第15题。。。。
1个回答
展开全部
f(x)= x^2 +|x-a| +1
case 1: x≥a
f(x)= x^2 +|x-a| +1
= x^2 +x-a +1
= (x+1/2)^2 - a + 3/4
min f(x) = f(-1/2) = -a+3/4
case 2: x<a
f(x)= x^2 +|x-a| +1
= x^2 -(x-a) +1
= (x-1/2)^2 + a + 3/4
min f(x) = f(1/2) = a + 3/4
case 1: x≥a
f(x)= x^2 +|x-a| +1
= x^2 +x-a +1
= (x+1/2)^2 - a + 3/4
min f(x) = f(-1/2) = -a+3/4
case 2: x<a
f(x)= x^2 +|x-a| +1
= x^2 -(x-a) +1
= (x-1/2)^2 + a + 3/4
min f(x) = f(1/2) = a + 3/4
本回答由提问者推荐
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
