非羊的天空
春芽
共回答了25个问题采纳率:92% 举报
f(x)=ax+b ; x>=0
=arctan(1/x)+e^(1/x) ; x0+)f(x)
= lim(x->0+) ax+b
= b
lim(x->0-)f(x)
= lim(x->0-)arctan(1/x)+e^(1/x)
=-π/2
f(0+)=f(0-)
=>b= -π/2
f'(0+)=a
f'(x-) = -1/[x^2(x^2+1)] -(1/x^2)e^(1/x)
f'(0-) = -[1- (x^2+1)e^(1/x)]/[x^2(x^2+1)]
f'(0+) = -1
a=-1
1年前
6