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

CALCULUSTHOMAS'CALCULUSARLYTRANSCENDChapter 4JAMESSTEWARTApplications of DerivativesAAStCiHA4.11.Absolute(Global)ExtremeValuesExtreme Values of FunctionsSlide 4-4ofDEFINITIONSLet f be a finction with domain D. Then f has an absolute (orLocal maximuglobal) maximum valuc on D at a point c ifNogresocatintf(x) ≤ f(c)for allx in Drvapoffnearbyand an absolute (or global) minimum value on D at c irAbsolute minimu(x)≥ (c)for all x in DNosmallervalueofLocal minimunFanywhere,AlboaNo smaler value oflocal minimmnearbyFIGURE 4.5How to classify maxima and minimaSlide 4- 5Slide 4-

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 4 Applications of Derivatives Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.1 Extreme Values of Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 4 1. Absolute (Global) Extreme Values Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 6

Exploring Extreme ValuesFunction ruleDomain DAbsolute extrema on Dmsin(a)y=r(00,00)No absolute maximum.Absolute minimum of 0 at x = 0.(b) y=x2[0, 2]Absolute maximum of 4 at x = 2.Absoluteminimum of 0 atx=0(c) y = x2(0, 2)Absolute maximum of 4 at x = 2.No absolute minimum(d) y = 2(0, 2)No absolute extrema.FIGURE4.1AbsoluteextremaforFunctionswiththesamedefiningrulecanhavedifferentthe sine and cosine functions onextrema,dependingonthedomain.[w/2, w/2]. These values can dependon the domain of a function.lide 4-7Slide4-8=107THEOREM1The Extreme Value Theorem(a) abs min cely(b) abs max and minIf f isContinuougon aClosed intervaD(a,b],then J attains both an absolute max-imum value M and an absolute minimum value m in [a, b]. That is, there arenumbers , and xg in [a, b] with f(x,) = m, f(x) M, and m ≤ f(x) ≤ M forevery other.x in [a, b] (Figure 4.3)The requirements in Theorem 1 that the interval be closed andfinite,and the functionconbinuous arekeyingredients.(c) abs mux oely(d) nomax or minFIGURE4.2Graphs for Example 1alide 4-9Slide 4- 10"Fu)E shows the requirements in Theorem-1that the interval be closed and trefunctioncontinuous arekeyingredientsWithoutthem,the conctusion oftheNo largest valuetheoremneed not hold.OFIGURE4.4Even a single point ofy=xdiscontinuity cankeepafunctionfrom0≤x<1having either a maximum or minimumvalue on a closed interval, The function0Smallest value0≤x<:1Mainrn'siMinismm at intorior poiris continuous at every point of[0, 1]except x = 1. yet its graph over [0, ]]FIGURE 4.3Some possibilities fora continusfiunction'smaximumanddoes not have a highest point.minimum ona csed ierval [a,Slide 4.12Slide 4-11

2016/11/15 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 7 Exploring Extreme Values Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 8 Functions with the same defining rule can have different extrema, depending on the domain. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 10 The requirements in Theorem 1 that the interval be closed and finite, and the function continuous are key ingredients. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 12 It shows the requirements in Theorem 1 that the interval be closed and the function continuous are key ingredients. Without them, the conclusion of the theorem need not hold

ExerciseExercisesFinding Extrema from Graphs3In Exercises 1-6, determine from the graph whether the function hasay absolute extreme values on [o, b], Thexplain how your answris consistent with Theorem 1.3. No absolute minimum, An absolute maximum at x = c.Since the function's domain is an open intervsl, tbeextreme values because f is continuous on [a,b]L. An absolute minimum at x = Co an absolute maximum at x = b.4, No abelute extrema.The function is neither costisuossTheorem I guarantees the existence of suchBor defined ona closed interval, soit neednot fullextreme values because h is continuous on [a,b].the conclusions of Theorem 1.2. An absolute minimum at x = b, an absolute maximum at x =c.Theorem I guarantees the existence of suchextremaluesbcausescontinuouson [b]Slide4-13Slide 4-14In Exerczses 7-10, find the absolute extreme values and where they occurExercisesExercises2R5. An sbeolute minimum at x = a and an absolute maximum at x = e.num st (1,0), local maximum at (1,0)7. Local miNote that y = g(x) is mot continuous butstil has extrema.When the hypothesis of Theorem18. Minima at (2,0) and (2,0), maximum at (0,2)is satisfied then extrema are guaranteed, but when thehypotbesis is aot satiafied, absolute extrema may or may not occuz.6. Absolute minimum at x = c and an abeolute maximum at x = aNote that y=g(x) is not continuous but stilhas absolute extrema, When the hypothesisof Theorem lissatiafied then extrema sre guaranteed, but whenthe hypothesis is not satisfied, absolute extrema may ot may not occurSlide 4-15Slide 4- 18ExercisesExercisesf'r'(t)doesdoes tnot exist9. Maximum at (0,5). Note that there is no minimusince the endpoint (2,0) is excluded from the graph.10, Local maximum at (3,0), local minimum st (2,0),maximum at (1,2), minimum at (0, 1)Slide 4-17

2016/11/15 3 Slide 4 - 13 Exercises Slide 4 - 14 Exercises Slide 4 - 15 Exercises Slide 4 - 16 Exercises Slide 4 - 17 Exercises Slide 4 - 18 Exercises

DEFINITIONAn interior point of the domain ofa function J where is zero orundefined is a critical point of f.2.Method forFinding AbsoluteExtreme Values on a ClosedHow to Find the Absolute Extrema ofa Continuous FunctionfonaFinite Closed IntervalInterval1. Evaluate f at all critical points and endpoints.2. Take the largest and smallest of these values.lide 4-19Slide 4- 20EXAMPLE2Find the absolute maximum and minimum values of j(x) = x on[2, 1].Solution The function is differentiable over its entire domain, so the only critical point isCricalPointswhere J'(x) Zx 0, namely x 0. We need to check the function's values at x = (Need Not Giveand at the endpoints x =2 and x = 1:Extreme ValuesCritical point value: J(0) = 0J(2) = 4Endpoint values:Lf(1) -1The function hasan absolute maximum value of 4 at x-2 and an absolute minimuvalue of 0 atx 0FIGURE 4.7Critical pointswithouextreme values. (a) y = 3r is 0 atx = 0.but y = x has no extremum there.(b) y = (1/3)x-/ is undefined at x = 0,but y =x//has no extremum thereSlide 4- 22Slide 4-21EXAMPLE4Find thec absolute maximum and minimum values of f(x) = x2/3 on theEXAMPLE3Find the absolute maximum and minimum values of f(x) = 10x(2 In x)interval [2, 3]on the interval [1, e']SolutionSolutionThe first derivativeThe first derivative is)-r3VF() 10(2 Inx) 1010(1 Inx):has no zeros but is undefined at the interior point x = 0, Thbe values of J at this one criti.Theonly critical point in the domain [,']is the poinbhere In x 1,The valuescal point and at the endpoints areof J at this one critical point and at the endpoints areCritical poinr value: J(0) = 0Critical point value:f(e) 10e(2) = (2) = V4Endpoint values:Endpoint values:/(1) = 10(2 In 1) = 20J(3) = (3)p = V9ftel) = 10e (2 2 In c) = 0.Wecanscefrom thislisthatthe functionabolutemaximumvalueisV2.08andiWe can see from this list that the function's absolute maximum value is 10e 27.2; it oc-occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at thecurs at the critical interior point x = e. The absolute minimum value is O and occurs at theimteriorpoint0 where thegraph hasa cusp (Figure4.9).right endpoint x = -Slide 4- 23slide 4- 24

2016/11/15 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 19 2. Method for Finding Absolute Extreme Values on a Closed Interval Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 21 Critical Points Need Not Give Extreme Values Slide 4 - 22 Slide 4 - 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 24

ExercisesFind the absolute maximum and minimum values ofeach function20,2≤153on the given interval.Absolute maximum:28. h(x) = -3r2/3, -1 ≤x≤1Localalso a local maximummaximum230. g(x) = V5 x, V5 ≤x ≤ 037. g(x) = xe, 1 ≤x ≤ 1739. f(o) ↓ + Inx, 0.5 ≤ x 4Absolute minimum;also a local minimumFIGURE 4.9The extreme values ofJ(c) =on[2, 3] ourat x = 0 andx = 3 (Example 3)lide 4- 25Slide 4- 28DEFINITIONSA function J has a local maximum value at an interior point cof its domain if(x) ≤ f(c)for all x in some open interval containing eA function / has a local minimum value at an interior point c of its domain if(x) ≥ f(e)for all x in some open interval containing c3.Local (Relative)ExtremeValuesDEFINITIONSLet y be a funection with domain D.Then f has an absolute (orglobal) maximum value on D at a point c iff(x) ≤ f(c)for all x in Dand an absolute(or global) minimum value on D at ciffor all x in Df(x) ≥ J(c)extremevalueinitsimmediateneighborhood.Slide 4- 27Slide 4- 28Local maximum valueTHEOREM2The First Derivative Theorem for Local Extreme ValuesIfyhasalocalmaximum orminimum value at an interiorpointcofits domainy=fx)and if f' is defined at c, thenI'(c) = 0.TheonlyplaceswhereafunctionfcanpossiblyhaveanSecant slopes 2 0Secant slopes s 0extreme value (local or global) are(never positive)(never negative)1 interior points where f=0 interior points where f"is undefined3 endpoints of the domain offFIGURE4.6A curve with a localmaximum value.The slope at e,simultaneously the limit of nonpositivenumbers and nonnegative mumbers, is zero,Slide 4- 29Slide 4-30

2016/11/15 5 Slide 4 - 25 Slide 4 - 26 Exercises Find the absolute maximum and minimum values of each function on the given interval. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 27 3. Local (Relative) Extreme Values Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 28 Local extrema are also called relative extrema. A function f has a local extrema at an endpoint c if the appropriate inequality holds for all x in some half-open interval containing c. An absolute extremum is also a local extremum, because being an extreme value overall makes it an extreme value in its immediate neighborhood. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 30 The only places where a function f can possibly have an extreme value (local or global) are interior points where f ' =0 interior points where f ' is undefined endpoints of the domain of f

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≤1x1derivative 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, 0yat each point xe (a, b), then fis increasing on [a,b]Let J(x)=0, hem x =2 ad x=220If y'(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  