已知函數(shù)f(x+1)為奇函數(shù),函數(shù)f(x-1)為偶函數(shù),且f(0)=2,則f(4)= .
【答案】分析:先根據(jù)函數(shù)f(x+1)為奇函數(shù)得到f(x+1)=-f(-x+1)⇒f(-2)=-f(4);再結(jié)合函數(shù)f(x-1)是偶函數(shù)得到f(x-1)=f(-x-1),進(jìn)而根據(jù)f(0)=f(-2)=-f(4)即可得到答案.
解答:解:因為函數(shù)f(x+1)為奇函數(shù)
所以有:f(x+1)=-f(-x+1)①
∵函數(shù)f(x-1)是偶函數(shù)
∴f(x-1)=f(-x-1)②
在②中令x=1得:f(0)=f(-2)
在①中令x=-3得:f(-2)=-f(4)
∴f(0)=f(-2)=-f(4)=2.
∴f(4)=-2
故答案為:-2.
點評:本題主要考查函數(shù)奇偶性的應(yīng)用.解決問題的關(guān)鍵在于根據(jù)函數(shù)f(x+1)為奇函數(shù)得到f(x+1)=-f(-x+1)⇒f(-2)=-f(4);再結(jié)合函數(shù)f(x-1)是偶函數(shù)得到f(x-1)=f(-x-1)⇒f(0)=f(-2).