According to the differential mean value theorem, there is c 1 at (0, x2) and c2 at (x 1, x 1+x2), so,

f(x2)-f(0)=f'(c 1)x2

f(x 1+x2)-f(x 1)= f '(C2)x2，

Note that f'' (x) < 0, so f'(x) is a subtraction function, so

F'(c 1)>F'(c2), so there is.

f(x 1+x2)-f(x 1)= f '(C2)x2 & lt； f'(c 1)x2=f(x2)

The conclusion of the mobile project.