求导数大神
2个回答
展开全部
1) f'(e) = [lnx + (x+a)/x - 1], for x = e, = 1
0+(e+a)/e - 1 = 1 ==> a = e
2) a = 0
f(x) = xlnx - x + 1
f'(x) = lnx = 0 ==> x = 1
By the first derivative test, fmin = f(1) = 0
Therefore, f(x) > or = 0
3) f'(x) = lnx + a/x = 0
a = -xlnx
a' = -lnx - 1 = 0
x = 1/e
a < (-1/e)ln(1/e) = 1/e
0+(e+a)/e - 1 = 1 ==> a = e
2) a = 0
f(x) = xlnx - x + 1
f'(x) = lnx = 0 ==> x = 1
By the first derivative test, fmin = f(1) = 0
Therefore, f(x) > or = 0
3) f'(x) = lnx + a/x = 0
a = -xlnx
a' = -lnx - 1 = 0
x = 1/e
a < (-1/e)ln(1/e) = 1/e
更多追问追答
追问
第三问,导函数等于0,a的值应该少个负号吧,后面能详细一点吗,有点不明白诶
追答
lnx + 1 = 0
lnx = -1
x = e^(-1) = 1/e > 0
在x = 1/e处,a取得最小值(1/e)ln(1/e) = -1/e。
所以,在其它点处有:
a > -1/e
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询