Functions of One Real Variable. A Survival Guide. Arindama Singh Department of Mathematics Indian Institute of Technology Madras

Transcription

1 Functions of One Rel Vrible A Survivl Guide Arindm Singh Deprtment of Mthemtics Indin Institute of Technology Mdrs

2 Contents Limit nd Continuity 3. Preliminries Limit One-sided Limits nd Asymptotes Continuity Continuous functions on Closed Intervls Review Problems Differentition Derivtive Mxim Minim Tests for Mxim-Minim L Hospitl s Rule Curvture Review Problems Integrtion The definite integrl The Indefinite Integrl Substitution nd Integrtion by Prts Logrithm nd Exponentil Lengths of Plne Curves Review Problems Bibliogrphy 73 2

3 Chpter Limit nd Continuity. Preliminries We use the following nottion: = the empty set. N = {, 2, 3,...}, the set of nturl numbers. Z = {..., 2,,,, 2,...}, the set of integers. Q = { p q : p Z, q N}, the set of rtionl numbers. R = the set of rel numbers. R + = the set of ll positive rel numbers. N Z Q R. The numbers in R Q is the set of irrtionl numbers. Exmples re 2, 3. etc. Along with the usul lws of +,, <, R stisfies the Archimedin property: If > nd b >, then there exists n n N such tht n b. Also R stisfies the completeness property: Every nonempty subset of R hving n upper bound hs lest upper bound (lub) in R. Explntion: Let A be nonempty subset of R. A rel number u is clled n upper bound of A if ech element of A is less thn or equl to u. An upper bound l of A is clled lest upper bound if ll upper bounds of A re greter thn or equl to l. Notice tht Q does not stisfy the completeness property. For exmple, the nonempty set A = {x Q : x 2 < 2} hs n upper bound, sy, 2. But its lest upper bound is 2, which is not in Q. Similr to lub, we hve the notion of glb, the gretest lower bound of subset of R. Let A be nonempty subset of R. A rel number v is clled lower bound of A if ech element of A is greter thn or equl to v. A lower bound m of A is clled gretest lower bound if ll lower bounds of A re less thn or equl to m. The completeness property of R implies tht Every nonempty subset of R hving lower bound hs gretest lower bound (glb) in R. 3

4 The lub cts s mximum of nonempty set nd the glb cts s minimum of the set. In fct, when the lub(a) A, this lub is defined s the mximum of A nd is denoted s mx(a). Similrly, if the glb(a) A, this glb is defined s the minimum of A nd is denoted by min(a). Moreover, both Q nd R Q re dense in R. Tht is, if x < y re rel numbers then there exist rtionl number nd n irrtionl number b such tht x < < y nd x < b < y. We my not explicitly use these properties of R but some theorems, whose proofs we will omit, cn be proved using these properties. These properties llow R to be visulized s number line: Let, b R, < b. [, b] = {x R : x b}, the closed intervl [, b]. (, b] = {x R : < x b}, the semi-open intervl (, b]. [, b) = {x R : x < b}, the semi-open intervl [, b). (, b) = {x R : < x < b}, the open intervl (, b). (, b] = {x R : x b}, the closed infinite intervl (, b]. (, b) = {x R : x < b}, the open infinite intervl (, b). [, ) = {x R : x }, the closed infinite intervl [, ). (, ) = {x R : x b}, the open infinite intervl (, ). (, ) = R, both open nd closed infinite intervl. We lso write R + for (, ) nd R for (, ). These re, respectively, the sets of ll positive rel numbers, nd the set of ll negtive rel numbers. A neighborhood of point c is n open intervl (c δ, c + δ) for some δ >. A deleted neighborhood of point c is (c δ, c + δ) {c} = (c δ, c) (c, c + δ) for some δ >. Let D R. Let c D. We sy tht c is n interior point of D if there exist n open intervl (, b) such tht c (, b) D. Notice tht c in interior point of D if there exists neighborhood of c which is contined in D. For exmple,. is n interior point of [, ). The point is not n interior point of [, ). In contrst, we sy tht is left end-point of the intervls [, b) nd of [, b]. Similrly, b is right end-point of the intervls (, b] nd of [, b]. { x if x The bsolute vlue of x R is defined s x = x if x < Thus x = x 2. And = or ; x y is the distnce between rel numbers x nd y. 4

5 Moreover, if, b R, then =, b = b, = b b if b, + b + b, b b. Let x R nd let >. The following re true:. x = iff x = ±. 2. x < iff < x < iff x (, ). 3. x iff x iff x [, ]. 4. x > iff < x or x > iff x (, ) (, ) iff x R [, ]. 5. x iff x or x iff x (, ] [, ) iff x R (, ). Therefore, for R, δ >, x < δ iff δ < x < + δ. The following sttements re useful in proving equlities from inequlities: Let, b R.. If for ech ɛ >, < ɛ, then =. 2. If for ech ɛ >, < b + ɛ, then b. Let A, B R. A function f from A to B is rule tht ssigns ech point in A to some point in B in unique wy. The unique here mens tht if f() f, then b; tht is, f() cnnot be sometimes number b nd sometimes number c. The unique does not necessrily men tht ll points in A re ssocited to the sme point in B. We write such function f s f : A B. If x is generic point in A nd if f ssigns this x to y in B, then we write f(x) = y. Here, y is the imge of x under f nd x is one of the pre-imges of y. Often we sy tht f(x) is function, loosely, or even s y = f(x) is function. The set A is clled the domin nd the set B is clled the co-domin of f. The rnge of f is the set of ctul vlues tht hve been ssigned to in B. Tht is, the rnge of f is the set {f(x) : x A}. Mostly, we will be interested in functions with domins s intervls. When function is given in the form y = f(x), we do not sy explicitly wht its domin is. rther, we find out lrgest subset of R which my serve s its domin. For exmple, y = x 2 hs domin s R nd rnge s [, ). y = x hs domin nd rnge s [, ). y = /x hs domin nd rnge s R {}. y = 2 x hs domin s (, 2] nd rnge s [, ). y = 3 x 2 hs domin s [ 3, 3] nd rnge s [, 3]. The grph of function y = f(x) is the set {(x, f(x)) : x domin of f}. We plot the points (x, f(x)) in the plne R 2. In this plotting, the convention is to regrd x s the independent vrible 5

6 nd y = f(x) s the function. Since f() is uniquely defined for ech in the domin of f, if you drw verticl line on the grph of f, then it never crosses the grph in more thn one point. For exmple, the unit circle in the plne does not represent function but the semicircles bove the x-xis represent functions. The semicircles to the right or left of the y-xis re not functions. Grphs of some known functions re s follows: { x if x. y = x = x if x < x if x < 2. y = x 2 if x if x > 3. y = f(x) = { x if x 2 x if < x 2 4. y = x = n if n x < n + for n N. It is the lrgest integer less thn or equl to x. The lrgest integer function or the floor function. Sometimes we write s [ ]. 5. y = x = n + if n < x n + for n N. It is the smllest integer greter thn or equl to x. The smllest integer function or the ceiling function. 6

7 6. The power function y = x n for n =, 2, 3, 4, 5 look like 7. The power function y = x n for n = nd n = 2 look like 8. The grphs of the power function y = x for = 2, 3, 3 2 nd = 2 3 re 9. Polynomil functions re y = f(x) = + x + 2 x 2 + n x n for some n N {}. Here, the coefficients of powers of x re some given rel numbers,..., n nd n. The highest power n in the polynomil is clled the degree of the polynomil. Grphs of some polynomil functions re s follows: 7

8 . A rtionl function is rtio of two polynomils; f(x) = p(x), where p(x) nd q(x) re q(x) polynomils, my or my not be of the sme degree. Grphs of some rtionl functions re s follows:. Algebric functions re obtined by dding subtrcting, multiplying, dividing or tking roots of polynomil functions. Rtionl functions re specil cses of lgebric functions. Some grphs of lhebric functions: 2. Trigonometric functions come from the rtios of sides of right ngled tringle. The ngles re mesured in rdin. The trigonometric functions hve period. Tht is, f(x + p) = f(x) hppens for some p >. The period of f(x) is the minimum of such p. The period for sin x is 2π. The functions cos x nd sec x re even functions nd ll others re odd functions. Recll tht f(x) is even if f( x) = f(x) nd it is odd if f( x) = f(x) for ech x in the domin of the function. Some of the useful inequlities re x sin x x for ll x R. sin x, cos x for ll x R. cos x x for ll x R. sin x x tn x for ll x (, π/2). In fct, if x, then sin x < x. Some exmples: 8

9 3. Exponentil functions re in the form y = x for some > nd. All exponentil functions hve domin (, ) nd co-domin (, ). They never ssume the vlue. Grphs of some exponentil functions: 4. Logrithmic functions re inverse of exponentil functions. Tht is, log x = log ( x ) = x. Some exmples: Functions tht re not lgebric re clled trnscendentl functions. Trigonometric functions, exponentil functions nd logrithmic functions re exmples of trnscendentl functions. We will come bck to exponentil functions nd logrithmic functions lter. Two functions cn be combined to give nother function. If f(x) nd g(x) hve the sme domins, then f ± g(x) re new functions defined by (f ± g)(x) = f(x) ± g(x). 9

10 Similrly, (f g)(x) = f(x)g(x) nd (f/g)(x) = f(x)/g(x) provided tht g(x). Also, you cn hve h(x) = f(x) g(x) s nother function provided tht the ltter is meningful. Another wy is to obtin function by tking composition. If the rnge of f(x) is subset of the domin of g(x), then g(f(x)) is meningful. The composite function (g f) is then defined s (g f)(x) = g(f(x). Notice tht the inverse of function f(x) is defined through the composition s f is tht function which stisfies f(f (x)) = f (f(x)) = x. In tht cse, f must be one-one function onto its co-domin. Sometimes, we restrict the domin of function so tht it my be oneone nd onto so tht its inverse cn be defined. For exmple, restricting the domin of sin x to [ π/2, π/2], mkes it one-one nd onto the rnge [, ]. Then its inverse sin (x) is function from [, ] to [ π/2, π/2] nd this is lso one-one nd onto. This mens tht sin x is tht number y [ π/2, π/2] for which sin y = x. Similrly, for other trigonometric functions their inverse functions re defined. Wrning: sin x is not the sme s (sin x). Inverse trigonometric functions re given below:

11 .2 Limit The ide is to find how function behves ner point in its domin. Suppose y = f(x) is function, sy, defined on n open intervl (, b). Let c (, b). Under wht condition cn we sy tht when x (, b) is tken ner c, the vlues f(x) sty ner f(c)? To tckle the vgue notion of nerness, suppose I quntify it. Suppose ɛ >. I sy tht two numbers r nd s re ɛ-ner if their difference is bounded by ɛ; Tht is, when r s < ɛ. But the mount of nerness in x to c need not be the sme mount in f(x). However, there should be some connection between the two. I my sy tht whtever nerness I choose between f(x) nd f(), I find certin nerness in x nd c such tht if x lies ner c, then f(x) lies ner f(). If this condition is stisfied, we sy tht the function f is continuous t c. If f is not defined t c but t every point in n open intervl round c, then it is not meningful to write f(c). But still there my exist some rel number l such tht f(x) lies ner l. In such cse, we sy tht f hs limit l t c. We stte it formlly. Let < c < b. Let f : D R be function whose domin D contins the union (, c) (c, b). Let l R. We sy tht the limit of f(x) s x pproches c is l nd write it s lim f(x) = l iff for ech ɛ >, there exists δ > such tht for ech x (, c) (c, b) with < x c < δ, we hve f(x) l < ɛ. This is the forml definition of the notion of limit of function. In this definition, δ my depend on ɛ nd lso on the point c. Notice tht given function our definition does not led us to find l. it is guess work. Once guess hs been tken, we cn only verify whether the guess is correct. The definition sys tht ech ɛ-neighborhood of l must contin the f-imge of some δ-neighborhood of c. Writing D δ (c) = (c δ, c + δ) for δ-neighborhood of c, it demnds tht ech D ɛ (l) must contin the f-imge of some D δ (c). To put it nother wy, write the inverse imge of f of subset A of R s {x dom(f) : f(x) A}. Then lim f(x) = l iff inverse imge of ech neighborhood of l is contined in some neighborhood of c. We will not lwys use this forml definition; rther we will work with our intuitive understnding of the notion tht f(x) hs limit l t c provided the following condition holds: if x is ner c, then f(x) is ner l If certin cse poses problem, we my come bck to the forml definition.

12 Exmple.. Find lim x x. From the grph of x, we guess tht this limit is. To show this, let ɛ >. Choose δ = ɛ. (Why? Do not sk me; see whether it works.) If x = x < δ = ɛ, then f(x) = x < ɛ. Hence lim x x =. Exmple.2. Find lim x f(x), where f(x) = { 3 x if x < x 2 + if x >. On the left side of x = the function is 3 x. If x is ner, we see tht f(x) is ner 2. On the right side, if x is ner, then we see tht x 2 + lies ner 2. Therefore, lim x f(x) = 2. We cn check it formlly. Let ɛ >. Choose δ = min{ɛ, + ɛ }. (Agin, do not sk me how I got it; but see how it works.) Suppose δ < x < + δ; x. Brek into two cses. () Let x <. Tht is, δ < x <, Then f(x) 2 = 3 x 2 = x < δ ɛ. (b) Let x >. Tht is, < x < + δ, Then f(x) 2 = x = x 2 < ( + δ) 2 ( + + ɛ ) 2 = + ɛ = ɛ. Therefore, x < δ implies f(x) 2 < ɛ. We see tht it is often difficult to follow the definition. However, the definition cn be used to prove certin properties of limits which help us in the intuitive understnding nd lso in determining the limit of function t point. Before stting the theorem, let us consider some more exmples. Remember tht the nerness in f(x) to l includes f(x) being equl to l wheres the nerness in x to c, x c. Exmple.3. Let f(x) = 5 for x ; nd f() =. Wht is lim x f(x)? When x is ner but not equl to, f(x) = 5 or tht f(x) is ner 5. Therefore, lim x f(x) = 5. It does not mtter whether f is defined t or not. It lso does not mtter wht vlue f() hs. Exmple.4. Let f(x) = x2 25 x 5 for x 5. Wht is lim f(x)? x 5 When x is ner 5, the numertor x 2 25 is ner. So is the denomintor x 5. It looks tht the limit is undefined. However, it is wrong, since for x 5, f(x) = x2 25 = x + 5. Now, x + 5 stys x 5 ner when x is ner 5. Hence lim f(x) =. x 5 2

13 Theorem.. (The Limit Properties) Let k be constnt; or constnt function.. lim k = k nd lim x = c. 2. lim (f(x) ± g(x)) = lim f(x) ± lim g(x). 3. lim kf(x) = k lim f(x). 4. lim [f(x)g(x)] = lim f(x) lim g(x). 5. lim [f(x)/g(x)] = [ lim f(x) ] / [ lim g(x) ] if 6. lim (f(x)) r = (lim f(x)) r lim g(x). if tking powers re meningful. 7. lim f(x) is unique rel number if it exists. 8. If lim g(x) =, nd lim [f(x)/g(x)] exists, then lim f(x) =. 9. (Sndwich) Let f, g, h be functions whose domin include (, c) (c, b) for < c < b. Suppose tht g(x) f(x) h(x) for ll x (, c) (c, b). If lim g(x) = l = lim h(x), then lim f(x) = l.. (Domintion Limit) Let f, g be functions whose domins include (, c) (c, b) for < c < b. Suppose tht both lim f(x) nd lim g(x) exist. If f(x) g(x) for ll x (, c) (c, b), then lim f(x) lim g(x). It follows from Theorem. tht if p(x) = + x+ n x n is polynomil, then lim p(x) = p(c). Also, if q(x) is nother such polynomil with q(c), then lim (p(x)/q(x)) = p(c)/q(c). Also, the (8) implies tht If lim f(x) nd lim g(x) =, then lim (f(x)/g(x)) does not exist. x For exmple, lim does not exist. x 3 x + 3 Cution: In (), if f(x) < g(x) hppens for ll x, then we my not be ble to conclude tht lim f(x) < lim g(x). The first limit my become equl to the second limit. For exmple, consider the function f(x) = x with the domin R {}. We hve < f(x) for ll x in this domin. Then lim = = lim x. x x 3

14 Exmple.5. Evlute () lim x sin x (b) lim sin x x x (c) lim x cos x (d) lim f(x), given tht lim f(x) =. () Since sin x, we hve x x sin x x. We know tht lim x = lim x =. By Sndwich theorem, lim x sin x =. x x x (b) Since x sin x x, Sndwich theorem gives lim x sin x =. (c) Since cos x x, it follows tht lim x ( cos x) =. This gives lim x cos x =. (d) f(x) f(x) f(x). Since lim f(x) =, by Sndwich theorem, lim f(x) =. Functions my not hve limit t given point becuse there my be jump t tht point, or it my be growing too lrge ner tht point, or even it my be oscillting too often ner the point..3 One-sided Limits nd Asymptotes One-sided limits cn be defined in n nlogous mnner. We write lim f(x) = l iff whenever x lies ner c nd remins less thn c, we hve tht f(x) lies ner l. Tht is, in the ɛ δ definition, we simply consider ll x stisfying ll those conditions long with x being less thn c. This is clled the left hnd limit of f(x) t x = c. Similrly, we write lim f(x) = l iff whenever x lies ner c + nd remins greter thn c, we hve tht f(x) lies ner l. This is the right hnd limit of f(x) t x = c. It thus follows tht lim f(x) = l iff lim f(x) = lim f(x) = l. + Exmple.6. Evlute the left hnd nd right hnd limits if they exist: () f(x) = x/ x t x = (b) g(x) = 4 x 2 t ll the points in its domin [ 2, 2]. () For the left hnd limit, x <. It gives f(x) = x/ x = x/( x) =. Thus lim f(x) =. x () =. For the right hnd limit, x >. Then lim (x/ x ) = lim (x/x) = lim x + x + x + Thus lim x (x/ x ) does not exist. (b) We cnnot spek of the left hnd limit of g(x) t x = 2 since the domin does not include n open intervl to the left of x = 2. Similrly, we cnnot spek of right hnd limit of g(x) t x = 2. But lim x 2+ 4 x2 = = lim 4 x2. Hence lim 4 x2 =. x 2 x 2 4

15 At ll (other) points c ( 2, 2), lim g(x) = 4 c 2. All the properties in Theorem. hold true for one-sided limits. Exmple.7. Show tht lim x sin x x =. Using the inequlity sin x < x < tn x for < x < π/2, we hve < x sin x < cos x Or cos x < sin x x <. By Sndwich theorem, sin x lim x + x =. When x <, write x = t for t >. Then sin x = sin t. sin x lim x x = lim sin t t + t sin x Hence lim x x =. We lso bring in in discussing limits. These re referred to s limits t. The forml definitions follow: Let I be (, ) or [, ) for some R. Let f : I R. Let l R. We sy tht lim x f(x) = l if for ech ɛ >, there exists n m > such tht if x is ny rel number greter then m, then f(x) l < ɛ. Let I be (, ) or (, ] for some R. Let f : I R. Let l R. We sy tht f(x) = l if for ech ɛ >, there exists n m > such tht if x is ny rel number less lim x thn m, then f(x) l < ɛ. =. Limits t or t hve ll the sme properties s the limit t ny point c. Exmple.8. Show tht lim x x = lim x x =. Let ɛ >. Choose m = /ɛ. If x > m, then x > nd /x = /x < ɛ. Hence lim x x =. For the other limit, let ɛ >. Choose m = /ɛ. If x < m, then > /x > /m = ɛ. So, /x = /x < ɛ. Hence lim x x =. As erlier, we will not use this ɛ δ definition often in our clcultions. These definitions re not new provided tht you consider ny neighborhood of s ny intervl (, ) nd ny neighborhood of is of the form (, ) for >. We rther depend on our intuitive understnding. For the limit t to be equl to l, we understnd tht s x increses without bound, f(x) remins ner l. Similrly, for the limit t to be equl to l mens tht s x decreses without bound, f(x) lies ner l. Geometriclly, it mens tht t or t, the line y = l merges with the curve y = f(x). 5

16 We sy tht the line y = l is n symptote of y = f(x) t if lim x f(x) = l. Similrly, the line y = l is n symptote of y = f(x) t if lim f(x) = l. x Both the symptotes t nd t re clled symptotes of y = f(x). An symptote y = l is lso clled horizontl symptote. An symptote cn intersect the curve s the following exmple shows. Exmple.9. Find the symptotes of y = f(x) = 2 + sin x x. Since sin x sin x, by Sndwich theorem, lim x x x ± x The line y = 2 is the symptote to y = 2 + sin x x =. Hence lim x ± t nd lso t. ( sin x) 2 + = 2. x In generl, stright line y = x+b is n symptote to y = f(x) if lim (f(x) (x + b)) = or x if lim (f(x) (x + b)) =, respectively t nd t. Such symptotes re clled slnted x symptotes. Usully, when rtionl function p(x)/q(x) hs the degree of the numertor p(x) one more thn the degree of the denomintor q(x), then slnted symptote exists. Exmple.. Find n symptote of y = f(x) = 2x2 3 ( Using long division write f(x) = 2x x + 4 = 7 x 8 ) (7x + 4). { ( 2 Notice tht lim f(x) x ± 7 x 8 ) } { } 5 = lim =. 49 x ± 49(7x + 4) Hence the line y = 2x 8 is n symptote to f(x) both t nd t x+4. 6

17 Sometimes it is esier to clculte the limits t infinity by substituting the independent vrible by its reciprocl. We see tht if the limits t infinity exist, then lim f(x) = lim f(/x) nd lim f(x) = lim f(/x). x x + x x Consider two functions f(x) nd g(x) with the sme domin (, ). Let f(x) = for ll positive rtionls nd f(x) = for ll positive irrtionls. In contrst, tke g(x) = /x for ll positive x. Limits for both s x pproches from the right hnd side, do not exist. For f(x), the limit does not exist becuse the vlues f(x) when x is ner fluctutes between nd, nd never remins ner either of them. Wheres s x pproches, g(x) increses without bound, so does not remin ner ny rel number. We wnt to seprte the cse of g(x) from the rest. We sy tht g(x) hs n infinite limit s x pproches. Forml definitions in vrious cses follow.. Let f(x) hve domin contining (c, b). Then lim f(x) = iff for ech m >, there + exists δ > such tht for every x with c < x < c + δ, we hve f(x) > m. Agin we work with n intuitive understnding tht lim f(x) = iff, s x decreses to c, f(x) increses without bound Let f(x) hve domin contining (, c). Then lim f(x) = iff for ech m >, there exists δ > such tht for every x with c δ < x < c, we hve f(x) > m. Tht is, lim f(x) = iff, s x increses to c, f(x) increses without bound. 3. Let f(x) hve domin contining (c, b). Then lim f(x) = iff for ech m >, there + exists δ > such tht for every x with c < x < c + δ, we hve f(x) < m. Tht is, lim f(x) = iff, s x decreses to c, f(x) decreses without bound Let f(x) hve domin contining (, c). Then lim f(x) = iff for ech m >, there exists δ > such tht for every x with c δ < x < c, we hve f(x) < m. Tht is, lim f(x) = iff, s x increses to c, f(x) decreses without bound. As erlier, we write lim f(x) = if Similrly, we write lim f(x) = if lim f(x) = = lim f(x). + lim f(x) = = lim f(x) Let f(x) hve domin contining (, ). Then lim x f(x) = iff for ech m >, there exists δ > such tht for every x with x > δ, we hve f(x) > m. Tht is, lim f(x) = iff, s x increses without bound, f(x) lso increses without x bound. Notice tht lim f(x) = iff lim f(/x) =. x x + 6. Let f(x) hve domin contining (, ). Then lim x f(x) = iff for ech m >, there exists δ > such tht for every x with x > δ, we hve f(x) < m. 7

18 Tht is, lim f(x) = iff, s x increses without bound, f(x) decreses without bound. x Agin, lim f(x) = iff lim f(/x) =. x x + 7. Let f(x) hve domin contining (, ). Then lim f(x) = iff for ech m >, x there exists δ > such tht for every x with x < δ, we hve f(x) > m. Tht is, lim f(x) = iff, s x decreses without bound, f(x) increses without bound. x Here, f(x) = iff lim lim x f(/x) =. x 8. Let f(x) hve domin contining (, ). Then lim f(x) = iff for ech m >, x there exists δ > such tht for every x with x < δ, we hve f(x) < m. Tht is, lim f(x) = iff, s x decreses without bound, f(x) lso decreses without x bound. Once more, lim f(x) = iff lim x f(/x) =. x If you understnd the principle behind limits t infinity nd infinite limits, then you will not hve to remember ech one seprtely. Cution: When limit of function t point, or t infinite, becomes infinite, it does not men tht the limit exists. It only sys tht this non-existence of limit is specil, in the sense of growing too lrge or becoming smller nd smller without bound. In fct, if the limit of function exists t x = c, then it is necessrily rel number, which is never equl to or. The lgebr with the symbols nd re s follows: Let R, b > nd c <. Then + =, = + ( ) =, ( ) =, / = /( ) =. b =, b ( ) =, c =, c ( ) =. + ( ),, ( ), ± / ± re indeterminte. All the properties of limits hold provided we do not end up t indeterminte forms. For exmple, if lim f(x) = d > nd lim g(x) =, then lim f(x)g(x) =. Exmple.. Let p(x) = + x + + n x n, where n. Let b >. Show the following; p(x) () lim x x = n. (b) lim n p(x) () lim x x n = lim x p(x) = if n > x ( x + n x + + n n x + n (c) lim p(x) = if n <. x ) = n. p(x) (b-c) From (), lim p(x) = lim lim x n = n. Now the nswers follow. x x x x n This exmple sys tht s x pproches, p(x) behves s its leding term. Infinite limits llow us to define verticl symptotes. A line x = c is clled verticl symptote of the function y = f(x) if one of the following hppens: lim f(x) = ± or lim f(x) = ±. + 8

19 Exmple.2. Find the horizontl nd verticl symptotes of y = f(x) = x + 3 x + 2. We see tht lim f(x) = Also, lim =. And, for no other point c, limit of f(x) is ± x 2 x 2+ s x pproches c. So, x = 2 is verticl symptote. Next, lim f(x) =. Also, lim f(x) =. So, y = is the horizontl symptote. x x Alterntively, write f(x) s f(x) = + x + 2. Now, both horizontl nd verticl symptotes re visible from this expression! As x 2, from left or from right, f(x). And s x ±, f(x). There re curves with infinitely mny symptotes. For exmple, y = sec x nd y = tn x hve verticl symptotes t points (2n + )π/2 for ech n N..4 Continuity Let f : D R be function. Let c be n interior point of D. We sy tht f(x) is continuous t c if lim f(x) = f(). If D = [, b) or D = [, b], then f(x) is clled continuous t the left end-point if lim f(x) = f(). x + If D = (, b] or D = [, b], then f(x) is clled continuous t the right end-point b if lim f(x) = f(). x b f(x) is clled continuous if it is continuous t ech point of its domin D. For domins of the type (, ] [2, 3), continuity cn be defined similrly t x = nd t x = 2, but we will not need them. 9

20 A function with domin s in intervl if found to be continuous t ll points, then you cn drw its grph without lifting the pen. In the following figure, the functions in () is continuous t x =. Others re not continuous t x =, though the types of discontinuity re different. Write out wht kinds of discontinuity you observe. Exmple.3. Where is the function defined below continuous? { if x < 4 x 3 f(x) = 2x + 3 if x 4 At ny point c < 3, f(x) = /(x 3) is continuous since lim f(x) = /(c 3). At x = 3, f(x) is not defined; so we cnnot spek of its continuity t x = 3. For 3 < c < 4, f(x) = /(x 3). It is gin continuous. At x = 4, lim f(x) = lim {/(x 3)} =. And lim f(x) = lim (2x + 3) =. Hence limit x 4 x 4 x 4+ x 4+ of f(x) does not exist t x = 4. Tht is, f(x) is not continuous t x = 4. For c > 4, clerly f(x) = 2x + 4 is continuous t x = c. Due to the properties of limits, the following theorem is obvious. Theorem.2. Let f(x) nd g(x) be functions continuous t x = c. Let k R. Then. f(x) + g(x), f(x) g(x), f(x) g(x), k f(x) re continuous t x = c. 2. f(x)/g(x) is continuous t x = c provided tht g(c). 3. [f(x)] k is continuous t x = c provided [f(x)] k is defined in n intervl round c. 4. In ddition, if h(x) is continuous t f(c), then (h f)(x) is continuous t c. Every polynomil nd every rtionl function is continuous. The trigonometric functions re continuous on their respective domins. The bsolute vlue function f(x) = x is continuous. The 2

21 function f(x) = /x is continuous on R {}, which is its domin. So, f(x) = /x is continuous function. Exmple.4. Is the function f(x) = x sin x continuous? x f (x) = x is continuous, f 2 (x) = sin x is continuous. Thus f 3 (x) = x sin x is continuous. f 4 (x) = x is continuous nd is never. So, f 5 (x) = f 3 (x)/f 4 (x) is continuous. The function f 6 (x) = x is continuous. So, f(x) = (f 6 f 5 )(x) is continuous. Its grph looks like: The function f(x) = sin x x function g(x) = sin x x is not continuous t x = since it is not defined t x =. However, the for x ; nd g() = is continuous everywhere since lim x sin x x = = g()..5 Continuous functions on Closed Intervls The forml definition cn be used to prove nice fct bout continuous functions. Theorem.3. Let f(x) be continuous t x = c, where the domin of f(x) includes neighborhood of c. If f(c) >, then there exists neighborhood (c δ, c + δ) such tht f(x) > for ech point x (c δ, c + δ). Proof: Suppose tht f(c) >. Let ɛ = f(c)/2. Since f(x) is continuous t x = c, we hve δ > such tht for ech x (c δ, c + δ), f(x) f(c) < f(c)/2. Tht is, f(c)/2 < f(x) < 3f(c)/2. As f(c) >, we see tht for ech x (c δ, c + δ), we hve f(x) >. Continuous functions on intervls enjoy nother nice property: The imge of closed bounded intervl under continuous function is closed bounded intervl. We write it formlly s follows. Theorem.4. Let f(x) be continuous function, whose domin contins [, b] for < b. Then there exist α, β R such tht {f(x) : x [, b]} = [α, β]. Proof of this result requires the completeness property of R nd we omit it. Of course, the imge s n intervl my degenerte into single point, for exmple, when f(x) is constnt function. We will discuss two importnt corollries of Theorem.4. 2

22 If f(x) is not known to be continuous, then the rel numbers α nd β given by α = glb{f(x) : x [, b]} nd β = lub{f(x) : x [, b]} my or my not be in the imge of [, b] under f. However, if f(x) is continuous, then α nd β must be in the imge of [, b] under f. This is the content of our first corollry. Theorem.5. (Extreme Vlue Theorem) Let f(x) be continuous on closed bounded intervl [, b]. Then there exist numbers c, d [, b] such tht f(c) f(x) f(d) for ech x [, b]. Proof: By Theorem.4, we hve α, β R such tht {f(x) : x [, b]} = [α, β]. Then α f(x) for ech x [, b] nd α {f(x) : x [, b]}. Tht is, there exists c [, b] such tht α = f(c). Similrly, β f(x) for ech x [, b] nd there exists d [, b] such tht β = f(d). The Extreme Vlues Theorem (EVT) sys tht the quntities re well-defined. It is informlly stted s min{f(x) : x [, b]} nd mx{f(x) : x [, b]} A continuous function on closed bounded intervl chieves its minimum nd mximum. The other importnt corollry of Theorem.4 is s follows: Theorem.6. (Intermedite Vlue Theorem) Let f(x) be continuous on closed bounded intervl [, b]. Let d be number between f() nd f(b). Then there exists c [, b] such tht f(c) = d. Proof: By Theorem.4, we hve α, β R such tht {f(x) : x [, b]} = [α, β]. Then α f() β nd α f(b) β. If d lies between f() nd f(b), then α d β. Therefore, there exists c [, b] such tht f(c) = d. Intermedite Vlue Theorem (IVT) sys tht function continuous on closed intervl [, b] tkes on every vlue between f() nd f(b). It helps in brcketing zero of function. A number α is clled zero of function if f(α) =. Given continuous function f(x), if f() nd f(b) hve opposite signs, then the intervl (, b) must contin point c where f(c) = ; IVT confirms this conclusion. Remrk: We hve derived IVT nd EVT from Theorem.4. However, IVT nd EVT together prove Theorem.4. In such n pproch, we would require independent proofs of IVT nd EVT; nd these proofs use the completeness principle of R. 22

23 Exmple.5. Let ABCD be squre. Drw curve lying inside ABCD nd joining the digonlly opposite points A nd C. Drw curve joining B nd D nd gin lying inside ABCD. Prove tht the two curves intersect somewhere inside the squre. (Though curve is much more generl, here we ssume tht they re represented by continuous functions of x.) Let the vertices of the squre be A(, ), B(, ), C(, ), nd D(, ) in the plne. The curves re (ssumed to be) continuous functions f, g : [, ] [, ]. We see tht f() =, f() = nd g() =, g() =. Then h(x) = f(x) g(x) is continuous function from [, ] to [, ]. And h() =, h() =. Therefore, by IVT, there is point c [, ] such tht h(c) =. At this point, f(c) = g(c), tht is, the curves intersect. Exmple.6. Prove tht every polynomil of odd degree hs rel root. Without loss of generlity let p(x) = + x + + n x n + x n, where n is odd. Now, lim p(x) =. So, there exists point < such tht p() <. Also, lim p(x) =. So, x x there exists point b > such tht p(b) >. Consider p s function from [, b] to R. Since [p(), p(b)], by IVT, we conclude tht there exists point c [, b] such tht p(c) =. If function is not continuous, then the conclusion of IVT my or my not hold. Exmple.7. Define f : [, ] R by f(x) = x if x nd f() =. Then f(x) is not continuous t x =. But it stisfies the conclusion of IVT, even on ny closed subintervl of [, ]. On the other hnd, let g : [, 2] R be given by g(x) = x for x nd g(x) = x + 2 for < x 2. Here, g is not continuous t x =, g() =, g(2) = 4. For no c [, 2], g(c) = 2..6 Review Problems Problem : () Find l = lim x x sin x. x sin x x. By Sndwich theorem, l =. Sme rgument shows tht lim x x n sin x (b) Find l = lim x sin x. = for ny n N. When x is ner, we my tke x s (mπ/2) for lrge m or we my choose x s (nπ) for lrge n. Then sin in the first cse, will be equl to nd in the seond cse, will be equl to x. Notice tht in ny neighborhood of x =, there re numbers of the first form nd lso of the second form for some lrge m nd n. Therefore, this limit does not exist. Problem 2: Find the limits lim f(x), lim g(x) where x x { { if x Q x if x Q f(x) = g(x) = if x Q, if x Q. In ny neighborhood of x =, there re rtionl s well s irrtionl numbers. Thus, for ny rtionl number close to x =, f(x) = where s for irrtionl numbers close to x =, f(x) =. Therefore, the first limit does not exist. 23

24 For g(x), if =, nd x is ny rtionl number close to =, then f(x) = x which is close to. On the other hnd, when x is ny irrtionl number close to, f(x) is equl to. Therefore, g(x) =. lim x For x, if x is ny rtionl number close to, then f(x) = x is close to, nd not to. Wheres if x is n irrtionl number close to, then f(x) is equl to. Therefore, limit of g(x) s x pproches does not exist, for. + x if x < Problem 3: Discuss the continuity of f(x) = + x + sin x if x π/2 3 if x π/2. For x <, f(x) = + x, which is continuous function. At x =, lim f(x) = lim ( + x) =, lim x x f(x) = lim ( + x + sin x) =, f() =. x + x + Hence f(x) is continuous t x =. For < x < nd < x < π/2, f(x) = + x + sin x is continuous. At x =, lim ( + x + sin x) = lim ( + + sin x) = + sin. x x lim ( + x + sin x) = lim ( + + sin x) = 2 + sin = f(). x + x So, f(x) is not continuous t x = ; it is only right-continuous t x =. For x > π/2, f(x) = 3, which is continuous. At x = π/2, lim f(x) = x π/2 So, f(x) is continuous t x = π/2. lim ( + x + sin x) = 3 = lim x π/2 In summry, f(x) is continuous everywhere except t the point x =. f(x) = f(π/2) = 3. x π/2+ Problem 4: Let f(x) be n unbounded function, i.e., for ech m N, there exists n x such tht f(x) > m. Does it follow tht lim x f(x) = ±? No. For exmple, let f(x) = x sin x. Let m N. Consider x = mπ/2. Now, sin x =. Consequently, x sin x = mπ/2 > m. But lim x sin x ±. Reson: If x is lrge nd in the x form mπ, then x sin x =. Problem 5: Find the lrgest integer m such tht lim x [ ] +x m( x) + sin(x ) = x + sin(x ) 4. We re interested in finding the limit of the given function when x is ner but x is not equl to. sin(x ) If m =, then using lim =, we find tht the limit is equl to ( x 2 x )2 =. Thus m = 4 is possible vlue. There cn be others. 24

25 m( x) + sin(x ) If m nd x is ner, then <. Its power is not defined. x + sin(x ) { x 2n sin( x Therefore, m =. Exercise : Let f(x) = ) if x if x =. Show tht f (), f (),..., f (n) () exist, f (n) (x) is not continuous t x =. { x 2n+ sin( x Exercise 2: Let g(x) = ) if x if x =. Show tht g (), g (),..., g (n) () exist, g (n) (x) is continuous t x =, but g (n) (x) is not differentible t x =. 25

26 Chpter 2 Differentition 2. Derivtive Let f(x) be function whose domin includes n open intervl (, b). Let c (, b). If the limit f(c + h) f(c) lim h h exists, we sy tht f(x) is differentible t x = c; nd we write the limit s f (c) nd cll it the derivtive of f(x) t x = c. We lso write f (c) s df, nd lso s df dx x=c dx (c). Throughout, we will use the prime ( ) nottion for differentition with respect to the concerned independent vrible. For exmple, if y = f(x), then both y nd f (x) denote dy dx, wheres if x = g(y), then both x nd g (y) will denote dx dy. Exmple 2.. Is f(x) = x 3 x 2 + x 2 differentible t x =? {( + h) 3 ( + h) 2 + ( + h) 2} { } lim h h Hence f(x) is differentible t x = nd f () = 2. Exmple 2.2. Is f(x) = x differentible t x =? = 2. + h lim h h h = lim h h does not exist since the left hnd limit is wheres the right hnd limit is. Therefore, x is not differentible t x =. Exmple 2.3. Is f(x) = x /3 differentible t x =? h /3 lim h h = lim h h 2/3 does not exist. Hence x /3 is not differentible t x =. 26

27 Let y = f(x) be curve. Let (x, f(x )) nd (x + h, f(x + h)) be two points on the curve. The line joining these points is secnt to the curve. The slope of the secnt is f(x + h) f(x ) = f(x + h) f(x ). x + h x h Then its limit s h gives the slope of the tngent to y = f(x) t the point x = x. Tht is, f (x ) is the slope of the tngent to the curve y = f(x) t x = x. We should be cutious in using the word tngent. A tngent to curve my be defined cripticlly s unique stright line tht hs exctly one point in common with the curve t identiclly two points. Tht is, t point, tngent is unique nd if we chnge its slope little bit, then it hs more thn one point in common with the curve. For exmple, there re mny stright lines tht touch the curve y = x t the point x = nd therefore, this curve hs no tngent t x =. Leving out the horizontl nd verticl tngents, we my tke the following view: if both dy dx nd do not exist, dx dy or if both dy dx = dx = t point, then tngent to the curve is not defined t tht point. If both dy dy dx nd re nonzero t point, then tngent to the curve exists t tht point. dx dy If curve is given prmetriclly by x = x(t), y = y(t), nd t point t = t, both x (t ) nd y (t ) re nonzero, then the stright line (y y(t ))x (t ) (x x(t ))y (t ) = is the tngent to the curve t t = t. Similrly, if the curve is given implicitly by f(x, y) = nd both m = dy dx (x, y ) nd m = dx dy (x, y ) re nonzero t the point (x, y ), then the tngent t (x, y ) is given by y y = m(x x ) nd lso by x x = m(y y ). However, differentibility sys something more thn the existence of tngent t point. For exmple, the tngent to the curve y = x /3 t x = is the y-xis wheres y = x /3 is not differentible t x =. Becuse, h /3 lim h h = lim h h 2/3 =. The curve hs verticl tngent t x =, which intersects the curve only t tht point. 27

28 We thus sy tht curve y = f(x) hs verticl tngent t x = c if f (c) is either equl to or equl to. Cution: Verticl tngents re not covered in the usul interprettion of derivtive s the slope of the tngent. Wht bout the curve y = x 2/3? Compute its derivtive: h 2/3 lim h h = lim h h /3. We get the left hnd limit s nd the right hnd limit s. Tht is, this curve does not even hve verticl tngent. We sy tht function f(x) is differentible if it is differentible t every point of its domin. Leving the pthologicl cses side, we will write f (x) s new function obtined from f(x). Then f (c) is the vlue of f (x) t x = c. Let φ(t) be differentible function on n open intervl (, b). Suppose tht φ(t) represents the position of moving body t time t. Let c (, b). Then φ (c) = instntneous velocity of the moving body t t = c. Theorem 2.. Ech function differentible t x = c is continuous t x = c. Proof: Let c be ny point in the domin of f(x). Given tht f(x) is differentible t x = c, we hve rel number f (c) nd f f(c + h) f(c) (c) = lim. Now, h h [ ] f(c + h) f(c) lim{f(x) f(c)} = lim{f(c + h) f(c)} = lim h h h h f(c + h) f(c) = lim lim h = f (c) =. h h h Therefore, lim f(x) = f(c). And the function f(x) is continuous t x = c. It only sys tht differentible function is continuous on its domin; it does not sy tht the derivtive of function is continuous. { x 2 sin(/x) if x Exmple 2.4. Let f(x) = if x =. We see tht f (x) = g(x) = 2x sin(/x) cos(/x) for x. Also, lim x g(x) does not exist. Hence f (x) is not continuous t x =. The derivtive of function need not be continuous, but it stisfies the conclusion of IVT. Its proof gin uses the completeness principle of R. 28

29 Theorem 2.2. (Drboux Theorem) Let f be differentible function whose domin includes n intervl [, b]. If d is between f () nd f (b), then there exists point c [, b] such tht d = f (c). Tht is, f (x) tkes on every vlue between f () nd f (b). Since derivtives re defined through the limiting process, the usul lws of limits pply to derivtives, with some twists. Theorem 2.3. Let f(x), g(x) be differentible functions. Let k R. Then. (f(x) + g(x)) = f (x) + g (x). 2. (k f(x)) = k f (x). 3. (f(x)g(x)) = f (x) g(x) + f(x) g (x). 4. (g(f(x))) = g (f(x)) f (x). 5. (f(x)/g(x)) = [f (x)g(x) f(x)g (x)]/(g(x) 2 ). 6. (f (x)) = f (f (x)). The fourth one, clled the Chin Rule is written more suggestively s dg(f(x)) dx = dg(f(x)) df(x) df(x) dx. We prove (3), (4) nd (6). Proof: (3) f(x + h)g(x + h) f(x)g(x) h Tking limit s h proves (3). For (4), consider x = ; define = f(x + h) f(x) g(x + h) g(x) g(x + h) + f(x). h h φ(h) = { g(f(+h)) g(f()) f(+h) f() g (f()) if f( + h) f() if f( + h) = f(). Now, φ is continuous t. And lim h φ(h) = φ() = g (f()). Also, g(f( + h)) g(f()) h = φ(h) Tking limit s h gives (g f) () = g (f()) f (). f( + h) f(). h (6) Let g(x) = f (x) be the inverse of the function f(x). We ssume tht f (x) is not zero in the domin of f(x) which is n intervl I. Differentiting f(g(x)) = x we hve This proves (6) f (g(x))g (x) = df dx = g (b) = (f (x)) x=f x=b (b). 29

30 We will not use the formul in (6) explicitly. We will rther differentite the equtions g(f(x)) = x or f(g(x)) = x s pproprite. For exmple, differentiting sin(sin x) = x, we obtin cos(sin x)(sin x) =. Since sin(sin x) = x, we hve cos(sin x) = (sin(sin x)) = x 2. Therefore, (sin x) = x 2. The derivtive of inverse of function in Theorem 2.3(6) is lso stted in different wy. For function y = f(x), we my write x = f (y) provided tht f exists. In this cse, dy dx dx dy =. This formultion helps in differentiting function given in prmetric form or even given implicitly. If function y = f(x) is given prmetriclly, sy, s x = g(t) nd y = h(t), then dy dx = dy dt dt dx = dy dt ( dx ). dt If y = f(x) is given implicitly by n eqution g(x, y) =, then we differentite g(x, y) s function of x, treting y lso s function of x. Since the derivtive of is, we get n eqution in y. Then we solve this new eqution in determining y. For exmple, if y = f(x) is given implicitly by x 2 + sin(xy) + y 2 =. Then differentiting the eqution we get: 2x + cos(xy)(y + xy ) + 2y y =. Solving this, we obtin y 2x + y cos(xy) = 2y + cos(xy). Anlogous to one-sided continuity t the end-points of semi-closed or closed intervl, we lso define one-sided derivtives. Let the domin of function f(x) be either [, b) or [, b]. If the limit f( + h) f() lim h + h exists, then this limit is the right hnd derivtive of f(x) t x = ; nd we denote it s f +(). Similrly, suppose the domin of function f(x) is either (, b] or [, b]. The left hnd derivtive of f(x) t x = b is defined s f (b) f(b + h) f(b) = lim h h provided this limit exists. Notice tht f (b) = lim h + f(b) f(b h) h. Also, notice tht f +() need not be the sme s lim f ( + h), nd f () need not be equl to h + lim f ( + h). These two equlities sy something bout the continuity of f (x) t the end-points. h + 3

31 To extend the vocbulry bit, suppose f(x) is differentible on (, b) nd it hs right hnd derivtive t x =, then we sy tht f(x) is differentible on [, b). Similrly, if f(x) is differentible on (, b) nd it hs left derivtive t x = b, then f(x) is clled differentible on (, b]. Suppose the domin of f(x) includes [, b]. If f(x) is differentible on (, b) nd hs both right derivtive t x = nd left derivtive t x = b, then we sy tht f(x) is differentible on [, b]. Obviously, t n interior point, if both the left hnd derivtive nd the right hnd derivtive re equl, then tht quntity is the derivtive t the point. The bsolute vlue function x is differentible on [, ). It is lso differentible on (, ]. But it is not differentible on (, ). Reson: the right hnd derivtive of x t x = is not equl to the left hnd derivtive t x =. The function f(x) = x for x, is not differentible t x =. Reson? lim h + + h The Right hnd derivtive t x = does not exist. Common functions nd their derivtives h =. Function Derivtive Where k x r rx r r sin x cos x cos x sin x tn x sec 2 x x nπ/2, n Z sec x sec x tn x x nπ/2, n Z cosec x cosec x cot x x nπ, n Z cot x cosec 2 x x nπ, n Z sin x x 2 x < tn x +x 2 sec x x x 2 x > Use the formuls: cos x = π/2 sin x, cot x = π/2 tn x, cosec x = π/2 sec x for their derivtives. 2.2 Mxim Minim Let function f(x) hve domin D. The function f(x) hs n bsolute mximum t point d D if f(x) f(d) for every x D. in such cse, we lso sy tht the point x = d is point of bsolute mximum of the function f(x). 3

32 Similrly, f(x) hs n bsolute minimum t b D if f(b) f(x) for every x D. In this cse, we sy tht the point x = b is point of bsolute minimum of the function f(x). The points of bsolute mximum nd bsolute minimum re commonly referred to s bsolute extremum points; nd the function is sid to hve bsolute extrem t those points. In the figure, x = d is point of bsolute mximum nd x = is point of bsolute minimum. But the point x = c is lso some sort of mximum. It is point where f(x) chieves its mximum compred with vlues of f(x) t ll the neighboring points. Let function f(x) hve domin D. The function f(x) hs locl mximum t point d D if f(x) f(d) for every x in some neighborhood of d contined in D. in such cse, we lso sy tht the point x = d is point of locl mximum of the function f(x). Similrly, f(x) hs n locl minimum t b D if f(b) f(x) for every x in some neighborhood of b contined in D. In this cse, we sy tht the point x = b is point of locl minimum of the function f(x). The points of locl mximum nd locl minimum re commonly referred to s locl extremum points; nd the function is sid to hve locl extrem t those points. As we know from Theorem.5, continuous function defined on closed intervl chieves its extremum points. These re the bsolute extrem. If differentibility of f(x) is ssumed, then something cn be sid bout locl extremum points. Theorem 2.4. Let f(x) with domin D hve locl extremum t x = c. Let c be n interior point of D. Suppose tht f(x) is differentible t x = c. Then f (c) =. Proof: Suppose tht f(x) hs locl minimum t x = c. There exists neighborhood (c δ, c + δ) of c such tht for every x (c δ, c + δ), we hve f(c) f(x). Now, for ll such points x we obtin: f(x) f(c) x c for x < c, nd f(x) f(c) x c for x > c. Tking the limit s x c, the first inequlity sys tht f (c) nd the second inequlity sys tht f (c). Therefore, f (c) =. When f(x) hs locl mximum, the inequlities bove get chnged. Then lso we conclude tht f (c) =. Notice tht there cn be points where function cn hve its extreme vlue where f(x) is not differentible or such point is not n interior point. In such scenrio, Theorem.5 is not pplicble. 32

33 Let f(x) hve domin D. A point c D is clled criticl point of f(x) if c is not n interior point of D, or if f(x) is not differentible t x = c, or if f (c) =. Then Theorem 2.4 sys tht if f(x) hs n extremum t x = c, then c is criticl point of f(x). A criticl point need not be n extremum point. For exmple, the function f(x) = x 3 defined on R hs criticl point t x = but this function does not chieve n extreme vlue t x =. Reson: to the left of x =, the vlues of x 3 re less thn = f() nd to the right of x =, the vlues of x 3 re greter thn. When is criticl point n extremum point? To nswer this question we discuss some results which re geometriclly ppeling. Theorem 2.5. (Rolle s Theorem) Suppose tht f : [, b] R is continuous, f(x) is differentible on (, b), nd f() = f(b). Then f (c) = for some c (, b). Proof: If f(x) is constnt function, then there is nothing to prove. Assume tht f(x) is not constnt function. By Theorem.5, f(x) hs mximum nd minimum in [, b]. Since f(x) is not constnt, t lest one of these extreme vlues is different from f(), nd lso different from f(b). Suppose such n extreme vlue is chieved t x = c. Then c (, b), is n interior point of [, b]. By Theorem 2.4, f (c) =. Exmple 2.5. Show tht x 3 + x + b hs unique rel root if >. Since f(x) = x 3 + x + b is polynomil of odd degree, it hs rel root. Suppose c < d re two rl roots of f(x). Then by Rolle s theorem, we hve r (c, d) such tht f (r) =. But f (x) = 3x 2 + >. Rolle s theorem cn be generlized. Theorem 2.6. (Men vlue Theorem) Suppose tht f : [, b] R is continuous nd f(x) is differentible on (, b). Then there exists c (, b) such tht f(b) f() = f (c)(b ). f(b) f() Proof: Define g : [, b] R by g(x) = f(x) f() (x ). b Then g(x) is continuous on [, b], differentible on (, b), nd g() = = g(b). By Rolle s theorem, we hve c (, b) such tht g (c) =. Tht is, f(b) f() = f (c)(b ). 33

34 The conclusion of the Men vlue theorem is stted in nother wy: f( + h) = f() + hf ( + θh) for some θ (, ). It follows from Theorem 2.6 by writing b = + h. We know tht the derivtive of constnt function is zero. Its converse follows from the Men vlue theorem provided the domin is n intervl. We do not consider degenerte intervls of the form [, ] etc. Theorem 2.7. Let I be n intervl contining t lest two points. Let f : I R be differentible. If f (x) = for ech x I, then f(x) is constnt function. Proof: Let s < t I. Then f(x) is continuous on [s, t] nd is differentible on (s, t). By the Men vlue theorem, f(t) f(s) = f (c)(t s). If f (x) = for ll x I, then f (c) = ; consequently, f(b) = f(). It thus follows tht functions whose derivtives re the sme cn only differ by constnt. In ddition to the ssumptions of the Men vlue theorem if m f (x) M for some rel numbers m nd M, then It follows tht m(b ) f(x) M(b ). This is sometimes clled s the Men vlue inequlity. MVT helps in determining whether function is incresing or decresing on n intervl, nd consequently it leds to test for mximum/minimum. Let f(x) be function defined on n intervl I. We sy tht f(x) is incresing on I if for ll s < t I, f(s) < f(t). Similrly, we sy tht f(x) is decresing on I if for ll s < t I, f(s) > f(t). A monotonic function on I is one which either increses on I or decreses on I. For exmple, f(x) = x 2 increses on [, ) nd decreses on (, ]. It is monotonic on both the intervls (, ] nd on [, ). But it is not monotonic on R, since it neither increses on R nor decreses on R. Theorem 2.8. Let f(x) be continuous on [, b] nd differentible on (, b).. If f (x) > on (, b), then f(x) is incresing on [, b]. 2. If f (x) < on (, b), then f(x) is decresing on [, b]. Proof: Let s < t [, b]. By MVT, f(t) f(s) = f (c)(t s) for some c (s, t). Since f (c) >, f(t) > f(s). Therefore, f(x) is incresing on [, b]. This proves (). Proof of (2) is similr. The point x = c is point of locl mximum for the function f(x) iff in the left neighborhood of c, f(x) is incresing nd in the right neighborhood of c, f(x) is decresing. The incresing nd decresing nture of f(x) re determined from the sign of f (x). If f (c) =, f (x) > on n intervl to the immedite left of x = c, nd f (x) < on n intervl 34

35 to the immedite right of x = c, then we sy tht f (x) chnges sign from + to t x = c. Similrly, f (x) chnges sign from to + mens tht f (c) =, f (x) < on n intervl to the immedite left of x = c, nd f (x) > on n intervl to the immedite right of x = c. Thus the following re true: If f (x) chnges sign from to + t x = c then f(x) hs locl minimum. If f (x) chnges sign from + to t x = c then f(x) hs locl mximum. Look t f (x) s function nd we cn tke its derivtive once more. Suppose f (c) <. Then f (x) is decresing on neighborhood of x = c. Since f (c) =, we see tht this decrese must be from + to. In tht cse, x = c is point of locl mximum. Anlogous rguments hold for point of locl minimum where f (x) chnges sign from to + t x = c. 2.3 Tests for Mxim-Minim We summrize the discussion. Test for Locl Extrem: Let c be n interior point of the domin of f(x) with f (c) =. f (x) chnges sign from + to t x = c iff x = c is point of locl mximum of f(x). If f (c) <, then x = c is point of locl mximum of f(x). f (x) chnges sign from to + t x = c iff x = c is point of locl minimum of f(x). If f (c) >, then x = c is point of locl minimum of f(x). Let x = c be left end-point of the domin of f(x). f (x) < on the immedite right of x = c iff x = c is point of locl mximum of f(x). f (x) > on the immedite right of x = c iff x = c is point of locl minimum of f(x). Let x = c be right end-point of the domin of f(x). f (x) > on the immedite left of x = c iff x = c is point of locl mximum of f(x). f (x) < on the left of x = c iff x = c is point of locl minimum of f(x). 35

36 Exmple 2.6. Find ll points of locl extrem of the function f(x) = x 3 3x 2 24x + 5. The domin of f(x) is (, ). Here, f (x) = 3x 2 6x 24 = 3(x + 2)(x 4). The criticl points re x = 2 nd x = 4. Now, f (x) = 6x 6. Then f ( 2) = 8 < sys tht x = 2 is locl mximum point nd the locl mximum vlue of f(x) t x = 2 is f( 2) = 33. f (4) = 8 >. Then x = 4 is locl minimum point nd the locl minimum vlue of f(x) t x = 4 is f(4) = 75. Exmple 2.7. Find ll locl mxim/minim for the function g(x) = x + sin x. g (x) = + cos x vnishes for x = (2n + )π, where n Z. Now, g (x) = sin x is lso zero t ll these criticl points. So, it does not help. We look for chnge of sign of g (x). Now, g (x) > for ll vlues of x except the criticl points. So, there is no chnge in sign of g (x) t the criticl points. Therefore, g(x) hs no locl mxim/minim. Exmple 2.8. Find the criticl points of f(x) = x /3 (x 4) defined on the intervl [, 2]. Identify the intervls in which f(x) is incresing or decresing. Find the bsolute nd locl extreme points nd vlues of f(x). f (x) = 3 x 2/3 (x 4) + x /3 () = x 4 + 3x 4(x ) =. 3x 2/3 3x 2/3 It is zero t x = nd is undefined t x =. The end-points of the domin of f(x) re nd 2. Thus x =,,, 2 re its criticl points. () For bsolute extrem: f( ) = 5, f() =, f() = 3, f(2) = 2 4/3. Compring these vlues, we see tht the bsolute minimum vlue of f(x) is 3 nd the point of bsolute minimum is x =. The point of bsolute mximum is x = nd the bsolute mximum vlue is 5. (b) For the incresing nd decresing nture of f(x), look for the signs of f (x). For x <, f (x) is ve. Thus f(x) is decresing on [, ]. For < x <, f (x) is ve. Thus f(x) is decresing on [, ]. For x >, f (x) is +ve. Thus f(x) is incresing on [, 2]. 36

37 (c) The chnge of sign of f (x) sys tht f(x) does not hve locl extremum t x =. And f(x) hs locl minimum t x =. The locl minimum vlue t x = is f() = 3. At the end-points, since f(x) is decresing on [, ], x = is locl mximum point, the locl mximum vlue is f( ) = 5. Similrly, x = 2 is lso locl mximum point, nd the locl mximum vlue t x = 2 is f(2) = 2 4/3. In summry, x = is locl mximum point nd lso n bsolute mximum point. x = is not n extremum point. x = is locl minimum point s well s n bsolute minimum point. x = 2 is locl mximum point. Notice tht lim x f (x) =. Thus y = f(x) hs verticl tngent t the origin. Exmple 2.9. A box is to be mde from sheet of crdbord tht mesures 2 2. The construction will be chieved by cutting squre from ech corner of the sheet nd then folding up the sides. Wht is the box of gretest volume tht cn be constructed in this fshion? Let x be the side length of the squres tht re to be cut from the sheet of crdbord. Then the side length of the resulting box will be 2 2x; the height of the box will be x. As result, the volume of the box will be f(x) = x(2 2x) 2 = 44x 48x 2 + 4x 3. This function is to be mximized. The criticl points re f (x) = 44 96x 2x 2 = x = 2, 6. Now, f (x) = x. Since x 6; compring f(), f(2), f(6), we see tht f(2) is mximum. Also, notice tht f (2) = 48 < nd f (6) = 48 >. Therefore, f(x) hs locl mximum t x = 2 nd locl minimum t x = 6. The gretest volume is f(2) = 2(2 4) 2 = 28. The second derivtives help us to know the shpe of curve. 37

38 Here, the grph of the function f(x) = 3 + sin x is concve down on (, π) nd is concve up from (π, 2π). The concvity chnges t x = π. The grph of function y = f(x) is concve up on n open intervl I if f (x) is incresing on I. The grph of y = f(x) is concve down on n open intervl I if f (x) is decresing on I. A point of inflection is point where y = f(x) hs tngent nd the concvity chnges. Now, f (x) increses on I when f (x) > on I. Similrly, f (x) < on I implies tht f (x) decreses on I. We thus hve the following: Second derivtive test for concvity: Let y = f(x) be twice differentible on n intervl I. If f (x) > on I, then the grph of y = f(x) is concve up over I. If f (x) < on I, then the grph of y = f(x) is concve down over I. If f (x) is positive on one side of x = c nd negtive on the other side, then the point (c, f(c)) on the grph of y = f(x) is point of inflection. At point of inflection, either f does not exist or it vnishes. Notice tht in the ltter cse, f (x) hs locl extremum. Here is summry of how the derivtives give informtion regrding the shpe of curve: Exmple 2.. () The curve y = x 2 is concve up everywhere. Becuse, (x 2 ) = 2 > t ll points. (b) The curve y = x 3 is concve down on (, ) where f (x) = 6x <. It is concve up on (, ), where f (x) = 6x >. The point x = is its inflection point. 38

39 (c) The curve y = f(x) = x 4 hs f (x) = 2x 2 = t x =. But the point (, ) is not point of inflection since f (x) does not chnge sign t x =. (d) For the curve y = f(x) = x /3, f (x) = (2/9)x 5/3 for x. And f (x) does not exist t x =. However, f (x) < for x < nd f (x) > for x >. Tht is, f (x) chnges sign t x =. Therefore, the point (, ) is point of inflection. Exmple 2.. Give rough sketch of the grph of y = f(x) = ( + x)2 + x 2.. The domin of f(x) is (, ). There re no symmetries. 2. f(x) = ( + x)2 + x 2, f (x) = 2( x2 ) ( + x 2 ) 2, f (x) = 4x(x2 3) ( + x 2 ) The criticl points re x =,. At x =, f ( ) = >. So, y = f(x) hs locl minimum t x =. At x =, f () = <. So, y = f(x) hs locl mximum t x =. 4. On the intervl (, ), f (x) <, so, f(x) is decresing. On the intervl (, ), f (x) >, so, f(x) is incresing. On the intervl (, ), f (x) <, so, f(x) is decresing. 5. f (x) = for x = 3,, 3. On the intervl (, 3), f (x) <. On the intervl ( 3, ), f (x) >. On the intervl (, 3), f (x) <. And on the intervl ( 3, ), f (x) >. Thus, ech of these points is point of inflection. Moreover, on (, 3), y = f(x) is concve down; on ( 3, ), it is concve up; on (, 3) it is concve down, nd on ( 3, ), it is concve up. 6. Rewrite f(x) = + (2/x) + (/x2 ). Thus f(x) decreses to s x nd f(x) increses to s x. Thus y = is horizontl symptote to y = (/x 2 ) + f(x). For every c R, limit of f(x) s x c remins finite. Thus there re no verticl symptotes. Since the curve decreses to rech the point (, ) nd then increses up to the point (, 2), nd then decreses towrds its symptote, the locl minimum t x = is n bsolute minimum. And lso the locl mximum t x = is n bsolute mximum. 7. The curve thus looks like : 39

40 2.4 L Hospitl s Rule Throughout, we hve used the MVT for deciding bout the incresing nd decresing nture of functions. The MVT cn be generlized bit. Theorem 2.9. (Cuchy Men Vlue Theorem) Let f(x) nd g(x) be continuous on [, b] nd differentible on (, b). If g (x) on (, b), then there exists c (, b) such tht f (c) g (c) = f(b) f() g(b) g(). Proof: If g(b) g() =, then by MVT, g (d) = for some d (, b), contrdicting our ssumption. Hence g(b) g(). Define h : [, b] R by h(x) = f(x) f() f(b) f() {g(x) g()}. g(b) g() Now, h(x) is continuous on [, b] nd differentible on (, b). Also, h() = h(b). By Rolle s theorem, there exists c (, b) such tht h (c) =. Tht is, f (c) f(b) f() g(b) g() g (c) =. Theorem 2.. (L Hospitl s Rule) Let f(x) nd g(x) be differentible on neighborhood of point x =. Suppose f() = g() = but g(x), g (x) in the deleted neighborhood of f (x) f(x) x =. If lim exists, then lim x g (x) x g(x) = lim f (x) x g (x). Proof: First, consider the intervl [, x] to the right of, which is contined in the given neighborhood of. Both f nd g re continuous on [, x], nd differentible on (, x). Also, g on (, x). By Cuchy MVT, there exists point θ (, x) such tht f (θ) g (θ) When x, we hve θ. Therefore, = f(x) f() g(x) g() = f(x) g(x). f (θ) lim θ + g (θ) = lim f(x) x + g(x). 4

41 Similrly, considering n intervl [t, ] to the left of the point, we conclude tht Combining the two bove completes the proof. f (τ) lim τ g (τ) = lim f(x) x g(x). Of course, when only one sided limit is relevnt, our proof shows tht L Hospitl s Rule is still pplicble. The method cn lso be used for evluting limits in the indeterminte forms such s ±, ±, nd. ± + x x/2 Exmple 2.2. () lim in form. x = lim x (/2)( + x) /2 /2 2x = lim x (/4)( + x) 3/2 2 (b) lim x cos x x + x 2 But lim x cos x x + x 2 = 8. x 2 cos x = lim x x = lim x sin x + 2x = lim x + x = lim sin x x cos x 2 = 2. =. in form. Wht is wrong? The second limit in the second clcultion is not in form. In fct, the limit there is equl to s it should be. sin x (c) lim x + x 2 cos x = lim x + 2x =. sin x cos x (d) lim = lim x x 2 x 2x =. sec x (e) For the limit lim, notice tht both the numertor nd denomintor re discontinuous x π/2 + tn x t x = π/2. We consider only one-sided limits. sec x lim x (π/2) + tn x = lim sec x tn x = lim sin x =. Similrly, x (π/2) sec 2 x x (π/2) sec x sec x lim =. Hence lim x (π/2)+ + tn x x π/2 + tn x =. x 2x 2 (f) lim x 3x 2 + 5x = lim 4x x (g) lim x sin x x = lim t + t 6x + 5 = lim x sin t = lim ( sin x x (h) For the limit lim x it is in + form. We need lim x ( sin x x ) = lim x x sin x x sin x 4 6 = 2 3. ± ± cos t =. t + ), we see tht s x +, it is in form nd s x, ± or. So, we rerrnge. ± = lim x cos x sin x + cos x = lim x sin x cos x sin x =. Let c be n rbitrry point in the domin of differentible function y = f(x). Suppose x 4

42 represents the chnge in x ner c. Then the chnge in y ner c is f(c + x) f(c). The derivtive f (x) t c is f f(c + x) f(c) (c) = lim. Therefore, ner c, the chnge in y is pproximtely x x equl to f (c) x. In bstrct terms, we express this s differentils. The chnge dx t ny rbitrry point x is n independent vrible clled the differentil of x. It is the sme s x. The corresponding chnge in y is dependent vrible nd is clled the differentil of y. The differentil of y is denoted by dy. We define this differentil of y s dy = f (x) dx. Then the differentil dy is n pproximtion of the increment y in y ner c. Notice tht in this view, the differentils give proper nottion in viewing dy s rtio. dx This helps. For exmple, the derivtive of product (f(x)g(x)) = f (x)g(x) + f(x)g (x) cn be stted in terms of differentils s d(fg) = df g + f dg. For y = f(x), the differentil dy = f (c)dx gives n pproximtion to the increment in y ner c. Thus y = f(c + dx) f(c) dy f(c + dx) f(c) + dy. It helps in pproximtely computing the function vlue t neighboring point from the vlue t tht point nd the differentil ner the point. Exmple 2.3. The rdius of circle of rdius increses to.. Estimte the chnge in re. Since y = f(x) = πx 2, the differentil formul gives the estimte s y dy = 2πx dx = π()(.2) = 2π. Of course, the chnge in re is π{(.) 2 2 } = π(. 2.) = 2. π. 2.5 Curvture The curvture of curve t point quntifies how much the curve is bent t tht point. For exmple, consider two circles, C of rdius nd C 2 of rdius 2. Intuitively, the bent on ny circle is independent of the point we choose on it. So, choose point P on C nd point P 2 on C 2. Now, the bent of C t P is certinly stiffer or more thn tht t P 2 on C 2. Thus the curvture of circle my be tken s something tht vries inversely with the rdius. We lso use the term rdius of curvture for the reciprocl of the curvture. To compre the curvtures of these circles t those points, let us tke nother pir of points, sy Q ner P on C nd Q 2 ner P 2 on C 2. The ngle between the tngents t P nd Q is more thn the ngle between the tngents to C 2 t P 2 nd Q 2. It shows tht the rte of chnge of the 42

43 ngle between tngent nd the x-xis with respect to the length of the curve is fster when the corresponding ngles between the tngents is more. We thus define the curvture s follows. Curvture of curve t point P on the curve is the rte of chnge in direction with respect to the rc length of the curve t P. Writing κ for the curvture, our definition yields the formul κ = dψ ds where s is the rc length nd ψ is the ngle between the tngent to the curve t tht point nd the positive x-xis. Further, the rdius of curvture, denoted by ρ is given by ρ = κ = ds. dψ We now derive the formuls for clculting curvture in crtesin coordintes. As we know, Then sec 2 ψ dψ dx = d2 y dx 2. Consequently, tn ψ = dy dx, ds = dx 2 + dy 2, dx ds = cos ψ. κ = dψ = dψ dx = d2 y ds dx ds dx 2 cos3 x = y [ + (y ) 2 ] 3/2. ρ = κ = [ + (y ) 2 ] 3/2. y In these formuls, we should tke cre tht t the point where the curvture or rdius of curvture re computed, the quntities in the denomintor re not. Similrly, in polr coordintes, the formul for the rdius of curvture is given by ρ = [ κ = r 2 + ( dr ) ] 2 3/2 [r ( dr ) 2 d 2 r ] r. dθ dθ dθ 2 Notice tht the curvture of stright line is t every point since ψ, the ngle between ny tngent to it (tngent is the stright line itself) nd the x-xis never chnges. Exmple 2.4. Find the rdius of curvture for circle of rdius, sy, centred t the origin. In polr coordintes, the circle is given by r =. We hve dr dθ =, d2 r dθ =. 2 ρ = κ = [ 2 ] 3/2 [ 2 ] = 3 =. 2 If C is curve hving rdius of curvture ρ t point P on the curve, then imgine circle of rdius ρ touching the curve t P nd lying on the convex side of the curve. 43

44 Then the centre of such circle, now clled n osculting circle is clled the centre of curvture of the curve t tht point. Clerly, the centre of curvture of circle is the centre of the circle, whtever point P on the circle we choose. If (, b) is the centre of curvture of curve y = y(x) t the point (x, y), then = x ρ sin ψ = x y [ + (y ) 2 ] y, b = y + ρ cos ψ = y + + y y. Exmple 2.5. Find the rdius nd centre of curvture for the prbol y 2 = 4x. Let (x, y) be ny point on the prbol y 2 = 4x. t P, we hve dy dx = 2 y, d 2 y dx 2 = 42 y 3. ρ = If (, b) is the centre of curvture t (x, y), then [ + 42 y 3 ] 3/2( 42 y 3 ) = 2(x + ) 3/2. = x y [ + (y ) 2 ] = x + 42 y 2, b = y + + y y 2 y = y y3 + 2y Review Problems Problem : Wht is the derivtive of f(x) = x? For x >, x = x. So, f (x) = 2. x For x <, x = x. So, f (x) = For the derivtive of f(x) t x =, f(h) f() So, lim h + h 2. x f(h) f() h = f(h) f() = nd lim h h { / h if h > / h if h <. =. Therefore, the derivtive of f(x) t x = does not exist. Even the left hnd nd right hnd derivtives do not exist t x =. { x 2 if x Problem 2: Find f (x) if f(x) = x 2 if x <. 44

45 If x >, then f (x) = 2x. If x <, then f (x) = 2x. For x =, f(h) f() lim h + h Thus f () =. Tht is, f (x) = 2 x. h 2 = lim h + h =. lim f(h) f() h h h 2 = lim h h =. Problem 3: Find the derivtive of y = f(x) where x nd y re given prmetriclly: () x = cos t, y = sin t, t π. Upper semi-circle y = 2 x 2. (b) x = cos t, y = b sin t, t π. Upper prt of n ellipse y = b x 2 / 2. (c) x = (t sin t), y = ( cos t), t 2π. The cycloid. (d) x = cos 3 t, y = sin 3 t, t 2π. The steroid x 2/3 + y 2/3 = 2/3. If function y = f(x) is given prmetriclly by x = φ(t), y = ψ(t), then ssuming tht φ (t), we hve dy dx = dy dt dt dx = ψ (t) φ (t). () dy dx d( sin t) / d( cos t) = dt dt / d( cos t) = cos t sin t = x y. (b) dy d(b sin t) = = b cos t dx dt dt sin t = b2 cos t 2 b sin t = b2 x 2 y. (c) dy d((t sin t)) / ( ) d(( cos t)) ( cos t) t = = = cot. dx dt dt sin t 2 (d) dy dx = d( sin3 t) / d( cos 3 t) = 3 sin2 t cos t dt dt 3 cos 2 t sin t = sin t ( y ) /3 cos t =. x Problem 4: If curve is given in polr coordintes, show tht the derivtive of the rdius vector with respect to the polr ngle is equl to the rdius vector multiplied by the cotngent of the ngle between the rdius vector nd the tngent to the curve t the given point. In polr coordintes, x = ρ cos θ, y = ρ sin θ, where the curve is given by ρ = ρ(θ). Let φ be the ngle formed by the tngent to the curve t some point M(ρ, θ) with the positive x-xis. Then tn φ = dy dx = dy/dθ dx/dθ = ρ (θ) sin θ + ρ cos θ ρ (θ) cos θ ρ sin θ. Let µ be the ngle between the rdius vector ρ nd the tngent t tht point M. Then µ = φ θ, nd then tn µ = tn φ tn θ + tn φ tn θ = (ρ sin θ + ρ cos θ) cos θ (ρ cos θ ρ sin θ) sin θ (ρ cos θ ρ sin θ) cos θ + (ρ sin θ + ρ cos θ) sin θ = ρ ρ. This gives ρ (θ)θ = ρ(θ) cot µ. Problem 5: Consider the function f(x) = x 2/3 defined on the intervl [, ]. Its derivtive is f (x) = 2 3 x /3. Notice tht f( ) = f() = but f (x) on [, ]. Why? 45

46 Reson: Rolle s theorem is not pplicble becuse, t x =, f(x) is not differentible. Also, notice tht t x =, f(x) ttins its mximum vlue. Problem 6: Find the mximum nd minimum vlues of ( x 2 ) in (, ). Let f(x) = ( x 2 ). Now, f (x) = 2x. It is only for x =. ( x 2 ) 2 When x is close to, f(x) becomes rbitrrily lrge. So, f(x) hs no mximum in (, ). When x <, f (x) < nd when x >, f (x) >. Tht is, t x =, f (x) chnges sign from to +. Therefore, f() = is minimum vlue of f(x). This is the only minimum vlue since f (x) is never for ny other point. Problem 7: Exmine f(x) = 2 sin x + cos 2x for mxim/minim for x R. Since f(x) hs period 2π, we consider it on (, 2π). Now, f (x) = 2 cos x 2 sin 2x = 2(cos x 2 sin x cos x) = 2 cos x( 2 sin x). The criticl points re = π/6, b = π/2, c = 5π/6, d = 3π/2. f (x) = 2 sin x 4 cos 2x. f (π/6) = 2(/2) 4(/2) = 3 <. Hence f(x) hs locl mximum t x = π/6 nd this mximum vlue is f(π/6) = 3/2. f (π/2) = = 2 >. Thus f(x) hs locl minimum t x = π/2 nd this minimum vlue is f(π/2) =. f (5π/6) = 2(/2) 4(/2) = 3 <. So, f(x) hs locl mximum t x = 5π/6) nd this mximum vlue is f(5π/6) = 3/2. f (3π/2) = 2( ) 4( ) = 6 >. Then f(x) hs locl minimum t x = 3π/2 nd this minimum vlue is f(3π/2) = 3. Wht bout the points x = nd x = 2π? Due to periodicity we need to consider x = only. f () = 2. Hence x = is not point of mxim/minim. In summry, the points of locl mxim re 2nπ + π/6 nd 2nπ + 5π/6 for n Z. The points of locl minim re 2nπ + π/2 nd 2nπ + 3π/2 for n Z. The bsolute mximum vlue of f(x) is 3/2 nd the bsolute minimum vlue of f(x) is 3. Problem 8: Determine the points of inflection of y = (x ) /3. y = 3 (x ) 2/3, y = 2 9 (x ) 5/3. The second derivtive does not vnish nywhere. But t x =, y does not exist. There is possibility of point of inflection t x =. 46

47 For x <, y >. So, the curve is concve up in left neighborhood of x =. For x >, y <. So, the curve is concve down in right neighborhood of x =. Therefore, x = is point of inflection. Notice tht t x = the curve hs verticl tngent since y (x) ± s x. Problem 9: Investigte the curve x(t) = 3t 3t2, y(t) = for give > for its decresing/incresing behvior. Sketch the + t3 + t3 curve. Both the functions re defined for ll vlues of t except for t =. We see tht lim x(t) = lim t t lim y(t) = lim t t x() = y() =, lim t x(t) = lim t y(t) =, 3t =, + t3 lim x(t) = t + 3t 2 =, + t3 lim t + lim x(t) =, lim y(t) =. t t Also, dx dt = 6(/2 t3 ) dy, ( + t 3 ) 2 dt = 3t(2 t3 ) ( + t 3 ) ; dy 2 dx = t(2 t3 ). For the prmeter t, we hve the 2t3 four criticl vlues (when dx/dt = = dy/dt): t =, t 2 =, t 3 = 2 /3, t 4 = 2 /3. At t =, x =, y =, we hve dy dy = nd when t, x =, y =, we hve. dx dx Thus the curve cuts the origin twice: with the tngent prllel to x-xis nd with the tngent prllel to y-xis. Also, when t = 2 /3, x = 4 /3, y = 2 /3, dy. dx At this point the tngent to the curve is verticl. When t = 2 /3, x = 2 /3, y = 4 /3, dy =. dt At this point the curve hs horizontl tngent. The informtion obtined from the derivtive re s follows: Rnge of t Rnge of x Rnge of y sign of vrition dy/dx in y = f(x) < t < < x < > y > decreses < t < < x < > y > decreses < t < 2 /3 < x < 4 /3 < y < 2 /3 + increses 2 /3 < t < 2 /3 4 /3 > x > 2 /3 2 /3 < y < 4 /3 decreses 2 /3 < t < 2 /3 > x > 4 /3 > y > + increses y lim x x = lim 3t 2 ( + t 3 ) t 3t( + t 3 ) =. 3t(t + ) lim (y ( )x) = lim =. x t + t 3 47

48 Hence the stright line y = x is n symptote to brnch of the curve s x. Similrly, y lim x x =, lim (y ( )x) =. x Therefore, the stright line y = x is lso n symptote to brnch of the curve s x. Now, we cn hve rough sketch of the curve, clled Descrtes Folium. Problem : Let f : (, b) R; c (, b). Suppose tht f is differentible t x = c nd f is continuous t x = c. Show tht f is differentible t x = c. Notice tht f my not be continuous but f cn be differentible. For exmple, the function tht tkes vlue for ll rtionl numbers nd tkes vlue for ll irrtionl numbers. If f(c) >, then f(x) remins positive in neighborhood of x = c. Tht is, in tht neighborhood, f(x) = f(x). Then f is differentible t x = c. Similrly, if f(c) <, then in neighborhood of x = c, f(x) = f(x) nd then f(x) is differentible t x = c. If f(c) =, then x = c is locl minimum point of f(x). So, f (c) =. This mens = lim h f(c + h) f(c) h = lim h f(c + h) h f (c) =. Problem : Find the rdius nd centre of curvture for y = x 2 t (, ). Here, y = 2x, y = 2. Tht is, t the origin, y =, y = 2. Thus ρ = (y ) [ + (y ) 2 ] 3/2 = (/2) () = 2 The centre of curvture t the origin is (, b), where = x y [ + (y ) 2 ](y ) =, b = y + ( + y )(y ) = 2. 48

49 Chpter 3 Integrtion 3. The definite integrl Integrtion is reverse process of integrtion. Look t the figure: Imgine h is very smll. Then f(x) cn be pproximted by the shded re divided by h. Write the re s F (x), sy, strting from some point x = on the left. Then the shded re is F (x + h) F (x). Consequently, f(x) Or, in the limit, we hve f(x) = F (x). This mens: F (x + h) F (x). h If F (x) denotes the re of the region bounded by y = f(x) nd the x-xis, strting from some point x = up to the point x = c, then f(c) = F (c). Tht is, the function which my serve s n inverse process of differentition is this re. How do we find this re? We cn pproximte the re by dividing the x-intervl, the domin of f(x) into smller prts, nd then tke the sum of ll smller rectngles of width s the smll prt on x-xis nd the height s the function vlue t some point inside the smller intervl. Tht is, suppose tht f : [, b] R. Divide [, b] into smller sub-intervls by choosing the brek points s = x < x <... < x n = b. The set P = {x, x,..., x n } is clled prtition of [, b]. Now P divides [, b] into n sub-intervls: [x, x ],, [x n, x n ]. Here, the kth sub-intervl is [x k, x k ]. The re under the curve y = f(x) rised over the kth sub-intervl is pproximted by f(c k )(x k x k ) for some choice of the point c k [x k, x k ]. 49

50 Write the choice points (lso clled smple points) s set C = {c,..., c n }. Then the Riemnn sum S(f, P, C) = n f(c k )(x k x k ) k= is n pproximtion to the whole re rised over [, b] nd lying between the curve y = f(x) nd the x-xis. If the sub-intervls become smller nd smller, we expect better nd better pproximtion of the re by the Riemnn sum. This hints t limit process. Tht is, by tking the norm of the prtition s P = mx (x k x k ), we would sy tht when the norm of the prtition pproches, the k Riemnn sum would pproch the required re. Thus, we define the re of the region bounded by the lines x =, x = b, y =, nd y = f(x) s lim P n f(c k )(x k x k ) k= provided tht this limit exists. Notice tht we cn choose c k from ech sub-intervl in mny wys. We cn choose n in mny wys. Thus the limit of he Riemnn sum must exist for ll such choices, nd ll those limits must be equl. If the limit of ll such Riemnn sums exist, nd re equl, we cll this limit s the integrl of f(x) from to b. Our heuristics sys tht the integrl is nothing but the re of the bounded region we just considered. In fct, when n re of such region is required, we simply compute this integrl. We write the integrl s in the following: b f(x) dx = lim S(f, P, C) = lim P P n f(c k )(x k x k ) We sy tht the integrl exists if for ll such choices of P nd of c k, the right hnd side limit exists. In tht cse, the integrl cn be computed by choosing prticulr type of P, or even prticulr wy of choosing these c k from the sub-intervls. The nottion demnds tht the limit, when exists, must not depend on the prticulr choice points in C. Since the bove limit looks different from lim f(x), we must stte it correctly. It mens the following: 5 k=

51 There exists l R so tht for ll ɛ >, there exists δ > such tht if P < δ, then for ll choices of C, S(f, P, C) l < ɛ. Of course, we will not use this definition lwys. The key to its ppliction is the following bsic fct, which we ccept without proof: Theorem 3.. Let f : [, b] R be continuous function. Then b f(x) dx exists. Since the integrl is the re bounded by the curve nd the x-xis between the two verticl lines tht re obtined from the limits of integrtion, prerequisite for function to be integrble is tht it must be bounded. Due to Theorem.4, continuous function on closed bounded intervl is bounded. Exmple 3.. Clculte the re of the region below the curve y = x 2 bove x-xis nd between the lines x = nd x =. Tht is, we compute x2 dx. By Theorem 3., this integrl exists. We thus choose prticulr type of prtition, sy, uniform prtition P : = x = n < x = 2 n < < x n = n n < x n = n n =. We lso choose the smple points C : c = x =, c n n = x n = n, the left end-points of n the prtition sub-intervls. Tht is, The Riemnn sum is S(f, P, C) = n ( k ) 2 f(c k )(x k x k ) = n n = n 3 k= n k 2 = k= n(n + )(2n + ) 6n 3. Notice tht s P, we hve n. Then Exmple 3.2. Show tht x 2 dx = lim n n(n + )(2n + ) 6n 3 dx does not exist. x = ( 6 lim + ) ( 2 + ) = n n n 3. The function /x is not defined t. As such (/x) is not defined. Suppose we redefine the function by tking its vlue t s. Tht is, we sk for f(x)dx, where f() = nd f(x) = /x for x (, ]. Here, f(x) is n unbounded function since lim x + (/x) =. Therefore, f(x) is not integrble. 5

52 Now, you see tht n integrble function must be bounded in its domin of definition. If f(x) becomes unbounded, then the limit of the Riemnn sum will not exist. Even though function is bounded, it need not be integrble. { if x Q Exmple 3.3. Show tht f(x)dx does not exist, where f(x) = if x Q. Agin, choose uniform prtition of [, ] s P : < < < n n n seprte sets of smple points =. Let us choose two C : x < x 2 < < x n, C 2 : t < t 2 < < t n, where ech x k is rtionl number in the k-th subintervl nd t k is n irrtionl number in the sme k-th subintervl. The Riemnn sums re the following: S(P, f, C ) = n k= f(x k ) n =, S(P, f, C 2 ) = n k= f(t k ) n n = n. The limits of the first Riemnn sum s n is. The limit of the second Riemnn sum if exists, will remin positive. Therefore, f(x) is not integrble. In fct, limit of the second Riemnn sum does not exist. Form the definition of the definite integrl we observe n interesting fct. Suppose we tke ny prtition P of the intervl [, b] on which the function f(x) is defined, sy, into n subintervls. Assume tht f(x) is bounded. Choose three different sets of smple points. Choose x k s such point tht f(x k ) is minimum in tht k-th subintervl. Tke t k be ny point in the subintervl. Finlly, choose y k s point where f(y k ) is mximum in the subintervl. Cll C s the set of smple points x k, C 2 for the t k s nd C 3 for the y k s. The Riemnn sums will stisfy the following: S(P, f, C ) S(P, f, C 2 ) S(P, f, C 3 ). If the function f(x) is integrble, then s P, ll of them will pproch the limit b f(x)dx. It is lso cler tht if lim S(P, f, C ) = S(P, f, C 3 ), then f(x) is integrble. The sum S(P, f, C ) P is clled the lower sum nd S(P, f, C 3 ) is clled the upper sum; their limits if exist, re respectively clled the lower integrl nd the upper integrl. Integrbility is equivlent to the sttement tht s the norm of prtition pproches, the difference between the lower nd upper sums pproch. Sometimes this chrcteriztion helps in proving tht given function is integrble. We will not use this in usul circumstnces due to Theorem 3.. In fct, Theorem 3. is proved using this technique! Since b f(x) dx is number, it is clled definite integrl. Exmple 3.4. Evlute the definite integrl cos x dx. Since cos x is continuous on [, ], the integrl exists. To evlute it, choose prtition P : = x < x = n < x 2 = 2 n < < x n = n n 52 k= < x n =.

53 Consider the sub-intervl [x k, x k ]. Since (sin x) = cos x. By MVT, there exists c k [x k, x k ] such tht sin x k sin x k = (cos c k )(x k x k ). Choose these c k s s the smple points in the sub-intervls. Then cos x dx = lim n n k= cos c k (x k x k ) = lim n n [sin x k sin x k ] k= = lim n (sin x n sin x ) = lim n (sin sin ) = sin. The following properties of the definite integrl re obvious from the definition. Theorem 3.2. (Properties of Definite Integrl). Let f(x) hve domin [, b]. Let c (, b). Then f(x) is integrble on [, b] iff f(x) is integrble on both [, c] nd [c, b]. In this cse, b f(x)dx = c f(x)dx + b c f(x)dx. 2. Let f(x) nd g(x) be integrble on [, b]. Then (f + g)(x) is integrble on [, b] nd b (f + g)(x)dx = b (f(x) + g(x)) dx = b f(x) dx + b g(x) dx. 3. Let f(x) be integrble on [, b]. Let c R. Then (cf)(x) is integrble on [.b] nd b (cf)(x) dx = b cf(x) dx = c b f(x) dx. 4. Let f(x) nd g(x) be integrble on [, b]. If for ech x [, b], f(x) g(x), then b f(x) dx b g(x) dx. 5. Let f(x) be integrble on [, b]. If m f(x) M for ll x [, b], then m(b ) b f(x) dx M(b ). 6. (Averge Vlue Theorem) Let f(x) be continuous on [, b]. Then there exists c [, b] such tht f(c) = b f(x) dx. b 7. Let f(x) be continuous on [, b]. If f(x) hs the sme sign on [, b] nd then f(x) is the zero function, i.e., f(x) = for ech x [, b]. 53 b f(x) dx =,

54 Notice tht (6) follows from (5) nd the intermedite vlue theorem for continuous functions. The vlue f(c) = b f(x) dx. is clled the verge vlue of the function f(x) nd (6) sys tht b continuous function chieves its verge vlue. It lso follows from (6) tht if then f(x) hs zero in [, b]. Convention: following: b If = b, then we tke If > b, then we tke b f(x)dx =, f(x) dx mkes sense when < b. We extend the integrl even when b by the b b f(x) dx =. f(x) dx = b f(x) dx. Notice tht this convention goes well with the properties listed in Theorem 3.2. Moreover, we lso extend Theorem 3.2() for ny rel number c; even when c is outside the intervl (, b) by b f(x) dx = c f(x) dx + In ll these extensions, we ssume tht the definite integrls exist. b c f(x) dx 3.2 The Indefinite Integrl If F (x) = f(x), then we sy tht the function F (x) is n ntiderivtive of f(x). Exmple 3.4 shows tht if we know n ntiderivtive of function, then we cn redily get its integrl. Theorem 3.3. (Fundmentl Theorem of Clculus ) Let f(x) be continuous on [, b]. Let F (x) be n ntiderivtive of f(x). Then b f(x) dx = F (b) F (). Proof: Since f(x) is continuous on [, b], the definite integrl exists. To evlute it, choose prtition P : = x < x = + b n < < x (n )(b ) n = + n < x n = b. Consider the sub-intervl [x k, x k ]. Since F (x) = f(x), By MVT, there exists c k [x k, x k ] such tht F (x k ) F (x k ) = f(c k )(x k x k ). Choose these c k s s the smple points in the sub-intervls. Then b f(x) dx = lim n n k= f(c k )(x k x k ) = lim n n [F (x k ) F (x k )] k= = lim n (F (x n ) F (x )) = F (b) F (). We my remember the conclusion of this theorem s b g (x) dx = g(x) b = g(b) g(). 54

55 FTC- mkes evlution of definite integrls esy, which you re fmilir with. For exmple, sec x tn x dx = (sec x) dx = sec x = sec sec( π/4) = 2. π/4 π/4 π/4 Its use is limited, for, we require n ntiderivtive of the integrnd in closed form. FTC- considers tking the integrl of the derivtive of f(x); it is f(x) f(). Wht bout tking the derivtive of the integrl? For this, we need the integrl to be function. Define g(x) = x f (t) dt for ech x [, b]. We then show tht g (x) = f(x). Theorem 3.4. (Fundmentl Theorem of Clculus 2) Let f(x) be continuous on [, b]. Then g(x) = x f(t) dt is continuous on [, b] nd differentible on (, b). Moreover, g (x) = d dx x f(t) dt = f(x). Proof: Let x (, b) nd let h > such tht x + h [, b]. Now, g(x + h) g(x) = [ x+h x ] f(t) dt f(t) dt = h h h x+h x f(t) dt = f(c) for some c [x, x + h], by the Averge Vlue theorem. We see tht if h, then c x. Thus g(x + h) g(x) lim h h = lim c x f(c) = f(x). Tht is, g(x) is differentible nd g (x) = f(x). Also, continuity of g(x) follows from its differentibility. When x is either equl to or b, we consider the pproprite one-sided limits: [ ] lim g( + h) g() = lim h + h + +h f(x) dx = lim hf(d) =. h + Here d is some point in [, + h] which exists due to Theorem 3.2(5). This proves tht g(x) is continuous t x =. Similr rgument proves continuity of g(x) t x = b. The two fundmentl theorems sy tht integrtion nd differentition re inverse processes. The reson behind introducing the definite integrl s the re is to obtin these two theorems. In fct, strting from inverse processes, we hve no wy of relting the integrl to the re. Using the Riemnn integrtion, we get both! Moreover, the fundmentl theorems sy tht the differentil eqution y = f(x) hs solution provided tht f(x) is continuous. Exmple 3.5. Clculte the re bounded by x-xis nd the prbol y = 6 x x 2. We get the points of intersection of the curve with x-xis by setting y = = 6 x x 2 = (x + 3)(2 x). Tht is, x = 2 or x = 3. The curve is non-negtive on [ 3, 2]. Therefore, the required re is 2 3 (6 x x 2 ) dx =

56 Cution: The ctul re is b f(x) dx. The signed re is given by b f(x) dx. If f(x) is not of the sme sign throughout [, b], then the integrl gives the sum totl of the signed res only. In Exmple 3.5, the function f(x) = 6 x x 2 for ech x [ 3, 2]. Therefore, f(x) = f(x). Hence the re is equl to the integrl s computed there. Exmple 3.6. Clculte the re of the region bounded by the x-xis, the lines x =, x = 2π, nd the curve y = f(x) = sin x. 2π f(x) dx = cos x 2π =. This gives the signed re, where the portions below nd bove the x-xis get cnceled. The true re is given by 2π f(x) dx = π = cos x π + cos x 2π π sin x dx + = 4. 2π π sin x dx Exmple 3.7. Find the re of the region bounded by the lines x =, x = 2, y =, nd the curve y = x 3 x 2 2x. f(x) = x 3 x 2 2x = x(x + )(x 2). On [, ], f(x). On [, 2], f(x). Thus the re of the given region is (x 3 x 2 2x) dx = = Exmple 3.8. Wht is wrong with (x 3 x 2 2x) dx x dx = = 2? 2 x For ech x [, ],. So, x2 x dx, if exists, must be non-negtive. The problem is 2 x 2 is not integrble on [, ], since it is not defined t x =. Moreover, it becomes unbounded s x pproches. Therefore, dx does not exist. x2 Due to the Fundmentl theorems, we introduce the indefinite integrl: x f(x)dx = f(t) dt for ny R whenever the definite integrl mkes sense. Notice tht if f(x)dx = g(x), then g (x) = f(x). Conversely, if g (x) = f(x), then x x f(x)dx = f(t) dt = g (t)dt = g(x) g(). 56

57 We write this by tking g() s n rbitrry constnt C; nd we cll it s constnt of integrtion. Tht is, g (x) = f(x) if nd only if f(x) dx = g(x) + C. The differentition of usul functions give rise to their integrtion. For exmple, (x 2 ) = 2x. Therefore, (2x)dx = x 2 + C Below, we list integrls of some functions which we use often. Common functions nd their integrls f(x) f(x) dx C x r x r+ r + + C for r sin x cos x + C cos x sin x + C sec 2 x tn x sec x tn x sec x + C cosec x cot x cosec x + C cosec 2 x cot x + C x 2 sin x + C for x < tn x + C + x 2 x sec x for x > x Substitution nd Integrtion by Prts The chin rule for differentition: (g(f(x)) = g(f(x))f (x) is trnslted to integrtion s in the following. Theorem 3.5. (Substitution). Let u = g(x) be differentible function whose rnge is n intervl I. Let f(x) be continuous on I. Then f(g(x))g (x) dx = f(u) du. 2. Let u = g(x) be continuously differentible function on [, b] whose rnge is n intervl I. Let f(x) be continuous on I. Then b f(g(x))g (x) dx = g(b) g() f(u) du. 57

58 Proof: Since f(x) is continuous, F (x) = f(x) dx is n ntiderivtive of f(x). Tht is, F (x) = f(x). By Chin rule for differentition, Then F (g(x)) = F (g(x))g (x) = f(g(x))g (x). f(g(x))g (x) dx = F (g(x)) dx = F (g(x)) + C = F (u) + C = F (u) du + C = f(u) du. This proves (). Similrly, for (2) we see tht b f(g(x))g (x) dx = F (g(x)) = F (g(b)) F (g()) = F (u) x=b x= u=g(b) u=g() = g(b) g() f(u) du. When u = g(x), du = g (x)du. Then the substitution rule looks like f(u)u dx = f(u)du. Similr comments pply for the definite integrl, where we tke cre of the limits of integrtion in terms of u. cos(2x) Exmple 3.9. sin 2 x dx = dx 2 = dx cos(2x) d(2x) = x 2 sin(2x) + C. 4 2π [ x Therefore, sin 2 xdx = 2 sin(2x) ] 2π = π. 4 Exmple 3.. Evlute 3x 2 x 3 + dx. Put u = x 3 +. Then du = 3x 2 dx; x 3 + = u. When x =, u =. When x =, u = 2. Thus 3x 2 x 3 + dx = u du = 3 u3/2 = Sometimes for integrtion, you my require to use breking rtionl function into simple form using decomposition into prtil frctions. dx Exmple 3.. Evlute (x + )(x 2 + 9). 2 Here, we try to determine constnts A, B, C, D nd E from (x + )(x 2 + 9) 2 = A x + + Bx + C x Dx + E (x 2 + 9) 2. 58

59 By simplifying nd compring, we obtin: A = /, B = /, C = /, D = nd E =. Therefore, dx (x + )(x 2 + 9) 2 = dx (x + ) xdx (x 2 + 9) + dx (x 2 + 9) xdx (x 2 + 9) 2 + dx (x 2 + 9) 2. For the second nd fourth integrls use the substitution u = x nd for the lst integrl, use the substitution x = 3 tn t nd integrte to obtin: Exmple 3.2. Evlute d x 4 sec t dt. dx Substitute u = x 4. Using FTC nd the Chin rule, d x 4 sec t dt = d dx du u sec t dt du dx = sec u du dx = 4x3 sec(x 4 ). Recll tht function f(x) is clled even if f( x) = f(x). In tht cse, f(x) dx = f( x)d( x) = f( x)dx = And when f(x) is n odd function, i.e., when f( x) = f(x), we see tht f(x) dx = f( x)d( x) = f( x)dx = In summry, let f(x) be continuous function on [, ]. Then f(x)dx = { 2 f(x)dx if f is even if f is odd. f(x)dx. f(x)dx. 59

60 The Fundmentl theorems re used to trnslte the product rule for differentition to integrtion. Recll tht the product rule for differentition sys tht (f(x)g(x)) = f (x)g(x) + f(x)g (x). Write h(x) = g (x). Then g(x) = h(x) dx, nd we get f(x)h(x) = (f(x)g(x)) f (x) h(x) dx. Integrte both the sides to get f(x)h(x) dx = f(x) h(x) dx [ f (x) ] h(x) dx dx + C. We remember it s follows (Red F s first nd S s second): Integrl of F S = F integrl of S integrl of (derivtive of F integrl of S). This is clled the integrtion by prts formul. We lso write it in the form u dv = uv v du + C. Exmple 3.3. Find x cos x dx. Integrting by prts, we hve [(x x cos x dx = x cos x dx ) cos x dx ] dx = x sin x sin x dx = x sin x+cos x+c. Since definite integrl uses re s its definition, it is firly strightforwrd to clculte the res bounded by curves. Of course, we use the fundmentl theorems for this purpose. Let f(x) g(x) be continuous functions with f(x) g(x) on [, b]. The re between the curves y = f(x) nd y = g(x) is b [f(x) g(x)] dx. In pplying this result, you must consider where the curves intersect. It my turn out tht inside the intervl [, b] the curves intersect more thn once. In tht cse, ll the integrls re to be computed seprtely since there will be sign chnge in f(x) g(x). Exmple 3.4. Find the re of the region enclosed by the line y = x nd the prbol y = 2 x 2. 6

61 The curves here intersect to give bounded region. The points of intersection re obtined from x = 2 x 2. They re x = nd x = 2. The line y = x lies below the prbol. Therefore, the re between them is 2 [2 x 2 ( x)] dx = ] [2x + x2 2 x = ( ) = = 9 2. Exmple 3.5. Find the re of the region in the first octnt bounded by the lines y =, y = x 2, nd the prbol x = y 2. The prbol y = x is the upper curve. But there re two lower lines. Their brek point is t x = 2. We divide the region into two prts by drwing the verticl line x = 2. Then we find the res of the regions mrked A nd B seprtely nd tke their sum. The required re is 2 [ x ] dx + 4 [ 2 ] 2 [ 2 = 3 x3/2 + 3 x3/2 2 [ x (x 2)] dx (x ] 2) = / / /2 = = 3. Alternte Method: If you integrte with respect to y, then single integrl gives the result. Viewed from y-xis, the upper curve is the line x = y + 2 nd the lower curve is x = y 2. The limits of integrtion re y = to y = 2. Therefore, the required re is 2 [ y (y + 2 y 2 2 ] y3 2 ) dy = + 2y = = Logrithm nd Exponentil The function f(t) = t is continuous on (, ). Therefore for ny x (, ), the integrl x exists. Notice tht when x (, ), we look t the integrl s s the vlue of function depending on x. This is clled the nturl logrithmic function ln x. Tht is, ln x = x dt for x >. t x t t dt dt. So, we define its vlue 6

62 We see tht ln : (, ) R. Since > for t >, ln is strictly incresing function. Also, t ln x > for x >, ln x < for < x <, nd ln =. Moreover, (ln x) =. Since x (ln x) = <, we conclude tht the grph of ln x is concve down. Also, ln x s x 2 x nd ln x s x +. All but the limit properties re esy to see. We prove tht ln x =. lim x For this, suppose k t m. Then /t /m. Consequently, m k t dt m k m dt = m m Let n N. Let x = 2 2n. Using the bove inequlity, we hve ln x = x k dt = m k m. 2 2n t dt = 2n 2 j t dt = 2n j= 2 t dt 2 j 2 j = 2 j j j= 2n j= 2 = n. Tht is, for ech n N, there exists x > such tht ln x n. This proves tht lim x ln x =. For the other limit, observe tht by substituting s = /t, we hve ds = (/s 2 ) ds. Then Therefore, ln x = /x t dt = x s s 2 x ds = ds = ln x. s lim ln x = lim ln x + y y = lim ln y =. y Due to its strict incresing nture, the limit properties imply tht ln : (, ) R is one-one nd onto function. Then ln x hs n inverse function. We cll this inverse function s the nturl exponentil function nd denote it by exp. Tht is, exp : R (, ); y = exp(x) iff x = ln y. Hence exp(ln x) = x for x (, ) nd ln(exp(x)) = x for x R. Differentiting the ltter eqution, we get exp(x) exp (x) =. Tht is, exp (x) = exp(x). Similr to ln, we hve the limit properties: lim exp(x) = ; nd lim exp(x) =. x x Moreover, exp(x) is strictly incresing nd is lwys positive. Let, b >. Consider f(x) = ln(xb) ln x. Now, f (x) = b =. So, f(x) is constnt. xb x But f() = ln b ln = ln b. Therefore, for ech x (, ), f(x) = ln b. In prticulr, f() = ln(b) ln = ln b. Tht is, ln(b) = ln + ln b, for ll, b >. 62

63 Similrly, if c, d R, let = exp(c), b = exp(d). Then ln = c nd ln b = d. Now, the previous eqution gives exp(c + d) = exp(ln + ln b) = exp(ln(b)) = b = exp(c) exp(d) for ll c, d R. Arbitrry powers of positive rel numbers cn then be defined using exp nd ln. Let > We define the exponentil function x = exp(x ln ). The number e is defined s the unique positive number stisfying ln e =. Then e x = exp(x ln e) = exp(x). We dispense with the symbol exp(x) nd write it henceforth s e x. The exponent symbolism is consistent with the usul powers like x 3 x 2 = x 2+3, etc. The definitions, the limit properties nd the differentition properties tell us tht for >, ln e = = e, e ln x = x, ln(e x ) = x, x = e x ln, lim ln x =, lim ln x =, lim x x + x ex =, lim x ex =, (ln x) = e x, (ex ) = e x, ( x ) = (ln ) x, dt =, e x dx = e x. t ln( + xh) Exmple 3.6. () lim = for x. h xh As h, both ln( + xh) nd xh pproch. By L Hospitl s rule, (with respect to h) (b) lim h e h h =. ln( + xh) lim h xh Using L Hospitl s rule gin, we hve lim h e h h (c) lim h ( + xh) /h = e x. x = lim h x( + xh) =. = lim h e h =. If x =, then ( + xh) /h = /h = e (/h) ln = e =. For x, lim ln( + ln( + xh) h xh)/h = lim h h = x lim h ln( + xh) xh = x. Since exp is continuous function, we hve ( ) lim ( + h xh)/h = exp lim ln( + h xh)/h = exp(x) = e x. In prticulr, e = lim h ( + h) /h. ln x (d) lim = for ny k N. x x/k The limit is in form; pply L Hospitl s rule: lim x 63 ln x = lim x/k x k =. x x/k

64 (e) lim = for ny k N. ex Agin, the limit is in x k form. So, lim x e = lim kx k x x e x x x k (f) If >, then lim = for ny k N. x For >, x s x. Therefore, x x k x k lim x = lim x x kx k ln() x = = lim x k! = = lim x e =. x k! (ln ) k x =. Exercise: In proving tht ln x s x, we hve shown tht ln(2 2n ) n. Use this nd the limit for e in (3) to prove tht 2 e 4. The following properties of the exponentil function cn esily be proved: If >, then lim x x = nd If < <, then lim x x = nd lim x x =. lim x x =. If, then the x-xis is horizontl symptote to the grph of y = x. The hyperbolic functions re defined through the exponentil nd the logrithmic functions: sinh x = ex e x, cosh x = ex + e x, tnh x = ex e x 2 2 e x + e. x sinh x = ln(x + x 2 + ), cosh x = ln(x + x 2 ), tnh x = ( ) + x 2 ln. x Notice tht cosh hs domin s x nd tnh hs domin s < x <. The derivtives of these functions cn be found out in the usul mnner. (sinh x) = cosh x, (cosh x) = sinh x, (tnh x) = (cosh x) 2. (sinh x) = + x 2, (cosh x) = x2, (tnh x) = x. 2 These formuls cn be used for evluting integrls s usul. For exmple, dx = + x 2 sinh x = sinh () sinh () = ln( + 2). 3.5 Lengths of Plne Curves Let C be curve given prmetriclly by x = f(t), y = g(t), t b. We ssume tht both the functions f(t) nd g(t) re continuously differentible nd the derivtives re not simultneously zero on [, b]. (Smoothness) We lso ssume tht the curve is trversed exctly once s t increses from to b. 64

65 Tke prtition of [, b] into n subintervls [t k, t k ] of equl length t. nd drw the secnts on the curve by using the prtition points. The length of the curve is defined s the limit of the sum of lengths of the secnts s n pproches. Now, the length of single secnt joining the points P k to P k is ( xk ) 2 + ( y k ) 2 = [f(t k ) f(t k )] 2 + [g(t k ) g(t k )] 2. By the MVT, there exist k, b k [t k, t k ] such tht f(t k ) f(t k ) = f ( k ) t, g(t k ) g(t k ) = g (b k ) t. Then the sum of ll the secnts corresponding to the subintervls is n [f ( k )] 2 + [g (b k )] 2 t. k= Since the length of the curve is the limit of this sum, we hve Length of the curve = L = b [f (t)] 2 + [g (t)] 2 dt. If the curve is given s function y = f(x), x b, then tke x = t nd y = f(t) s its prmeteriztion. We then hve the length s L = b b + [f (x)] 2 dx = + (y ) 2 dx. Notice tht this formul is pplicble when f (x) is continuous on [, b]. Similrly, if curve is given by function x = g(y), c y d, where g (y) is continuous on [c, d], the length of the curve is L = d c d + [g (y)] 2 dy = + (x ) 2 dy. In the length formul, the integrnd long with the differentil is written s ds. Here, s stnds for the length element on given curve. We write L = b ds with limits nd b for the vrible of integrtion, which my be x, y or t. Here, ds = [x (t)] 2 + [y (t)] 2 dt = ( dy ) 2 ( dx ) 2 (dx) 2 + (dy) 2 = + dx = + dy. dx dy 65 c

66 This view mkes it possible to see the rc length s function: s(x) = x + [f (t)] 2 dt for the curve y = f(x) where the curve is mesured from the point (, f()). 66

67 Exmple 3.7. Find the length of the curve y = x3/2 trced for x. Here, y = 2 2 x /2. The length of the curve is Exmple 3.8. L = + (y ) 2 dx = Find the length of the steroid x = cos 3 t, y = sin 3 t, t 2π. Due to symmetry of the curve, its length is four times the rc trced when t π/2. (cos 3 t) = 3 cos 2 t sin t, (sin 3 t) = 3 sin 2 t cos t. Therefore, the length of the steroid is L = 4 π/2 [(cos 3 t) ] 2 + [(sin 3 t) ] 2 dt = x dx = 3 8 ( + 8x)3/2 = 3 6. π/2 3 cos t sin t dt = 6. Exmple 3.9. Find the length of the rc of the curve y = (x/2) 2/3 for x 2. Here, y = 3 ( ) /3. It is not defined t x =. We express the sme curve s x x = g(y) = 2y 3/2, y. Now, x = g (y) = 3 y, which is continuous on [, ]. The length L of the rc of the curve is L = + [g (y)] 2 dy = 2 + 9y dy = 27 ( + 9y)3/2 = 2( ). 27 Exmple 3.2. Find the rc length function for the curve y = x 2 (ln x)/8, where the rc is mesured from the point (, ). Write f(x) = x 2 (ln x)/8. Then f (x) = 2x /(8x), nd + [f (x)] 2 = 2x + /(8x). Then s(x) = x + [f (t)] 2 dt = x ( 2x + ) dt = x 2 + ln x. 8x 8 Notice tht for x <, the vlues of s(x) re negtive, since s is mesured from the point (, ). 67

68 The clcultion of re in polr coordintes is similr to the Crtesin coordintes; but it uses the bsic re s circulr sector insted of rectngle. Recll tht circulr sector of ngle θ while the circle hs rdius r is given by θr 2 /2, where θ is mesured in rdins. Suppose tht curve is given in polr coordintes by r = f(θ) for continuous function f(θ), where α θ β. Divide the intervl [α, β] into n sub-intervls [θ k, θ k ], sy of equl length θ = θ k θ k. Approximte the re of the sector of the region bounded by the curve r = f(θ) nd the rys θ = θ k nd θ = θ k by circulr sector of the sme ngle θ nd rdius f(r k ), where r k is some point in the subintervl [θ k, θ k ]. The pproximte re of the sector is [f(r k )] 2 θ/2. Form the Riemnn sum s n pproximtion to the required re. Then tke the limit s the norm of the prtition, i.e., mx(θ k θ k ) pproches, tht is, s n. This gives the re s A = lim n n k= 2 [f(r k)] 2 θ = 2 β α [f(θ)] 2 dθ = β α r 2 dθ. Notice tht our ssumptions for the derivtion of this formul re tht the function f(θ) is continuous on the intervl [α, β], where β α 2π, tht is, when the curve is trversed only once. Moreover, we ssume tht f(θ) remins non-negtive. The rc-length of the curve r = f(θ) for α θ β cn be obtined directly from the prmetric form. We prmeterize the curve in Crtesin coordintes s Here, the prmeter is θ. Then x = r cos θ = f(θ) cos θ, y = r sin θ = f(θ) sin θ, α θ β. x (θ) = f (θ) cos θ f(θ) sin θ, y (θ) = f (θ) sin θ + f(θ) cos θ. Consequently, [(x (θ)] 2 + [y (θ)] 2 = [f(θ)] 2 + [f (θ)] 2. Therefore, the required length of the rc of the curve is L = β α β [(x (θ)] 2 + [y (θ)] 2 dθ = [(f(θ)]2 + [f (θ)] 2 dθ. Agin, our ssumption for this formul is tht f (θ) is continuous on the intervl [α, β]. In summry, for r = r(θ), α θ β, the re of the sector nd the rc length of the curve re Are = β α r 2 dθ, Length = α β α r2 + (r ) 2 dθ 68

69 Exmple 3.2. Find the re of one petl of the rose curve r = 3 cos(3θ). One petl of the rose curve is trversed when θ vries from π/6 to π/6. Thus, the re enclosed by it is A = 2 = 9 4 π/6 π/6 π/6 π/6 [3 cos(3θ)] 2 dθ ( + cos(6θ)) dθ = 3π 4. Exmple Find the re of the region lying between the inner nd the outer loops of the limcon r = 2 sin θ. The inner loop is trced when θ vries from π/6 to 5π/6. The re of the region enclosed by the inner loop is A = 2 = 2 5π/6 π/6 5π/6 π/6 ( 2 sin θ) 2 dθ [3 4 sin θ 2 cos(2θ)] dθ = π The outer loop is trced when θ vries from 5π/6 to 3π/6. Thus the re enclosed by the outer loop is A 2 = 3π/6 ( 2 sin θ) 2 dθ = 2π π/6 Therefore, the required re is A 2 A = π Exmple Find the re of the region common to the interiors of the circle r = 6 cos θ nd the crdioid r = 2 2 cos θ. Both the curves re symmetric bout the x-xis. We work with the upper-hlf plne. The gry shded region lies between the circle nd the rdil line θ = 2π/3. The circle hs the coordinte (, π/2) t the pole. Thus this region hs the re 2 2π/3 π/2 ( 6 cos 2 θ) dθ. The region shded pink is bounded by the rdil lines θ = 2π/3 nd θ = π. Thus it hs the re 2 π 3π/6 (2 2 cos θ) 2 dθ. 69