Clculus I-II Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing or decresing on its entire domin. Exmple 1.1 e x is incresing on (, ) nd ln x is incresing on (0, ). f(x) = 1/x is decresing on (, 0) nd on (0, ). The sine function is incresing on [ π/, π/], [3π/, 5π/], etc. A function f(x) is proportionl to nother function g(x) if there is nonzero constnt k such tht f(x) = kg(x). Exmple 1. The force of grvity F between two msses M nd m is proportionl to the inverse squre function of the distnce between them, r, nmely Here k = GMm nd g(r) = 1/r. F = GMm r The grph of function f is concve up if it bends upwrd s we move from left to right, tht is if, when we pick two points (x 1, y 1 ) nd (x, y ) on the grph of f, the grph lies below or on the line segment joining those points on tht intevl. Similrly, the grph of f is concve down if it bends downwrd, or if the grph of f lies bove or on ny line segment joining two points on its grph. An inflection point is point (x, f(x)) on the grph of f t which the grph chnges concvity (or equivlently just the x vlue of tht point). Exmple 1.3 The cosine function is concve up on [ π, 3π ] nd concve down on [ π, π ], etc. It hs inflection points t odd multiples of π/. A function f is sid to be even if f( x) = f(x) nd odd if f( x) = f(x). Even functions re symmetric bout the y-xis, while odd ones re symmetric bout the origin. Exmple 1.4 The cosine function is even, f(x) = x is even, nd f(x) = e 3x is even, while the sine function is odd, f(x) = x 3 is odd, nd f(x) = rctn(x) is odd. 1. Limits nd Continuity Suppose f is defined on n intervl round given rel number c, except perhps t x = c itself. The limit of f s x pproches given rel number c is rel number L stisfying the following condition: for ll ɛ > 0 there is δ > 0 such tht x c < δ nd x c 1
implies In this cse we write f(x) L < ɛ lim f(x) = L Remrk 1.5 In plin English, this mens tht the limit L exists provided the following holds: however smll I wnt to mke my ɛ-window on the y-xis round L, I cn find smll enough δ-window on the x-xis round c which is mpped by f entirely into the ɛ-window. Remrk 1.6 Note the logicl implictions: I hve freedom to choose ɛ. I cn choose it to be one in billion, whtever I wnt. But once chosen, I m typiclly restricted in my choice of δ. For exmple, if f(x) = x nd L = 100, then clerly c = 10, nd to show tht the limit s x pproches 10 is 100, I need to find δ for ny given ɛ. Suppose I ve chosen my ɛ, sy ɛ = 1/1000. Then you cn check tht good enough δ is δ = 1/1000. It s not the only one, since I cn pick it to be smller thn this, but I m restricted by how big I cn mke it. It cn t be much bigger thn tht. Remrk 1.7 Note lso tht the definition of limit includes bsolute vlues round x c. We write x c nd this implies the existence nd equlity of the left nd right limits, lim f(x) nd lim f(x) + respectively. Exmple 1.8 The function f(x) = x 1 x 1 hs left nd right limits s x pproches 1, but the two limits re not equl. Hence f hs no limit t x = 1. We cn nlogously define lim f(x) x ± The difference is we cn t use the δ prt of the definition. The ɛ prt is the sme, we cn choose ny ɛ > 0 such tht f(x) L < ɛ under some pproprite condition on x. It s just tht condition cn t be the existence of δ such tht x c < δ (becuse c = nd x =, which is never less thn δ), so insted we demnd the existence of n M > 0 such tht x M implies f(x) L < δ. A function f is continuous t point x = c if the limit lim f(x) = L exists nd moreover L = f(c), i.e. if lim f(x) = f(c) We sy f is continuous on n intervl [, b] or (, b), etc., if it is continuous t ll points in tht intervl. Exmple 1.9 All polynomils re continuous on (, ). All rtionl functions p(x)/q(x) re continuous except t zeros of q(x). For exmple, x +3 x 5 is continuous everywhere except t ± 5. The exponentil function e x is continuous on (, ), the log function ln x is continuous on (0, ). The sine nd consine functions re continuous on (, ). The bsolute vlue function f(x) = x is continuous everywhere on (, ).
Discontinuities, or points of discontinuity of f, re x-vlues where f is not continuous. For exmple 0 is point of discontinuity of csc x nd of f(x) = 1/x. A removble discontinuity is one tht cn be plugged, e.g. x = is removble discontinuity of f(x) = x 4 x+. An essentil discontinuity is one tht cnnot be plugged, typiclly becuse f goes to ± ner tht point, for exmple x = 1 is n essentil discontinuity of f(x) = /(x 1). A jump discontinuity is discontinuity where f jumps finite mount ner it, for exmple x = 1 for f(x) = x 1 x 1. 1.3 Rte of Chnge nd the Derivtive The verge rte of chnge of function f between x = nd x = b is defined to be the slope of the line connecting the points (, f()) nd (b, f(b)) on the grph of f: verge rte of chnge = f(b) f() b If we define h = b, then b = + h, nd the verge rte of chnge between nd b = + h becomes the difference quotient: f( + h) f() h If the limit of the difference quotient s h pproches 0 exists, then we sy tht f is differentible t x = nd we cll this limit the derivtive of f t. It is rel number, denoted equivlently by f () df dy () f( + h) f() dx dx = lim x= h 0 h If f is differentible on n entire intervl, then vrying the point, in other words letting x vry, we get function, the derivtive function f (x). One of our tsks is to find vrious derivtive functions for well-known functions. Exmple 1.10 Common derivtive functions re the following: 1. (x n ) = nx n 1 (power rule). (sin x) = cos x 3. (cos x) = sin x 4. (tn x) = sec x 5. (e x ) = e x 6. (ln x) = 1 x 7. ( x ) = (ln ) x 8. (rctn x) = 1 1+x 9. (rcsin x) = 1 1 x 3
Remrk 1.11 To compute more complicted derivtives, such s those of xe x, for exmple, we will need to know how to brek up the process of tking derivtive into severl steps involving only derivtives of things we know how to compute, such s (1)-(6) bove. For exmple, we know how to tke the derivtive of x, x nd e x, nd we will develop methods to compute (xe x ) in terms of these esier ones. This is the content of the sum, product, quotient nd chin rules below. The derivtive f () t x = cn be interpreted s the slope of the tngent line to the grph of f t the point (, f()). Since we hve slope, f (), nd point, (, f()), we cn find the eqution of the tngent line using point-slope: y f() = f ()(x ) Adding f() to both sides gives the eqution for the tngent line, which we lso cll the liner pproximtion to f ner x = or locl lineriztion ner x = : y = f() + f ()(x ) This is the key ide behind the ide of derivtive, to pproximte complicted function f(x) with the simplest one possible, line, t lest loclly. The tngent line is n honest-togoodness pproximtion of f ner x =, becuse we know tht s x, f(x) f() x f (), i.e. from the definition of limit we get f(x) f() x f () f(x) f() f ()(x ) multiply both sides by (x ) f(x) f() + f ()(x ) dd f() to both sides Exmple 1.1 Let us pproximte the vlue of f(x) = e sin x ner x = π using locl lineriztion, sy t x = 3 which is ner x = π. Since f (x) = e sin x cos x, we hve f (π) = e sin π cos π = 1, nd becuse f(π) = e sin π = 1, we get f(x) f(π) + f (π)(x π) = 1 (x π) = 1 + π x Therefore, f(3) 1 + π 3 = π 1.14 The error in the tngent/liner pproximtion is the difference between the ctul vlue f(x) t x nd the pproximte vlue obtined from the tngent line t x ner x = : E(x) = f(x) [ f() + f ()(x ) ] It is theorem, not hrd to prove, tht E(x) lim x x = 0 It is lso fct, not proven in the book until the section on Tylor s theorem, tht E(x) f () (x ) 4
Exmple 1.13 The error in the previous exmple is E(x) = f(x) (1 + π x) = e sin x + x 1 π so, for exmple our pproximtion t x = 3 is off by bout E(3) = e sin 3 + 3 1 π 0.00997 Not bd! And, to complete this exmple, note tht f (π) (3 π) = 1 (cos 3e sin 3 (sin 3)e sin 3 )(3 π) 0.00968 which is indeed close to E(3). 1.4 Optimiztion A criticl point of f is point x = where f () = 0, tht is criticl points re zeros of the derivtive of f. Hence, when looking for them you hve to solve the eqution f (x) = 0 for x. The y-vlue of f t criticl point y = f() is clled criticl vlue. Note tht f() 0 in generl it s f () tht equls 0! A locl minimum of f is point x = such tht f(x) f() ner, nd locl mximum of f is point x = such tht f(x) f() ner. These re upgrded to globl minimum nd globl mximum if f(x) f() or f(x) f(), respectively, for ll x in the domin of f under considertion, not just those nerby. 1.5 The Riemnn Integrl Let [, b] be nonempty closed intervl in (, ) nd let f : [, b] (, ) be bounded nd continuous. Let P = ( = x 0 < x 1 < < x n = b) be prtition of [, b], tht is n incresing finite incresing sequence of points in the intervl. For given prtition P choose, for ech i = 0, 1,..., n, n rbitrry t i in [x i 1, x i ]. The Riemnn Sum of f for given prtition P is then defined s S P = n f(t i ) (x i x i 1 ) i=1 If we choose the x i such tht x n x n 1 = x n 1 x n = = x 1 x 0, tht is if x = b n then the Riemnn sum is written simply s nd x i = + i x n S P = f(t i ) x i=1 If we choose the left endpoint every time, t i = x i, then we get the left Riemnn sum, n 1 S L = f(x i ) x i=0 5
nd if we choose the right endpoint, t i = x i + 1, then we get the right Riemnn sum, S R = n f(x i ) x i=1 Similrly, we hve the upper nd lower Riemnn sums, U nd L, obtined by choosing t i in [x i, x i+1 ] such tht f(t i ) = mx xi t x i+1 f(t) nd f(t i ) = min xi t x i+1 f(t), respectively (since f is continous, we cn find such t i by the Extreme Vlue Theorem!). We cn lso choose t i to be the midpoint of [x i, x i+1 ], in which cse we obtin the midpoint Riemnn sum. Remrk 1.14 Note tht U does not in generl equl S L or S R, nd likewise with L. A good exmple is the upper hlf of the circle, f(x) = 1 x on [ 1, 1]. Try to compute S L, S R, U nd L. Exmple 1.15 Let us compute the Riemnn sum for f(x) = 1 x on [ 1, 1], nd so obtin severl pproximtions of the upper hlf of the unit circle. Let us choose n = 4 for convenience, which mens x 0 = 1, x 1 = 1, x = 0, x 3 = 1, nd x 4 = 1, nd so P = ( 1, 1, 0, 1, 1) is our prtition of [ 1, 1]. Let us compute the left Riemnn sum first: this mens choosing t i = x i, nd since x = 1 ( 1) 4 = 1, we hve S L = 4 1 f(x i ) x x=0 = f( 1) 1 + f( 1 ) 1 + f(0) 1 + f(1 ) 1 ( 1 = ( 1) + 1 ( 1/) + 1 0 + ) 1 (1/) 1 ( ) 3 3 = 0 + + 1 + 1 3 + 1 = Similrly, by choosing t i = x i+1, we get the right sum, S R = 4 1 f(x i+1 ) x x=0 = f( 1/) 1 + f(0) 1 + f(1/) 1 + f(1) 1 ( 1 = ( 1/) + 1 0 + 1 (1/) + ) 1 1 1 ( ) 3 3 = + 1 + 1 + 1 3 + 1 = The upper sum requires little bit of extr work. We hve to figure out where f is lrgest on 6
ech subintervl [x i, x i+1 ], nd this will vry depending on i. Let us look t picture: 1 x Thus, for exmple, on the intervl [x 0, x 1 ] = [ 1, 1 ], it is cler tht f is lrgest when x = 1, sowe must choose t 0 = 1. Proceeding nlogously, we get t 1 = 0, t = 0, nd t 3 = 1, nd so U = n 1 f(t i ) x i=0 = f( 1 ) 1 + f(0) 1 + f(0) 1 + f(1 ) 1 ( 1 = ( 1/) + 1 0 + 1 0 + ) 1 (1/) 1 ( ) 3 3 = + 1 + + 0 1 3 + = Similrly, to compute the lower Riemnn sum, we see from the picture 1 x tht, for exmple on [x 0, x 1 ] = [ 1, 1 ], f is smllest t x = 1, so we will need t 0 = 1, nd proceeding nlogously we get tht t 1 = 1, t = 1 nd t 3 = 1. Thus, L = n 1 f(t i ) x i=0 = f( 1) 1 + f( 1 ) 1 + f(1 ) 1 + f(1) 1 ( 1 = ( 1) + 1 ( 1/) + 1 (1/) + ) 1 1 1 ( ) 3 3 = 0 + + + 0 1 3 = As finl observtion, we note tht L S L = S R U. 7
The next step is, of course, refining the prtition of [, b] nd summing over more nd more rectngles, thus, hopefully, getting more ccurte pproximtion to the ctul re. A function f is sid to be Riemnn integrble on [, b] if, no mtter how the choice of prtitions P of [, b] is mde, the limit of these Riemnn sums s we dd more nd more rectngles, lim n S P n f(t i ) (x i+1 x i ) i=1 exists s rel number. It is fct, proven lter in undergrdute nlysis, tht the choice of prtition doesn t mtter, so you my choose, if you wnt, the left Riemnn sum every time. In this cse, if the limit exists, f is sid to be Riemnn integrble on [, b] nd the vlue of this limit is clled the Riemnn integrl or definite Riemnn integrl of f on [, b], nd this limit is denoted n 1 f(x) dx = lim S P = lim f(t i )(x i+1 x i ) n n If we choose x = x i+1 x i to be constnt nd equl to b n, then we my write i=0 n 1 f(x) dx = lim f(t i ) x n i=0 8
Theorems Theorem.1 (Limit Lws) Let k be constnt rel number. Then, 1. lim (kf(x)) = k lim f(x). lim ( f(x) + g(x) ) = lim f(x) + lim g(x) ( ) ( 3. lim f(x)g(x) = lim f(x) )( lim g(x) ) f(x) 4. lim g(x) = lim f(x) lim g(x) if g(x) 0 5. lim k = k (we think of k s constnt function) 6. lim x = c (we think of x s the function f(x) = x) Theorem. (Continuity Lws) Let k be constnt rel number nd suppose f nd g re continuous on n intervl [, b] (or n open or hlf-open intervl, it doesn t mtter). Then, the following functions re continuous: 1. kf(x). f(x) + g(x) 3. f(x)g(x) 4. f(x) g(x) if g(x) 0 5. (f g)(x) Theorem.3 (Intermedite Vlue Theorem) If f is continuous on closed nd bounded intervl [, b] nd k is y-vlue between f() nd f(b) (whether f() k f(b) or f(b) k f()), then there is t lest one number c between nd b such tht f(c) = k. Remrk.4 This mens tht the entire rnge of y-vlues between f() nd f(b) is hit by f on the intervl [, b], possibly more thn once. Exmple.5 Since f(x) = x is continuous on [4, 100], nd since f(4) = nd f(100) = 10, ll numbers between nd 10 re hit. For exmple 5: there is number, nmely c = 5, in between 4 nd 100, such tht f(c) = 5. Exmple.6 Consider the polynomil p(x) = 5x 3 + πx + 1. It hs t lest one root between 1 nd 0, becuse p( 1) = 4 π nd p(0) = 1, nd since 0 lies in between these two y vlues nd p(x) is continuous on [ 1, 0], we know there is number c between 1 nd 0 such tht p(c) = 0. Tht s our root. 9
Exmple.7 Consider f(x) = 1/x. Then f( 1) = 1 nd f(1) = 1, but the Intermedite Vlue Theorem doesn t pply to give c between 1 nd 1 such tht f(c) = 0, becuse f is not continuous on [ 1, 1]. Indeed, f(x) never equls 0. Theorem.8 (Extreme Vlue Theorem) If f is continuous function on closed nd bounded intervl [, b], then f ttins (globl) minimum nd (globl) mximum on tht intervl, tht is mx x b f(x) nd min x b f(x) exist s rel numbers. This mens there re points x 1 nd x in the intervl [, b] such tht f(x 1 ) = mx x b f(x) nd f(x ) = min x b f(x) Exmple.9 The prbol f(x) = x + 6x 1, being polynomil, is continuous everywhere, so for exmple on [0, 4] it must chieve its minimum nd its mximum. In fct, s cn be seen from the grph: f(0) = min f(x) = 1 nd f(3) = mx f(x) = 8 0 x 4 0 x 4 Of course, we don t need the grph, becuse (using the first nd second derivtive tests) we cn ctully find the mximum nd minimum nlyticlly s follows: 1. First, find ll criticl points, which mens ompute the derivtive, set it equl to zero nd solve for x: f (x) = x + 6 = 0, so x = 3 is criticl point.. Use, e.g. the second derivtive test to clssify it: f (3) = < 0, so it s locl mx. 3. Check the vlue of f t the criticl point (which lies in the intervl [0, 4]) ginst the vlue of f t the endpoints 0 nd 4: f(0) = 1, f(3) = 8 nd f(4) = 7, so it looks like 0 is the globl min nd 3 is the globl mx here, i.e. f(0) = min 0 x 4 f(x) = 1 nd f(3) = mx 0 x 4 f(x) = 8. Exmple.10 The function f(x) = 1 x is continuous on (, 0) (0, ), but on (0, ) the Extreme Vlue Theorem doesn t pply, becuse (0, 1) is not closed intervl. As it turns out, f(x) hs neither (locl or globl) mx or min on (0, 1) (it just gets bigger nd bigger without bound s x pproches 0 nd it gets closer nd closer to 1/ s x pproches, but it never reches 1/, nd so it never reches smllest vlue. 10
Theorem.11 (Differentition Rules) If two functions f nd g re differentible t x = nd k is constnt, then the functions c (the constnt function), cf(x), f(x) + g(x), f(x)g(x), f(x)/g(x) (when g(x) 0), nd (f g)() (when f is differentible t b = g()) re ll differentible t x =, nd moreover stisfy the following formuls: (1) c = 0 () (cf) () = cf () (constnt fctor rule) } (linerity) (3) (f + g) () = f () + g () (sum rule) (4) (fg) () = f()g () + f ()g() (product rule) (5) ( ) f () = f ()g() f()g () g g () (quotient rule) (6) (f g) () = f ( g() ) f () (chin rule) Theorem.1 (Differentibility Implies Continuity) If f is differentible t point x =, then it is continuous t x =. Consequently, if f is differentible on n open intervl (, b), then it is continuous on tht intervl. Remrk.13 Of course the converse is not true. You cn hve continuous function which is not differentible. For exmple f(x) = x is continuous t x = 0 but not differentible there (the left nd right limits of the difference quotient s h 0 don t gree). Theorem.14 (Men Vlue Theorem) If f is continuous on [, b] nd differentible on (, b), then there is some c between nd b such tht f(b) f() = f (c)(b ) Exmple.15 The Men Vlue Theorem is used for mny things, but the simplest of its uses is in proving inequlities. For exmple, let us prove the inequlity sin x x on [0, ). The trick is to define new function, h(x), s the difference of the two functions under considertion, tht is let h(x) = x sin x 11
Then we need only show tht h(x) 0 to prove our ssertion. Now, since for ny x > 0 we know tht h is continous on [0, x] nd differentible on (0, x), the MVT pplies to give the existence of c between 0 nd x such tht h(x) h(0) = h (c)(x 0) Now, h(0) = 0 sin 0 = 0, nd moreover h (x) = 1 cos x 0, which, long with the fct tht x > 0, implies h (x)(x 0) 0, so we hve x sin x = h(x) 0 on [0, ), which is wht we set out to prove. Corollry.16 (Constnt Function Theorem) Let f be continuous on [, b] nd differentible on (, b). Then f (x) = 0 on ll of (, b) if nd only if f is constnt on the intervl. Corollry.17 Let f nd g be continuous on [, b] nd differentible on (, b). If f (x) = g (x) on (, b), then f nd g differ by constnt, i.e. for some constnt c. f(x) = g(x) + c Remrk.18 This is importnt for integrls nd nti-derivtives. Nmely, ll nti-derivtives differ by constnt, nd so the most generl ntiderivtive of function is written f(x)dx = F (x) + C where F (x) is some specific ntiderivtive nd C is n rbitrry constnt, tht constnt showing up in the lst corollry. Corollry.19 Let f be differentible on n open intervl (, b). (1) If f (x) 0 on (, b), then f is incresing on (, b). () If f (x) 0 on (, b), then f is decresing on (, b). Moreover, if the inequlities re strict, then f is strictly incresing or strictly decresing, respectively. Corollry.0 (First Derivtive Test) Let f be differentible on n open intervl contining criticl point x 0 of f, i.e. one for which f (x 0 ) = 0. (1) If f (x) > 0 for x < x 0 nd f (x) < 0 for x > x 0, then x 0 is locl mximum of f. () If f (x) < 0 for x < x 0 nd f (x) > 0 for x > x 0, then x 0 is locl minimum of f. 1
Corollry.1 (Second Derivtive Test) Let f be twice differentible on n open intervl contining criticl point x 0 of f, i.e. one for which f (x 0 ) = 0. (1) If f (x 0 ) > 0, then x 0 is locl minimum of f. () If f (x 0 ) < 0, then x 0 is locl mximum of f. (3) If f (x 0 ) = 0, then no conclusion cn be drwn. x 0 my be locl minimum or locl mximum or point of inflection. Theorem. (Inverse Fuction Theorem) Let f be continuous nd invertible on [, b] nd differentible t some point x 0 in [, b]. If f (x 0 ) 0, then f 1 is differentible t f(x 0 ) nd if we let y 0 = f(x 0 ), then (f 1 ) ( ) 1 y 0 = f (x 0 ) Theorem.3 (L Hôspitl s Rule) Let f nd g be differentible on (, b) nd let g (x) 0 on (, b). We llow = nd b =. If for some x 0 in [, b] we hve lim x x0 f(x) = lim x x0 g(x) = 0 or lim x x0 f(x) = lim x x0 g(x) = ±, nd if the limit f (x) lim x x 0 g (x) = L f(x) g(x) exists s rel number or s ±, then the limit lim x x0 nd f(x) lim x x 0 g(x) = lim f (x) x x 0 g (x) = L exists s rel number or ±, Theorem.4 If f nd g nd continuous on [, b] nd k is constnt, then (linerity) (f(x) + g(x)) dx = nd if f(x) g(x) on [, b], then kf(x) dx = k f(x) dx As consequence, if m f(x) M on [, b], then f(x) dx + f(x) dx g(x) dx g(x) dx m(b ) f(x) dx M(b ) 13
Theorem.5 Let f be Riemnn continuous on [, b]. Then for ll c in [, b] we hve f(x) dx = c f(x) dx + c f(x) dx Theorem.6 f f is continous on [, b], then f(x) dx = b f(x) dx Remrk.7 This is more mtter of convention thn theorem. Theorem.8 (Averge/Men Vlue Theorem for Integrls) [, b], then there is number c in [, b] such tht If f is continuous on f(c) = 1 b f(x)dx Theorem.9 (Fundmentl Theorem of Clculus I) differentible on (, b), nd if f is continuous on (, b), then f (x) dx = f(b) f() If f is continuous on [, b] nd Theorem.30 (Fundmentl Theorem of Clculus II) then the function F : [, b] R given by F (x) = x f(t) dt is continuous on [, b] nd differentible on (, b), with derivtive given by If f is continuous on [, b], F (x) = f(x) Remrk.31 This theorem, FTC II, gives the existence of ntiderivtives for ny Riemnn integrble function on [, b]. However, this result is only of theoreticl interest, s the form of the ntiderivtive, F (x) = x f(t) dt, isn t relly computtionlly helpful. However, it is useful for fmilirizing yourself with the nottion nd definitions. For exmple, if we wnt to tke the x-derivtive of F (sin x), this is strightforwrd, just pply the chin rule: letting y = sin x, d df dy F (sin x) = = f(x) cos x dx dy dx But it s slightly hrder to see in the form d sin x dx f(t) dt. However, the two expressions re the sme! d sin x f(t) dt = df dy = f(x) cos x dx dy dx 14
Theorem.3 (Integrtion by Prts) If f nd g re continuous on [, b] nd differentible on (, b), nd if f nd g re Riemnn integrble on [, b], then f(x)g (x) dx = [ ] b f(x)g(x) x= f (x)g(x) dx (.1) Theorem.33 (Chnge of Vrible) [, b] nd g is continuous on [α, β], then If f : [, b] [α, β] is continously differentible on g ( f(x) ) f (x) dx = f(b) f() g(y) dy (.) 15