Question

请问这道高数题怎么解？

Answer 1

x<0，导数为1/(1+x)
x>=0，导数为b
1、x=0，左右导数相等，有1/1=b
2、函数连续，得出a+ln1=2，得出a=2

Answer 2

左导数是h->0-: [a+ln(1+h)-2]/h, 因a+ln(1+h)-2=0, 所以a=2,
即h->0-: [a+ln(1+h)-2]/h=1.
所以,右导数是h->0+: (bh+2-2)/h=b=1.

Answer 3

f(x)
=a+ln(1+x) ; x<0
=bx+2 ; x≥0
f(0-)=lim(x->0-) [a+ln(1+x) ] = a
f(0)=f(0+)=lim(x->0+) (bx+2) = 2
f(0)=f(0+)=f(0-)
=>a =2
f'(0-)
=lim(h->0) [ a+ln(1+h) -f(0) ]/h
=lim(h->0) [ 2+ln(1+h) -2 ]/h
=lim(h->0) ln(1+h) /h
=1
f'(0+)
=lim(h->0) [ bh+2 -f(0) ]/h
=lim(h->0) [ bh +2 -2 ]/h
=b
f'(0-)=f'(0+)
=> b=1
(a,b)=(2,1)

Answer 4