Question

设函数f(x){e^x,x<0 a-x，x≥0在(-∞,+∞)上连续,则常数的值为？

Answer 1

lim[x-->0-]f(x)=lim[x-->0+]e^x=1
f(0)=a
lim[x-->0+]f(x)=lim[x-->0+](a-x)=a
因为函数f(x){e^x,x<0 a-x，x≥0在(-∞,+∞)上连续，
所以常数的值为：a=1 。

Answer 2

f(x)
=e^x ; x<0
=a-x ; x≥0
f(0-) = lim(x->0) e^x =1
f(0)=f(0+)=lim(x->0) (a-x) = a
f(0)=f(0+)=f(0-)
a=1