Only = is in (2 32), so satisfies the conclusion of Rolle’s Theorem. √ SECTION 4.2 THE MEAN VALUE THEOREM ¤ 317 2−0 ⇔ 4 is a rational function that is continuous on its domain, with , then () = () = 0. Since is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that 1 1 √ 1 1 0 (−∞ 2 . line has thethe same √ =0) ∪ (0 ⇔ ∞), so=it1is continuous ⇔ = 1.on tangent 0 The secant 0and 2 domain line and is differentiable on 2 2 . Also, that the given equation 2there is a2number in ( ) such that ()2= 0. But () = 3 + 0. This contradiction shows 2 1 5 − 1 0 can’t have root.2 − 1 = 0 ⇔ = ±1. Only 1 is in 1 2 so 1 satisfies the are 2parallel. = 2 two = distinct (2). real () roots, = 0 so ⇔it has exactly = one 0 ⇔ 2 2 (4) − (0) 1 2 = 15. () = , [0 4]. 1 0 () ⇔ √ −= 1 8. () = + 1, 2 2 . 0 ()4= −10− 12 = 2 2 . Section 4.2 The Mean Value Theorem 21. conclusion Let () =of3Rolle’s − 15Theorem. + for in [−2 2]. If has two real roots and in [−2 2], with , then () = () = 0. Since (2) − (0) −2 − 1 2−0 2 0 0 0 2 ¤ not 316 does CHAPTER APPLICATIONS OF DIFFERENTIATION −24= 0. −2 contradict Rolle’s Theorem, since (0) does not exist, and so isiscontained not differentiable on we (−1 1).|| 2, so 2 4. () Now () = 3 − 15. Since is in ( ), which in [−2 2], have such that 1− 1− ⇔ − = ln ⇔ − = 2 32 − 15 3 · 4 − 152 = −3 0. This contradicts () − () 2() = 0, so the 0on can’t ⇔ have two realnot roots follows that 4].= tan is 0continuous on[1 and differentiable 14. = 10. It () () = tan ., [1 (0) = 0 = tan = 4] (). 0 () = sec (1 ≥ 4). 1, so 0 ()given = 0 equation has no solution. This does 1+−2−2 − line and the tangent line are in =[−2 − ln2]. Rolle’s ≈has 08386. The secant 0 real Hence, it at most one root in [−2 2]. on (0 ). (Also, () is not continuous contradict 22 − 1Theorem, since 2 does not exist, and so√ is not differentiable √ 2 2 3 3 ⇔ ( + 2) = = 18 ⇔ = −2 ± 3 2. −2 + 3 2 ≈ 224 is in (1 4). 2 parallel. ( +[0 2)].)4 4 − 1 on() 22. = + 4 + . Suppose that () = 0 has three distinct real roots , , where . Then − 23 0 − 23 differentiable the polynomial 2]. ] and on( implies is ano number inThis ( ) ⇔ =Theorem 16.9. () = () = () = 1 −, [0 .is continuous (−1) = 1on −[ (−1) =⇔ 1 − − 1 = 0= = (1).), Rolle’s 0 () − 23 −13 , so 0that () there = 0 has solution. 2 differentiable − 0 and (4) Theorem (0) 1 are 1polynomials −2 () = √ () () 0. 2]. By there numbers onwith and 1 are 0 [0 11. () 22 −=3 + 1,= is−continuous on [0 2] and (0 2) since continuous and −3 0 Rolle’s 15. 4]. 17. () = (,−[0 3)−2 ⇒()= () = −2 ( − 3)⇔ . √ (4)=− (1) =1⇔ 0 ()(42 − 1) ⇒ 1 2 − =2 ·3 ⇒ 2 3 4−0 4 2 1 (−2) ( − 3) 0 () − () (2) (0) 3However −1 )= 0 ( = 0 must ⇔ have4 at − least two real−solutions. and 0 = 0 (1on 02 ), so () differentiable R. 3 = = = 1 ⇔ 4 = 4 ⇔ = 1, which () = √ 1 1 3√ =−1.The secant line and the tangent 2 − 0 line 2 3 =0 −6 ⇔ 3⇒ = 1− 3)⇔ 3 =−8 2 −2 ⇒ = 1, which = ( ⇒ − 3 = is not in the open interval (1 does at not ()This can have 2 =3 4 + 4 = 4( + 1) = 4( + 1)( − + 1) has as its only real solution = −1. Thus, 4). 20 = (() 4 is in (0− 2)3) are parallel. most two real roots. Value Theorem since is not continuous at = 3. contradict the Mean 12. () = 3 − 3 + 2, [−2 2]. is continuous on [−2 2] and differentiable on (−2 2) since polynomials are continuous and (4 ). 23. (a) Suppose that a cubic polynomial () has roots 1 2 3 4 , so (1 ) =1 (2 ) = (3 ) = 2 −(2 − 1)() if 2 − 1 ≥ 0 (2) − 3− 2 if 4 − ≥ 02 −2 if 12 () − (−2) 0 2 2 0 −2 on 3 with = =and 1 ⇔ 3 = are − 3==− =4 and ⇔ 18. differentiable ()By = Rolle’s 2− − |2 −R.1|=0() ⇒ () = (2) − (0)1⇔ 1 2 , 2 − , , Theorem there numbers 1 2 3 1 1 2 3 3 3 4 − − 1)] ⇔ 4 16. () = , [0 2]. ()2=− [−(2 = 0 2 −1(−2) 2 if 12 if 2− −1 +⇔ 2 if 2 2 − 0 2 318 ¤ 0CHAPTER 40 APPLICATIONS OF DIFFERENTIATION (2 ) =2 0 (3 ) = 0. Thus, the second-degree polynomial 0 () has three distinct real roots, which is 2 4(1 ) =−2 −2 in (−20 2). 0 √ ⇔ = ± = , which 1are − −3 both − = ()(3 − 0) ⇒ − (3) −31 (0) − 1 = () · 3 ⇒ 0 () = − 43 [not ± 2]. This does not contradict the Mean ⇔ − ⇔ = 0 3 = ln 26. If 3 impossible. ≤ () 2 ≤ 5 for all , then by the 2 Mean Value Theorem, (8) − (2) = 0 () · (8 − 2) for some in [2 8]. Value Theorem since is not differentiable at = 12 . −2 − ,by 13. ( (b) () =prove 1−2 [0 3]. all,isthat continuous and differentiable on at R,on so(2 it is continuous on [0 3][2 and 3). is differentiable for particular, isdegree differentiable androots. continuous 8].differentiable Thus, thehypotheses We induction ainpolynomial has most This ison certainly true for =on 1.(0 Suppose = − ln ≈ 08386.so,The secant lineof and the tangent line are 8)real 2 −6 19. of Let that () = 2 . Then (−) − (8) 10 0− (0) = 01() and is the sum of1the 2 and the () −+cos () Since − 0.−6Since − polynomial −6 the Mean Value Theorem satisfied.) and (2) =6 3 ≤ 0 () ≤ degree 5, it follows that for are all−2 polynomials and let of −2 = −2 −2 ⇔ −2 = ln + 1. Suppose ⇔ that () has =the result is true ⇔ = of degree ⇔ =() be a polynomial 0 () parallel. − 3 − 0 6 6 0 function cos , is continuous and differentiable for all . By the Intermediate Value Theorem, there is a number trignometric () ≤+6 1· 5real 18say ≤ (8) 30 6 · 3more ≤ 6than roots, 2(2) ≤ · · · +1 +2 . Then (1 ) = (2 ) = · · · = (+2 ) = 0. 1 − 3 ⇒ −6 1 1− 1 1 −2 −2 −3 0 =By − ln ≈ which in3)(0 3). () in (− 0)−such that⇒ () =() 0. Thus, the given has at equation has distinct real roots and there are= real 1equation (4) +1 1there 0 ()(4 1real root. If ⇒ the +2 and 17. =Rolle’s ( 3)Theorem 0897, −2numbers (is− . Value − with (1)least = one −2+1 = · 3 ⇒ 21) 2 27. Suppose Theorem isa−number 0+1 12 with 2 that such6 a function exists. By the Mean (−2) ( − 3)3 with , () =+1 Sincethe isth continuous on [ ] and differentiable Rolle’s Theorem implies that =then · · ·= 0 ( ) ==0.0.Thus, degree polynomial 0 () has at leaston( + 1),roots. This contradiction shows 0( 1 5 () )(2) − (0) 0 5 −6 4.32 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 319 3 this is impossible sinceSECTION = . But 30 () = () ≤ for all , so no such function can exist. = ( −2 3) = −8 0 ⇒ − 3 = −2 ⇒ = 1,2which is not in the open interval (1 4). This does not 23 − 0⇒ 4 that (a− 3)2016 number has + real roots. there is atin ( ) such thatReserved. () May = 0. 0 () = 2or− sin 0 since sin accessible ≤ 1. This contradiction shows that the most c() ° Cengage Learning. All1 Rights not But be scanned, copied, duplicated, or posted to a publicly website, in whole or in part. √ −1 − 2 arctan + 2 . Note that the domain of is [0 ∞). Thus, 35. Let () = arcsin contradict roots, is() notsocontinuous one ==3. can’t have two distinct it has exactly root. 28. given Let equation = the − .Mean NoteValue that ()since =real (), = () − at () 0. Then since and are continuous on [ ] and +since 1Theorem 24. (a) Suppose that ()= () = 0 where . By Rolle’s Theorem applied to on [ ] there is a number such that is , there is differentiable on ( satisfies Value1Theorem. Therefore, ( and + 1)thus − ( − 1) the2assumptions 1 1 1 of the Mean 1 ), so =20.− 3 and (2= − −1 1) + 1 (0) ≥ 3√ − 20. Let () +.0 () Then (−1) 1if − 20 − and 0.2 Sinceif is−≥ the and if thenatural √2 sum of a polynomial = 0. · 0√= 1= == 0 () −2 12 2 0 2 1 + 2 ( + 1) ( + 1) 18. a () = 2 −with |2 − ⇒ 0 0 0 () = ( + 1) = () − () = = 1 1|− 1 2such ()( − ). Since () 0, ()( − ) 0, so number = that () 1 2 if in − [−(2 − differentiable 1)] if 2 −for 1 0 1 +Intermediate 2 if Value 1that − () = 2 exponential is continous and the Theorem, a number 2 (b) Supposefunction, all..ByBy Rolle’s Theorem applied to () onthere [ ]isand [ ] there are + 1 () = () = 0 where SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 319 () − () = () 0 and hence () (). 0 0 0 4 0 0 0 (−1 0) such that ()(3 () 0.0) Thus, least one real root. If the has distinct real roots and is a Then (3)numbers − (0)== on − − By 1equation () 3 at =() [not ±equation 2].find This notcontradict Mean = and ⇒ the −3 given with =() = has 0· and 0. By Rolle’s Theorem applied () [the ] there 3 [0 ∞). () (0 ∞) continuity of⇒ ,() () == − on To ,does we to let = 0on ⇒ − 1 by Theorem 5.√ 35. Consider Let () the = arcsin + 2 1. Noteand thatdifferentiable the domain of is Let [0 ∞). 29. function () = − sin2,arctan which00 iscontinuous onR. be Thus, a number such that 0 2. number −with not differentiable such + 1+ Value Theorem is at − ==0. arcsin(−1) 2since arctan(0) = that ⇒() − .0 + 2 = 0 = . Thus, () = 0 ⇒ 2 2 2 Then is continuous differentiable on (0 ). By the Mean Value Theorem, there is a number in (0 ) such that 1 on [0 ](and + 1) − ( − 1) 1 1() 1 2 0 Suppose √ 1 +0cos that is . times differentiable on R− · (0) +√ 1 distinct roots.−Then sum has at least one real2 root. = −2 −2 = 0. ()() = = (c) 19. Let Then (−) 1and 0has and ==cos 1√ real 0. Since is√ the of sin the polynomial andtook the 2 − 0− 2Learning. arctan ( − . 0); = 2 Allthat ()( − is, sin = (cos )(). Now 1 for 0 2, so 1or·inpart. = . We arcsin () − ° (0) = c 2016 1 + Cengage Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible 2 ( + 1) ( + website, 1) inwhole 2 + 1) +1 −1 1function − 0 trignometric cos continuous and all . By(1 the Intermediate Value Theorem, + 1 , 25. to Bybethe Value 2), (4) −sosin (1)= ()(4 −1)satisfying forfor some ∈ But for every ∈Theorem, (1 4) we there have is a number anMean arbitrary number in (0is differentiable for all 0 4). 2. 10 1 ( 1 ) − (0) 65 − 50 6 0)= such that = Thus, the given equation has real root. the equation has distinct and 0 (− 0 () = . If Then == 0 real 36. Let () be the of≥the car the hours 2:00 PMand . At 2:10 PM, = one = = 90. By () 2. Putting (0 () 20. into above substituting (1) 10, get in Then ≥() velocity on ∞) by Theorem 5.after Byequation continuity of ,atleast () on = [0 ∞). To find , ⇒ roots 1 we let 60 6 we − 0 () − 16(−) 0 6 = 30. satisfies the conditions for the Mean Value Theorem, so we use this theorem on the interval [− ]: () arcsin(−1) with ,−then on [ and differentiable ), Rolle’s 0 () = + 0 − is−16. (−)implies that =0.+ Since −0+ 2= =16. 0 =]So . Thus, ()possible = 0on ( ⇒ ()(4 − 1)() == 10 3⇒ () is ≥ continuous the smallest value of (4)Theorem (4) = (1) +2arctan(0) 210 + 3 · 2 the Mean Value Theorem, there2 is a number such that 0 16 with 0 () = 90. Since 0 () is the acceleration at time , 0 0 there is a in ( such () (−) = 0. But (). () = 2 − sin 0 since sinabove ≤ 1.equation, This contradiction shows that the for some number ∈ (− But)since is odd, = − Substituting this into the we get √ that − 1 ). 2 c 2016 Cengage ° Rights Reserved. May not90 be scanned, 2Learning. arctan arcsin −PM .is exactly the acceleration = hours afterAll2:00 kmhcopied, . or duplicated, or posted to a publicly accessible website, in whole or in part. 2 + 1 can’t have twodistinct given equation real roots, so it has exactly one root. () () + () = 0 () ⇒ = 0 (). 2 and () be the position functions of the two runners and let () = () − (). By hypothesis, 37. Let () 1 (of ) − (0) 1 65 − 50 10 6 a polynomial 20. () be = the 3 + . Then (−1) −1 + 12:00 0PM and (0) = PM 1 sum the = natural Then = and 36. Let Let () velocity of the car =hours after . At 2:10 , 0.=Since= is. the 90. By 1Theorem, (0) = (0) −(0) () = () = 0, where the finishing Then by1 the Mean Value 31. Let () = sin and=let0and . Then ()− is () continuous on []is and differentiable ( ). By the Mean Value Theorem, 60 time. 6 on − 0 6 6 exponential function, is continous and differentiable for all . By0 the Intermediate Value Theorem, there is a number in 0 ) = (cos )( − ()(− ). isThus, there is a number ∈ ( ) withissin − sin 0 such = () () − − () (0) = 1 with the Mean Value,Theorem, a number ()==(0) 90. Since 0 () the acceleration at time , () = 0, so 0 () = 0. Since there is a time with 0 there , such that () =that 0 6. But (−1 0) such that () = 0. Thus, the given equation hasat−least 0 one real root. If the equation has distinct real roots and |sin − sin | ≤ |cos | | − | ≤ | − |. If , then |sin − sin | = |sin − sin | ≤ | − | = | − |. If = , both the acceleration hours after 2:00 PM is exactly 90 kmh2 . 0 () = 0 () − 0 () = 0, we have 0 () = 0 (). So at time , both runners have the same speed 0 () = 0 (). sides of the inequality are 0. 37. Let () and () be the position functions of the two runners and let () = () − (). By hypothesis, 38. Assume that 0 is differentiable (and hence continuous) on R and that 0 () 6= 1 for all . Suppose has more than one fixed 0 32. Suppose that Cengage () = . Let All () =Reserved. , so () = . Then, by Corollary 7, () = () + , where isorainconstant, so c 2016 ° not be scanned, copied, orduplicated, or posted to atime. publiclyThen accessible in whole (0) = (0) − (0) Learning. = 0 and Rights () = ()May − () = 0, where is the finishing bywebsite, the Mean Valuepart. Theorem, point. Then there are numbers and such that , () = , and () = . Applying the Mean Value Theorem to the () = + . () − (0) . But () = (0) 0, so 0 () = 0. Since there is a time , with 0 , such that 0 () = () = − () − function on [ ], we find that there is a number in ( )0such that 0 0 () = 2 . But then 0 () = = 1, 0 − 0 0 1 −0 () = (1 + 1)0 = −1 − 2, so 33. For 0, () = (), so () = (). For 0, () = (1) = −1 and 0 0 0 0 0 0 0 () =0 () −0 () = 0, we have0 () = (). So at time , both runners have the same speed () = (). contradicting assumption that () 6= 1offor every realannumber shows0)that our∞)] supposition was wrong, thatthat is, that (). However, thedomain () is not interval.[itThis is (−∞ ∪ (0 so we cannot conclude again () =our there is a time , with 0 , such that 0 () = () − (0) . But () = (0) = 0, so 0 () = 0. Since −0 0 () = 0 () − 0 () = 0, we have 0 () = 0 (). So at time , both runners have the same speed 0 () = 0 (). 38. Assume that is differentiable (and hence continuous) on R and that 0 () 6= 1 for all . Suppose has more than one fixed point. Then there are numbers and such that , () = , and () = . Applying the Mean Value Theorem to the function on [ ], we find that there is a number in ( ) such that 0 () = () − () − . But then 0 () = = 1, − − contradicting our assumption that 0 () 6= 1 for every real number . This shows that our supposition was wrong, that is, that cannot have more than one fixed point. 4.3 How Derivatives Affect the Shape of a Graph 1. (a) is increasing on (1 3) and (4 6). (b) is decreasing on (0 1) and (3 4). (c) is concave upward on (0 2). (d) is concave downward on (2 4) and (4 6). (e) The point of inflection is (2 3). 2. (a) is increasing on (0 1) and (3 7). (b) is decreasing on (1 3). (c) is concave upward on (2 4) and (5 7). (d) is concave downward on (0 2) and (4 5). (e) The points of inflection are (2 2), (4 3), and (5 4). 3. (a) Use the Increasing/Decreasing (I/D) Test. (b) Use the Concavity Test. (c) At any value of where the concavity changes, we have an inflection point at ( ()). 4. (a) See the First Derivative Test. (b) See the Second Derivative Test and the note that precedes Example 7. c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. ° 2
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