已知f(x)是奇函数,且f(2-x)=f(x),当x∈[2,3]时,f(x)=log2(x-1),则f[1/3]= 要过程
2个回答
展开全部
x属于[2,3]时,f(x)=log2(x-1),则:
f(2+1/3)=log2(2+1/3-1)=log2(4/3)=log2(4)-log2(3)=2-log2(3)
又f(2-x)=f(x),故:
f(2+1/3)=f(2-(-1/3))=f(-1/3)
则f(-1/3)=2-log2(3)
又f(x)为奇函数,即f(-x)=-f(x)故:
f(-1/3)=-f(1/3)
则f(1/3)=-f(-1/3)=-(2-log2(3)=-2+log2(3)
f(2+1/3)=log2(2+1/3-1)=log2(4/3)=log2(4)-log2(3)=2-log2(3)
又f(2-x)=f(x),故:
f(2+1/3)=f(2-(-1/3))=f(-1/3)
则f(-1/3)=2-log2(3)
又f(x)为奇函数,即f(-x)=-f(x)故:
f(-1/3)=-f(1/3)
则f(1/3)=-f(-1/3)=-(2-log2(3)=-2+log2(3)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询