函數f(x)=√(x+3)+√(1-x)的最大值和最小值為何?
回答 (2)
2020-03-27 6:24 pm
f(x) = √[√(x+3) + √(1-x)]^2 (-3 ≦ x ≦ 1)
= √{4+2√[(3+x)(1-x)]}
= √{4+2√[4-(x+1)^2]}
最大值發生在 x+1 = 2, 即 f(-1) = √8 = 2√2.
最小值發生在端點 x = -3, 1. f(-3) = f(1) = 2.
= √{4+2√[(3+x)(1-x)]}
= √{4+2√[4-(x+1)^2]}
最大值發生在 x+1 = 2, 即 f(-1) = √8 = 2√2.
最小值發生在端點 x = -3, 1. f(-3) = f(1) = 2.
2020-03-30 12:01 pm
f(x)=√(x+3)+√(1-x)
f'(x)=0.5(x+3)^-0.5 +(-1)0.5(1-x)^-0.5
f'(x)=0.5(x+3)^-0.5 -0.5(1-x)^-0.5
f'(x)=[(1-x)^0.5 -(x+3)^0.5]/{2[(x+3)(1-x)]^0.5}
f'(x)=0
(1-x)^0.5 -(x+3)^0.5=0
x+3=1-x
2x=-2
x=-1
f(1)=√(1+3)+√(1-1)
f(1)=2
最小值=2
f'(x)=0.5(x+3)^-0.5 +(-1)0.5(1-x)^-0.5
f'(x)=0.5(x+3)^-0.5 -0.5(1-x)^-0.5
f'(x)=[(1-x)^0.5 -(x+3)^0.5]/{2[(x+3)(1-x)]^0.5}
f'(x)=0
(1-x)^0.5 -(x+3)^0.5=0
x+3=1-x
2x=-2
x=-1
f(1)=√(1+3)+√(1-1)
f(1)=2
最小值=2
收錄日期: 2021-04-18 18:26:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200327065242AAyJux0