30 is close to t 5 = 15.

Transcription

1 Limits Definition 1 (Limit). If te values f(x) of a function f : A B get very close to a specific, unique number L wen x is very close to, but not necessarily equal to, a limit point c, we say te limit of f(x), as x approaces c, is L an write lim x c f(x) = L. Te number c is referre to as te limit point. Note: Tis is not a rigorous efinition. It is really a preliminary attempt at a efinition an will be replace later by a more rigorous efinition. One obvious problem is tat te concept of closeness is subjective. Some limits are relatively easy to calculate. Example: lim x 5 (x 2 + 3). It s obvious tat x 2 will be close to 25 an x will be close to 28 wen x is close to 5, so lim x 5 (x 2 + 3) = 28. 3t + 9 Example: lim t 7. It s obvious tat 3t is close to 21, 3t + 9 is close to 30, t 5 is t 5 close to 2 an tus 3t is close to 3t + 9 t 5 = 15. t 5 2 = 15 wen t is close to 7, so lim t 7 Most of te limits tat come up naturally are not quite so obvious an easy to calculate, but often yiel to a little algebraic manipulation an common sense. x 2 + x 12 Example: lim x 3. If one examines te numerator an enominator of x2 + x 12 x 3 x 3 for values of x close to 3, it is obvious bot are close to 0 an one cannot immeiately estimate te quotient. Suc limits are referre to as ineterminate. Tis particular type of ineterminate is sometimes referre to symbolically as te 0 0 case. We say symbolically because, of course, tere is no quotient 0. It is a sort way of referring 0 to te fact tat we are trying to fin te limit of a quotient were bot te numerator an te enominator are getting very close to 0. However, one may simplify x2 + x 12, fining x2 + x 12 x 3 x 3 = (x 3)(x + 4) x 3 wen x 3. Since x + 4 is clearly close to 7 wen x is close to 3, so must x2 + x 12 x 3 x 2 + x 12 it follows tat lim x 3 = 7. = x + 4 x 3 We ave implicitly mae use of te following teorem, wic is an almost immeiate consequence of te preliminary efinition of a limit. Teorem 1. If f(x) = g(x) for x c an eiter f or g as a limit at c, ten bot must ave a limit at c an teir limits must be te same. In effect, tis is a reiteration of te fact tat te value of a function at a limit point as no effect on a limit. We use tis teorem in te calculation of te limit almost every time we are face wit an ineterminate. Te calculation generally looks like te following: lim x c f(x) = = lim x c g(x) = L. 1 an

2 2 Te iea is tat we start wit an ineterminate f (x) an keep simplifying, eac time getting eiter a new form of f (x) or anoter function tat is equal to f (x) except at c, until we ave foun some function g(x) wose limit is easy to fin. For example, te calculation in te previous example migt look like te following: x2 + x 12 (x 3)(x + 4) limx 3 = limx 3 = limx 3 (x + 4) = 7. x 3 x 3 To repeat, in a nutsell, we usually win up calculating limits of ineterminates by simplifying until we ve foun a function, equal to te original except at te limit point, wic is not ineterminate. Most of te time, te algebraic manipulation we nee to o to simplify is in one of te following tree categories. (1) Factor an cancel. (2) Rationalize an cancel. (3) Simplifying a complex fractional expression Te example we saw fell in te first category, factor an cancel. Consier te following examplee T[((1TJ-284r)1(e)-t ctt c(nce)-1-157n tfraa.955tf59(1wing treelo prncel.

3 Tecnique 2: lim z 4 1 z 1 4 z 4 = lim z 4 4 z = lim z 4 4z(z 4) = lim 1 z 4 4z = z 4 z 4 4z 4z Here, we observe te complications arose from te enominators of 4 an z in te numerator an simply multiplie by 1 in te form 4z 4z. Tere will be oter manipulations tat will come up, an sometimes te simplication involve using tese tecniques may not be quite as straigtforwar, but most of te examples we run across will fall into one of tese tree cases. Different Types of Limits Besies orinary, two-sie limits, tere are one-sie limits (left-an limits an rigtan limits), infinite limits an limits at infinity. One-Sie Limits lim x 5 x2 4x 5. One migt tink tat since x 2 4x 5 0 as x 5, it woul follow tat lim x 5 x2 4x 5 = 0. But since x 2 4x 5 = (x 5)(x + 1) < 0 wen x is close to 5 but x < 5, x 2 4x 5 is unefine for some values of x very close to 5 an te limit as x 5 oesn t exist. But we woul still like a way of saying x 2 4x 5 is close to 0 wen x is close to 5 an x > 5, so we say te Rigt-Han Limit exists, write lim x 5 + x2 4x 5 = 0 an say x 2 4x 5 approaces 0 as x approaces 5 from te rigt. Sometimes we ave a Left-Han Limit but not a Rigt-Han Limit. Sometimes we ave bot Left-Han an Rigt-Han Limits an tey re not te same. Sometimes we ave bot Left-Han an Rigt-Han Limits an tey re equal, in wic case te orinary limit exists an is te same. x 2 if x < 1 f(x) = x 3 if 1 < x < 2 x 2 if x > 2. lim x 1 f(x) = lim x 1 + f(x) = 1, so te left an rigt an limits are equal an lim x 1 f(x)1. lim x 2 f(x) = 8 wile lim x 2 + f(x) = 4, so te left an rigt an limits are ifferent an lim x 2 f(x) oesn t exist. Limits at Infinity 2x Suppose we re intereste in estimating about ow big is wen x is very big. It s x + 1 2x easy to see tat x + 1 = 2x x(1 + 1) = 2 2x if x 1 an tus will be very close to 2 x + 1 x x if x is very big. We write 2x lim x x + 1 = 2 an say te limit of 2x x + 1 is 2 as x approaces. 3

4 2x Similarly, will be very close to 2 if x is very small an we write x + 1 2x lim x x + 1 = 2 2x an say te limit of is 2 as x approaces. Here, small oes not mean close to x + 1 0, but it means tat x is a negative number wit a large magnitue (absolute value). A convenient way to fin a limit of a quotient at infinity (or minus infinity) is to factor out te largest term in te numerator an te largest term in te enominator an cancel wat one can. 5x 2 3 lim x 8x 2 2x + 1 = x 2 (5 3 ) x lim 2 x x 2 ( ) = x x 2 lim x 5 3 x x + 1 x 2 = 5 8 lim x 5x 3 8x 2 2x + 1 = lim x x(5 3 x ) x 2 (8 2 x + 1 x 2 ) = 5 3 x x(8 2 x + 1 x 2 ) = 0 Infinite Limits If x is close to 1, it s obvious tat lim x lim x 1 1 (x 1) 2 = 1 is very big. We write (x 1) 2 1 an say te limit of is as x approaces 1. (x 1) 2 1 Similarly, lim x 1 (x 1) =. 2 Tecnically, a function wit an infinite limit oesn t actually ave a limit. Saying a function as an infinite limit is a way of saying it oesn t ave a limit in a very specific way. Infinite limits are inferre fairly intuitively. If one as a quotient f(x), one may look at g(x) ow big f(x) an g(x) are. For example, If f(x) is close to some positive number an g(x) is close to 0 an positive, ten te limit will be. If f(x) is close to some positive number an g(x) is close to 0 an negative, ten te limit will be. If f(x) is close to some negative number an g(x) is close to 0 an positive, ten te limit will be. If f(x) is close to some negative number an g(x) is close to 0 an negative, ten te limit will be. 4

5 Variations of Limits One can also ave one-sie infinite limits, or infinite limits at infinity. 1 lim x 1 + x 1 = 1 lim x 1 x 1 = Asymptotes If lim x f(x) = L ten y = L is a orizontal asymptote. If lim x f(x) = L ten y = L is a orizontal asymptote. If lim x c + f(x) = ± ten x = c is a vertical asymptote. If lim x c f(x) = ± ten x = c is a vertical asymptote. Properties of Limits Wen calculating limits, we intuitively make use of some basic properties it s wort noting. Eac can be proven using a formal efinition of a limit. We list some of tem, usually bot using matematical notation an using plain language. It is te plain language tat soul be remembere. In eac case, unless note oterwise, we assume te limits written own actually exist. As is usually te case, x, y, z, u, v, s, t will represent variables an a, b, c, k, L will represent constants. lim x c k = k Te limit of a constant is tat constant. lim x c x = c Wen x gets close to c, x gets close to c. lim x c [kf(x)] = k lim x c f(x) Te limit of a constant times a function is equal to te constant times te limit. lim x c [f(x) ± g(x)] = lim x c f(x) ± lim x c g(x) Te limit of a sum or ifference is te sum or ifference of te limits. lim x c [f(x) g(x)] = [lim x c f(x)] [lim x c g(x)] Te limit of a prouct is te prouct of te limits. f(x) lim x c g(x) = lim x c f(x) lim x c g(x) Te limit of a quotient is equal to te quotient of te limits. Of course, tis epens on te quotient existing, so te enominator on te rigt must not equal 0. lim x c f g(x) = lim x c f(g(x)) = f(lim x c g(x)) if f is continuous at lim x c g(x). Te limit of a composition is te composition of te limits, provie te outsie function is continuous at te limit of te insie function. Example: lim x 3 5x + 1 = 16 = 4. (Squeeze Teorem) If f(x) g(x) (x) in some open interval containing c an lim x c f(x) = lim x c (x) = L, ten lim x c g(x) = L. If te values of a function lie between te values of two functions wic ave te same limit, ten tat function must sare tat limit. Continuity 5

6 Definition 2 (Continuity). A function f is sai to be continuous at c if lim x c f(x) = f(c). Goemetrically, tis correspons to te absence of any breaks in te grap of f at c. Wen we ve calculate limits, most of te time we starte wit a function tat was not continuous at te limit point, simplifie to get anoter function wic was equal to te original function except at te limit point but was continuous at te limit point, an ten it was easy to fin te limit of te latter function. Rule of Tumb: Most functions we run across will be continuous except at points were tere is an obvious reason for tem to fail to be continuous. Examples of Continuous Functions: Polynomial Functions Rational Functions (Quotients of Polynomial Functions) except were te enominator is 0. Te exponential function Te natural logaritm function sin an cos tan except at o multiples of π/2, were it obviously isn t since tan = sin an cos cos takes on te value 0 at o multiples of π/2. Wen we perform most algebraic manipulations involving continuous functions, we win up wit continuous functions. Again, te exception is if tere s an obvious reason wy te new function wouln t be continuous somewere. Te sum of continuous functions is a continuous function. Te ifference of continuous functions is a continuous function. Te prouct of continuous functions is a continuous function. Te quotient of continuous functions is a continuous function except were te enominator is 0. Te composition of continuous functions is a continuous function. Teorem 2 (Intermeiate Value Teorem). If a function is continuous on a close interval [a, b], ten te function must take on every value between f(a) an f(b). Corollary 3 (Zero Teorem). If a function is continuous on a close interval [a, b] an takes on values wit opposite sign at a an at b, ten it must take on te value 0 somewere between a an b. Te Zero Teorem leas to te Bisection Meto, wic is a foolproof way of estimate a zero of a continuous function to any esire precision provie we are able to fin bot positive an negative values of te function. Te Bisection Meto works as follows. We wis to estimate a zero for a continuous function f. We start by fining points a 0, b 0 at wic te function as opposite sign. Witout loss of generality (WLOG), let us assume a 0 < b 0. We evaluate te function at te mipoint a 0 + b 0 m 0. Unless f(m 0 ) = 0, in wic case we ave foun a zero for f an can stop, f(m 0 ) 2 must ave a ifferent sign tan eiter f(a 0 ) or f(b 0 ). 6

7 If f(m 0 ) iffers in sign from f(a 0 ), ten we know f as a zero on te interval [a 0, m 0 ]; oterwise, we know f as a zero on te interval [m 0, b 0 ]. Eiter way, we now ave an interval we may enote by [a 1, b 1 ] wic is alf te wit of te original interval but wic also contains a zero of f. We can repeat tis process until we ave an interval as small as we esire. Number of Iterations We can also etermine, in avance, ow many iterations we nee in orer to obtain an interval of wit smaller tan any preetermine number ɛ > 0. Clearly, after te first iteration we will ave an interval of wit b 0 a 0 2. After te next iteration, we will ave an interval of wit b 0 a 0 = b 0 a 0 4 After te tir iteration, we will ave an interval of wit b 0 a = b 0 a Continuing, it soul be clear tat after n iterations we will ave an interval of wit b 0 a 0. 2 n We can easily fin a value of n for wic tis wit is less tan ɛ by solving as follows. b 0 a 0 < ɛ 2 n b 0 a 0 ɛ < 2 n 2 n > b 0 a 0 ɛ n ln 2 > ln ( b 0 a 0 ) ɛ ) n > ln ( b 0 a 0 ɛ. ln 2 Te number of iterations we will nee is tus te smallest integer greater tan or equal to ln ( b 0 a 0 ) ɛ. ln 2 One soul not memorize tis formula; te calculations are routine enoug so tat you soul be able to carry tem out fairly easily an quickly wen you nee to an oing so is goo practice. Definition of a Limit Definition 3 (Limit). We say lim x c f(x) = L if for every ɛ > 0 tere is some δ > 0 suc tat f(x) L < ɛ wenever 0 x c < δ. 7 Tis gives a precise meaning to te vague iea of being close to. Essentially, regarless of ow close (ɛ) someone insists f(x) be to te limit L in orer to be convince f(x) is close to L, it must be possible to sow tat f(x) is at least tat close provie x is close enoug (witin δ of) to c. Tere are variations of tis efinition tat give rigorous meaning to eac of te variations: one-sie limits, infinite limits an limits at infinity. Spee Suppose an object is moving along a straigt line an te istance it as travele at a given time is given by te formula s = f(t), were t represents time an s represents istance. Te average spee of te object over a time perio t 0 t t 1 woul be equal to s t = f(t 1) f(t 0 ) t 1 t 0.

8 If t 1 was close to t 0, tis woul likely be close to te instantaneous spee of te object f(t 1 ) f(t 0 ) wen t = t 0, suggesting tat te instantaneous spee is equal to lim t1 t 0, wic t 1 t 0 is equal to te simpler looking expression lim t t0 f(t) f(t 0 ) t t 0. Slope of a Tangent Line Consier te grap of a function y = f(x). Te slope of a line connecting two points (x 0, y 0 ) an (x 1, y 1 ) on its grap woul be equal to y x = y 1 y 0 = f(x 1) f(x 0 ). x 1 x 0 x 1 x 0 If te two points are close togeter on te grap, wic will be te case if x 1 is close to x 0, tis woul likely be close to te slope of te line tangent to te grap at te point (x 0, y 0 ), f(x 1 ) f(x 0 ) suggesting tat te slope of te tangent is equal to lim x1 x 0, wic is equal x 1 x 0 f(x) f(x 0 ) to te simpler looking expression lim x x0. x x 0 Te Derivative f(t) f(t 0 ) f(x) f(x 0 ) Te similarity between lim t t0 an lim x x0 suggest someting t t 0 x x 0 more general is going on an tat it woul be wortwile to stuy tat expression, leaing to te efinition of a erivative. Definition (Te Derivative). f f(x) f(c) (c) = lim x c is calle te erivative of f at c if te limit exists. x c Alternative Form: f f(c + ) f(c) (c) = lim 0. If f (c) exists, ten f is sai to be ifferentiable at c; oterwise, it is sai f fails to be ifferentiable at c. Te Derivative Function Te corresponence c f (c) wic associates f (c) wit c for every c D f effectively efines a function, wic we also call te erivative an enote by f. We may use te alternate form of te erivative at a point to create te following efinition. Definition (Te Derivative Function). Te erivative of a function f is te function f given by te formula f f(x + ) f(x) (x) = lim 0 werever te limit exists. Variations f (x) = lim x 0 f(x + x) f(x) x f (x) = lim z x f(z) f(x) z x. f (x) = lim x 0 y x 8

9 Te process of fining a erivative is calle ifferentiation. Notation Te erivative function may be enote by eac of te following equivalent notations: f (x) = y = y x = x (f(x)). Te latter two versions are calle Leibniz Notation. y soul be rea te erivative of y (wit respect to x). x Te value of te erivative at a point c may be enote by eac of te following equivalent notations: f (c) = y x=c = y x=c = x (f(x)) x=c. x Often, te part x=c inicating te erivative is evaluate at c will be omitte an must be inferre from te context. Differentiable Functions Are Continuous Suppose f is ifferentiable at c. Ten f f(x) f(c) (c) = lim x c exists. Intuitively, it x c woul appear tat f(x) woul ave to be close to f(c) if x is close to c (oterwise, te limit wouln t exist), wic is essentially wat it means for f to be continuous at c. Tis can be proven as follows. Proof. ( Suppose f is ifferentiable at) c. Ten lim x c f(x) = f(x) f(c) lim x c f(c) + (x c) = x c f(x) f(c) f(c) + lim x c lim x c (x c) = x c f(c)+f (c) 0 = f(c), were eac calculation is justifiable from eiter te properties of limits or te efinition of a erivative. Properties an Formulas k = 0 Te erivative of a constant function is ientically 0. x x (kx) = k (ax + b) = a x Power Rule: x (xn ) = nx n 1 Te erivative of a constant times a function is equal to te constant times te erivative. Tis may be written as u (ku) = k x x or x (kf(x)) = kf (x). Te erivative of a sum is equal to te sum of te erivatives. Tis may be written as u (u + v) = x x + v x or (f + g) (x) = f (x) + g (x). Te erivative of a ifference is equal to te ifference of te erivatives. Tis may be written as 9

10 u (u v) = x x v x or (f g) (x) = f (x) g (x). Tese tree properties, taken togeter, imply tat we may ifferentiate term-by-term. Tis means tat if we want to calculate te erivative of a function containing several terms, we may fin te erivative of eac term separately an a te erivatives togeter. In particular, combine wit te Power Rule, tis makes te ifferentiation of polynomials routine. Example: x (5x8 + 3x 5 8x x + 4) = 5 x (x8 ) + 3 x (x5 ) 8 x (x2 ) + (11x + 4) = x 5 8x x 4 8 2x + 11 = 40x x 4 16x Eac of tese properties may be proven using te efinition of a erivative. For example, we may prove te erivative of a sum is equal to te sum of te erivatives wit te following routine calculation. Proof. Let φ(x) = f(x) + g(x). By efinition, φ φ(z) φ(x) (x) = lim z x = z x lim z x [f(z) + g(z)] [f(x) + g(x)] z x = f(z) + g(z) f(x) g(x) lim z x = z x f(z) f(x) + g(z) g(x) lim z x = ( z x ) f(z) f(x) g(z) g(x) lim z x + z x z x Since te limit of a sum is equal to te sum of te limits, we get φ f(z) f(x) g(z) g(x) (x) = lim z x + lim z x. z x z x Once again by te efinition of a erivative, te expression on te rigt is equal to f (x) + g (x). Te Prouct Rule Te formula for te erivative of a prouct turns out to be more complicate tan te formulas for erivatives of sums an ifferences. In plain language, Te erivative of a prouct is equal to te first factor times te erivative of te secon plus te secon factor times te erivative of te first. Symbolically, tis may be written several ways. (fg) = fg + gf x (f(x)g(x)) = f(x)g (x) + g(x)f (x) v (uv) = u x x + v u x 10

11 11 Example: x (x2 (5x 3)) = x 2 x (5x 3) + (5x 3) x (x2 ) = x 2 (5) + (5x 3) 2x Tis may be proven wit te following calculation. Proof. Let φ(x) = f(x)g(x). By te efinition of a erivative, φ φ(z) φ(x) (x) = lim z x = z x lim z x f(z)g(z) f(x)g(x) z x f(z)g(z) f(z)g(x) + f(z)g(x) f(x)g(x) lim z x z x f(z)[g(z) g(x)] + [f(z) f(x)]g(x) lim z x = = ( z x ) g(z) g(x) f(z) f(x) lim z x f(z) + g(x). z x z x From te properties of limits, te expression on te rigt may be written as φ g(z) g(x) (x) = lim z x f(z) lim z x z x f(z) f(x) lim z x g(x) z x From te efinition of a erivative, f(z) f(x) lim z x = f (x) z x an g(z) g(x) lim z x = g (x). z x + Since f is continuous (ifferentiable functions are continuous), it also follows tat lim z x f(z) = f(x), so we get φ (x) = f(x)g (x) + f (x)g(x). = Te Quotient Rule Te erivative of a quotient is equal to te enominator times te erivative of te numerator minus te numerator times te erivative of te enominator, all ivie by te square of te enominator. Symbolically, ( ) tis may be written several ways. f = gf fg g( ) g 2 f(x) = g(x)f (x) f(x)g (x) x g(x) (g(x)) 2 ( u ) v u = x uv x ( x v v 2 u ) vu uv = v v 2

12 Example: ( ) sin x = x x (x 2 + 3) x (sin x) sin x x (x2 + 3) (x 2 + 3) 2 = (x 2 + 3) cos x sin x(2x) (x 2 + 3) 2 = (x 2 + 3) cos x 2x sin x (x 2 + 3) 2. Te Quotient Rule may be proven wit a calculation similar to tat use to prove te Prouct Rule. Proof. Let φ(x) = f(x) g(x). By te efinition of a erivative, φ(z) φ(x) φ (x) = lim z x = z x f(z) lim f(x) f(z)g(x) f(x)g(z) g(z) g(x) g(x)g(z) z x = lim z x = z x z x f(z)g(x) f(x)g(x) f(x)g(z) + f(x)g(x) lim z x g(x)g(z)(z x) g(x)[f(z) f(x)] f(x)[g(z) g(x)] = lim z x g(x)g(z)(z x) f(z) f(x) g(z) g(x) g(x) f(x) = lim z x z x z x. g(x)g(z) Using te properties of limits, we get φ (x) = f(z) f(x) g(x) lim z x f(x) lim z x z x f(x) g(z) g(x) z x g(x) lim z x g(z) = g(x)f (x) f(x)g (x). (g(x)) 2 Derivatives of Trigonometric Functions Let f(x) = sin x. From te efinition of a erivative, f (x) = lim 0 f(x + ) f(x) sin(x + ) sin x = lim 0. Conveniently, we ave a trigonometric ientity tat enables us to rewrite sin(x + ) as sin x cos + cos x sin, so we ave f (x) = lim 0 sin x cos + cos x sin sin x lim 0 sin x cos sin x + cos x sin lim 0 sin x(cos 1) + sin cos x = = = 12

13 ( lim 0 sin x cos 1 + cos x sin ) = sin x lim 0 cos 1 + cos x lim 0 sin. sin We will sow tat lim 0 1 cos lim 0 = 1 an = 0, from wic it will follow tat f (x) = (sin x) 0 + (cos x) 1 = cos x. We tus ave te formula (sin x) = cos x subject to proving te claims about te limits x of sin 1 cos an. sin Claim 1. lim 0 = 1. Proof. First consier 0 < < π/2, raw te unit circle wit center at te origin, an consier te sector wit central angle were one sie lies along te x-axis an te oter sie lies in te first quarant. Since te area of te circle is π an te ratio of te area of te sector to te area of te circle is, te area of te sector is 2π 2π π = 2. Now consier te rigt triangle were te ypotenuse coincies wit te sie of te sector lying in te first quarant an te base lies along te x-axis. Te vertices will be (0, 0), 1 (cos, 0), (cos, sin ), so its legs will be of lengt cos, sin an its area will be sin. 2 cos Since te triangle is containe witin te sector, its area will be smaller tan te area of te sector. Hence 1 2 cos sin < 2. 2 sin Multiplying bot sies by yiels te inequality cos < 1 cos. Now consier te rigt triangle wit one leg coinciing wit te sie of te sector lying along te x-axis an te ypotenuse making an angle wit tat leg. Its vertices are (0, 0), (1, 0), (1, tan ), so its legs will be of lengt 1, tan an its area will be 1 tan. 2 Since te sector is containe witin tis triangle, its area will be smaller tan te area of te triangle. Hence 2 < 1 tan. 2 Multiplying bot sies by 2 cos an making use of te ientity tan cos = sin yiels te inequality cos < sin. Combining te two inequalities we ave obtaine yiels (1) cos < sin < 1 cos if 0 < < π/2. Now, suppose π/2 < < 0. Ten 0 < < π/2 an te ouble inequality (1) yiels 13

14 14 (2) cos( ) < sin( ) < 1 cos( ). Since cos( ) = cos an sin( ) = sin, it follows tat sin( ) an (2) becomes = sin = sin (3) cos < sin < 1 cos. We tus see (1) ols bot for 0 < < π/2 an for π/2 < < 0. 1 Since lim 0 cos = lim 0 cos = 1, by te Squeeze Teorem it follows tat lim sin 0 = 1 1 cos Claim 2. lim 0 = 0. Proof. We make use of te ientity involving sin an an algebraic manipulation reminiscent of rationalization, enabling us to prove te claim wit a fairly routine calculation. 1 cos 1 cos lim 0 = lim cos 1 + cos = lim 0 lim 0 sin lim 0 sin 1 cos 2 (1 + cos ) = lim 0 sin 1 + cos = lim 0 sin 2 (1 + cos ) = sin 1 + cos = 1 0 = 0. Derivatives of Logaritmic an Exponential Functions We will ultimately go troug a far more elegant evelopment ten wat we can o now. Consier first an exponential function of te form f(x) = a x for some constant a > 0. Note te ifference between a power function x x n an an exponential function x a x. For a power function, te variable is raise to a power; for an exponential function, a constant is raise to a variable power. Let s try to calculate te erivative for f. Using te efinition of a erivative, we may write f (x) = lim 0 f(x + ) f(x) = a x+ a x lim 0 = a x (a 1) lim 0 = a x a 1 lim 0. We may write x (ax ) = k a x a 1, were k = lim 0 epens on a.

15 We are assuming k exists! It oes, but tis is not so easy to sow. Evaluating 2 1 for values of close to 0 yiels values close to 0.69, wile evaluating 3 1 for values of close to 0 yiels values close to 1.1. k a x is obviously simplest if k = 1. Te numerical calculations suggest tere is some 2 < a < 3 for wic a = 1. Tat number is calle e, yieling te formula x (ex ) = e x. Te function x e x is calle te exponential function an is often enote by exp. Te exponential function is essentially unique in aving te property tat it s its own erivative! Te ajective essentially is use because every constant multiple of te exponential function as te same property, but no oter function as tat property! If we interpret te erivative as a measure of rate of cange, te fact tat te exponential function is its own erivative may be interprete to mean tat te rate at wic te exponential function canges is equal to te magnitue of te exponential function. It turns out tat all functions wose rates of cange are proportional to teir sizes are exponential functions. Note te omission of te efinite article. Te Natural Logaritm Function Recall te efinition of a logaritm function: log b x is te power wic b must be raise to in orer to obtain x. In oter wors, l = log b x if b l = x. Te logaritm wit base e is known as te natural logaritm function an is enote by ln. Tus, l = ln x if an only e l = x. We ll try to figure out te erivative of te natural logaritm function ln. Our calculations will not be rigorous; we will obtain te correct formula, but a legitimate erivation will ave to wait until we learn about te efinite integral. Let f(x) = ln x. Let s start calculating f (x). Accoring to te efinition of a erivative, f (x) = lim z x f(z) f(x) z x lim z x ln z ln x z x. We nee to estimate te ifference quotient ln z ln x z x = 15 wen z is close to x. We ll o it in a rater strange way. Let Z = ln z an X = ln x. ln z ln x Ten we know z = e Z an x = e X an we may write as Z X e Z e X. z x Now, let s go back an take anoter look at te erivative of te exponential function, but from a ifferent perspective an wit sligtly ifferent notation. Sometimes it pays to write someting a few ifferent ways! Let g(x) = e X. By te efinition of a erivative, g e Z e X (X) = lim Z X Z X.

16 e X. But we know g (X) = e X, so tis suggests tat wen Z is close to X, ez e X Z X 16 is close to But ez e X Z X is te reciprocal of Z X Z X 1, suggesting tat is close to e Z ex e Z ex e. X On te oter an, wen trying to fin te erivative of te natural log function we came up someting suggesting Z X was close to te erivative. Tis suggests tat te e Z e X erivative is equal to 1 e. X Recall e X = x, so 1, wic suggests e X = 1 x x (ln x) = 1 x. Inee, tis is te erivative, altoug we re not reay for a rigorous erivation. We will, owever, make use of tis formula. Logaritms to Oter Bases Te key properties of logaritms are: log b (xy) = log b x + log b y (Te log of a prouct equals te sum of te logs.) log b (x/y) = log b x log b y (Te log of a quotient equals te ifference of te logs.) log b (x r ) = r log b x (Te log of someting to a power is te power times te log.) We can use tese properties to sow tat, in a very real sense, natural logaritms suffice an we can always write any logaritm in terms of a natural logaritm. Suppose l = log b x. It follows tat b l = x an tus ln(b l ) = ln x. Using te tir property of logaritms, we see l ln b = ln x an, solving for l, we get l = ln x. Tis gives us te crucial ientity ln b log b x = ln x ln b. Tis enables us to calculate erivatives involving logaritms to any base, as sown in te following general example. Example: Fin x (log b x). Solution: x (log b x) = ( ) ln x = 1 x ln b ln b x (ln x) = 1 ln b 1 x = 1 x ln b. Oter Exponential Functions A calculation similar to te erivation of te ientity log b x = ln x involving exponential functions. Let a x = y. Ten: ln(a x ) = ln y x ln a = ln y e x ln a = e ln y. Since e ln y = y an y = a x, it follows tat a x = e x ln a. Caution: ln b yiels a useful ientity

17 In ig scool algebra, a meaning was given to rational exponents: If a > 0, m, n Z, n > 0, ten a m/n = n a m. However, no meaning was given to a x if x is irrational. Tat can be one an te ientity a x = e x ln a will play a key role. Let s take anoter look at te erivative of orinary exponential functions. We foun x (ax ) = k a x a 1, were k = lim 0. Let s play te same sort of game we playe wen trying to calculate te erivative of te natural log function an let a = H, noting tat a 1 will be close to k wen is close to 0, an ence wen H is close to 1. a 1 H 1 may be written as log a H log a 1, since log a 1 = 0. If we let f(x) = log a x an trie to calculate f (1), we migt write f f(h) f(1) (1) = lim H 1 = H 1 log lim a H log a 1 H 1. H 1 But we earlier sowe tat f (x) = 1 x ln a, so f (1) = 1 ln a. Tis suggests tat log a H log a 1 1 is close to if H is close to 1, wic suggests its H 1 ln a H 1 reciprocal log a H log a 1, an ence a 1 as well, is close to ln a if H is close to 1 an is close to 0. We tus expect k = ln a an x (ax ) = a x ln a. Te Cain Rule Suppose y = f g(x). Assuming f an g ave erivatives were appropriate, te Cain Rule says tat (f g) = (f g) g. In more practical language, if we write y = f(u) an u = g(x), it comes out as y x = y u u x. To see wy, go back to te efinition of a erivative an write y x = (f g) (x) = f g(x + ) f g(x) lim 0 = f(g(x + )) f(g(x)) lim 0. We like to write tis as f(g(x + )) f(g(x)) g(x + ) g(x) lim 0, but it s possible tat g(x+) = g(x), an g(x + ) g(x) ence te enominator g(x + ) g(x) = 0, for some 0, so we consier two separate cases. If tere are arbitrarily small values of for wic g(x + ) = g(x), ten bot (f g) (x) an g (x) will ave to equal 0 an te Cain Rule certainly ols in a trivial fasion. So we nee only verify te Cain Rule in te remaining case were g(x + ) g(x) is is close to 0. For tis case, we write 17

18 u = g(x), g(x + ) = g(x) + k, k = g(x + ) g(x). We can ten write y x = f(g(x + )) f(g(x)) g(x + ) g(x) lim 0 = g(x + ) g(x) f(u + k) f(u) g(x + ) g(x) lim 0. k Since te limit of a prouct is equal to te prouct of limits, we ave y x = f(u + k) f(u) g(x + ) g(x) lim 0 lim 0. k Since k 0 wen 0, te limit f(u + k) f(u) lim 0 k will equal te limit f(u + k) f(u) lim k 0, wic is equal to f (u) or y k u, wile lim 0 by efinition, equal to g (x) = u x. Tis emonstrates tat y x = y u u x. g(x + ) g(x) Strategy For Calculating Derivatives Te following strategy will work witout fail to calculate te erivative of any function efine implicitly, provie te function is constructe from te basic elementary functions. Strategy Differentiate Term-By-Term. For eac term, etermine weter it is a basic, elementary function, a prouct, a quotient or a composite function an ten use te appropriate formula or rule. Tat s all tere is. Te Catc One must, of course, correctly recognize elementary functions: te power functions, sin, cos, tan, sec, csc, cot, ln, exp. One must also remember te Prouct an Quotient Rules an be able to use te Cain Rule. A Common Mistake: Not recognizing tat a term can be rewritten in te form of a basic elementary function, prouct or quotient an tus incorrectly jumping to te conclusion tat te term is a composite function. Implicit Differentiation Functions may be efine several ways. One of te most common ways is via an explicit formula, suc as f(x) = x 2 or y = 5 sin x + tan x. It is also possible to efine functions implicitly. One uses an equation involving an inepenent variable an a epenent variable, often x an y, an efines a function y = f(x) by saying y = f(x) for a particular value of x if tose values for x an y satisfy te equation. 18 is,

19 Sometimes aitional conitions must be place on te solutions in orer to ensure a function is efine, tat is, so tat for any value of x tere is only one value of y satisfying te equation along wit te conitions. Examples: xy = 1 (Te same function may be efine explicitly via te formula f(x) = 1. Its grap is an x yperbola.) x 2 + y 2 = 1, y 0 (Te same function may be efine explicitly via te formula f(x) = 1 x 2. Its grap is a unit semi-circle, centere at te origin, above te x-axis.) x 2 + y 2 = 1, y 0 (Te same function may be efine explicitly via te formula f(x) = 1 x 2. Its grap is a unit semi-circle, centere at te origin, below te x-axis.) Derivative of functions efine implicitly may be calculate using a tecnique calle Implicit Differentiation. Te process is iabolically simple. One starts wit an equation of te form F (x, y) = G(x, y), were F (x, y) an G(x, y) are algebraic expressions involving te variables x an y, efining a function y = f(x) implicitly an observes x (F (x, y)) = (G(x, y)). x If one ifferentiates correctly, one obtains a linear equation in y x easily by moving all te terms involving y x 19 wic may be solve to te left sie, all te oter terms to te rigt sie, an ten iviing bot sies by y x. Wat s te Catc? One must calculate x (F (x, y)) an (G(x, y)) correctly! x One must recognize te inepenent variable is x an remember tat y is a function of x. Example Consier te function y = f(x) efine implicitly by xy = 1. One calculates as follows: x (xy) = x (1) x y x + y x x x y x + y 1 = 0 x y x + y = 0 x y x = y y x = y x. Example = 0 Using te Prouct Rule on te left.

20 Consier te function y = f(x) efine implicitly by x 2 + y 2 = 1, y 0. One calculates as follows: x (x2 + y 2 ) = x (1) x (x2 ) + x (y2 ) = 0 2x + x (y2 ) = 0 To calculate x (y2 ), one nees to use te Cain Rule. Write z = y 2. We nee a ecomposition of te composite function z = y 2. It is eceptively simple: z = y 2 y = f(x). We may ten apply te Cain Rule as follows: z x = z y y x = 2y y x. Plugging tis erivative into te equation 2x + x (y2 ) = 0 yiels 2x + 2y y x = 0 2y y x = 2x y x = 2x 2y y x = x y. Relate Rates Relate rates problems eal wit situations in wic two or more variables are canging an te rates at wic tey re canging are relate ence te term Relate Rates. Steps Recognize te basic variables involve an represent tem by symbols. Remember tat te inepenent variable is usually time an it is common to represent time by t. Determine te relationsips between te unerlying variables. Every fact an every relationsip generally may be escribe by an equation, so one obtains one or more equations: 20 A 1 = B 1 A 2 = B 2 A 3 = B 3... A n = B n

21 For eac equation, ifferentiate eac sie wit respect to t, an equate te two sies to get one or more equations relating te rates of cange: t (A 1) = t (B 1) t (A 2) = t (B 2) t (A 3) = t (B 3)... t (A n) = t (B n) Solve te equations algebraically to obtain te rates esire. State te conclusion implie by te solution to te equations. Te Mean Value Teorem A stuent at te University of Connecticut appens to be travelling to Boston. He enters te Massacussetts Turnpike at te Sturbrige Village entrance at 9:15 in te morning. Since e uses Fast Lane, e never actually picks up a ticket, but sensors recor tat e goes troug te toll lane at Newton at 9:53, just 38 minutes later. He soon receives a traffic summons in te mail inicating tat e violate te poste spee limit an ecies to appeal, since tere is no inication e was caugt by raar. Appearing in court, te prosecutor argues tat e travelle 44.8 miles in 38 minutes an terefore travelle at an average spee of miles per our, more tan 5 miles above te igest poste spee limit of 65 miles per our. Te stuent argues tat, altoug e may ave average more tan 65 miles per our, tere is no evience tat e actually ever travelle at an instantaneous spee of more tan 65 miles per our. Te juge consiers te arguments an quickly rejects te stuent s plea, noting tat te Mean Value Teorem implies tat since is average spee was miles per our, tere a to be at least one instant uring wic is instantaneous spee was miles per our. Rolle s Teorem Teorem 4 (Rolle s Teorem). Suppose a function f is continuous on te close interval [a, b], ifferentiable on te open interval (a, b) an f(a) = f(b). Ten tere is a number c (a, b) suc tat f (c) = 0. Geometrically, tis says tat if tere are two points on a smoot curve takes at te same eigt, tere must be a point in between were te tangent is orizontal. Rolle s Teorem is a special case of te Mean Value Teorem. Proof. Suppose f satisfies te ypoteses of Rolle s Teorem. By te Extreme Value Teorem for Continuous Function, tere must be some point in [a, b] at wic f attains a minimum an some point at wic f attains a maximum. One possibility is tat f is constant on te entire interval, in wic case f is ientically 0 on (a, b) an te conclusion is clearly true. 21

22 So let s consier te oter possibility, tat f is not constant. Ten eiter te minimum or te maximum must occur at some point c (a, b), tat is, at some point oter tan te enpoints. We will sow tat f (c) = 0 if f as a maximum at c. Similar reasoning woul sow f (c) = 0 if f a a minimum at c, sowing te conclusion is true an completing te proof. We know f f(x) f(c) (c) = lim x c. x c Since tis orinary limit exists, it follows tat bot te left an limit an te rigt an limit exist an are bot equal to f (c). We ll consier tem separately. Te Left Han Limit: f f(x) f(c) (c) = lim x c. x c Since f as a maximum at c, if x < c, ten f(x) f(c), so f(x) f(c) 0. But, if x < c, f(x) f(c) f(x) f(c) it s also true tat x c < 0 an it follows tat 0. We see lim x c x c x c is te limit of non-negative numbers an terefore can t be negative. It follows tat f (c) 0. Te line of reasoning for te rigt an limit is similar. Te Rigt Han Limit: f f(x) f(c) (c) = lim x c +. x c Since f as a maximum at c, if x > c, ten f(x) f(c), so f(x) f(c) 0. But, if x > c, f(x) f(c) f(x) f(c) it s also true tat x c > 0 an it follows tat 0. We see lim x c + x c x c is te limit of numbers less tan or equal to 0 an terefore can t be positive. It follows tat f (c) 0. Since f (c) 0 an f (c) 0, it follows tat f (c) = 0 Consequences of Rolle s Teorem Besies being a special case of te Mean Value Teorem an being a step in te pat to proving te Mean Value Teorem, Rolle s Teorem as some interesting applications of its own. Corollary 5. A polynomial equation of egree n as at most n solutions. Tis is equivalent to te statement tat a polynomial of egree n as at most n zeros. Since a polynomial may be ifferentiate as many times as necessary, wit eac erivative being a polynomial of lower egree, one immeiate consequence of Rolle s Teorem is tat te erivative of a polynomial as at least one zero between eac pair of istinct zeros of te original polynomial. Te erivative of a linear polynomial is a non-zero constant, aving no zeros, so a linear polynomial can t ave more tan 1 zero. Te erivative of a quaratic polynomial is linear, aving no more tan 1 zero, so a quaratic can t ave more tan 2 zeros. Te erivative of a cubic polynomial is quaratic, aving no more tan 2 zeros, so te cubic can t ave more tan 3 zeros. Tis clearly goes on forever. Matematical Inuction. Te Mean Value Teorem Te argument can be mae rigorous troug te use of 22

23 Teorem 6 (Te Mean Value Teorem). Suppose a function f is continuous on te close interval [a, b] an ifferentiable on te open interval (a, b). Ten tere is a number c (a, b) suc tat f(b) f(a) = f (c)(b a) or, equivalently, f f(b) f(a) (c) =. b a Geometrically, te Mean Value Teorem says tat if tere is a smoot curve connecting two points, tere must be some point in between at wic te tangent line is parallel to te line connecting tose two points. Analytically, te Mean Value Teorem says te rate of cange of a ifferentiable function must, at some point, take on its average, or mean value. Te proof epens on te fact tat te particular point at wic te tangent line is parallel to te line connecting te enpoints also appens to be te point at wic te curve is furters away from tat line. Te proof essentially consists of applying Rolle s Teorem to te function measuring te istance between te line an te curve. f(b) f(a) Since te line goes between te points (a, f(a)) an (b, f(b)), its slope will be b a an its equation may be written, in point-slope form, f(b) f(a) y f(a) = (x a). b a Solving for y, we may write te equation in te form f(b) f(a) y = f(a) + (x a). b a Since te secon coorinate of a point on te curve wit first coorinate x is f(x), te vertical istance [ between te line an te ] curve will equal f(b) f(a) f(x) f(a) + (x a) b a f(b) f(a) = f(x) f(a) (x a). b a We are now prepare to prove te Mean Value Teorem. f(b) f(a) Proof. Let φ(x) = f(x) f(a) (x a). b a It is easy to see tat φ satisfies te ypoteses of Rolle s Teorem on te interval [a, b]. Certainly, te fact tat φ is bot continuous an ifferentiable on [a, b] follows immeiately from te fact tat f is. In aition, f(b) f(a) φ(a) = f(a) f(a) (a a) = 0 b a an f(b) f(a) φ(b) = f(b) f(a) (b a) = f(b) f(a) [f(b) f(a)] = 0. b a Tus, tere must be some c (a, b) suc tat φ (c) = 0. We first obtain φ (c) as follows. φ (x) = f f(b) f(a) (x) b a φ (c) = f f(b) f(a) (c) b a Since φ (c) = 0, it follows tat 23

24 f f(b) f(a) (c) = 0. b a f f(b) f(a) (c) = b a f (c)(b a) = f(b) f(a) Consequences of te Mean Value Teorem Peraps te most important consequence of te Mean Value Teorem is tat it gives precise meaning to te most important single concept in elementary Calculus, Te Derivative Measures Rate of Cange. Teorem 7. a. If te erivative of a function is positive at all points on an interval, ten te function is increasing on tat interval. b. If te erivative of a function is negative at all points on an interval, ten te function is ecreasing on tat interval. To prove tis, we nee a precise efinition of wat it means to be increasing or ecreasing. Definition 4 (Strictly Increasing). A function f is sai to be strictly increasing on an open interval I if f(a) < f(b) wenever a, b I an a < b. Definition 5 (Nonecreasing). A function f is sai to be nonecreasing on an open interval I if f(a) f(b) wenever a, b I an a < b. Note te subtle ifference. Often, we will simply say a function is increasing. Generally, in tose cases, it will not really be important weter we mean strictly increasing or simply nonecreasing. Tere are similar efinitions of te terms strictly ecreasing an nonincreasing. Here, too, we will often use te ambiguous term ecreasing. Definition 6 (Strictly Decreasing). A function f is sai to be strictly ecreasing on an open interval I if f(a) > f(b) wenever a, b I an a < b. Definition 7 (Nonincreasing). A function f is sai to be nonincreasing on an open interval I if f(a) f(b) wenever a, b I an a < b. Wit tese efinitions, we are reay to prove te erivative measures rate of cange. We will prove just one of wat are really four ifferent parts, tat if te erivative is strictly positive ten te function is strictly increasing. Proof. Suppose f (x) > 0 x I an let a, b I, a < b. We nee to prove tat f(a) < f(b). Note: We ve introuce te notation to mean for all or for every. Since f (x) > 0 x I, it follows tat f is continuous on [a, b] an ifferentiable on (a, b), so by te Mean Value Teorem tere is some c (a, b) suc tat f(b) f(a) = f (c)(b a). Since f (x) > 0 x I, it follows tat f (c) is positive. Since a < b, it follows tat b a is also positive, so tat f (c)(b a) is also positive an ence f(b) f(a) must be positive. Clearly, f(a) must be smaller tan f(b). 24

25 Very Different Functions Can t Have te Same Derivative It s obvious tat if two functions iffer by only a constant term, ten tey will ave te same erivative. Example: x (x2 ) = x (x2 + 5) = 2x. A natural question is weter only functions iffering by a constant can sare te same erivative. Te Mean Value Teorem enables us to see tis is true. Teorem 8. If f (x) = g (x) for all x in some interval, ten tere is some constant k suc tat f(x) = g(x) + k. Proof. Let φ = f g an let a be some fixe point in te interval. Now let x be in te interval. Clearly, te φ is bot continuous an ifferentiable on te interval wit enpoints a an x an we can apply te Mean Value Teorem. We use tat language because it is possible tat a < x an te interval is [a, x] but also possible tat a > x an te interval is [x, a]. By te Mean Value Teorem, tere is some c in te interval suc tat φ(x) φ(a) = φ (c)(x a). Because φ = f g, it follows tat φ (c) = 0, so φ(x) φ(a) = 0. It follows tat φ(x) = φ(a), or φ(x) = k, were we let k = φ(a). Since φ = f g, we ave f(x) g(x) = k, or f(x) = g(x) + k. Corollary 9. Only constant functions ave erivatives wic are ientically 0. Tis teorem will prove useful wen we try to fin functions wit a given erivative. One application will be to etermine te acceleration ue to gravity.example An object is roppe from a eigt of 128 feet. How long oes it take to reac te groun? Solution Let: be te eigt of te object, measure in feet. v be te velocity of te object, measure in feet per secon. t be te time since te object was roppe, measure in secons. We know: = 128 wen t = 0 v = 0 wen t = 0 = v (Since velocity is te rate at wic te eigt canges.) t v = 32 (Since acceleration is te rate at wic te velocity canges an te t acceleration ue to gravity is 32 feet per secon per secon an is in te ownwar, or negative, irection.) We want: Te value of t wen = 0 Since ( 32t) = 32 an two functions can ave te same erivative only if tey iffer t by a constant, it follows tat 25

26 v = 32t + c for some constant c. Since v = 0 wen t = 0, it follows tat 0 = c c = 0 v = 32t Since t ( 16t2 ) = 32t, = v = 32t an two functions can ave te same erivative t only if tey iffer by a constant, it follows tat = 16t 2 + k for some constant k. Since = 128 wen t = 0, it follows tat 128 = k k = 128 = 16t Tus, wen = 0, 0 = 16t t 2 = 128 t 2 = = 8 t = 8 = It tus takes 2 2 secons for te object to fall to te groun. We can also figure out ow fast it was going wen it it te groun: Since v = 32t, wen t = 2 2 we ave v = = , so te object is travelling at a spee of 64 2 feet per secon wen it its te groun.example Determine te acceleration ue to gravity. Solution: We may perform te following experiment: We rop a coin from te ceiling an measure te time it takes to it te groun. Let: t be te time since te coin is roppe T be te amount of time it takes to it te groun be te eigt of te coin H be te eigt of te ceiling v be te spee of te coin g be te acceleration ue to gravity We know: t = v v t = g = H wen t = 0 v = 0 wen t = 0 = 0 wen t = T We want: Te value of g. 26

27 Since v (gt) = g, = g an two functions can ave te same erivative only if tey t t iffer by a constant, it follows tat v = gt + c for some constant c. Since v = 0 wen t = 0, it follows tat 0 = g 0 + c c = 0 v = gt Since ( 1 2 t gt2) = gt, = v = gt an two functions can ave te same erivative only if t tey iffer by a constant, it follows tat = 1 2 gt2 + k for some constant k. Since = H wen t = 0, it follows tat H = 1g k k = H = 1 2 gt2 + H Since = 0 wen t = T, it follows tat 0 = 1gT 2 + H 2 1 gt 2 = H 2 g = H 1 T 2 2 g = 2H T 2 So te acceleration ue to gravity is 2H T 2 feet per secon per secon. Note tat g is negative, since te acceleration is ownwar. Linear Approximations Suppose we want to approximate te value of a function f for some value of x, say x 1, close to a number x 0 at wic we know te value of f. By its nature, te tangent to a curve ugs te curve fairly closely near te point of tangency, so it s natural to expect te 2 n coorinate of a point on te tangent line close to te point (x 0, f(x 0 )) will be fairly close to te actual value of f(x 1 ). Tat value is calle te Linear Approximation to f(x 1 ), or te Tangent Line Approximation. Since te tangent line goes troug (x 0, f(x 0 )) an as slope f (x 0 ), it will ave equation y f(x 0 ) = f (x 0 )(x x 0 ), wic may also be written as y = f(x 0 ) + f (x 0 )(x x 0 ). Definition 8 (Linear Approximation). Te linear approximation of f(x) near x = x 0 is L(x) = f(x 0 ) + f (x 0 )(x x 0 ). Example Fin an approximation to Solution: Let f(x) = 3 x. We use te linear approximation for f(x) near x = 8. 27

28 We nee f(8) an f (8) an fin tem as follows: f(x) = 3 x = x 1/3 f(8) = 3 8 = 2 f (x) = 1 3 x 2/3 = x = 1 2/3 3( 3 x) 2 f 1 (8) = 3( 3 8) = = Using te formula L(x) = f(x 0 ) + f (x 0 )(x x 0 ), we get L(x) = (x 8). 12 Our approximation to is L(8.01) = (8.01 8) = = Note a calculator approximation for is Differentials An equivalent meto of approximating values of functions is calle an approximation using ifferentials. Definition 9 (Increment of x). x = x 1 x 0 Definition 10 (Increment of y). y = y 1 y 0 = f(x 1 ) f(x 0 ) Wit tis notation, f(x 1 ) = f(x 0 ) + y. Te Linear Approximation assumes f(x 1 ) f(x 0 ) + f (x 0 )(x 1 x 0 ) = f(x 0 ) + f (x 0 ) x. Definition 11 (Differential of x). x = x. Te ifferential of x is synonymous wit te increment of x. Definition 12 (Differential of y). y = f (x 0 ) x = f (x 0 )x = y x x. Wit tis efinition, te Linear Approximation may be written f(x 1 ) f(x 0 ) + y. Inepenently, y may be tougt of as an approximation to te amount y, or f(x), canges. Example A layer of paint inces tick is applie to a sperical object 16 inces in iameter. Approximately ow muc paint is use? Solution: Te amount of paint is equal to te amount te volume of a spere will cange by if its raius increases from 8 inces to inces. Letting V be te volume of te spere an r be its raius, we know V = 4 3 πr3. 28

29 V r = 4πr2. Taking r 0 = 8, r 1 = 8.001, we fin: r = r = = V r r=8 = 4π 8 2 = 256π V = V r = 256π = 0.256π r So it will take 0.256π cubic inces of paint. Newton s Meto Newton s Meto is esigne to estimate a zero of a ifferentiable function. It generally works faster tan te Bisection Meto wen it works an it oes not require one to first fin two points at wic te function as opposite signs. Te rawback is tat it oesn t always work. Given a function f an an initial value x 0, one obtains a sequence of values x 0, x 1, x 2, x 3,... using te formula 29 x n = x n 1 f(x n 1) f (x n 1 ). If one is fortunate, te sequence quickly approaces a zero of f. It is generally relatively easy to set up a spreaseet to o te calculations, wic can also be one using a calculator. Stuents familiar wit any programming languages soul fin writing a program to implement Newton s Meto an easy yet eucational exercise. Geometrically, x n is te first coorinate of te point at wic te line tangent to te grap of f at te point (x n 1, f(x n 1 )) crosses te x axis. Tis may be seen as follows. Te equation of te tangent line is y f(x n 1 ) = f (x n 1 )(x x n 1 ). If x n is te first coorinate of te point were tis line crosses te x axis, since te secon coorinate of tat point is 0, we may plug x = x n 1, y = 0 in tat equation to get 0 f(x n 1 ) = f (x n 1 )(x n x n 1 ). We may ten solve for x x as follows: f(x n 1 ) = f (x n 1 )(x n x n 1 ) f(x n 1) f (x n 1 ) = x n x n 1 x n = x n 1 f(x n 1) f (x n 1 ) L Hôpital s Rule L Hôpital s Rule provies a convenient way of fining limits of ineterminate quotients. Effectively, it states tat if one wises to fin a limit of a quotient f(x) an bot f(x) an g(x) g(x) eiter 0 or bot ±, ten te limit of te quotient f(x) is equal to te limit of g(x) te quotient f (x) of te erivatives if te latter limit exists. g (x)