Differentiable Functions

Transcription

1 Capter 8 Differentiable Functions A differentiable function is a function tat can be approximated locally by a linear function. 8.. Te derivative Definition 8.. Suppose tat f : (a, b) R and a < c < b. Ten f is differentiable at c wit derivative f (c) if f(c + ) f(c) = f (c). 0 Te domain of f is te set of points c (a, b) for wic tis it exists. If te it exists for every c (a, b) ten we say tat f is differentiable on (a, b). Grapically, tis definition says tat te derivative of f at c is te slope of te tangent line to y = f(x) at c, wic is te it as 0 of te slopes of te lines troug (c, f(c)) and (c +, f(c + )). We can also write f (c) = x c [ f(x) f(c) x c since if x = c +, te conditions 0 < x c < δ and 0 < < δ in te definitions of te its are equivalent. Te ratio f(x) f(c) x c is undefined (0/0) at x = c, but it doesn t ave to be defined in order for te it as x c to exist. Like continuity, differentiability is a local property. Tat is, te differentiability of a function f at c and te value of te derivative, if it exists, depend only te values of f in a arbitrarily small neigborood of c. In particular if f : A R ], 39

2 40 8. Differentiable Functions were A R, ten we can define te differentiability of f at any interior point c A since tere is an open interval (a, b) A wit c (a, b) Examples of derivatives. Let us give a number of examples tat illustrate differentiable and non-differentiable functions. Example 8.2. Te function f : R R defined by f(x) = x 2 is differentiable on R wit derivative f (x) = 2x since [ (c + ) 2 c 2 ] (2c + ) = = (2c + ) = 2c Note tat in computing te derivative, we first cancel by, wic is valid since 0 in te definition of te it, and ten set = 0 to evaluate te it. Tis procedure would be inconsistent if we didn t use its. Example 8.3. Te function f : R R defined by { x 2 if x > 0, f(x) = 0 if x 0. is differentiable on R wit derivative f (x) = { 2x if x > 0, 0 if x 0. For x > 0, te derivative is f (x) = 2x as above, and for x < 0, we ave f (x) = 0. For 0, we consider te it f() f(0) f() = 0 0. Te rigt it is and te left it is f() 0 + = = 0, 0 f() 0 = 0. Since te left and rigt its exist and are equal, te it also exists, and f is differentiable at 0 wit f (0) = 0. Next, we consider some examples of non-differentiability at discontinuities, corners, and cusps. Example 8.4. Te function f : R R defined by { /x if x 0, f(x) = 0 if x = 0,

3 8.. Te derivative 4 is differentiable at x 0 wit derivative f (x) = /x 2 since f(c + ) f(c) /(c + ) /c = 0 0 c (c + ) = 0 c(c + ) = 0 c(c + ) = c 2. However, f is not differentiable at 0 since te it f() f(0) / 0 = 0 0 does not exist. = 0 2 Example 8.5. Te sign function f(x) = sgn x, defined in Example 6.8, is differentiable at x 0 wit f (x) = 0, since in tat case f(x + ) f(x) = 0 for all sufficiently small. Te sign function is not differentiable at 0 since and 0 sgn sgn 0 sgn = = 0 sgn { / if > 0 / if < 0 is unbounded in every neigborood of 0, so its it does not exist. Example 8.6. Te absolute value function f(x) = x is differentiable at x 0 wit derivative f (x) = sgn x. It is not differentiable at 0, owever, since f() f(0) = 0 0 = sgn 0 does not exist. (Te rigt it is and te left it is.) Example 8.7. Te function f : R R defined by f(x) = x /2 is differentiable at x 0 wit f (x) = sgn x 2 x. /2 If c > 0, ten using te difference of two square to rationalize te numerator, we get 0 f(c + ) f(c) = 0 (c + ) /2 c /2 (c + ) c = 0 [ (c + ) /2 + c /2] = 0 (c + ) /2 + c /2 = 2c /2.

4 42 8. Differentiable Functions If c < 0, we get te analogous result wit a negative sign. differentiable at 0, since does not exist. f() f(0) = /2 However, f is not Example 8.8. Te function f : R R defined by f(x) = x /3 is differentiable at x 0 wit f (x) = 3x 2/3. To prove tis result, we use te identity for te difference of cubes, and get for c 0 tat f(c + ) f(c) 0 a 3 b 3 = (a b)(a 2 + ab + b 2 ), = 0 (c + ) /3 c /3 (c + ) c = 0 [ (c + ) 2/3 + (c + ) /3 c /3 + c 2/3] = 0 (c + ) 2/3 + (c + ) /3 c /3 + c 2/3 = 3c 2/3. However, f is not differentiable at 0, since does not exist. f() f(0) = 0 0 2/3 Finally, we consider some examples of igly oscillatory functions. Example 8.9. Define f : R R by { x sin(/x) if x 0, f(x) = 0 if x = 0. It follows from te product and cain rules proved below tat f is differentiable at x 0 wit derivative f (x) = sin x x cos x. However, f is not differentiable at 0, since wic does not exist. f() f(0) = sin 0 0, Example 8.0. Define f : R R by { x 2 sin(/x) if x 0, f(x) = 0 if x = 0.

5 8.. Te derivative Figure. A plot of te function y = x 2 sin(/x) and a detail near te origin wit te parabolas y = ±x 2 sown in red. Ten f is differentiable on R. (See Figure.) It follows from te product and cain rules proved below tat f is differentiable at x 0 wit derivative f (x) = 2x sin x cos x. Moreover, f is differentiable at 0 wit f (0) = 0, since f() f(0) = sin 0 0 = 0. In tis example, x 0 f (x) does not exist, so altoug f is differentiable on R, its derivative f is not continuous at Derivatives as linear approximations. Anoter way to view Definition 8. is to write f(c + ) = f(c) + f (c) + r() as te sum of a linear (or, strictly speaking, affine) approximation f(c) + f (c) of f(c + ) and a remainder r(). In general, te remainder also depends on c, but we don t sow tis explicitly since we re regarding c as fixed. As we prove in te following proposition, te differentiability of f at c is equivalent to te condition r() 0 = 0. Tat is, te remainder r() approaces 0 faster tan, so te linear terms in provide a leading order approximation to f(c + ) wen is small. We also write tis condition on te remainder as r() = o() as 0, pronounced r is little-o of as 0. Grapically, tis condition means tat te grap of f near c is close te line troug te point (c, f(c)) wit slope f (c). Analytically, it means tat te function f(c + ) f(c)

6 44 8. Differentiable Functions is approximated near c by te linear function f (c). Tus, f (c) may be interpreted as a scaling factor by wic a differentiable function f srinks or stretces lengts near c. If f (c) <, ten f srinks te lengt of a small interval about c by (approximately) tis factor; if f (c) >, ten f stretces te lengt of an interval by (approximately) tis factor; if f (c) > 0, ten f preserves te orientation of te interval, meaning tat it maps te left endpoint to te left endpoint of te image and te rigt endpoint to te rigt endpoints; if f (c) < 0, ten f reverses te orientation of te interval, meaning tat it maps te left endpoint to te rigt endpoint of te image and visa-versa. We can use tis description as a definition of te derivative. Proposition 8.. Suppose tat f : (a, b) R. Ten f is differentiable at c (a, b) if and only if tere exists a constant A R and a function r : (a c, b c) R suc tat r() f(c + ) = f(c) + A + r(), 0 = 0. In tat case, A = f (c). Proof. First suppose tat f is differentiable at c according to Definition 8., and define r() = f(c + ) f(c) f (c). Ten r() f(c + ) f(c) 0 = f (c) = 0, 0 so te condition in te proposition olds wit A = f (c). Conversely, suppose tat f(c + ) = f(c) + A + r() were r()/ 0 as 0. Ten [ f(c + ) f(c) = A + r() ] = A, 0 0 so f is differentiable at c wit f (c) = A. Example 8.2. For Example 8.2 wit f(x) = x 2, we get (c + ) 2 = c 2 + 2c + 2, and r() = 2, wic goes to zero at a quadratic rate as 0. Example 8.3. For Example 8.4 wit f(x) = /x, we get c + = c c 2 + r(), for c 0, were te quadratically small remainder is r() = 2 c 2 (c + ).

7 8.2. Properties of te derivative Left and rigt derivatives. For te most part, we will use derivatives tat are defined only at te interior points of te domain of a function. Sometimes, owever, it is convenient to use one-sided left or rigt derivatives tat are defined at te endpoint of an interval. Definition 8.4. Suppose tat f : [a, b] R. Ten f is rigt-differentiable at a c < b wit rigt derivative f (c + ) if f(c + ) f(c) = f (c + ) 0 + exists, and f is left-differentiable at a < c b wit left derivative f (c ) if f(c + ) f(c) f(c) f(c ) = = f (c ) A function is differentiable at a < c < b if and only if te left and rigt derivatives at c bot exist and are equal. Example 8.5. If f : [0, ] R is defined by f(x) = x 2, ten f (0 + ) = 0, f ( ) = 2. Tese left and rigt derivatives remain te same if f is extended to a function defined on a larger domain, say x 2 if 0 x, f(x) = if x >, /x if x < 0. For tis extended function we ave f ( + ) = 0, wic is not equal to f ( ), and f (0 ) does not exist, so te extended function is not differentiable at eiter 0 or. Example 8.6. Te absolute value function f(x) = x in Example 8.6 is left and rigt differentiable at 0 wit left and rigt derivatives f (0 + ) =, f (0 ) =. Tese are not equal, and f is not differentiable at Properties of te derivative In tis section, we prove some basic properties of differentiable functions Differentiability and continuity. First we discuss te relation between differentiability and continuity. Teorem 8.7. If f : (a, b) R is differentiable at at c (a, b), ten f is continuous at c.

8 46 8. Differentiable Functions Proof. If f is differentiable at c, ten f(c + ) f(c) = 0 0 = 0 = f (c) 0 = 0, [ f(c + ) f(c) [ f(c + ) f(c) ] ] 0 wic implies tat f is continuous at c. For example, te sign function in Example 8.5 as a jump discontinuity at 0 so it cannot be differentiable at 0. Te converse does not old, and a continuous function needn t be differentiable. Te functions in Examples 8.6, 8.8, 8.9 are continuous but not differentiable at 0. Example 9.24 describes a function tat is continuous on R but not differentiable anywere. In Example 8.0, te function is differentiable on R, but te derivative f is not continuous at 0. Tus, wile a function f as to be continuous to be differentiable, if f is differentiable its derivative f need not be continuous. Tis leads to te following definition. Definition 8.8. A function f : (a, b) R is continuously differentiable on (a, b), written f C (a, b), if it is differentiable on (a, b) and f : (a, b) R is continuous. For example, te function f(x) = x 2 wit derivative f (x) = 2x is continuously differentiable on R, wereas te function in Example 8.0 is not continuously differentiable at 0. As tis example illustrates, functions tat are differentiable but not continuously differentiable may beave in rater patological ways. On te oter and, te beavior of continuously differentiable functions, wose graps ave continuously varying tangent lines, is more-or-less consistent wit wat one expects Algebraic properties of te derivative. A fundamental property of te derivative is tat it is a linear operation. In addition, we ave te following product and quotient rules. Teorem 8.9. If f, g : (a, b) R are differentiable at c (a, b) and k R, ten kf, f + g, and fg are differentiable at c wit (kf) (c) = kf (c), (f + g) (c) = f (c) + g (c), (fg) (c) = f (c)g(c) + f(c)g (c). Furtermore, if g(c) 0, ten f/g is differentiable at c wit ( ) f (c) = f (c)g(c) f(c)g (c) g g 2. (c)

9 8.3. Te cain rule 47 Proof. Te first two properties follow immediately from te linearity of its stated in Teorem For te product rule, we write f(c + )g(c + ) f(c)g(c) (fg) (c) = 0 (f(c + ) f(c)) g(c + ) + f(c) (g(c + ) g(c)) = 0 f(c + ) f(c) g(c + ) g(c) = g(c + ) + f(c) = f (c)g(c) + f(c)g (c), were we ave used te properties of its in Teorem 6.34 and Teorem 8.9, wic implies tat g is continuous at c. Te quotient rule follows by a similar argument, or by combining te product rule wit te cain rule, wic implies tat (/g) = g /g 2. (See Example 8.22 below.) Example We ave = 0 and x =. Repeated application of te product rule implies tat x n is differentiable on R for every n N wit (x n ) = nx n. Alternatively, we can prove tis result by induction: Te formula olds for n =. Assuming tat it olds for some n N, we get from te product rule tat (x n+ ) = (x x n ) = x n + x nx n = (n + )x n, and te result follows. It also follows by linearity tat every polynomial function is differentiable on R, and from te quotient rule tat every rational function is differentiable at every point were its denominator is nonzero. Te derivatives are given by teir usual formulae Te cain rule Te cain rule states tat te composition of differentiable functions is differentiable. Te result is quite natural if one tinks in terms of derivatives as linear maps. If f is differentiable at c, it scales lengts by a factor f (c), and if g is differentiable at f(c), it scales lengts by a factor g (f(c)). Tus, te composition g f scales lengts at c by a factor g (f(c)) f (c). Equivalently, te derivative of a composition is te composition of te derivatives (regarded as linear maps). We will prove te cain rule by sowing tat te composition of remainder terms in te linear approximations of f and g leads to a similar remainder term in te linear approximation of g f. Te argument is complicated by te fact tat we ave to evaluate te remainder of g at a point tat depends on te remainder of f, but tis complication sould not obscure te simplicity of te final result. Teorem 8.2 (Cain rule). Let f : A R and g : B R were A R and f (A) B, and suppose tat c is an interior point of A and f(c) is an interior point of B. If f is differentiable at c and g is differentiable at f(c), ten g f : A R is differentiable at c and (g f) (c) = g (f(c)) f (c).

10 48 8. Differentiable Functions Proof. Since f is differentiable at c, tere is a function r() suc tat f(c + ) = f(c) + f (c) + r(), r() 0 = 0, and since g is differentiable at f(c), tere is a function s(k) suc tat It follows tat were g (f(c) + k) = g (f(c)) + g (f(c)) k + s(k), (g f)(c + ) = g (f(c) + f (c) + r()) s(k) = 0. k 0 k = g (f(c)) + g (f(c)) (f (c) + r()) + s (f (c) + r()) = g (f(c)) + g (f(c)) f (c) + t() t() = g (f(c)) r() + s (φ()), Since r()/ 0 as 0, we ave t() 0 = 0 φ() = f (c) + r(). s (φ()). We claim tat tis it exists and is zero, and ten it follows from Proposition 8. tat g f is differentiable at c wit (g f) (c) = g (f(c)) f (c). To prove te claim, we use te facts tat φ() f (c) as 0, s(k) k 0 as k 0. Rougly speaking, we ave φ() f (c) wen is small and terefore s (φ()) s (f (c)) 0 as 0. In detail, let ɛ > 0 be given. We want to sow tat tere exists δ > 0 suc tat s (φ()) < ɛ if 0 < < δ. First, coose δ > 0 suc tat r() < f (c) + if 0 < < δ. If 0 < < δ, ten φ() f (c) + r() < f (c) + ( f (c) + ) < (2 f (c) + ). Next, coose η > 0 so tat s(k) k < ɛ 2 f (c) + if 0 < k < η.

11 8.3. Te cain rule 49 (We include a in te denominator on te rigt-and side to avoid a division by zero if f (c) = 0.) Finally, define δ 2 > 0 by η δ 2 = 2 f (c) +, and let δ = min(δ, δ 2 ) > 0. If 0 < < δ and φ() 0, ten 0 < φ() < η, so s (φ()) ɛ φ() 2 f (c) + < ɛ. If φ() = 0, ten s(φ()) = 0, so te inequality olds in tat case also. Tis proves tat s (φ()) = 0. 0 Example Suppose tat f is differentiable at c and f(c) 0. Ten g(y) = /y is differentiable at f(c), wit g (y) = /y 2 (see Example 8.4). It follows tat te reciprocal function /f = g f is differentiable at c wit ( f ) (c) = g (f(c))f (c) = f (c) f(c) 2. Te cain rule gives an expression for te derivative of an inverse function. In terms of linear approximations, it states tat if f scales lengts at c by a nonzero factor f (c), ten f scales lengts at f(c) by te factor /f (c). Proposition Suppose tat f : A R is a one-to-one function on A R wit inverse f : B R were B = f (A). Assume tat f is differentiable at an interior point c A and f is differentiable at f(c), were f(c) is an interior point of B. Ten f (c) 0 and (f ) (f(c)) = f (c). Proof. Te definition of te inverse implies tat f (f(x)) = x. Since f is differentiable at c and f is differentiable at f(c), te cain rule implies tat ( f ) (f(c)) f (c) =. Dividing tis equation by f (c) 0, we get te result. Moreover, it follows tat f cannot be differentiable at f(c) if f (c) = 0. Alternatively, setting d = f(c), we can write te result as (f ) (d) = f (f (d)). Proposition 8.23 is not entirely satisfactory because it assumes te existence and differentiability of an inverse function. We will return to tis question in Section 8.7 below, but we end tis section wit some examples tat illustrate te

12 50 8. Differentiable Functions necessity of te condition f (c) 0 for te existence and differentiability of te inverse. Example Define f : R R by f(x) = x 2. Ten f (0) = 0 and f is not invertible on any neigborood of te origin, since it is non-monotone and not oneto-one. On te oter and, if f : (0, ) (0, ) is defined by f(x) = x 2, ten f (x) = 2x 0 and te inverse function f : (0, ) (0, ) is given by f (y) = y. Te formula for te inverse of te derivative gives (f ) (x 2 ) = f (x) = 2x, or, writing x = f (y), (f ) (y) = 2 y, in agreement wit Example 8.7. Example Define f : R R by f(x) = x 3. Ten f is strictly increasing, one-to-one, and onto. Te inverse function f : R R is given by f (y) = y /3. Ten f (0) = 0 and f is not differentiable at f(0)= 0. On te oter and, f is differentiable at non-zero points of R, wit (f ) (x 3 ) = f (x) = 3x 2, or, writing x = y /3, (f ) (y) = 3y, 2/3 in agreement wit Example Extreme values One of te most useful applications of te derivative is in locating te maxima and minima of functions. Definition Suppose tat f : A R. Ten f as a global (or absolute) maximum at c A if f(x) f(c) for all x A, and f as a local (or relative) maximum at c A if tere is a neigborood U of c suc tat f(x) f(c) for all x A U. Similarly, f as a global (or absolute) minimum at c A if f(x) f(c) for all x A, and f as a local (or relative) minimum at c A if tere is a neigborood U of c suc tat f(x) f(c) for all x A U.

13 8.4. Extreme values 5 If f as a (local or global) maximum or minimum at c A, ten f is said to ave a (local or global) extreme value at c. Teorem 7.37 states tat a continuous function on a compact set as a global maximum and minimum but does not say ow to find tem. Te following fundamental result goes back to Fermat. Teorem If f : A R R as a local extreme value at an interior point c A and f is differentiable at c, ten f (c) = 0. Proof. If f as a local maximum at c, ten f(x) f(c) for all x in a δ-neigborood (c δ, c + δ) of c, so wic implies tat Moreover, wic implies tat f(c + ) f(c) f (c) = 0 + f(c + ) f(c) f (c) = 0 0 for all 0 < < δ, f(c + ) f(c) 0. 0 for all δ < < 0, f(c + ) f(c) 0. It follows tat f (c) = 0. If f as a local minimum at c, ten te signs in tese inequalities are reversed, and we also conclude tat f (c) = 0. For tis result to old, it is crucial tat c is an interior point, since we look at te sign of te difference quotient of f on bot sides of c. At an endpoint, we get te following inequality condition on te derivative. (Draw a grap!) Proposition Let f : [a, b] R. If te rigt derivative of f exists at a, ten: f (a + ) 0 if f as a local maximum at a; and f (a + ) 0 if f as a local minimum at a. Similarly, if te left derivative of f exists at b, ten: f (b ) 0 if f as a local maximum at b; and f (b ) 0 if f as a local minimum at b. Proof. If te rigt derivative of f exists at a, and f as a local maximum at a, ten tere exists δ > 0 suc tat f(x) f(a) for a x < a + δ, so f(a + ) f(a) f (a + ) = Similarly, if te left derivative of f exists at b, and f as a local maximum at b, ten f(x) f(b) for b δ < x b, so f (b ) 0. Te signs are reversed for local minima at te endpoints. In searcing for extreme values of a function, it is convenient to introduce te following classification of points in te domain of te function.

14 52 8. Differentiable Functions Definition Suppose tat f : A R R. An interior point c A suc tat f is not differentiable at c or f (c) = 0 is called a critical point of f. An interior point were f (c) = 0 is called a stationary point of f. Teorem 8.27 its te searc for maxima or minima of a function f on A to te following points. () Boundary points of A. (2) Critical points of f: (a) interior points were f is not differentiable; (b) stationary points were f (c) = 0. Additional tests are required to determine wic of tese points gives local or global extreme values of f. In particular, a function need not attain an extreme value at a critical point. Example If f : [, ] R is te function { x if x 0, f(x) = 2x if 0 < x, ten x = 0 is a critical point since f is not differentiable at 0, but f does not attain a local extreme value at 0. Te global maximum and minimum of f are attained at te endpoints x = and x =, respectively, and f as no oter local extreme values. Example 8.3. If f : [, ] R is te function f(x) = x 3, ten x = 0 is a critical point since f (0) = 0, but f does not attain a local extreme value at 0. Te global maximum and minimum of f are attained at te endpoints x = and x =, respectively, and f as no oter local extreme values Te mean value teorem Te mean value teorem is a key result tat connects te global beavior of a function f : [a, b] R, described by te difference f(b) f(a), to its local beavior, described by te derivative f : (a, b) R. We begin by proving a special case. Teorem 8.32 (Rolle). Suppose tat f : [a, b] R is continuous on te closed, bounded interval [a, b], differentiable on te open interval (a, b), and f(a) = f(b). Ten tere exists a < c < b suc tat f (c) = 0. Proof. By te Weierstrass extreme value teorem, Teorem 7.37, f attains its global maximum and minimum values on [a, b]. If tese are bot attained at te endpoints, ten f is constant, and f (c) = 0 for every a < c < b. Oterwise, f attains at least one of its global maximum or minimum values at an interior point a < c < b. Teorem 8.27 implies tat f (c) = 0. Note tat we require continuity on te closed interval [a, b] but differentiability only on te open interval (a, b). Tis proof is deceptively simple, but te result is not trivial. It relies on te extreme value teorem, wic in turn relies on te completeness of R. Te teorem would not be true if we restricted attention to functions defined on te rationals Q.

15 8.5. Te mean value teorem 53 Te mean value teorem is an immediate consequence of Rolle s teorem: for a general function f wit f(a) f(b), we subtract off a linear function to make te values of te resulting function equal at te endpoints. Teorem 8.33 (Mean value). Suppose tat f : [a, b] R is continuous on te closed, bounded interval [a, b] and differentiable on te open interval (a, b). Ten tere exists a < c < b suc tat f f(b) f(a) (c) =. b a Proof. Te function g : [a, b] R defined by f(b) f(a) g(x) = f(x) f(a) (x a) b a is continuous on [a, b] and differentiable on (a, b) wit g (x) = f f(b) f(a) (x). b a Moreover, g(a) = g(b) = 0. Rolle s Teorem implies tat tere exists a < c < b suc tat g (c) = 0, wic proves te result. Grapically, tis result says tat tere is point a < c < b at wic te slope of te tangent line to te grap y = f(x) is equal to te slope of te cord between te endpoints (a, f(a)) and (b, f(b)). As a first application, we prove a converse to te obvious fact tat te derivative of a constant functions is zero. Teorem If f : (a, b) R is differentiable on (a, b) and f (x) = 0 for every a < x < b, ten f is constant on (a, b). Proof. Fix x 0 (a, b). Te mean value teorem implies tat for all x (a, b) wit x x 0 f (c) = f(x) f(x 0) x x 0 for some c between x 0 and x. Since f (c) = 0, it follows tat f(x) = f(x 0 ) for all x (a, b), meaning tat f is constant on (a, b). Corollary If f, g : (a, b) R are differentiable on (a, b) and f (x) = g (x) for every a < x < b, ten f(x) = g(x) + C for some constant C. Proof. Tis follows from te previous teorem since (f g) = 0. We can also use te mean value teorem to relate te monotonicity of a differentiable function wit te sign of its derivative. (See Definition 7.54 for our terminology for increasing and decreasing functions.) Teorem Suppose tat f : (a, b) R is differentiable on (a, b). Ten f is increasing if and only if f (x) 0 for every a < x < b, and decreasing if and only if f (x) 0 for every a < x < b. Furtermore, if f (x) > 0 for every a < x < b ten f is strictly increasing, and if f (x) < 0 for every a < x < b ten f is strictly decreasing.

16 54 8. Differentiable Functions Proof. If f is increasing and a < x < b, ten f(x + ) f(x) 0 for all sufficiently small (positive or negative), so f(x + ) f(x) f (x) = 0. 0 Conversely if f 0 and a < x < y < b, ten by te mean value teorem tere exists x < c < y suc tat f(y) f(x) y x = f (c) 0, wic implies tat f(x) f(y), so f is increasing. Moreover, if f (c) > 0, we get f(x) < f(y), so f is strictly increasing. Te results for a decreasing function f follow in a similar way, or we can apply of te previous results to te increasing function f. Note tat altoug f > 0 implies tat f is strictly increasing, f is strictly increasing does not imply tat f > 0. Example Te function f : R R defined by f(x) = x 3 is strictly increasing on R, but f (0) = 0. If f is continuously differentiable and f (c) > 0, ten f (x) > 0 for all x in a neigborood of c and Teorem 8.36 implies tat f is strictly increasing near c. Tis conclusion may fail if f is not continuously differentiable at c. Example Define f : R R by { x/2 + x 2 sin(/x) if x 0, f(x) = 0 if x = 0. Ten f is differentiable on R wit { f /2 cos(/x) + 2x sin(/x) if x 0, (x) = /2 if x = 0. Every neigborood of 0 includes intervals were f < 0 or f > 0, in wic f is strictly decreasing or strictly increasing, respectively. Tus, despite te fact tat f (0) > 0, te function f is not strictly increasing in any neigborood of 0. As a result, no local inverse of te function f exists on any neigborood of Taylor s teorem If f : (a, b) R is differentiable on (a, b) and f : (a, b) R is differentiable, ten we define te second derivative f : (a, b) R of f as te derivative of f. We define iger-order derivatives similarly. If f as derivatives f (n) : (a, b) R of all orders n N, ten we say tat f is infinitely differentiable on (a, b). Taylor s teorem gives an approximation for an (n + )-times differentiable function in terms of its Taylor polynomial of degree n.

17 8.6. Taylor s teorem 55 Definition Let f : (a, b) R and suppose tat f as n derivatives f, f,... f (n) : (a, b) R on (a, b). Te Taylor polynomial of degree n of f at a < c < b is P n (x) = f(c) + f (c)(x c) + 2! f (c)(x c) n! f (n) (c)(x c) n. Equivalently, P n (x) = n a k (x c) k, k=0 a k = k! f (k) (c). We call a k te kt Taylor coefficient of f at c. Te computation of te Taylor polynomials in te following examples are left as an exercise. Example If P (x) is a polynomial of degree n, ten P n (x) = P (x). Example 8.4. Te Taylor polynomial of degree n of e x at x = 0 is P n (x) = + x + 2! x2 + n! xn. Example Te Taylor polynomial of degree 2n of cos x at x = 0 is P 2n (x) = 2! x2 + 4! x4 + ( ) n (2n)! x2n. We also ave P 2n+ = P 2n since te Tayor coefficients of odd order are zero. Example Te Taylor polynomial of degree 2n + of sin x at x = 0 is P 2n+ (x) = x 3! x3 + 5! x5 + ( ) n (2n + )! x2n+. We also ave P 2n+2 = P 2n+. Example Te Taylor polynomial of degree n of /x at x = is P n (x) = (x ) + (x ) 2 + ( ) n (x ) n. Example Te Taylor polynomial of degree n of log x at x = is P n (x) = (x ) 2 (x )2 + 3 (x )3 + ( ) n+ (x ) n. We write f(x) = P n (x) + R n (x). were R n is te error, or remainder, between f and its Taylor polynomial P n. Te next teorem is one version of Taylor s teorem, wic gives an expression for te remainder due to Lagrange. It can be regarded as a generalization of te mean value teorem, wic corresponds to te case n =. Te idea of te proof is to subtract a suitable polynomial from te function and apply Rolle s teorem, just as we proved te mean value teorem by subtracting a suitable linear function.

18 56 8. Differentiable Functions Teorem 8.46 (Taylor wit Lagrange Remainder). Suppose tat f : (a, b) R as n + derivatives on (a, b) and let a < c < b. For every a < x < b, tere exists ξ between c and x suc tat f(x) = f(c) + f (c)(x c) + 2! f (c)(x c) n! f (n) (c)(x c) n + R n (x) were R n (x) = (n + )! f (n+) (ξ)(x c) n+. Proof. Fix x, c (a, b). For t (a, b), let g(t) = f(x) f(t) f (t)(x t) 2! f (t)(x t) 2 n! f (n) (t)(x t) n. Ten g(x) = 0 and g (t) = n! f (n+) (t)(x t) n. Define ( ) n+ x t (t) = g(t) g(c). x c Ten (c) = (x) = 0, so by Rolle s teorem, tere exists a point ξ between c and x suc tat (ξ) = 0, wic implies tat It follows from te expression for g tat g (x ξ)n (ξ) + (n + ) g(c) = 0. (x c) n+ n! f (n+) (ξ)(x ξ) n (x ξ)n = (n + ) g(c), (x c) n+ and using te expression for g in tis equation, we get te result. Note tat te remainder term R n (x) = (n + )! f (n+) (ξ)(x c) n+ as te same form as te (n + )t-term in te Taylor polynomial of f, except tat te derivative is evaluated at a (typically unknown) intermediate point ξ between c and x, instead of at c. Example Let us prove tat ( ) cos x = 2. By Taylor s teorem, x 0 cos x = 2 x2 + (cos ξ)x4 4! for some ξ between 0 and x. It follows tat for x 0, cos x x 2 2 = 4! (cos ξ)x2. Since cos ξ, we get cos x x 2 2 4! x2, x 2

19 8.7. * Te inverse function teorem 57 wic implies tat cos x x 0 x 2 2 = 0. Note tat as well as proving te it, Taylor s teorem gives an explicit upper bound for te difference between ( cos x)/x 2 and its it /2. For example, cos(0.) (0.) Numerically, we ave 2 cos(0.) (0.) , In Section 2.7, we derive an alternative expression for te remainder R n as an integral * Te inverse function teorem Te inverse function teorem gives a sufficient condition for a differentiable function f to be locally invertible at a point c wit differentiable inverse: namely, tat f is continuously differentiable at c and f (c) 0. Example 8.24 sows tat one cannot expect te inverse of a differentiable function f to exist locally at c if f (c) = 0, wile Example 8.38 sows tat te condition f (c) 0 is not, on its own, sufficient to imply te existence of a local inverse. Before stating te teorem, we give a precise definition of local invertibility. Definition A function f : A R is locally invertible at an interior point c A if tere exist open neigboroods U of c and V of f(c) suc tat f U : U V is one-to-one and onto, in wic case f as a local inverse (f U ) : V U. Te following examples illustrate te definition. Example If f : R R is te square function f(x) = x 2, ten a local inverse at c = 2 wit U = (, 3) and V = (, 9) is defined by (f U ) (y) = y. Similarly, a local inverse at c = 2 wit U = ( 3, ) and V = (, 9) is defined by (f U ) (y) = y. In defining a local inverse at c, we require tat it maps an open neigborood V of f(c) onto an open neigborood U of c; tat is, we want (f U ) (y) to be close to c wen y is close to f(c), not some more distant point tat f also maps close to f(c). Tus, te one-to-one, onto function g defined by { y if < y < 4 g : (, 9) ( 2, ) [2, 3), g(y) = y if 4 y < 9 is not a local inverse of f at c = 2 in te sense of Definition 8.48, even toug g(f(2)) = 2 and bot compositions f g : (, 9) (, 9), g f : ( 2, ) [2, 3) ( 2, ) [2, 3) are identity maps, since U = ( 2, ) [2, 3) is not a neigborood of 2.

20 58 8. Differentiable Functions Example Te function f : R R defined by { cos (/x) if x 0 f(x) = 0 if x = 0 is locally invertible at every c R wit c 0 or c /(nπ) for some n Z. Teorem 8.5 (Inverse function). Suppose tat f : A R R and c A is an interior point of A. If f is differentiable in a neigborood of c, f (c) 0, and f is continuous at c, ten tere are open neigboroods U of c and V of f(c) suc tat f as a local inverse (f U ) : V U. Furtermore, te local inverse function is differentiable at f(c) wit derivative [(f U ) ] (f(c)) = f (c).. Proof. Suppose, for definiteness, tat f (c) > 0 (oterwise, consider f). By te continuity of f, tere exists an open interval U = (a, b) containing c on wic f > 0. It follows from Teorem 8.36 tat f is strictly increasing on U. Writing V = f(u) = (f(a), f(b)), we see tat f U : U V is one-to-one and onto, so f as a local inverse on V, wic proves te first part of te teorem. It remains to prove tat te local inverse (f U ), wic we denote by f for sort, is differentiable. First, since f is differentiable at c, we ave f(c + ) = f(c) + f (c) + r() were te remainder r satisfies r() 0 = 0. Since f (c) > 0, tere exists δ > 0 suc tat r() 2 f (c) for < δ. It follows from te differentiability of f tat, if < δ, f (c) = f(c + ) f(c) r() f(c + ) f(c) + r() f(c + ) f(c) + 2 f (c). Absorbing te term proportional to on te rigt and side of tis inequality into te left and side and writing f(c + ) = f(c) + k, we find tat 2 f (c) k for < δ. Coosing δ > 0 small enoug tat (c δ, c + δ) U, we can express in terms of k as = f (f(c) + k) f (f(c)).

21 8.7. * Te inverse function teorem 59 Using tis expression in te expansion of f evaluated at c +, we get tat f(c + ) = f(c) + f (c) + r(), f(c) + k = f(c) + f (c) [ f (f(c) + k) f (f(c)) ] + r(). Simplifying and rearranging tis equation, we obtain te corresponding expansion for f evaluated at f(c) + k, were te remainder s is given by f (f(c) + k) = f (f(c)) + f (c) k + s(k), s(k) = f (c) r () = f (c) r ( f (f(c) + k) f (f(c)) ). Since f (c) /2 k, it follows tat s(k) k 2 r() f (c) 2. Terefore, by te sandwic teorem and te fact tat 0 as k 0, s(k) = 0. k 0 k Tis result proves tat f is differentiable at f(c) wit [ f (f(c)) ] = f (c). Te expression for te derivative of te inverse also follows from Proposition 8.23, but only once we know tat f is dfferentiable at f(c). One can sow tat Teorem 8.5 remains true under te weaker ypotesis tat te derivative exists and is nonzero in an open neigborood of c, but in practise, we almost always apply te teorem to continuously differentiable functions. Te inverse function teorem generalizes to functions of several variables, f : A R n R n, wit a suitable generalization of te derivative of f at c as te linear map f (c) : R n R n tat approximates f near c. A different proof of te existence of a local inverse is required in tat case, since one cannot use monotonicity arguments. As an example of te application of te inverse function teorem, we consider a simple problem from bifurcation teory. Example Consider te transcendental equation y = x k (e x ) were k R is a constant parameter. Suppose tat we want to solve for x R given y R. If y = 0, ten an obvious solution is x = 0. Te inverse function teorem applied to te continuously differentiable function f(x; k) = x k(e x ) implies tat tere are neigboroods U, V of 0 (depending on k) suc tat te equation as a unique solution x U for every y V provided tat te derivative

22 60 8. Differentiable Functions y x Figure 2. Grap of y = f(x; k) for te function in Example 8.52: (a) k = 0.5 (green); (b) k = (blue); (c) k =.5 (red). Wen y is sufficiently close to zero, tere is a unique solution for x in some neigborood of zero unless k =. of f wit respect to x at 0, given by f x (0; k) = k is non-zero i.e., provided tat k (see Figure 2).

23 8.7. * Te inverse function teorem x k Figure 3. Plot of te solutions for x of te nonlinear equation x = k(e x ) as a function of te parameter k (see Example 8.52). Te point (x, k) = (0, ) were te two solution brances cross is called a bifurcation point. Alternatively, we can fix a value of y, say y = 0, and ask ow te solutions of te corresponding equation for x, x k (e x ) = 0,

24 62 8. Differentiable Functions depend on te parameter k. Figure 2 plots te solutions for x as a function of k for 0.2 k 2. Te equation as two different solutions for x unless k =. Te branc of nonzero solutions crosses te branc of zero solution at te point (x, k) = (0, ), called a bifurcation point. Te implicit function teorem, wic is a generalization of te inverse function teorem, implies tat a necessary condition for a solution (x 0, k 0 ) of te equation f(x; k) = 0 to be a bifurcation point, meaning tat te equation fails to ave a unique solution branc x = g(k) in some neigborood of (x 0, k 0 ), is tat f x (x 0 ; k 0 ) = * L Hôspital s rule In tis section, we prove a rule (muc beloved by calculus students) for te evaluation of inderminate its of te form 0/0 or /. Our proof uses te following generalization of te mean value teorem. Teorem 8.53 (Caucy mean value). Suppose tat f, g : [a, b] R are continuous on te closed, bounded interval [a, b] and differentiable on te open interval (a, b). Ten tere exists a < c < b suc tat f (c) [g(b) g(a)] = [f(b) f(a)] g (c). Proof. Te function : [a, b] R defined by (x) = [f(x) f(a)] [g(b) g(a)] [f(b) f(a)] [g(x) g(a)] is continuous on [a, b] and differentiable on (a, b) wit (x) = f (x) [g(b) g(a)] [f(b) f(a)] g (x). Moreover, (a) = (b) = 0. Rolle s Teorem implies tat tere exists a < c < b suc tat (c) = 0, wic proves te result. If g(x) = x, ten tis teorem reduces to te usual mean value teorem (Teorem 8.33). Next, we state one form of l Hôspital s rule. Teorem 8.54 (l Hôspital s rule: 0/0). Suppose tat f, g : (a, b) R are differentiable functions on a bounded open interval (a, b) suc tat g (x) 0 for x (a, b) and Ten f(x) = 0, x a + g(x) = 0. x a + f (x) f(x) x a + g = L implies tat (x) x a + g(x) = L. Proof. We may extend f, g : [a, b) R to continuous functions on [a, b) by defining f(a) = g(a) = 0. If a < x < b, ten by te mean value teorem, tere exists a < c < x suc tat g(x) = g(x) g(a) = g (c)(x a) 0, so g 0 on (a, b). Moreover, by te Caucy mean value teorem (Teorem 8.53), tere exists a < c < x suc tat f(x) f(x) f(a) = g(x) g(x) g(a) = f (c) g (c).

25 8.8. * L Hôspital s rule 63 Since c a + as x a +, te result follows. (In fact, since a < c < x, te δ tat works for f /g also works for f/g.) Example Using l Hôspital s rule twice (verify tat all of te ypoteses are satisfied!), we find tat cos x sin x x 0 + x 2 = x 0 + 2x = cos x = x Analogous results and proofs apply to left its (x a ), two-sided its (x a), and infinite its (x or x ). Alternatively, one can reduce tese its to te left it considered in Teorem For example, suppose tat f, g : (a, ) R are differentiable, g 0, and f(x) 0, g(x) 0 as x. Assuming tat a > 0 witout loss of generality, we define F, G : (0, /a) R by Te cain rule implies tat F (t) = f F (t) = t 2 f ( t ( ), G(t) = g t ( ). t ), G (t) = ( ) t 2 g. t Replacing its as x by equivalent its as t 0 + and applying Teorem 8.54 to F, G, all of wose ypotesis are satisfied if te it of f (x)/g (x) as x exists, we get f(x) x g(x) = F (t) t 0 + G(t) = F (t) t 0 + G (t) = f (x) x g (x). A less straigtforward generalization is to te case wen g and possibly f ave infinite its as x a +. In tat case, we cannot simply extend f and g by continuity to te point a. Instead, we introduce two points a < x < y < b and consider te its x a + followed by y a +. Teorem 8.56 (l Hôspital s rule: / ). Suppose tat f, g : (a, b) R are differentiable functions on a bounded open interval (a, b) suc tat g (x) 0 for x (a, b) and g(x) =. x a + Ten f (x) f(x) x a + g = L implies tat (x) x a + g(x) = L. Proof. Since g(x) as x a +, we ave g 0 near a, and we may assume witout loss of generality tat g 0 on (a, b). If a < x < y < b, ten te mean value teorem implies tat g(x) g(y) 0, since g 0, and te Caucy mean value teorem implies tat tere exists x < c < y suc tat f(x) f(y) g(x) g(y) = f (c) g (c).

26 64 8. Differentiable Functions We may terefore write f(x) g(x) = [ f(x) f(y) g(x) g(y) g(x) g(y) g(x) [ g(y) ] + f(y) g(x) g(x). = f (c) g (c) ] + f(y) g(x) It follows tat f(x) g(x) L f (c) g (c) L + f (c) g(y) g (c) g(x) + f(y) g(x). Given ɛ > 0, coose δ > 0 suc tat f (c) g (c) L < ɛ for a < c < a + δ. Ten, since a < c < y, we ave for all a < x < y < a + δ tat f(x) g(x) L < ɛ + ( L + ɛ) g(y) g(x) + f(y) g(x). Fixing y, taking te sup of tis inequality as x a +, and using te assumption tat g(x), we find tat sup f(x) x a g(x) L ɛ. + Since ɛ > 0 is arbitrary, we ave wic proves te result. sup x a + f(x) g(x) L = 0, Alternatively, instead of using te sup, we can verify te it explicitly by an ɛ/3 -argument. Given ɛ > 0, coose η > 0 suc tat f (c) g (c) L < ɛ for a < c < a + η, 3 coose a < y < a + η, and let δ = y a > 0. Next, coose δ 2 > 0 suc tat g(x) > 3 ( L + ɛ ) g(y) for a < x < a + δ 2, ɛ 3 and coose δ 3 > 0 suc tat g(x) > 3 ɛ f(y) for a < x < a + δ 3. Let δ = min(δ, δ 2, δ 3 ) > 0. Ten for a < x < a + δ, we ave f(x) g(x) L f (c) g (c) L + f (c) g(y) g (c) g(x) + f(y) g(x) < ɛ 3 + ɛ 3 + ɛ 3, wic proves te result.

27 8.8. * L Hôspital s rule 65 We often use tis result wen bot f(x) and g(x) diverge to infinity as x a +, but no assumption on te beavior of f(x) is required. As for te previous teorem, analogous results and proofs apply to oter its (x a, x a, or x ± ). Tere are also versions of l Hôspital s rule tat imply te divergence of f(x)/g(x) to ±, but we consider ere only te case of a finite it L. Example Since e x as x, we get by l Hôspital s rule tat x x e x = x e x = 0. Similarly, since x as as x, we get by l Hôspital s rule tat log x x x = /x x = 0. Tat is, e x grows faster tan x and log x grows slower tan x as x. We also write tese its using little o notation as x = o(e x ) and log x = o(x) as x. Finally, we note tat one cannot use l Hôspital s rule in reverse to deduce tat f /g as a it if f/g as a it. Example Let f(x) = x + sin x and g(x) = x. Ten f(x), g(x) as x and ( f(x) x g(x) = + sin x ) =, x x but te it f (x) x g (x) = ( + cos x) x does not exist.

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