求极值？
1个回答
展开全部
f(x) = x+√(1-x^2)
1-x^2≥0
-1≤x≤1
定义域=[-1,1]
f(x) = x+√(1-x^2)
f'(x) = 1 - x/√(1-x^2)
f'(x) =0
1 - x/√(1-x^2) =0
√(1-x^2) =x
1-x^2 =x^2
x=√2/2 or - √2/2
f'(x) | x=√2/2+ <0 , f'(x) | x=√2/2+ >0
x=√2/2 ( max)
f'(x) | x=√2/2- >0 , f'(x) | x=√2/2+ <0
x=-√2/2 ( min)
f(x) = x+√(1-x^2)
max f(x)
=f(√2/2)
=√2/2 + √2/2
=√2
min f(x)
=f(-√2/2)
=-√2/2 + √2/2
=0
1-x^2≥0
-1≤x≤1
定义域=[-1,1]
f(x) = x+√(1-x^2)
f'(x) = 1 - x/√(1-x^2)
f'(x) =0
1 - x/√(1-x^2) =0
√(1-x^2) =x
1-x^2 =x^2
x=√2/2 or - √2/2
f'(x) | x=√2/2+ <0 , f'(x) | x=√2/2+ >0
x=√2/2 ( max)
f'(x) | x=√2/2- >0 , f'(x) | x=√2/2+ <0
x=-√2/2 ( min)
f(x) = x+√(1-x^2)
max f(x)
=f(√2/2)
=√2/2 + √2/2
=√2
min f(x)
=f(-√2/2)
=-√2/2 + √2/2
=0
本回答被网友采纳
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
