应用拉格朗日定理求证|arctanx1-arctanx2|小于或等于|x1-x2|

Question

应用拉格朗日定理求证|arctanx1-arctanx2|小于或等于|x1-x2|

Answer 1

令f(t)=arctant
根据拉格朗日中值定理,在(min{x1,x2},max{x1,x2})中存在一点P,使得
f'(P)(max{x1,x2}-min{x1,x2})=f(max{x1,x2})-f(min{x1,x2})
1/(1+P^2)*(max{x1,x2}-min{x1,x2})=f(max{x1,x2})-f(min{x1,x2})
f(max{x1,x2})-f(min{x1,x2})