Question

高一函数

已知函数f（x)对任意实数a,b均有f（a)+f（b)=f（a+b)，且当x＞0时，f（x)＜0,f（1)=－三分之二。
1判断并证明在f（x)R上的单调性。2求f（x)在【-3,3】上的最大值与最小值

已知函数f（x)对任意实数a,b均有f（a)×f（b)=f（a×b)，定义域为R，且f（27）=3，写出一个满足条件的函数

Answer 1

(1)
for y> x
y= x+ c ( c > 0)
f(y) = f(x+c)
= f(c) + f(x)
< f(x)
f is decreasing on R
(2)
put
a = b =0
f(0) + f(0) = f(0)
=> f(0) =0
put a = b =1
f(1) + f(1) = f(2)
=> f(2) = -4/3
put a=1 , b =2
f(1) + f(2) = f(3)
=> f(3) = -2

put a= -1, b=1
f(0) = f(1) + f(-1)
0 = -2/3 + f(-1)
=> f(-1) = 2/3
put a = b =-1
=> f(-2) = 4/3
put a=-2, b=-1
=> f(-3) = 2
f（x)在【-3,3】上
最大值 = f(-3) =2
最小值 = f(3) = -2

(3)
f(x) = log(3) x ( log with base 3 )