《微积分》课程教学课件（Calculus）04. Applications of Derivatives

ExercisesStrategy for finding critical pointsFind the critical points, domain endpoints, and absoluteextreme values for each function.1. Let y'=0y = p - 4)69. y = x3P(x + 2)71. y=+V4 yV-2. Let the denominator of y' be 04-2x,x≤1x<074. 3-(3-43. Check the lef-hand derivative and right-hand73. y =(3 + 2 , x ≥ 0(x + 1, x>1derivative at dividing point, If they are not similar, the2x+ 4,X≤?.derivative at that point doesn'texist, and the dividingx2 + 6r - 4, X>1point is critical point.++-1++4*5176.y6r+&x,>1Slide 4-31Slide 4-32Exercises70. y=2/(2x)+号x-1/(2-4) 8元Exercises77, Let f(x) = (x 2)0,3/a., Does '(2) exist?derivativecrit. pt.valueb, Show that the only local extreme value of j occurs at x = 2.e. Does the result in part (b) contradict the Extreme Value Theorem?3x=-1minimurundefine0X=0ca77. (a) No, since f(x) = (x 2)-1/3, which is undefined at x = 2.3X= 1minim(b) Thederivativisdeined andnoroforall2. Also, [2) =andf(x) >forall275.(c) No, f(x) need not bave a global maximumThe lkf-hand derivative at x=1 is [-2r-2]., =-428 =2, x ≤介becausets domainallreal numbeAy retrictioY:RThe rgh-hand deriatiog x=l i [-2+6]., =42#+6, x>1-offtoaclosed intevalfthfom [ab] wouldhaveestemum-raloeerit, pl, derivativeNot similarboth maximum vale and a misimum value on the interval.maximum5= 1andeiaedlocal min1Slide 4- 33Slide 4- 344.21.Rolle'sTheoremTheMeanValueTheorem andDifferentialEquationsSlide 4-38

2016/11/15 6 Slide 4 - 31 Strategy for finding critical points 1. Let y’=0 2. Let the denominator of y’ be 0 3. Check the left-hand derivative and right-hand derivative at dividing point. If they are not similar, the derivative at that point doesn’t exist, and the dividing point is critical point. Slide 4 - 32 Exercises Find the critical points, domain endpoints, and absolute extreme values for each function. Slide 4 - 33 Exercises 70. 75. The left-hand derivative at x=1 is   1 2 2 4 x x      The right-hand derivative at x=1 is   1 2 6 4 x x     Not similar Slide 4 - 34 Exercises 77. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.2 The Mean Value Theorem and Differential Equations Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 36 1. Rolle’s Theorem

f(c)=(two linesaretwokeyparallelingredientsTHEOREM3Rolle's TheoremOSuppose that y - y(x) is Continuouut every point of the closed interval [a, b]anddiferentiablot everyponrorits interior (a,b).f(a) = (b),nthen there is at least one number cin (a, b) at whichI'(c) 0.Tes)-0eaFIGURE4.10Rolle's Theorem says that= f(x)a differentiable curve has at least onehorizontal tangent between any two pointswhere it crosses a borizontal line. It mayE1have just one (a), or itmay have more (b)(b)Slide 4-37Slide 4-38CounterexampleThe hypotheses of Rolle'stheoremare essential iftheyfail at evenone point, thegraph may not haveahorizontaltangent2.Mean ValueTheorem(b) Dicor(e) Connss on fa,b] but no(a) Diseatarinteriorpoistof [a, b]difSerentiableatan intericmdpoinl of [a, b]poinrThere may be no horizontal tangeot if the hypotheses of Rolle>Theorem do not holdFIGURE4.11Slide 4- 39Slide 4- 40A Physical InterpretationTangent parllel to chordWe can think of the mumber (f(b) f(a)/(b a) as the average change in f over [a, b)and f(c) as an instantaneous change.Thenthe Mean ValueTheorem saysthat at some inteoThe Mean Value Theorem isrior point the instantaneous change must equal the average change over the entire interval.Slopef(c)BRolle' Theorem on a slant.The Mean Value Theoremf(b) f(a)THEOREM4The Mean Value TheoremSlopecormects the average rale.b-aSuppose y f(x) is continuous on a closed interyal [a, b] and dfferentiable onof change of a function overthe interval's interior (a, b), Then there is at least one point c in (a, b) at whichan interval with thef(b) f(a)Xinstantaneous rate of=f'(c)(1)00change of the function at ay-fu)point.FIGURE4.13Geometrically, the MeanThe Mean Value Theorem do not require f to beValue Theorem says that somewheredifferentiable at eithera or b, continuity at a and bbetween A and B the curve has at least oneis enough.tangent parallel to chord AB.Slide 4- 41Slide 4- 42

2016/11/15 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 37 two lines are parallel Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 38 two key ingredients Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 39 Counterexample The hypotheses of Rolle’s theorem are essential. If they fail at even one point, the graph may not have a horizontal tangent. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 40 2. Mean Value Theorem Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 41 The Mean Value Theorem is Rolle’ Theorem on a slant. The Mean Value Theorem connects the average rate of change of a function over an interval with the instantaneous rate of change of the function at a point. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 42 The Mean Value Theorem do not require f to be differentiable at either a or b, continuity at a and b is enough

Which ofthe functions in Exercises 9-14 satisfy the hypotheses of theExercisesMean Value Theorem on the given interval, and which do not? GiveAPhysical Interpretationreasons for your answerss"f(n)9. (x) = x2/3, [-1, 8]400TheMeanValueTheorem(8, 352)10, (x) -x45,[0, 1]says that at some interiorsatisfy3320point the instantaneous① fx) = V(I -x),[0, 1]240change must equal thesin160x<0average change over themf(x)=1+f(0)?)=At this point,entire interval.Othe car's speedx= 080Discontinuous atx(was 30 mph.16. For what values of a, m, and b does the function5Time (sec)=3,b=4,3X=0melf(s) =x2 + 3r + a,0<x<1FIGURE4.18Distance versus elapsed(mr +b,1 ≤x≤2time for the car in Example 3.satisfy the hypotheses of the Mean Value Theorem on the interval[0, 2]?lide 4- 4310e 4- 44+y=r+cC2.Corollary2 says thatfunctions have identicalderivatives on an intervalCOROLLARY1If y'(x) = 0 at each point x of an open interval (a, b), thenC=(only if their values on theJ(a) =C for all xe (a, b), where Cis a constant.interval have a constantdifference.COROLLARY2If f(z) = g (x) at each point x in an open interval (a, b), thenthereexists a constantCsuch that J() g(x)+ Cforall e(a,b),That isFIGURE4.19From a goomtetrie poing is a constanton (a,b)of view, Corollary 2 ofthe Mean ValucTheorem says that the graphs of functionswith identical derivatives on an interval candiffer only by a vertical shift there. Thegraphs of the funections with derivative 2rare the parabolas y = x + C, shoiwn herefor selected values of Cslide 4- 45Slide 4- 48Ercis8,fidal psbefuconswitegivendrivtiExercisesh.y'-r33.aJ1e y'-r'c y =32 + 2 134. a. y' = 2xb.y-2x-11ey-s+#35. ay1ay-l-1b.y-36. 8. )c. y'.V2VVx3.Differential Equationsmy=f+c09=f+33 () y=f+c(6) y =2Vi +C36 (a) y=↓-1/ =y=/n+Cny= Vi+c(0) y=23-2V/h+c34 (a) y=+c() y=-$+C(0) y=+x+CSlide 4- 47Slide 4.48

2016/11/15 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 43 A Physical Interpretation The Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval. Slide 4 - 44 Exercises satisfy 0 lim ( ) 1 (0) x f x f     Discontinuous at x=0 a=3,b=4, m=1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 46 Corollary 2 says that functions have identical derivatives on an interval only if their values on the interval have a constant difference. Slide 4 - 47 Exercises 33 34 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 48 3. Differential Equations

In Exercises 39-42, find the function with the given derivative whoscExercisesgraph pases through the pointP.39. (r) = 2r 1,:P(0, 0)A differential equation is an equation relating anunknown function and one or more of its derivatives. + 2, P(1, 1)40. g (x) =39.(x)=2x+C;0=(0)=020+CC=0→x)=xAfunctionwhosederivativessatisfyadifferentialequationiscalledasolutiontothedifferentialequation.40,g(x)=-=+r+Exampless(0)=4.9t2g(1)=1+1+c=1e=-1 g(r)=-'+-11.d's/dr=9.8m/sec2dy/dx=sinxy=-COSx+3lide 4- 49Slide 4- 50Finding Postion from Velocity or AccelerationExercisesExercises 4750 give the aceeleration a ds/dr, initial velocily,ExercisesExercises 43-46 give the velocity w = ds/dr and initial position ofaand initial position of a body moving on a coordinate line. Find thebody moving along a coordinate line. Find the body's position atbody's position at timet.timer48. a = 9.8, (0) = 3,(0) = 043. # = 9.8r + 5, (0) = 1049, a = 4 sin 2), (0) = 2, s(0) = 344. w = 32t 2, s(0.5) = 445. = sin #,(0)= 050, 4cos(0) = 0, s(0) = 146cos(）148&=9.8=→V=9.8t+CiatV=-3and1=0wehaveCy=-39y=9.8t3#1=4.913+Cyj 1#=0and t=0we bareCg=0→$=4.912_3t = 4.3+ 5t + 1043,v== 9.8t+5 =m4.9+5t+C at#=10 and t=0 we bave4=10498 =4 sin (2t) = Y=2 cos(2t) +Cji at y= 2 and t = 0 we bave Cy = 0.y =2 cos(2t)44Y==32120=1022++C:81*=4and1haveC=148=16132++1 s = sin (2t) + Cgi at s= 3 and 1 = 0 we bave Cy = 3 s =sla(2t) 345i(nt)--+co(n)+= and twea -()5e -()→-()+Cjat y=0 nd t= we haneC,=0=V=ain()ds_22eos(二0)=(0)=sin(二0)+Cs=co()+Cg; at's=1 and t= 0 we have Cg =0+==cos()46T :dt()= sin2#+=1c=1s(0)=si20+1Slide 4- 51Slide 4-524.31.FirstDerivative Test forIncreasing Functions andDecreasing FunctionsThe Shape of a GraphSlide 4-54

2016/11/15 9 Slide 4 - 49 A differential equation is an equation relating an unknown function and one or more of its derivatives. A function whose derivatives satisfy a differential equation is called a solution to the differential equation. 2 2 2 d s dt m / 9.8 / sec  2 s t t ( ) 4.9  dy dx x / sin  y x    cos 3 Slide 4 - 50 Exercises 2 1 2 1 40. ( ) ; ( 1) 1 1 1 1 ( ) 1 g x x c x g c c g x x x                   39. Slide 4 - 51 Exercises 43. 44 . 45 46 2 2 2 2 cos( ) ( ) sin( ) 2 ( ) 1 sin 2 1 1 ( ) sin( ) 1 ds v t s t t C dt s c c s t t                     Slide 4 - 52 Exercises 48 49 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.3 The Shape of a Graph Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 54 1. First Derivative Test for Increasing Functions and Decreasing Functions

>4++2DEFINITIONSLet f be a function defined oe an interval Iand let xj and x, be1hFunctionFunctionany two points in I.decreasingincreasingIf (x2) > f(xi) whenever x, <x. then J is said to be inereasing on /.1.y'>0y<02.9If f(x2) < f(xi) whenever xy < x2, then f is said to be decreasing on ,A fiunction that is increasing or decreasing on / is called monotonic on J.0y'-0The function f(x) = x° isFIGURE4.20monotonic on the intervals (oo, o) and[0, co], but it is not monotonic on(00, 00).lide 4-55Slide 4-58EXAMPLE1Find the critical points of f(x) =x- 12x 5 and identify the inter-COROLLARY3Suppose that J is contimuous on [o, b] and diferenitiable onvals on which fis increasing and on which Jis decreasing(a,b)Solution J(x)= 3x 12 = 3(x 4)= 3(x+2)(x2)Y=12x-I(x) >at each point xe (a, b), then fis increasing on [a,b]Let J(x)=0, hem x =2 ad x=220If y'(x) < 0 at each point xe (a, b), then / is decreasing on [a, b],They partition the x-axs into intervaks as fblows.(-2.11)1 Weapplythefirstderivativetestto findwhere aIntervak(0-2)(-2,2)(2,+)Sinofr+functionisincreasinganddecreasingBehavior offincreasingdecreasingincreasing2Thecritical points ofa functionf partitionthedomainintointervalsonwhichfiseitherpositiveornegativeSo the increasing intervak of (x) are (-ag,-2)and (2,+ ao), the decreasing interval is (-2,2)3Wedetemine thesign off in each interval by212.-21evaluating f for one value of x in the interval.FIGURE4.21 Thefunction f(r)=12r 5 is monotonic on threeseparate intervals (Example 1),Slide 4- s7Slide 4-58Howto Find the Increasing Interval andExercisesDecreasing Interval byFirst Derivative TestFind the open itervalson which the function is increasingand decreasing,Method.131. f(x) = r - 6Vx 124. F(0) = 60 0Step 1. Determine the domain of f(x)+241. J)=e+e"35. f(r) =Step 2. Find the critical points of the function on the domain,and identify the intervals where the critical points partition thedomain.24.The decreasing intervals are(n,2)and (2,+o)Step 3. Detemine the sign offon each intervalthe increasing interval is (-2,V2)Step 4. Identify the monotonic interval off.31.The increasing interval is (10, +00)the decreasing interval is (L,10)[Method2]35.The increasing intervals are (-,I) and (3,+00)Step 1. Determine the domain of f(x)the decreasing intervals are (,2) and (2,3)Step 2. Solve the inequalities f(x)>0 and f(x)<0 in the41.The increasing interval is (jln2,+)domain,the solutions ofthesetwo inequalities are thethe decreasing interval is (-a,_In2)increasing interval and decreasing interval respectivelySide 4- 55Slide 4-60

2016/11/15 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 55 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 56 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 57 We apply the first derivative test to find where a function is increasing and decreasing. The critical points of a function f partition the domain into intervals on which f' is either positive or negative. We determine the sign of f' in each interval by evaluating f' for one value of x in the interval. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 58 2 2 f x x x x x '( ) 3 12 3( 4) 3( 2)( 2)        Let f x then x and x '( ) 0, 2 2     Intervals (-∞,-2) (-2,2) (2,+∞) Sign of f’ + - + Behavior of f increasing decreasing increasing They partition the x-axis into intervals as follows. So the increasing intervals of f(x) are (- ∞,-2) and (2,+ ∞), the decreasing interval is (-2,2). Slide 4 - 59 How to Find the Increasing Interval and Decreasing Interval by First Derivative Test Step 2. Find the critical points of the function on the domain, and identify the intervals where the critical points partition the domain. Step 3. Determine the sign of f ' on each interval. Step 4. Identify the monotonic interval of f. Step 2. Solve the inequalities f ’(x)>0 and f ’(x)<0 in the domain, the solutions of these two inequalities are the increasing interval and decreasing interval respectively. Method 1 Method 2 Step 1. Determine the domain of f(x). Step 1. Determine the domain of f(x). Slide 4 - 60 Exercises 24.The decreasing intervals are ( , 2) ( 2, )    and the increasing interval is ( 2, 2)  31.The increasing interval is (10, )  the decreasing interval is (1,10) 35.The increasing intervals are ( ,1) (3, )   and the decreasing intervals are (1,2) (2,3) and 1 ( ln 2, ) 3 41.The increasing interval is   the decreasing interval is 1 ( , ln 2) 3  