第1个回答 2019-08-08
f(x)
=e^x ; x≤1
=ax+b ; x>1
f(1+)= lim(x->1) (ax+b) =a+b
f(1)=f(1-)=lim(x->1) e^x = e
f(1+)=f(1)=f(1-)
a+b=e (1)
f'(1-) = e
f'(1+)
=lim(h->0) [ a(h+1) +b -f(1)]/h
=lim(h->0) [ a(h+1) +b -e ]/h
=lim(h->0) [ (a+b-e) +ah ]/h
=a
f'(1+) =f'(1-)
=> a=e
from (1)
a+b=e
e+b=e
b=0
ie
(a,b)=(e,0)本回答被网友采纳