www.07xjs.com:高中数学课时跟踪检测(七)函数的最大小值与导数新人教A版选修2_2

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课时跟踪检测(七) 函数的最大(小)值与导数 层级一 学业水平达标 1.设 M,m 分别是函数 f(x)在[a,b]上的最大值和最小值,若 M=m,则 f′(x)( A.等于 0 C.等于 1 B.小于 0 D.不确定 ) 解析: 选 A 因为 M=m,所以 f(x)为常数函数,故 f′(x)=0,故选 A. 2.函数 y=2x -3x -12x+5 在[-2,1]上的最大值、最小值分别是( A.12,-8 C.12,-15 解析:选 A y′=6x -6x-12, 由 y′=0? x=-1 或 x=2(舍去). 2 3 2 ) B.1,-8 D.5,-16 x=-2 时,y=1;x=-1 时,y=12;x=1 时,y=-8. ∴ymax=12,ymin=-8.故选 A. 3.函数 f(x)=x -4x(|x|<1)( A.有最大值,无最小值 B.有最大值,也有最小值 C.无最大值,有最小值 D.既无最大值,也无最小值 解析:选 D f′(x)=4x -4=4(x-1)(x +x+1). 令 f′(x)=0,得 x=1.又 x∈(-1,1)且 1?(-1,1), ∴该方程无解,故函数 f(x)在(-1,1)上既无极值也无最值.故选 D. 1 4.函数 f(x)=2 x+ ,x∈(0,5]的最小值为( 3 2 4 ) x ) A.2 C. 17 4 1 1 3 2 B.3 D.2 2+ 1 2 x -1 x2 =0,得 x=1, 解析:选 B 由 f′(x)= 2 x x - = 且 x∈(0,1)时,f′(x)<0,x∈(1,5]时,f′(x)>0, ∴x=1 时,f(x)最小,最小值为 f(1)=3. ln x 5.函数 y= 的最大值为( x ) B.e D.10 1 A.e C.e -1 2 解析:选 A 令 y′= x x-ln x 1-ln x = =0? x=e.当 x>e 时,y′<0;当 x x2 2 -1 -1 0<x<e 时,y′>0,所以 y 极大值=f(e)=e ,在定义域内只有一个极值,所以 ymax=e . 6.函数 y= x-x(x≥0)的最大值为__________. 1 1-2 x 1 解析: y′= -1= ,令 y′=0 得 x= . 4 2 x 2 x 1 1 ∵0<x< 时,y′>0;x> 时,y′<0. 4 4 1 ∴x= 时,ymax= 4 1 答案: 4 7.函数 f(x)=xe ,x∈[0 ,4]的最小值为________. 解析:f′(x)=e -x e =e (1-x). 令 f′(x)=0,得 x=1(e >0), 1 4 ∴f(1)= >0,f(0)=0,f(4)= 4>0, e e 所以 f(x)的最小值为 0. 答案:0 8.若函数 f(x)=x -3x-a 在区间[0,3]上的最大值、最小值分别为 m,n,则 m-n= ________. 解析:∵f′(x)=3x -3, ∴当 x>1 或 x<-1 时,f′(x)>0; 当-1<x<1 时,f′(x)<0. ∴f(x)在[0,1]上单调递减,在[1,3]上单调递增. ∴f(x)min=f(1)=1-3-a=-2-a=n. 又∵f(0)=-a,f(3)=18-a,∴f(0)<f(3). ∴f(x)max=f(3)=18-a=m, ∴m-n=18-a-(-2-a)=20. 答案:20 9.设函数 f(x)=e - x -x. 2 (1)若 k=0,求 f(x)的最小值; (2)若 k=1,讨论函数 f(x)的单调性. 解:(1)k=0 时,f(x)=e -x,f′(x)=e -1. 当 x∈(-∞,0)时,f′(x)<0;当 x∈( 0,+∞)时,f′(x)>0,所以 f(x)在(-∞, 2 x x x 2 3 -x -x -x -x -x 1 1 1 - = . 4 4 4 k 2 0)上单调递减,在(0,+∞)上单调递增,故 f(x)的最小值为 f(0)=1. 1 2 x (2)若 k=1,则 f(x)=e - x -x,定义域为 R. 2 ∴f′(x)=e -x-1,令 g(x)=e -x-1, 则 g′(x)=e -1, 由 g′(x) ≥0 得 x≥0,所以 g(x)在[0,+∞)上单调递增, 由 g′(x)<0 得 x<0,所以 g(x)在(- ∞,0)上单调递减, ∴g(x)min=g(0)=0,即 f′(x)min=0,故 f′(x)≥0. 所以 f(x)在 R 上单调递增. 10.已知函数 f(x)=x +ax +bx+5,曲线 y=f(x)在点 P(1,f(1))处的切线方程为 y =3x+1. (1)求 a,b 的值; (2)求 y=f(x)在[-3,1]上的最大值. 解:(1)依题意可知点 P(1,f(1))为切点,代入切线方程 y=3x+1 可得,f(1)=3×1 +1=4, ∴f(1)=1+a+b+5=4,即 a+b=-2, 又由 f(x)=x +ax +bx+5 得, 又 f′(x)=3x +2ax+b, 而由切线 y=3x+1 的斜率可知 f′(1)=3, ∴3+2a+b=3,即 2a+b=0, ? ?a+b=-2, 由? ?2a+b=0. ? 2 3 2 3 2 x x x 解得? ? ?a=2, ?b=-4, ? ∴a=2,b=-4. (2)由(1)知 f(x) =x +2x -4x+5, 3 2 f′(x)=3x2+4x-4=(3x-2)(x+2), 2 令 f′(x)=0,得 x= 或 x=-2. 3 当 x 变化时,f(x),f′(x)的变化情况如下表: x f′(x) f(x) -3 (-3,-2) + -2 0 ?-2,2? ? 3? ? ? - 2 3 0 ?2,1? ?3 ? ? ? + 1 8 极大值 极小值 4 3 ?2? 95 ∴f(x)的极

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