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51问答网 > 已知函数f（x）=lnx， g(x)= a x ，设F（x）=f（x）+g（x）．（Ⅰ）当a=1时，求函数F（x）的单

已知函数f（x）=lnx， g(x)= a x ，设F（x）=f（x）+g（x）．（Ⅰ）当a=1时，求函数F（x）的单

2024-12-28 09:02:42

推荐回答（1个）

回答1：

（Ⅰ）由已知a=1，可得 F(x)=f(x)+g(x)=lnx+

1

x

，函数的定义域为（0，+∞），
则 F′(x)=

1

x

-

1

x 2

=

x-1

x 2

由 F′(x)=

1

x

-

1

x 2

=

x-1

x 2

＞0 可得F（x）在区间（1，+∞）上单调递增，
F′(x)=

1

x

-

1

x 2

=

x-1

x 2

＜0 得F（x）在（0，1）上单调递减；
（Ⅱ）由题意可知 k=F′( x 0 )=

x 0 -a

x

20

≤

1

2

对任意0＜x₀ ≤3恒成立，
即有 x 0 -

1

2

x

20

≤a 对任意0＜x₀ ≤3恒成立，即 ( x 0 -

1

2

x

20

) max ≤a ，
令 t= x 0 -

1

2

x

20

=-

1

2

(

x

20

-2 x 0 )=-

1

2

( x 0 -1 ) 2 +

1

2

≤

1

2

，
则 a≥

1

2

，即实数a的最小值为

1

2

．

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