Question

求解高数问题

Answer 1

简单计算一下即可，答案如图所示

Answer 2

(1)
y'=3ax^2+2bx+c

y'(0)=0
=>c=0
y'(1)=0
3a+2b+c=0
3a+2b=0 (1)
y=ax^3+bx^2+cx+d
过点 (0,1)
=> d =1
过点 (1,0)
a+b+c +d =0
a+b=-1 (2)
(1)-2(2)
a=2
from (1)
3a+2b=0
2b=-6
b=-3
ie
y=ax^3+bx^2+cx+d = 2x^3-3x^2+1
(2)
lim(x->a) [e^(-x) -e^(-a)]/(x-a)
洛必达
=lim(x->a) -e^(-x)
=-e^(-a)

Answer 3

1.水平切线，1阶导数为0
2.0/0型，用洛必达法则。