Question

第四，五题都不会，你能为我解答吗？

Answer 1

(4)
lim(x->0) x/[f(x0-2x) -f(x0)] = 1/4 (0/0)
lim(x->0) 1/[-2f'(x0-2x) ] = 1/4
1/[-2f'(x0)] = 1/4
-2f'(x0) = 4
f'(x0)= -1/2

(5)
lim(x->1) (x^2+ax+b)/(1-x) =5 (0/0)
=>
1^2+a(1) +b =0
a+b = -1 (1)

lim(x->1) (x^2+ax+b)/(1-x) =5 (0/0)
lim(x->1) (2x+a)/(-1) =5
-(2+a) =5
2+a=-5
a=-7

from (1)
a+b=-1
-7+b=-1
b=6

追问

第四题答案是负二

追答

(4)
lim(x->0) x/[f(x0-2x) -f(x0)] = 1/4 (0/0)
lim(x->0) 1/[-2f'(x0-2x) ] = 1/4
1/[-2f'(x0)] = 1/4
-2f'(x0) = 4
f'(x0)= -2

Answer 2