已知函数y=f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1),且在[0,1]上单调递减,则:f(7/2),f(7/3),f(7/5)的大小顺序?
热心网友
因函数y=f(x)对任意实数都有f(-x)=f(x),f(x)=-f(x+1)所以f(7/2)=f[-7/2] =-f[-(7/2)+1] =-f(-5/2) =f[-(5/2)+1] =f(-3/2) =-f[-(3/2)+1] =-f(-1/2) =f(1/2)同理f(7/3)=f(1/3)f(7/5)=f(3/5)因f(x)在[0,1]上单调递减,而（3/5)(1/2)(1/3)所以f(7/3)f(7/2)f(7/5)
热心网友
f(x)在[0,1]上单调递减所以将所有的数按照f(x)=-f(x+1),f(-x)=f(x)化到[0,1]比较f(7/2)=-f(5/2)=f(3/2)=-f(1/2)=f(-1/2)=f(1/2)f(7/3)=-f(4/3)=f(1/3)f(7/5)=-f(2/5)=f(-3/5)=f(3/5)因为f(1/3)f(1/2)f(3/5)所以f(7/3)f(7/2)f(7/5)