高中数学,第三题求解
展开全部
4/3>0
f(4/3)=2×(4/3)=8/3
-4/3<0
f(-4/3)=f(-4/3+1)
=f(-1/3)
=f(-1/3+1)
=f(2/3)
=2×(2/3)
=4/3
提示:只要<0,就+1,直到自变量>0为止,再乘以2
f(4/3)+f(-4/3)=8/3+4/3=4
综合起来写:
f(4/3)+f(-4/3)
=2×(4/3)+f(-4/3+1)
=8/3+f(-1/3)
=8/3+f(-1/3+1)
=8/3+f(2/3)
=8/3+2×(2/3)
=8/3+4/3
=4
f(4/3)=2×(4/3)=8/3
-4/3<0
f(-4/3)=f(-4/3+1)
=f(-1/3)
=f(-1/3+1)
=f(2/3)
=2×(2/3)
=4/3
提示:只要<0,就+1,直到自变量>0为止,再乘以2
f(4/3)+f(-4/3)=8/3+4/3=4
综合起来写:
f(4/3)+f(-4/3)
=2×(4/3)+f(-4/3+1)
=8/3+f(-1/3)
=8/3+f(-1/3+1)
=8/3+f(2/3)
=8/3+2×(2/3)
=8/3+4/3
=4
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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