第15题。。。。
答案:1 悬赏:30 手机版
解决时间 2021-11-25 20:24
- 提问者网友：贪了杯
- 2021-11-25 03:07
第15题。。。。
最佳答案
- 五星知识达人网友：不如潦草
- 2021-11-25 03:30
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: xf(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: xf(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
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