已知函数f(x)=e∧x+ax,g(x)=ax-lnx,其中a

(i) 先考虑a = 0
f(x) = e^x,f'(x) = e^x > 0
g(x) = -lnx,g'(x) = -1/x < 0 (在定义域x > 0内)
单调性不可能相同
(2) a < 0
f(x) = ax + e^x,
f'(x) = a + e^x = 0,x = ln(-a)
0 < x < ln(-a)时:f'(x) < 0
x > ln(-a)时:f'(x) > 0
g'(x) = a - 1/x < 0 (a < 0; x > 0时,-1/x < 0)
即a < 0时,f(x)和g(x)在0 < x < ln(-a)时均为减函数