证明(2)vx1,x,∈(a,b),x,< x1十 记 0 2 0-X1 对f(x)在x1,x,x0,x2l上分别应用L定理,得 f(x)-f(x1)=f(51(x1<51<x0) f(x2)-f(x)=f(2)h(x<52<x2) 两式相减,得 2f(x0)-[f(x1)+f(x2)=f(41)-f(2 由假设∫"(x)<0→f(x)在a,b内单调减 由51<2→f(1)-f(2)>0 →2f(x0)-Lf(x1)+f(x2)>0证明 1 2 1 2 (2)x , x (a,b), x x 0 1 2 0 1 2 0 , 2 h x x x x x x x = − = − + 记 = 对f (x)在[x1 , x0 ],[x0 , x2 ]上 分别应用L—定理,得 f (x0 ) − f (x1 ) = f ( 1 )h ( ) x1 1 x0 f (x2 ) − f (x0 ) = f ( 2 )h ( ) x0 2 x2 两式相减,得 2 f (x0 ) −[ f (x1 ) + f (x2 )] = [ f ( 1 ) − f ( 2 )]h 由假设 f (x) 0 f (x)在[a,b]内单调减 由 1 2 f ( 1 ) − f ( 2 ) 0 2 f (x0 ) −[ f (x1 ) + f (x2 )] 0