Question

函数f(X)=2x³-6x²+7在（0,2）是减函数 写具体的过程 注意是具体 少一步我就会看不太懂

Answer 1

不妨设2>x1>x2>0

f(x1)-f(x2)=2x1^3-6x1^2+7-2x2^3+6x2^2-7

=2(x1^3-x2^3)-6(x1^2-x2^2)

=2(x1-x2)(x1^2+x1x2+x2^2-3x1-3x2)

=2(x1-x2)[(x1^2-2x1)+(x2^2-2x2)+1/2(x1x2-2x1)+1/2(x1x2-2x2)]

=2(x1-x2)[x1(x1-2)+x2(x2-2)+x1/2(x2-2)+x2/2(x1-2)]

x1>x2,x1>0,x2>0,x1-2