若函数
f
(
x
)
f\left(x\right)
f(x)在
x
=
x
0
x=x_0
x=x0处取得极值，
且函数
y
=
f
(
x
)
y=f\left(x\right)
y=f(x)图像与直线
y
=
b
y=b
y=b交于
A
(
x
1
,
b
)
,
B
(
x
2
,
b
)
A\left(x_1,b\right),B\left(x_2,b\right)
A(x1,b),B(x2,b)亮点，
则
A
B
AB
AB的中点坐标
M
(
x
1
+
x
2
2
,
b
)
M\left(\frac{x_1+x_2}{2},b\right)
M(2x1+x2,b),但是
x
0
x_0
x0不一定等于
x
1
+
x
2
2
\frac{x_1+x_2}{2}
2x1+x2
一般：给你 f ( x 1 ) = f ( x 2 ) f\left(x_1\right)=f\left(x_2\right) f(x1)=f(x2)，然后证明 x 1 + x 2 > 2 x 0 x_1+x_2>2x_0 x1+x2>2x0之类的
假设要证明
x
1
+
x
2
>
2
x
0
x_1+x_2>2x_0
x1+x2>2x0
设
x
1
<
x
0
<
x
2
x_1
不妨假设
(
−
∞
,
x
0
)
\left(-\infty, x_0\right)
(−∞,x0)单调递减，
(
x
0
,
+
∞
)
\left(x_0,+\infty\right)
(x0,+∞)单调递增
那么
2
x
0
−
x
2
<
x
1
<
x
0
2x_0-x_2
也就是
f
(
2
x
0
−
x
2
)
>
f
(
x
1
)
=
f
(
x
2
)
f\left(2x_0-x_2\right)>f\left(x_1\right)=f\left(x_2\right)
f(2x0−x2)>f(x1)=f(x2)
令
F
(
x
)
=
f
(
x
)
−
f
(
2
x
0
−
x
)
(
x
>
x
0
)
F\left(x\right)=f\left(x\right)-f\left(2x_0-x\right)\left(x>x_0\right)
F(x)=f(x)−f(2x0−x)(x>x0)
要证明
F
(
x
)
<
0
F\left(x\right)<0
F(x)<0
也可以构造函数
G
(
x
)
=
f
(
1
+
x
)
−
f
(
1
−
x
)
G\left(x\right)=f\left(1+x\right)-f\left(1-x\right)
G(x)=f(1+x)−f(1−x)
f
(
x
)
=
x
e
−
x
,
x
1
≠
x
2
f\left(x\right)=xe^{-x},x_1\neq x_2
f(x)=xe−x,x1=x2
f
(
x
1
)
=
f
(
x
2
)
f\left(x_1\right) = f\left(x_2\right)
f(x1)=f(x2)
证明：
x
1
+
x
2
>
2
x_1+x_2>2
x1+x2>2
证明：
f
′
(
x
)
=
(
1
−
x
)
e
−
x
f^{\prime}\left(x\right)=\left(1-x\right)e^{-x}
f′(x)=(1−x)e−x
x
<
1
x<1
x<1单调递增，
x
>
1
x>1
x>1单调递减
x
=
1
x=1
x=1极值点
不妨假设
x
1
<
1
<
x
2
x_1<1
x
1
+
x
2
>
2
2
−
x
2
<
x
1
<
1
f
(
2
−
x
2
)
<
f
(
x
1
)
f
(
2
−
x
2
)
<
f
(
x
2
)
\begin{aligned} x_1+x_2 &>2 \\ 2-x_2 &
因此构造
F
(
x
)
=
f
(
x
)
−
f
(
2
−
x
)
(
x
>
1
)
F\left(x\right)=f\left(x\right)-f\left(2-x\right)\left(x>1\right)
F(x)=f(x)−f(2−x)(x>1)
F
′
(
x
)
=
(
x
−
1
)
(
e
2
(
x
−
1
)
−
1
)
e
x
>
0
F^{\prime}\left(x\right)=\frac{\left(x-1\right)\left(e^{2\left(x-1\right)}-1\right)}{e^{x}}>0
F′(x)=ex(x−1)(e2(x−1)−1)>0
F
(
x
)
>
F
(
1
)
=
0
F\left(x\right)>F\left(1\right)=0
F(x)>F(1)=0
因此
f
(
x
2
)
>
f
(
2
−
x
2
)
f\left(x_2\right)>f\left(2-x_2\right)
f(x2)>f(2−x2)
即
x
1
+
x
2
>
2
x_1+x_2>2
x1+x2>2
f
(
x
)
=
x
−
a
e
x
f\left(x\right) = x-ae^x
f(x)=x−aex
x
1
≠
x
2
x_1\neq x_2
x1=x2
f
(
x
1
)
=
f
(
x
2
)
=
0
f\left(x_1\right)=f\left(x_2\right)=0
f(x1)=f(x2)=0
证明：
x
1
+
x
2
>
2
x_1+x_2>2
x1+x2>2
证明：
f
(
x
1
)
=
f
(
x
2
)
=
0
f\left(x_1\right)=f\left(x_2\right)=0
f(x1)=f(x2)=0
则
x
1
=
a
e
x
1
,
x
2
=
a
e
x
2
x_1=ae^{x_1}, x_2=ae^{x_2}
x1=aex1,x2=aex2
相减得到
x
1
−
x
2
=
a
(
e
x
1
−
e
x
2
)
x_1-x_2=a\left(e^{x_1}-e^{x_2}\right)
x1−x2=a(ex1−ex2)
x
1
+
x
2
=
a
(
e
x
1
+
e
x
2
)
=
x
1
−
x
2
e
x
1
−
e
x
2
(
e
x
1
+
e
x
2
)
=
x
1
−
x
2
e
x
1
−
x
2
−
1
(
e
x
1
−
x
2
+
1
)
\begin{aligned} x_1+x_2 &=a\left(e^{x_1}+e^{x_2}\right)\\ &=\frac{x_1-x_2}{e^{x_1}-e^{x_2}}\left(e^{x_1}+e^{x_2}\right)\\ &=\frac{x_1-x_2}{e^{x_1-x_2}-1}\left(e^{x_1-x_2}+1\right) \end{aligned}
x1+x2=a(ex1+ex2)=ex1−ex2x1−x2(ex1+ex2)=ex1−x2−1x1−x2(ex1−x2+1)
不妨假设
x
1
>
x
2
x_1>x_2
x1>x2
令 t = x 1 − x 2 > 0 t=x_1-x_2>0 t=x1−x2>0
要证明
t
e
t
−
1
(
e
t
+
1
)
>
2
\frac{t}{e^t - 1}\left(e^t+1\right)>2
et−1t(et+1)>2
即
t
(
e
t
+
1
)
−
2
(
e
t
−
1
)
>
0
t\left(e^t+1\right)-2\left(e^t-1\right)>0
t(et+1)−2(et−1)>0
令
F
(
t
)
=
t
(
e
t
+
1
)
−
2
(
e
t
−
1
)
(
t
>
0
)
F\left(t\right)=t\left(e^t+1\right)-2\left(e^t-1\right)\left(t>0\right)
F(t)=t(et+1)−2(et−1)(t>0)
F
′
(
t
)
=
t
e
t
−
e
t
−
1
F^{\prime}\left(t\right)=te^{t}-e^t-1
F′(t)=tet−et−1
F
′
′
(
t
)
=
t
e
t
>
0
F^{\prime\prime}\left(t\right)=te^t>0
F′′(t)=tet>0
则
F
′
(
t
)
>
F
′
(
0
)
=
0
F^{\prime}\left(t\right)>F^{\prime}\left(0\right)=0
F′(t)>F′(0)=0
则
F
(
t
)
>
F
(
0
)
=
0
F\left(t\right)>F\left(0\right)=0
F(t)>F(0)=0