求y=x+(1-x)½极值点和极值

2023-07-26 11:02发布

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y=x+(1-x)^(1/2)1-x≥0-1≤x≤1y=x+(1-x)^(1/2)y = 1 - (1/2)(1-x)^(-1/2)y=02(1-x)^(1/
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1楼 · 2023-07-26 11:33.采纳回答

y=x+(1-x)^(1/2)
1-x≥0
-1≤x≤1
y=x+(1-x)^(1/2)
y' = 1 - (1/2)(1-x)^(-1/2)
y'=0
2(1-x)^(1/2) -1 =0
1-x =1/4
x= 3/4
y'|x=3/4+ <0 , y'|x=3/4- >0
x=3/4 ( max)
max y = y(3/4) =3/4+(1-3/4)^(1/2) =3/4 +1/2 = 5/4追问

2(1-x)^(1/2) -1 =0这部怎么来的？看不懂
追答
y'=0
1 - (1/2)(1-x)^(-1/2)=0
[2(1-x)^(1/2) -1]/[2(1-x)^(1/2)] =0

2(1-x)^(1/2) -1 =0
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