极值点偏移练习1
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
构造对称函数
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)
例1
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
换元
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