最新微积分英文版4PPT课件.ppt
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1、微积分英文版微积分英文版4三、其他未定式三、其他未定式 二、二、型未定式型未定式一、一、型未定式型未定式机动 目录 上页 下页 返回 结束 洛必达法则 第三三章 1)的情形从而机动 目录 上页 下页 返回 结束 2)的情形.取常数可用 1)中结论机动 目录 上页 下页 返回 结束 3)时,结论仍然成立.(证明略)说明说明:定理中换为之一,条件 2)作相应的修改,定理仍然成立.定理2 目录 上页 下页 返回 结束 例例3.求解解:原式机动 目录 上页 下页 返回 结束 例例3.求解解:原式机动 目录 上页 下页 返回 结束 例3.例4.说明说明:例如,而用洛必达法则在满足定理条件的某些情况下洛必
2、达法则不能解决 计算问题.机动 目录 上页 下页 返回 结束 3.1Maxima&MinimaMaxima:point whose function value is greater than or equal to function value of any other point in the intervalMinima:point whose function value is less than or equal to function value of any other point in the intervalExtrema:Either a maxima or a minim
3、aWhere do extrema occur?Peaks or valleys(either on a smooth curve,or at a cusp or corner)f(c)=0orf(c)isundefinedDiscontinutiesEndpoints of an intervalThese are known as the criticalpoints of the functionOnce you know you have a critical point,you can test a point on either side to determine if its a
4、 max or min(or maybe neitherjust a leveling off point)3.2Monotonicity and ConcavityLet f be defined on an interval I(open,closed,or neither).Then f isa)INCREASING on I if,b)DECREASING on I if,c)MONTONIC on I if it is ether increasing or decreasingMonotonicity TheoremLet f be continuous on an interva
5、l I and differentiable at every interior point of I.a)If f(x)0 for all x interior to I,then f is increasing on Ib)If f(x)0 for all x in(1,c)and f(x)0 for all x in(c,b),then f(c)is a local max.value.b)If f(x)0 for all x in(c,b),then f(c)is a local min.value.c)If f(x)has the same sign on both sides of
6、 c,then f(c)is not a local extreme value.Second Derivative TestLet f and f exist at every point in an open interval(a,b)containing c,and suppose that f(c)=0.a)If f(c)0,then f is a local min.value of f.3.4Practical ProblemsOptimization problems finding the“best”or“least”of“most cost effective”,etc.of
7、ten involves finding the extrema of the functionUse either 1st or 2nd derivative testExampleA fence is to be constructed using three lengths of fence(the 4th side of the enclosure will be the side of the barn).I have 120 yd.of fencing and the barn is 150 long.In order to enclose the largest possible
8、 area,what dimensions of fence should be used?(continued on next slide)Example continuedArea is to be optimized:A=l x wPerimeter=120 yd=360=2l+ww=360 2lSo,2 lengths of 90 and a width of 180.HOWEVER,the barn is only 150 wide,so in order to enclose the greatest area,we wont use a critical point of the
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