第2个回答 2021-05-21
+-×÷﹢﹣±/=≈≡≠∧∨∑∏∪∩∈⊙⌒⊥∥∠∽≌<>≤≥≮≯∧∨√﹙﹚[]﹛﹜∫∮∝∞⊙∏
f(x)
=ax+b ; x>1
=x^2 ; x≤1
f(1+)=lim(x->1+) (ax+b) = a+b
f(1)=f(1-)=lim(x->1-) x^2 = 1
x=1, f(x)连续
=>
f(1+)=f(1-)=f(1)
a+b=1 (1)
f'(1+)
=lim(h->0+) [ f(1+h) -f(1) ]/ h
=lim(h->0+) [ a(1+h)+b -1 ]/ h
=lim(h->0+) [ a+b -1 +ah ]/ h
=lim(h->0+) [0 +ah ]/ h
=a
f'(1-)
=lim(h->0-) [ f(1+h) -f(1) ]/ h
=lim(h->0-) [ (1+h)^2 -1 ]/ h
=lim(h->0-) [ 2h+h^2 ]/ h
=lim(h->0-) (2+h)
=2
x=1, f(x) 可导
=>
f'(1+)=f'(1-)
a=2
from (1)
a+b=1
2+b=1
b=-1
(a,b)=(1,-1)