函数g(X)周期为1,已知f(X)=X+g(X)定义域为【0,1】时,值域为【-2,5】,求定义域为【0,3】时值域

题目

函数g(X)周期为1,已知f(X)=X+g(X)定义域为【0,1】时,值域为【-2,5】,求定义域为【0,3】时值域

答案

g(x)周期为1,因此有g(x+1)=g(x)
f(x+1)=(x+1)+g(x+1)
=x+g(x)+1
=f(x)+1
同理,
f(x+2)=f(x+1)+1
=>f(x+2)=f(x)+2
当0<=x<=1时,-2<=f(x)<=5
=>-1<=f(x)+1<=6
=>-1<=f(x+1)<=6 ,则1<=x<=2时,-1<=f(x)<=6
=>0<=f(x+2)<=7,则2<=x<=3时,0<=f(x)<=7
因此有0<=x<=3时,-2<=f(x)<=7
所以f(x)的值域是[-2,7]