2022年人教A版高中数学选修1-1课时提升作业十八3.1.1变化率问题3.1.2导数的概念精讲优练课型含答案 .pdf

温馨提示：此套题为Word版，请按住 Ctrl,滑动鼠标滚轴，调节合适的观看比例，答案解析附后。关闭 Word文档返回原板块。课时提升作业十八变化率问题导数的概念一、选择题(每小题 4 分,共 12 分)1.(2016 杭州高二检测)设函数 y=f(x)=x2-1,当自变量x 由 1 变为 1.1 时,函数的平均变化率为()A.2.1 B.1.1 C.2 D.0【解析】选A.=2.1.2.(2016 洛阳高二检测)一质点运动的方程为s=5-3t2,若该质点在时间段内相应的平均速度为-3 t-6,则该质点在t=1 时的瞬时速度是()A.-3 B.3 C.6 D.-6【解析】选D.=-3 t-6,当 t 0 时,-3 t-6-6,所以瞬时速度为-6.3.设函数 f(x)在 x0处可导,则=()A.f(x0)B.-f(x0)C.f(x0)D.-f(x0)【解析】选B.=-f(x0).二、填空题(每小题 4 分,共 8 分)4.某质点的运动方程为s=-2t2+1,则该质点从t=1 到 t=2 时的平均速度为.【解析】=-6.5.(2016 佛山高二检测)一物体的运动方程为s=7t2+8,则其在t=时的瞬时速度为1.【解析】=7t+14t0,当(7 t+14t0)=1 时,t=t0=.答案:三、解答题6.(10 分)(2016 石家庄高二检测)一辆汽车按规律s=3t2+1 做直线运动(时间单位:s,位移单位:m),求这辆汽车在t=3s 时的瞬时速度.【解析】设这辆汽车在3s 到(3+t)s 这段时间内的位移的增量为s,则s=3(3+t)2+1-28=3(t)2+18t,所以=3 t+18,所以(3 t+18)=18.故这辆汽车在t=3s 时的瞬时速度为18m/s.【补偿训练】1.甲、乙两人走过的路程s1(t),s2(t)与时间 t 的关系如图所示,试比较两人的平均速度哪个大?文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3文档编码:CK4T9M9Y9H9 HT1S9O5E7A2 ZL3B6F9U1L3【解题指南】欲比较两人的平均速度,其实就是比较两人走过的路程对时间的平均变化率,通过平均变化率的大小关系得出结论.【解析】由图象可知s1(t0)=s2(t0),s1(0)s2(0),所以,所以在从0 到 t0这段时间内乙的平均速度大.2.(1)计算函数f(x)=x2从 x=1 到 x=1+x 的平均变化率,其中 x 的值为:2;1;0.1;0.01.(2)思考:当x 越来越小时,函数 f(x)在区间上的平均变化率有怎样的变化趋势?【解析】(1)因为 y=f(1+x)-f(1)=(1+x)2-12=(x)2+2x,所以=x+2.当 x=2 时,=x+2=4;当 x=1 时,=x+2=3;当 x=0.1 时,=x+2=2.1;当 x=0.01 时,=x+2=2.01.(2)当x 越来越小时,函数 f(x)在区间上的平均变化率逐渐变小,并接近于 2.一、选择题(每小题 5 分,共 10 分)1.(2016 太原高二检测)物体的运动方程是s=-4t2+16t,在某一时刻的速度为零,则相应时刻为()A.t=1 B.t=2 C.t=3 D.t=4 文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4【解析】选B.=-8t+16,令-8t+16=0,得 t=2.2.(2016 菏泽高二检测)若 f(x0)=1,则=()A.B.-C.1 D.-1【解题指南】根据导数的定义求解.【解析】选B.=-=-f(x0)=-1=-.【误区警示】本题易对导数的概念不理解而误选成D.二、填空题(每小题 5 分,共 10 分)3.(2016 烟台高二检测)汽车行驶的路程s 和时间 t 之间的函数图象如图,在时间段,上的平均速度分别为,则三者的大小关系为.【解析】=kOA,=kAB,=kBC,由图象知kOAkABkBC.答案:4.如图所示,水波的半径以1m/s 的速度向外扩张,当半径为5m时,这水波面的圆面积的膨胀率是m2/s.文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4【解题指南】求出在时刻t 的水波面积,进而求出在时刻t0的瞬时膨胀率,代入半径求膨胀率.【解析】因为水波的半径以v=1m/s 的速度向外扩张,水波面积S=r2=(vt)2=t2,所以水波面积在时刻t0时的瞬时膨胀率S(t0)=2 t.当半径为5m时,t=5s,所以 S(5)=2 5=10,即半径为5m时,这水波面积的膨胀率是10,答案:10 三、解答题5.(10 分)建造一栋面积为xm2的房屋需要成本y 万元,y 是 x 的函数,y=f(x)=+0.3,求 f(100),并解释它的实际意义.【解题指南】根据导数的定义,求出函数值y 在 x=100 时的瞬时变化率即可,最后由瞬时变化率解释 f(100)的意义.【解析】根据导数的定义,得f(100)=0.105.f(100)=0.105表示当建筑面积为100m2时,成本增加的速度为1050 元/m2,也就是说当建筑面积为 100m2时,每增加 1m2的建筑面积,成本就要增加1050 元.【补偿训练】如果一个质点从起点A开始运动,在时间 t 的位移函数为y=f(t)=t3+3.文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 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时,求的值.【解析】(1)y=f(4+t)-f(4)=(4+t)3+3-43-3=(t)3+48t+12(t)2=(0.01)3+48(0.01)+12(0.01)2=0.481201.所以=48.1201.(2)当t 0 时,=48.关闭 Word 文档返回原板块文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 ZG5W9E8K5Q4文档编码:CY5N5Q8E2O10 HU7P2L2F8M5 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