Question

已知函数f（x）=-x²+2ex+m-1，g（x）=x+e²/x（x＞0）若g（x）=m有实数根求m的取值范

Answer 1

(1)
这相此山当于求g(x)的值域(x＞0)
g'(x)
=
1
-
e²/x²
=
0,
x
=
e
0
<
x
<森洞中
e:
g'(x)
<
0
x
>
e:
g'(x)
>
0
最小值g(e)
=
2e
m
≥
2e即可
(2)h(x)
=
g(x)
-
f(x)
h'(x)
=
1
-
e²/x²-(-2x
+
2e)
=
2x
-
e²/x²-
2e
+
1
=
0
显然其根为x
=
e
(一元三次方程另二根小于0)
0
<
x
<
e:
h'颤滚(x)
<
0
x
>
e:
h'(x)
>
0
最小值h(e)=
1
+
2e
-
e²
-m
g(x)-f(x)=0有两个相异实根,
h(e)=
1
+
2e
-
e²
-m
<
0
m
>
1
+
2e
-
e²