Question

设函数f（x）处处连续求导 求 a，b

Answer 1

f(x)
=[1-√(1-x)]/x ; x<0
=ax+b ; x≥0
f(0-)
= lim(x->0) [1-√(1-x)]/x (0/0)
= lim(x->0) 1/[2√(1-x)]
=1/2
f(0+) = b
f(0-)=f(0+)
=> b =1/2
f'(0+)
= lim(h->0) ( ah + 1/2 - f(0) ) /h
=lim(h->0) (ah+ 1/2 - 1/2 )/h
= a
f'(0-)
= lim(h->0) [ f(0) - [1-√(1-h)]/h ] /h
= lim(h->0) [ 1/2 - [1-√(1-h)]/h ] /h
= lim(h->0) [ h - 2[1-√(1-h)] ] /(2h^2)
= lim(h->0) [（1/4) h^2 ] /(2h^2) ( 等价无穷小 ）
=1/8
=> a= 1/8
h->0
分子：
√(1-h) ~ 1 - (1/2)h +(1/8)h^2
2[1-√(1-h)] ~ h - (1/4)h^2
h- 2[1-√(1-h)] ～ (1/4)h^2

追问

为啥呀