设f(x)=｛ex,x≤1 ax+b,x＞1｝在x=1可导，试求a与b 大神求解。
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f(x)
=e^x ,x≤1
=ax+b ,x＞1
f(0)=f(0-)=lim(x->0) e^x = 1
f(0+) =lim(x->0) ( ax+b) =b
f(0)=f(0+)=f(0-) => b=1
f'(0-)
=lim(h->0) [e^h - f(0)] /h
=lim(h->0) [e^h - 1] /h
=lim(h->0) h/h
=1
f'(0+)
=lim(h->0) [(ah+1) - f(0)] /h
=lim(h->0) ah /h
=a
f'(0+)=f'(0-) => a=1
ie
(a,b)=(1,1)
=e^x ,x≤1
=ax+b ,x＞1
f(0)=f(0-)=lim(x->0) e^x = 1
f(0+) =lim(x->0) ( ax+b) =b
f(0)=f(0+)=f(0-) => b=1
f'(0-)
=lim(h->0) [e^h - f(0)] /h
=lim(h->0) [e^h - 1] /h
=lim(h->0) h/h
=1
f'(0+)
=lim(h->0) [(ah+1) - f(0)] /h
=lim(h->0) ah /h
=a
f'(0+)=f'(0-) => a=1
ie
(a,b)=(1,1)
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追问
在x=1可导，怎么带的0哦？？不懂
追答
看错
f(x)
=e^x ,x≤1
=ax+b ,x＞1
f(1)=f(1-)=lim(x->1) e^x = e
f(1+) =lim(x->1) ( ax+b) =a+b
f(1)=f(1+)=f(1-)
a+b = e (1)
f'(1-)
=lim(h->0) [e^(1+h) - f(1)] /h
=lim(h->0) [e^(1+h) - e] /h
= e
f'(1+)
=lim(h->0) [a(h+1)+b - f(1)] /h
=lim(h->0) (ah +a+b -e)/h
=a
f'(1+)=f'(1-) => a=e
from (1)
a+b=e
b=0
(a,b) = (e,0)
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