数学题,在线等回答,满意采纳

设f(x) = x^x ,首先f(0)不存在.如果考虑x<0,则要考虑复数情况.
设y=f(x)
y = x^x
ln(y) = x * ln(x)
dy / y = x * dx / x + ln(x) * dx
dy / y = dx * (1 + ln(x))
dy / x^x = dx * (1 + ln(x))
dy = dx * x^x * (1 + ln(x))
dy / dx = x^x * (1 + ln(x))
dy/dx = 0
0 = x^x * (1 + ln(x))
而x^x ≠ 0
0 = 1 + ln(x)
-1 = ln(x)
e^(-1) = x
1 / e = x
所以f(x)在x>0的最小值为
f(1/e) = (1/e)^(1/e) = 0.69220062755535
既然如此,怎么来会有a^a+b^b=1呢?c＝d
a＝－b
则abcd＝－1