University Mathematics 2

Transcription

1 University Matematics 2 1 Differentiability In tis section, we discuss te differentiability of functions. Definition 1.1 Differentiable function). Let f) be a function. We say tat f is differentiable at = a if fa + ) fa) 0 eists. Te value of tis it is called te derivative of f at = a and is denoted by f a). We say tat f is differentiable on a, b) if f is differentiable at every point on a, b). Te first property of differentiable function is tat it must be continuous. Let s recall te definition of continuous function. Definition 1.2 Continuous function). Let f) be a function. We say tat f is continuous at = a if te it of f at = a eists and f) = fa). a We say tat f is continuous on a, b) if f is continuous at every point on a, b). Teorem 1.3. If f is differentiable at = a, ten f is continuous at = a. Proof. Suppose f is differentiable at = a. Ten f) a = fa + ) 0 = fa + ) fa)) + fa)) 0 fa + ) fa) = 0 = f a) 0 + fa) = fa) Terefore f is continuous at = a. ) + fa) However te converse of te above teorem is false. Tere eists function wic is continuous at a point but not differentiable at tat point. Here is an eample. )

2 University Matematics 3 Eample 1.4. Let f) = = {, if < 0, if 0 Ten 1. f is continuous at = f is not differentiable at = 0. Proof. Te graps of f) = and its derivative are sown in Figure 1. Figure 1: f) = 1. Te left and rigt-and its of f) = at = 0 are Tus we ave f) = ) = 0 f) + = + f) = 0 = f0) Terefore f is continuous at = 0.

3 University Matematics 4 2. Observe tat f) f0) 0 f) f0) = = = = are not equal. Tus f) f0) 0 does not eist. Terefore f is not differentiable at = 0. Eample 1.5. Let Find f ). f) = sin = { sin, if < 0 sin, if 0 Solution. Te graps of f) = sin and its derivative are sown in Figure 2. For < 0, we ave For > 0, we ave At = 0, we ave f ) = d sin ) = cos sin d f ) = d sin ) = cos + sin d f 0) = f) f0) 0 = sin 0 0 = 0 = 0 sin )

4 University Matematics 5 Figure 2: f) = sin Terefore Eample 1.6. Let cos sin, if < 0 f ) = 0, if = 0 cos + sin, if > 0 ) 1 sin, if 0 f) = 0, if = 0 Determine weter f) is differentiable at = 0. Solution. Te graps of f) and its derivative are sown in Figure 3. Since ) 1 sin 0 f) f0) = 0 0 ) 1 = sin 0

5 University Matematics 6 Figure 3: f) = sin ) 1 does not eist. Terefore f) is not differentiable at = 0. Te following eample is important because it sows tat te derivative of a differentiable function can sometimes be discontinuous. Eample 1.7 Function wit discontinuous derivative). Let ) 1 2 sin, if 0 f) = 0, if = 0 1. Find f ) for Find f 0) 3. Sow tat f ) is not continuous at = 0. Solution. Te graps of f) and its derivative are sown in Figure Wen 0, f ) = 2 sin ) ) 1 1 cos

6 University Matematics 7 Figure 4: f) = 2 sin ) 1 2. f f) f0) 0) = 0 ) 1 2 sin 0 = 0 ) 1 = sin 0 Since 1 ) sin 1 is bounded and = 0, we ave 3. Te it 0 ) 1 f 0) = sin = 0 0 f ) = 2 sin ) 1 cos )) 1 does not eist since cos 1 ) does not eist. Terefore f ) is not continuous at = 0.

7 University Matematics 8 Te differentiability at = 0 of te functions in te above eamples are summarized in te table below. Eample f) f) is continuous at = 0 f) is differentiable at = 0 f ) is continuous at = 0 Grap 1.4 Yes No Not applicable Figure sin ) Yes Yes Yes Figure sin Yes No Not applicable Figure 3 ) sin Yes Yes No Figure 4 Note. In all of te eamples above, we define f0) = 0.

8 University Matematics 9 2 Mean Value Teorem Imagine a veicle traveling on a road. Suppose at time t = a and t = b, te displacements of te veicle are fa) and fb) respectively. Ten te average velocity of te veicle is fb) fa) b a One may ask weter tere always eists a time t = ξ suc tat te velocity of te veicle is eactly equal to te average velocity. Rougly speaking, te mean value teorem gives an affirmative answer to tis question if we assume tat velocity is defined at any time between a and b. Te rigorous statement of mean value teorem is stated below. Teorem 2.1 Lagrange s mean value teorem). Let a, b be two real numbers wit a < b. Suppose f is a function suc tat 1. f is continuous on [a, b]. 2. f is differentiable on a, b). Ten tere eists ξ a, b) suc tat f ξ) = fb) fa) b a In te veicle eample above, f) is te displacement of te veicle and is te time. Ten f ) is te velocity of te veicle. One may wonder weter te following situation gives a counter eample to te above teorem. Suppose from t = 0 to t = 1, te veicle remains at rest and from t = 1 to t = 2, te veicle travels wit a velocity of 2 units. Ten te average velocity of te veicle is 1 but te velocity of veicle is never equal to 1 from t = 0 to t = 2. Tis does not contradict te teorem because we assumed tat f is defined on a, b) but velocity is not defined at t = 1. Before we give te proof of te Lagrange s mean value teorem Teorem 2.1), we state two variants of mean value teorem. Te first one is a special case of te Lagrange s mean value teorem and te second one is a generalization of it. Teorem 2.2 Rolle s teorem). Let a, b be two real numbers wit a < b. Suppose f is a function suc tat

9 University Matematics f is continuous on [a, b]. 2. f is differentiable on a, b). 3. fa) = fb) Ten tere eists ξ a, b) suc tat f ξ) = 0 Teorem 2.3 Caucy s mean value teorem). Let a, b be two real numbers wit a < b. Suppose f and g are functions suc tat 1. f and g are continuous on [a, b]. 2. f and g are differentiable on a, b). 3. g ) 0 for any a, b) Ten tere eists ξ a, b) suc tat f ξ) g ξ) = fb) fa) gb) ga) Note tat Rolle s teorem is a special case of Lagrange s mean value teorem. If we take g) = in te Caucy s mean value teorem, we obtain Lagrange s mean value teorem. So Caucy s mean value teorem is a generalization of Lagrange s mean value teorem. First we prove Rolle s teorem. Te following teorem will be needed for tis purpose. Teorem 2.4 Etreme value teorem). Suppose f is a function wic is continuous on a closed and bounded interval [a, b]. Ten tere eists p, q [a, b] suc tat fp) f) fq) for any [a, b]. In oter words, f is bounded and attains bot its maimum and minimum values. Te rigorous proof of te Etreme value teorem requires some argument in analysis and is omitted. We also need te following teorem wose proof is very easy.

10 University Matematics 11 Teorem 2.5. Let a, b, ξ be real numbers suc tat a < ξ < b. Suppose f is differentiable at = ξ and eiter f) fξ) for any a, b), or Ten f ξ) = 0. f) fξ) for any a, b). Proof. Suppose f) fξ) for any a, b). Te proof for te oter case is more or less te same. For any < 0 wit a < ξ + < ξ, we ave Now f ξ) eists and we ave fξ + ) fξ) 0 f ξ) = 0 fξ + ) fξ) 0 On te oter and, for any > 0 wit ξ < ξ + < b, we ave Tus we also ave Terefore we ave f ξ) = 0. fξ + ) fξ) 0 f ξ) = 0 + fξ + ) fξ) Now we are ready to prove Rolle s teorem. Proof of Rolle s teorem. Suppose f is continuous on [a, b], differentiable on a, b) and fa) = fb). By Etreme value teorem Teorem 2.4) tere eists p, q [a, b] suc tat 0 fp) f) fq) for any [a, b]. If p a, b), i.e., p a, b, ten we take ξ = p. If q a, b), ten we take ξ = q. If bot p and q do not lie on a, b), ten f is a constant function

11 University Matematics 12 and we take ξ to be any point in a, b). In any of te above cases, we ave f ξ) = 0 by Teorem 2.5. Net we use Rolle s teorem to prove Lagrange s mean value teorem. Proof of Lagrange s mean value teorem. Let g) = f) Te function g) is constructed so tat fb) fa) gb) ga) = fb) b a = fb) fa)) = 0 fb) fa). b a ) b fa) fb) fa) b a) b a ) fb) fa) a b a Applying Rolle s teorem to g) on [a, b], tere eists ξ a, b) suc tat g ξ) = 0 wic means f ξ) fb) fa) b a = 0 and te proof of Lagrange s mean value teorem is complete. It is well known tat a function wit non-negative derivative is monotonic increasing. We may use Lagrange s mean value teorem to give a rigorous proof of tis statement. Teorem 2.6. Let f) be a function wic is differentiable on a, b). Suppose f ) 0 for any a, b). Ten for any, y a, b) wit < y, we ave f) fy). Proof. Suppose f ) 0 for any a, b) and, y a, b) wit < y. By Lagrange s mean value teorem, tere eists ξ, y) suc tat fy) f) = f ξ)y ) wic is non-negative since f ξ) 0 and y > 0. Tis completes te proof of te teorem.

12 University Matematics 13 Proof of Caucy s mean value teorem. Let f and g be functions wic are continuous on [a, b] and are differentiable on a, b). Suppose g ) 0 for any a, b). First of all ga) gb), for oterwise g ξ) = 0 for some ξ a, b) by Rolle s teorem wic violates our assumption on g. Tus we may let Ten b) a) = ) = f) fb) fa) gb) ga) g) ) ) fb) fa) fb) gb) ga) gb) fb) fa) fa) gb) ga) ga) = fb) fa)) = 0 fb) fa) gb) ga)) gb) ga) Applying Rolle s teorem to ) on [a, b], tere eists ξ a, b) suc tat ξ) = 0 f fb) fa) ξ) gb) ga) g ξ) = 0 f ξ) = f ξ) g ξ) = fb) fa) gb) ga) g ξ) fb) fa) gb) ga) Note tat g ξ) 0 since we assumed g ) 0 for any. Tis completes te proof of Caucy s mean value teorem.

13 University Matematics 14 3 L Hopital Rule In tis section, we study an application of Caucy s mean value teorem wic gives a powerful tool to evaluate its. Teorem 3.1 L Hopital rule). Let f) and g) be functions and a [, + ] Here a can be or + ) wic satisfy Ten 1. f) and g) are differentiable for any a. 2. a f) = a g) = 0, or a f), a g) = ± ). 3. g ) 0 for any a. 4. a f ) g ) = l f) a g) = l Proof. For simplicity, we assume a ± and f) = g) = 0. a a Redefine f and g, if necessary, so tat fa) = ga) = 0. Ten for any > a, we ave 1. f and g are continuous on [a, ]. 2. f and g are differentiable on a, ). 3. g y) 0 for any y a, ). By Caucy s mean value teorem Teorem 2.3), tere eists ξ a, ), ere ξ depends on ), suc tat Terefore f ξ) g ξ) f) fa) = g) ga) = f) g) f) a + g) f ξ) = a + g ξ) f ξ) = ξ a + g ξ) = l

14 University Matematics 15 Note tat ξ a as a since ξ a, ). Te same is true for te left-and it and te proof of L Hopital rule is complete. Let s sow ow L Hopital rule still works wen a = +. Define F y) = f 1) and Gy) = g 1 ). Ten y y F y) = f 1) y and G ) = g 1) y y 2 y 2 Applying L Hopital rule to F ) and G) at = 0, we ave f) + g) F y) = y 0 + Gy) F y) = y 0 + G y) f 1 y ) y 2 y 0 + g 1 y ) y 2 = f 1 = ) y y 0 + g 1) y f ) = + g ) Oter cases of L Hopital rule can be proved in similar ways. Eample 3.2. Evaluate te following its. 1. e 3 e sin 2. e 32 1 cos cos ln1 cos ) ln sin ln ) 4. + ln ) Solution.

15 University Matematics Since e3 e ) = sin = 0, we may apply L Hopital rule and get e 3 e sin = 3e 3 + e cos = = 4 2. Applying L Hopital rule two times, we get e 32 1 cos cos 2 = = = = 2 6e 32 sin + 2 sin e e 32 cos + 4 cos Since ln1 cos ) = ln sin =, + + we may apply L Hopital rule and get ln1 cos ) + ln sin = + = + sin 1 cos cos sin sin 2 1 cos ) cos 1 cos 2 = + 1 cos ) cos 1 + cos = + cos = 2 4. Since + ln ) = + ln42 + 1) = +

16 University Matematics 17 we may apply L Hopital rule and get ln ) + ln ) = + = )15 2 2) ) )15 2 ) = ) 2 3 = 3 2 Te its in te above eamples are of te forms 0 0 and. L Hopital rule can also be used to evaluate its of te forms 0, 0 0, 0 and 1. Eample 3.3. Evaluate te following its ln ) 1 ln 4. cos ) 1 2 Solution. 1. ln ln = = = + ) = 0

17 University Matematics Terefore Terefore Terefore ) ln + ) ln ) 1 ln ln = + ln ) = + ln = 0 = e 0 = 1 + = + ln2 + 1) 1 ln ) ln 2 + 1) = + ln = + = + = ln2 + 1) 1 ln ) = e 2 cos ) 1 2 ) = lncos ) 1 2 ) ln cos = 2 tan = 2 sec 2 = 2 = cos ) 2 = e 1 2