Question

f(x+y)=f(x)+f(y),f(-3)=a，如果x>0,f(x)<0,f(1)=-½，求f(x)在区间[-2,4]上的最大值和最小值

Answer 1

f(3+0)=f(3)+f(0) f(0)=0;
f(2)= -1 f(3)= -1.5 f(4)= -2
f(2)=f（4）+f(-2)= -1
f(-2)=1

最大f(-2)=1
最小 f(4)= -2

Answer 2

f(x+y) = f(x) + f(y), f(-3) = a
if x>0, f(x)<0, f(1) = -1/2
put x=y =1
f(2) = 2f(1) = -1
put x=y = 2
f(4) = 2f(2) = -2
put x=-3,y=1
f(-2) = f(-3)+ f(1)
= a -1/2

for x> y
x = y+c ( c>0)
f(x) = f(y+c)
= f(y) + f(c) (c> 0 , f(c) <0)
< f(y)
f is decreasing function
f(x)在区间[-2,4]上的
最大值 = f(-2)=a -1/2
最小值 = f(4)=-2