微积分第三章习题3.1答案
习题3.1=-cos 2x
x 2-4cos x sin x x +ln x (-2cos 2x +2sin 2x ) 1.
(1)∆x =0.1
(2)∆y =f (1.1)-f (1)=0.21
(3)平均变化率=∆y ∆x =2.1
(4)f '(x ) x =1=2x x =1=2
2. ∆lim ∆y -2x -2∆x +5+2x -5-x →0∆x =∆lim x →0∆x =∆lim 2∆x x →0∆x =-2 3.
解:y =-2x +7斜率为-2,抛物线上点的斜率必为-2,即:
(x 2) '=2x =-2→x =-1,抛物线上的点为(-1,1) 4
(1)12x 3(2)4πr 2(3)3x -1
2(4)-2u -3
2
53
(5)6x 5-6x 2+3(6)13
x 2-10x 2+3-4
3
2x -1
2+1-
2x 2
2(7)-8x 3+x
5. 解:
银行遭受损失的变化率为:f '(t ) =-2t +10
(1)2001对应t =3; f '(t ) t =3=-2t +10t =3=4
(2)2003对应t =5; f '(t )
t =5=-2t +10t =5=0
(2)2005对应t =7; f '(t ) t =7=-2t +10t =7=-4
6.
(1)f '(t ) 的含义为在t 时刻,温度的变化率.
(2)f '(20)的单位是C 0/min ,f '(20)含义为在食物放入 热烤箱20分钟时,时间增加一分钟,温度增加二度. 7. (1)C'(x ) =-20x +300
解:(2)C'(10)=-200+300=100
(3)C(10)-1000+3000+130
10=10=213
8. -4+3-3(8)12x 2x 2
2(1)P'(x ) =-x +73
(2)10000=10(千元) ,30000=30(千元)
广告支出为10000万元时,公司收入的变化率为:解:21P '(x ) x =10=-x +7=33x =10
2P '(x ) x =30=-x +7=-133x =30 广告支出为30000万元时,公司收入的变化率为:
9.
(1)f (5)=32.5, f (5.05)=32.3998, x ∈[5000,5050]
32.3998-32.5=-0.00200550
f (5)=32.5, f (5.01)=32.48, x ∈[5000,5050]
32.48-32.5=-0.00200150
(2)f '(x ) x =5=-0.2x -1=-2
10.
解:x →0
11. 解:
x →1+
x →1-lim f (x ) =1, f (0)=0, 所以函数不连续,也不可导。 lim f (x ) =lim 2-x =1, +x →1x →1lim f (x ) =lim x =1, -
f (1)=1所以函数在x=1处连续
f (1-∆x ) -f (1) -∆x =lim -=1∆x →0∆x →0-∆x -∆x
f (1+∆x ) -f (1) -∆x f (x ) 在x=1的右导数为:lim +=lim +=-1∆x →0∆x →0∆x ∆x
左右导数不等,所以f (x ) 在x=1的导数不存在。f (x ) 在x=1的左导数为:lim -
12. 解:
(1)f (0)=1, lim f (x ) =2+a -x →0
根据连续的性质知:2+a =1→a =-1, b 可取任意常数。
(2)f (x ) 在x =0的左导数为2
f (x ) 在x =0的右导数为b →b=2,a 可取任意常数