b 1 k /(1 k) fork<1 for k 1 Another exmple (using the clcultion from the prtil frctions exmple given bove) /[(x 1)(x 2)(x 3)] 4 ln( (b 1)(b 3)/ b 2 ) ln( (4 1)(4 3)/ 4 2 ) ln( 3/2), b becuse b ( (b 1)(b 3)/ b 2 ) 1ndln(1) 0. However, if we consider the improper integrl 3 /[(x 1)(x 2)(x 3)], then we need to pick some midpoint where there is no discontinuity, for exmple, x 4 nd consider its t both endpoints 4 b /[(x 1)(x 2)(x 3)] /[(x 1)(x 2)(x 3)]+ /[(x 1)(x 2)(x 3)] b 4 We just clculted he second it nd it is finite, but the first it 4 /[(x 1)(x 2)(x 3)] [ln( 3/2) ln( ( 1)( 3)/ 2 )] is infinite, so the improper integrl 3 /[(x 1)(x 2)(x 3)] diverges.