上述不等式成立是因为 fx)>0, 2.设x)在{ab上连线,在(ab内厂(x)>0,证明(x)≈(x)-(a)在(a,b)内单增 证明假设axb)f(x)-/(a=f(5)(a<5) x-a P f(x2)-f(a)f(x2)-f(x1)+f(x1)-f( ∫(52)x2-x)+f(51)(x1-a) f(51)(x2-x1+x1-a) 厂(5)=叭(x1) 不等式成立是因为51<x1<2:f"'(x)>0说明∫(x)单增,于是∫(2)>f(51) 3.设fx)在[a,b上连续,在(a,b)内可导且∫(x)≤0,求证 F(x) 在(a,b)内也F(x)≤0 证明:方法1:因为f(x)≤0,所以x)单减 f()d+-f(x) [f(x)-f(1)]d<0 所以F(x)单减 方法2:假设a<x1<x2<b F(x1) f(=/(5),(a<5≤x因为几单减5在区间ax)内部) F(1)=~1 ndt f(5)(x1-a)+∫(2)(x2-x) f(51)(x1-a)+f(51)(x2-x1)=f(51)=F( 因为fx)单减2在区间(x1,x2)内部,即a<51<x1<2<x2,且∫(2)<∫(51) 4证明方程tanx=1-x在(0,1)内有唯一实根 证明令F(x)=tanx-1+x.F(O)=-1<0,F(1)=tanl>0, 所以在(0,1)中存在ξ,使F(2)=0上述不等式成立是因为 f(x) > 0, t < x. 2. 设 f(x)在[a, b]上连续, 在(a, b)内 f ' ' (x ) > 0 , 证明 x a f x f a x - - = ( ) ( ) f ( ) 在(a, b)内单增. 证明. 假设 a < x1 < x2 < b, ' ( ) ( ) ( ) ( ) 1 1 1 1 f f x x a f x f a x = - - = (a < x1 <x1 ) x a f x f x f x f a x a f x f a x - - + - = - - = 2 2 1 1 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) f( ) x a f x x f x a - - + - = 2 2 2 1 1 1 '(x )( ) ' (x )( ) ' ( ) ( ) ' ( )( ) 1 1 2 1 2 1 1 f x x a f x x x a x f x = = - - + - > 不等式成立是因为x1 <x1 <x2. f ' ' (x ) > 0 说明 f '(x ) 单增, 于是 ' ( ) ( ) 2 1 f x > f x . 3. 设 f(x)在[a, b]上连续, 在(a, b)内可导且 f '(x ) £ 0 , 求证: Ú - = x a f t dt x a F x ( ) 1 ( ) 在(a, b)内也 F'(x ) £ 0 . 证明: 方法 1: 因为 f '(x ) £ 0 , 所以 f(x)单减. ( ) 1 ( ) ( ) 1 ' ( ) 2 f x x a f t dt x a F x x a - + - - = Ú = Ú - - x a f t dt x a ( ) ( ) 1 2 + Ú - x a f x dt x a ( ) ( ) 1 2 = Ú - < - x a f x f t dt x a [ ( ) ( )] 0 ( ) 1 2 所以 F(x)单减. 方法 2: 假设 a < x1 < x2 < b ( ) ( ) 1 ( ) 1 1 1 1 f t dt f x x a F x x a = - = Ú , (a < x1 < x1)(因为 f(x)单减, x1在区间(a, x1)内部). ˙ ˚ ˘ Í Î È + - = - = Ú Ú Ú 2 1 2 1 ( ) ( ) 1 ( ) 1 ( ) 2 2 2 x x x a x a f t dt f t dt x a f t dt x a F x = [ ( )( ) ( )( )] 1 1 1 2 2 1 2 f x a f x x x a - + - - x x < [ ( )( ) ( )( )] ( ) ( ) 1 1 1 1 2 1 1 1 2 f x a f x x f F x x a - + - = = - x x x 因为 f(x)单减, x2在区间(x1, x2)内部, 即 a < x1 < x1 < x2 <x2, 且 ( ) ( ) 2 1 f x < f x . 4.证明方程tan x = 1 - x 在(0, 1)内有唯一实根. 证明. 令 F(x ) = tan x -1 + x . F(0) =-1 <0, F(1) = tan1 >0, 所以在(0, 1)中存在x, 使 F(x) = 0