Question

f(x)=(X-1)e^X,若任意x>;1，都有f(X)>=x+m+ln(X-1)成立，求m取值范围

Answer 1

g(x)
=
f(x)
-
[
x
+
ln(x-1)
]
= xe^x
-
e^x
-
x
-
ln(x-1)；
g'(x)
= xe^x
-
1
-
1/(x-1)
=
0，
xe^x
=
1
+
1/(x-1)
=
x/(x-1)，e^x
=
1/(x-1)，x
=
-ln(x-1)；
g[-ln(x-1)]
=
-ln(x-1)/(x-1)
-
1/(x-1)
=
[
1
+
ln(x-1)
]/(1-x)
=
[
1
+
ln(x-1)
]/[
1
+
ln(x-1)
]
=
1，即
g(x) 最小值为
1；
故
1
-
m
>
0，m
<
1；m 的取值范围是 m
<
1
。