§4.1 Maximum and Minimum Values



§4.1 Maximum and Minimum ValuesA function f has an absolute maximum at a no. c in its domain D iffc≥f(x) for all x∈D.A function f has an absolute minimum at a no. c in its domain D iffc≤f(x) for all x∈D.The no. f(c) is the maximum value or minimum value, respectively, of f on D.Example. y=fx=x2, D=[-1,2)-2…0…2…x-, 0…4…y-, ff has an absolute minimum at x=0.f0=0 is the absolute minimum value.f has no absolute maximum.■The maximum and minimum values of f are the extreme values of f.Extreme Value TheoremIf f is continuous on a closed interval [a,b] then f attains an absolute maximum and an absolute minimum on [a,b].Example. y=fx=x2, D=[-1,2]-2…0…2…x-, 0…4…y-, ff has an absolute minimum at x=0.f has an absolute maximum at x=2.A function has a local maximum at a no. c if there is an open interval I containing c such thatfc≥f(x) for all x∈IA function has a local minimum at a no. c if there is an open interval I containing c such thatfc≤f(x) for all x∈IExample. …a…c…d…b…x-, ff has a local maximum at c.f has a local minimum at d.f has an absolute maximum at b.■?? Practice with maxima and minimaFermat’s TheoremIf f has a local maximum or minimum at a no. c, and if f'(c) exists, then f'c=0.Why. …c…, f, tangent at local max.A critical number of a function f is a no. c in the domain of f such that either f'c=0 or f'(c) does not exist.Rephrase Fermat’s Theorem.If f has a local maximum or minimum at a no. c, then c is a critical no.Example. Find the critical numbers of fx=4x3-9x2-12x+3Since f is a polynomial, f' exists at all real numbers.f'=12x2-18x-12=62x2-3x-2=6(2x+1)(x-2)Critical numbers are x=-12, x=2■Example. Find the critical numbers ofGx=3x2-xG'x=13x2-x-23(2x-1)=132x-1x2-x23G'=0 at x=12.G' does not exist for x=0 or x=1. …0…1…x-, …0…y-, G■Closed Interval MethodTo find absolute maximum and minimum values of a continuous function f on a closed interval [a,b].Find the values of f at the critical numbers in [a,b].Find f(a) and f(b).The largest of these is the absolute maximum and the smallest of these is the absolute min.Example. Find the absolute maximum and minimum values of fx=1-2x-x2 on [-4,1].Find the critical numbersf'x=-2-2x=-2(1+x)x=-1 is a critical number.f-1=1+2-1=2f-4=1+8-16=-7f1=1-2-1=-2The absolute max. value is 2 at x=-1. The absolute min. value is -7 at x=-4. ■Example. Find the absolute maximum and minimum values of fx=lnxx on [1, 3]f(x) is differentiable on 1,3].f'x=lnx'x-(lnx)x'x2 =1x x-lnxx2 =1-lnxx2 x=e makes the numerator zerofe=lnee=1e=0.367…f1=ln11=0f3=ln33=0.366…f has an absolute maximum value of 1e=0.367… at x=ef has an absolute minimum value of 0 at x=1■4.2 The Mean Value TheoremRolle’s Theorem Let f satisfyf is continuous on the closed interval [a,b].f is differentiable on (a,b)fa=f(b)Then there is a no. c on (a,b) such that f'c=0.Why? …a…b…, …,f, fa=fb, tangentExample. Verify that fx=sinx+cosx, 0≤x≤2πsatisfies the hypotheses of Rolle’s theorem. Find all numbers c satisfying the conclusions of Rolle’s theorm.f is continuous on [0,2π].f is differentiable on (0,2π).f0=0+1=f(2π)f does satisfy the hypotheses. Look for c such that f'c=cosc-sinc=0.axes, unit circle, radius at angle c, pt. (cosc,sinc) At c=π/4and c=5π/4 we have cosc=sinc.■The Mean Value TheoremLet f satisfyf is continuous on the closed interval a,b.f is differentiable on (a,b).Then there is a no. c on (a,b) such that f'c=fb-f(a)b-aPicture …a…b…,…,f,fa,fb, line segment g, c, tangent to f at c?? Estimate numbers c satisfying the Mean Value Theorem.Proof. Let h=f-h…a…b…, …,hBy Rolle’s theorem there is a c∈(a,b) such thath'c=0orf'c-g'c=0Let m be the slope of line segment g.m=fb-f(a)b-aThen f'c=m■Example. Verify that f'x=x, 1≤x≤4satisfies the hypotheses of the Mean Value Theorem. Find all numbers c that satisfy the conclusion.f(x) is continuous on [1,4]f(x) is differentiable on (1,4)f does satisfy the hypotheses. Find a no. c that satisfies the conclusions. We wantf'c=fb-f(a)b-a=4-14-1=13fx=x12f'x=12x-12=13x-12=23x12=32x=94=2.25∈(1,4)■Example. At 2pm you leave Pullman for the Spokane airport. After driving north 85 miles, you arrive at 3:30pm. Let s(t) be your position as a function of time t. Assume s(t) is continuous and differentiable. Recall thats't=vt, your instantaneous velocity.Notation of Mean Value theorema=0 hours sa=0 miles b=112hourssb=85 milesBy the Mean Value theorem there is a no. c such that s'c=sb-s(a)b-a=85-0112-0mileshour≈57 mileshourAt t=c instantaneous velocity = average velocity.■Theorem. If f'x=0 for all x∈(a,b), then f is constant on (a,b).Picture …a…b…x-, …, fProof. Let x1, x2 be any two numbers such that a<x1<x2<bBy the Mean Value theorem, there is a no. c in (x1,x2) such that f'c=fx2-f(x1)x2-x1By hypothesis, f'c=0.Therefore fx1=f(x2)■Corollary. If f'x=g'(x) for all x∈(a,b) then fx=gx+C on (a,b).Picture …a…b…x-, …, f,gProof. Let h=f-g. Then h'x=f'x-g'x=0 for x∈(a,b)Then by the previous theoremh is constant on (a,b).hx=Cfx-gx=Cfx=gx+C. ■§4.3 Derivatives and the Shapes of GraphsIncreasing/Decreasing TestIf f'x>0 on an interval, then f is increasing on that interval.If f'x<0 on an interval, then f is decreasing on that interval.Picture …x-,…,f,f'>0, f'<0Proof. Follows from the Mean Value Theorem.Example. Find the intervals where the function fx=x3-2x2+xis increasing or decreasing.Solution. Find the critical numbers. Since f is differentiable at every real no., need only find where f'=0.f'x=3x2-4x+1=3x-1(x-1)=3x-13x-1The critical numbers are x=1/3 and x=1.Construct a sign chart for f':Intervalx-13x-1f'fx<1/3--+increasing13<x<1 +--decreasingx>1+++increasing■The First Derivative TestSuppose c is a critical number of a continuous function f.If f' changes from positive to negative at c, then f has a local maximum at c.If f' changes from negative to positive at c, then f has a local minimum at c.If f' does not change sign at c, then f has no local extremum at c.Picture …,…, curve with f'>0, f'<0, f'>0,f'>0Previous Example (Continued). Find the x-coordinates of the local maxima and minima offx=x3-2x2+xBy the sign chart and first derivative testf has a local maximum at x=1/3f has a local minimum at x=1Graph f. Find the local maximum and minimum values.f13=127-29+13=1-6+927=427≈0.15f1=1-2+1=0Note also f0=0.…0…13…1…x-,…0…0.15…, curve■Example (Alternative Sign Chart) Find the local maximum and minimum values of gx=x-2sin(x)Solution. Find the critical numbers of g. Since g is differentiable at every real number, need only find where g'=0.g'x=1-2cos(x) cannot be factoredThen g'x=0 when cosx=1/2.45°-45°-90° triangle recall cosπ4=12. See also cos-π4=12.The critical numbers are x=-π4 and x=π4These divide the real line into 3 subintervals. Evaluate g' at test points in each subinterval.g'-π2=1>0, g'0=1-2<0,g'π2=1>0Alternate sign chart.-π…π…x-, g'>0, g'<0, g'>0 , segments with corresponding slopesSince g' is continuous, it cannot change sign within a subinterval.By the first derivative testg has a local minimum at x=-π/4g has a local maximum at x=π/4This can be read directly off the sign chart! ■Concavity…,…,f CU curve above tangentsf' increasingf is concave up (CU)…,…,f,CD curve below tangentsf' decreasingf is concave down (CD)Definitions. If the graph of f lies above its tangents on an interval I, then f is concave up on I.If the graph of f lies below its tangents on an interval I, then f is concave down on I.?? graph of gwhere is g CU and CD?Concavity TestIf f''x>0 on I, then f is CU on I.If f''x<0 on I, then f is CD on I.Why?f''>0 ?f' increasing ?f CUf''<0 ?f' decreasing ?f CD ■Example. Find the intervals where the function fx=x3-2x2+xis CU and CDSolution. f'x=3x2-4x+1f''x=6x-4=6x-23sign chart for f''intervalx-2/3 f''fx<2/3--CDx>2/3++CU■A point P on a curve is an inflection point if the curve changes from concave up to concave down, or from concave down to concave up at P.Example (continued).fx=x3-2x2+xf''x=6x-23f has an inflection point at x=23.f23=233-2232+23=827-89+23=8-24+1827=227The inflection point is 23,227■?? graph of g where are the inflection points?The Second Derivative TestSuppose f'' is continuous on an open interval containing c.If f'c=0 and f''c>0, then f has a local minimum at c.If f'c=0 and f''c<0, then f has a local maximum at c.Picture…c…,…,f CU, horizontal tangent at cf'c=0 f''(c)>0the slope is increasingf has a local minimum…c…,…,f CD, horizontal tangent at cf'c=0f''c<0 the slope is decreasingf has a local maximumExample. Apply the second derivative test to find the x-coordinates of the maxima and minima of fx=x3-2x2+xSolution. Start by finding the critical numbersf'x=3x2-4x+1=3x-1(x-1)=3x-13(x-1)f'x=0 for x=1/3 and x=1Find the sign of f'' at each critical no. where f'c=0f''x=6x-4=6x-23f''13<0 a local max. by the 2nd derivative testf''1>0 a local min. by the 2nd derivative testgraphf0=0f13=427≈0.15f1=0recall the inflection point 23,227…0…13…23….1…x-,…0…0.15…y-, fNote the local max, the local min & inflection point.■Example. Find the local maximum and minimum values of f using the first and second derivative tests.fx=xx2+4Solution. Start by finding the critical numbers (this is common to both tests).f'x=1x2+4-x(2x)x2+42=4-x2x2+42=2-x(2+x)x2+42critical numbers are x=-2 and x=2Apply the first derivative testConstruct a sign chart for f'Interval2-x2+xf'fx<-2+--decreasing-2<x<2+++increasingx>2-+-decreasingBy the first derivative testf has a local min at x=-2 and a local max at x=2Apply the second derivative testMust calculate the second derivativef''x=-2xx2+42-4-x22x2+4(2x)x2+44=-2xx2+4-4x(4-x2)x2+43=-2x3-8x-16x+4x3x2+43=2x3-24xx2+43=2x(x2-12)x2+43Thenf''-2=-4(4-12)4+43>0 local min at x=-2f''2=4(8-12)4+43<0 local max at x=2■Compare the first and second derivative testsThe first derivative test may be easier to apply.The first derivative tests handles more cases. You can apply it to a function that has a corner or cusp at x=c, where the 2nd derivative DNE.axes, curve, tangent 1st& 2nd deriv. tests OKfunction w/ corner 1st derivative test only The second derivative tests generalizes to calculus of several variables.§4.4 Curve SketchingCurve Sketching ChecklistDomainIntercepts: x and ySymmetry: odd, even or periodicAsymptotes: horizontal and verticalIntervals of increase and decreaseLocal maxima and minimaConcavity and points of inflectionSketch the curve!Example. Sketch the graph of fx=x3+6x2+9xA. Domainf is a polynomial. The domain is all real numbers.B. InterceptsWhat is x when y=0?x3+6x2+9x=0xx2+6x+9=0xx+32=0x intercept x=0 or x=-3 ***What is y when x=0? f0=0y intercept y=0 ***C. Symmetry. The polynomial involves both even and odd powers of x. No symmetry.D. Asymptotes.No horizontal or vertical asymptotes.E. Intervals of increase or decrease?f'x=3x2+12x+9 =3(x2+4x+3) =3x+1(x+3) The sign of f' changes when f'=0Critical numbers at x=-1 and x=-3Sign chart for f'…-3…-1…x-,f'>0,f'<0,f'>0, strokesf increases on (-∞,-3) and (-1,∞) ***f decreases on (-3,-1) ***F. Local max and min valuesBy the first derivative testf has a local max. at x=-3.f-3=-33+6-32+9-3 =-27+54-27=0Local max at -3,0 ***f has a local min at x=-1f-1=-13+6-12+9-1=-1+6-9=-4Local min at (-1,-4) ***G. Concavity and inflection pointsf''x=6x+12=6(x+2)sign chart for f''…-2…x-, CD f''<0, CU f''>0 ***f-2=-23+6-22+9(-2)=-8+64-18=-2Inflection point (-2,-2) ***Sketch f-4…1…x-,-5…0…y-,fWhew! ■Example. Sketch the graph of gx=xx2-9A. Domain x∈Rx≠-3 or 3}=-8,-3∪-3,3∪(3,∞)B. x-intercept: x=0y-intercept: y=0C. Symmetry: fx=odd functioneven function=odd functionD. Horizontal asymptote: a value approached as x→±∞?indeterminate form type ∞/∞limx→±∞xx2-9=limx→±∞12x=0 Vertical asymptotes: denom=0 and num≠0 atx=-3 and x=3limx→3+xx2-9=∞ limx→3-xx2-9=-∞E. Intervals of increase and decreasef'x=1(x2-9)-x(2x)x2-92=-x2-9x2-92<0f'(x) is always decreasing in its domain; no max or min values G. Concavity and points of inflectionf''x=-x2-9'x2-92--x2-9x2-92'x2-94=-2xx2-92-(-x2-9)2(x2-9)(2x)x2-94=-2x(x2-9)-(-x2-9)(4x)x2-93=-2x3+18x+4x3+36xx2-93=2x3+54xx2-93=2x(x2+27)x2-93sign chart for f''interval2xx2-9f''fx<-3-+-CD-3<x<0--+CU0>x>3+--CDx>3+++CUInflection point at (0,0)Sketch graph-4…-2…0…2…4…x-,-2…0…2…y-,fwhere f2=24-9=-25 and f4=416-9=47■Example Graph y=fx=ln?(cos(x)) A. Domain lnz is defined for z>0 Sketch cosx-2π…0…2π…x-,-1…0…1…y-, cosxcosx>0 for D=…∪–π2,π2 ∪3π2,5π2∪…D is the domain of f ***f(x) will have period 2π restrict this discussion to -π2<x<π2B. x-intercept when y=ln(cosx)=0, cosx=1x=…,-2π,0, 2π,… ***y- intercept y=f0=ln(cos0)=ln1=0 ***C. Symmetry cos(x) is periodic and eveny=ln(cosx) is periodic and even too ***D. Horizontal asymptotes: none Vertical asymptotes:lnz has a vertical asymptote at z=0ln(cos(x)) ………………………….. cosx=0 occurs for x=…,-3π2, -π2,π2,3π2, … ***E. Intervals of increase and decreasefx=ln?(cosx)f'x=-sinxcosx=-tanx-π/2…0…π/2…x-,…0…, tanx, -tanxf'>0 on (-π2, 0) and f'<0 on 0,π2F. Local max and min values Critical number at x=0. By the first derivative test, a local max at x=0f has a local max at (0, 0) ***G. Concavity and points of inflection From the graph of f' we see f''=f''<0f is CD ***Graph y=ln(cosx)-2π…0…2π…x-,-1…0…1…y-, ln(cosx)■§4.5Optimization ProblemsProblem Solving StrategyRead problem carefully.Draw a picture if relevant.Assign notation to the given information and unknown quantities.Express quantity to be optimized in terms of one variable.Find the domain of the variable.Find the absolute maximum or minimum (give a clear argument that it is an abs. max or min) and answer the question.Example. A farmer want to fence an area of 1.5 million square feet in a rectangular field and then divide it into half with a fence parallel to one of the sides of the rectangle. What are the dimensions of the field that minimize the length of the fence?2. fence, x, x, x, x, y, y, y3. Area of field A=1.5 106 square feet. Length of fence LA=2xyL=4x+3y4. Express L in terms of one variable.y=A/(2x)Lx=4x+3A/(2x)5. 0<x<∞6. Find a critical numberL'x=4+32A-x-2=4-32A1x2 At the critical no. c, L'c=0:4=32A1c2 or c2=38A c=38A12=38?3210612=34103=750 feet Is this an absolute minimum? WriteL'x= 1x2(4x2-32A)=4x2x2-38A =4x2x2-7502For x<750 ft, L'<0. For x>750 ft, L'>0.…750…x-, slopesL has an absolute minimum at x=750 feet.y=A2x=1.5 1061.5 103=1000 feet. ■First Derivative Test For Absolute Extreme ValuesSuppose c is a critical no. of a continuous function f defined on an interval.If f'x>0 for all x<c and f'x<0 for all x>c, then f(c) is an absolute max. value of f.If f'x<0 for all x<c and f'x>0 for all x>c, then f(c) is an absolute min. value of f.Sign chart for part (a) …c…, f' values, strokesSign chart for part (b) …c…, f' values, strokesExample. From a piece of cardboard of side 1 foot, cut out square corners of side x and fold up flaps to make an open top box. What x gives the maximum volume?2.square, corners of side x, middles 12-2x3. V= volume of box x= size of corners4. Vx=12-2x12-2xx =4x6-x2 5. 0≤x≤6 inches6. Find the critical numbersV'x=46-x2+4x26-x(-1)=46-x[6-x-2x]=4(6-x)(6-3x)=12(6-x)(2-x)x=2 and x=6 are critical numbers.Use the closed interval method.Evaluate V at the endpoints of the domain.V0=0.V6=0.Evaluate V at the critical pointV2=8?16=128The largest value gives the absolute max. of V. x=2 inches ■Example. A woman at a point A on the shore of a circular lake wants to arrive at a point C opposite A on the other side in the shortest possible time. She can walk at a rate of 4 miles per hour and row at a rate of 2 miles per hour. How should she proceed?2. circle, A…0…C, radius r, r, B, segment x, ∠0AB=θ, arc y3. x(θ) distance rowing, θ=0 row onlyy(θ) distance walking, θ=π/2 walk onlyTθ=12xθ+14y(θ)4. Express quantity to be optimized in terms of 1 variable add ∠0BA=θ, ∠AOB=π-2θ, ∠B0C=2θyθ=r(2θ) add segment BC Note ∠ABC is a right angle. Let ∠0BC=∠0CB=α… Obtain θ+α=π/2x2r=cos?(θ) xθ=2r cos?(θ) time spent travelingTθ=12xθ+14y(θ) 5. domain: 0≤θ≤π/26. Find the absolute min. valueUse the closed interval method.T0=12x(0)=12(2r)=r Tπ2=14yπ2=14πr≈0.79r Find the critical numbersT'θ=12x'θ+14y'θ =12(-2rsinθ+142r=r12-sinθ Then the critical value of θ givessinθ=12?θ=π/6Tπ6=12xπ6+14yπ6=12 2rcosπ6+14 r 2?π6=r32+rπ12≈1.13rThe shortest time corresponds to θ=π/2, or walking all the way.■STOP §4.7 AntiderivativesA function F is called an antiderivative of f on an interval I if F'x=f(x) for all x in IExample. Find an antiderivative of fx=xFx=12x2?? Gx=■Recall: If F'=G' for all x on an interval I, then Fx=Gx+Con I.Theorem. If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I isFx+C,where C is an arbitrary constant.Example. Find the most general antiderivative of fx=x.Fx=12x2+C■Example. Find the most general antiderivative of fx=xn.ddx1n+1ddxxn+1=1n+1n+1xn=xnThe most general antiderivative is Fx=1n+1xn+1+C■Table of AntiderivativesFunctionAntiderivativexn, n≠11n+1xn+1+C1xlnx+Ccos(x)sinx+Csin?(x)-cosx+Csec2(x)tanx+C11-x2sin-1x+C11+x2tan-1x+Ca f(x)aFx+Cfx+g(x)Fx+Gx+Cwhere F and G are functions satisfying F'x=f(x) and G'x=g(x).Example. Find the most general antiderivative of fx=x2+x+1xSolution. First simplify algebraicallyfx=x+1+x-1thenFx=12x2+x+lnx+C■Example. Find the most general antiderivative of ft=sint-2tSolutionFt=-cost-223t32+C=-cost-43t32+C■Geometrical Picture of AntiderivativesConsider fx=xgeneral form of antiderivativeFx=12x2+Cget a family of antiderivatives corresponding to different values of C-2…0…2…x-,-1…0…3…y-, several curvesDifferential EquationsA differential equation is an equation involving one or more derivatives of an unknown functionExample. Solvedydx=xwith the extra condition that y=1 when x=0.General solutiony=12x2+Cparticular solutiony=12x2+1■Example. Solve for f(x)f'x=4-31+x2-1with f1=0.General solution.fx=4x-3tan-1x+Cfind the particular solution0=f1=4-3tan-11+C=4-3π4+Cthen C=3π4-4.fx=4x-3tan-1x+3π4-4■Example. Solve for f(x)f''x=20x3-10with f1=1 and f'1=-5.General antiderivative of f''f'x=2014x4-10x+Cuse f'(1)=-5f'1=5-10+C=-5we see C=0. Thusf'x=5x4-10xgeneral antiderivative of f'fx=x5-5x2+Duse f1=1f1=1-5+D=1we see D=5concludefx=x5-5x2+5■Position, Velocity and AccelerationConsider motion along a straight lines(t) position at time t units are feetv(t) velocity at time t units are feet/seca(t)acceleration at time t units are feet/sec2wheres't=v(t)v't=a(t)Example. You are standing on the edge of a cliff 96 feet above the Snake river.cliff, water, 0…96…s- with origin at water(a) At t=0, throw a rock straight up at 80 ft/sec. Find v(t) and s(t) given a=-32 ft/sec2.First considerv't=a=-32general solutionvt=-32t+Cwe have the initial condition that v0=80 ft/secvt=-32t+80Next considers't=vt=-32t+80general solutionst=-16t2+80t+Dwe have a second initial condition s0=96 ft.st=-16t2+80t+96(b)When does the rock hit the water?Find t such that st=0-16t2+80t+96=0divide by 16-t2+5t+6=0factort2-5t-6=0t+1t-6=0The answer we want corresponds to the positive roott=6 seconds. ■ ................
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