对于函数f(x)=x^2-26x+1921,任取x1,x2∈[1,29]且x1≠x2,
证明:│f(x2)-f(x1)│<32│x2-x1│
对于函数f(x)=x^2-26x+1921,任取x1,x2∈[1,29]且x1≠x2,
证明:│f(x2)-f(x1)│<32│x2-x1│
f(x2)-f(x1)=x2²-26x2+1921-x1²+26x1-1921=x2²-x1²-26x2+26x1=(x2-x1)(x2+x1)-26(x2-x1)=(x2-x1)(x2+x1-26)
∵x1,x2∈[1,29],x1≠x2
∴x2+x1∈(2,58)
∴x2+x1-26∈(-24,32)
∴|x2+x1-26|<32
∴|f(x2)-f(x1)|=|(x2-x1)(x2+x1-26)|=|x2+x1-26|×|x2-x1|<32|x2-x1|