当x>-1时,求f(x)=x+1⼀(x+1)的最小值 已知0<x<1⼀3,求函数y=x(1-3x)的最大值

2025-01-05 16:40:17
推荐回答(2个)
回答1:

1.f(x)=x+1/(x+1)
=(x+1)+1/(x+1)-1
>=2-1=1,
当x=0时取等号,
∴它的最小值=1.
2.0当x=1/6时它取最大值1/12.

回答2:

f(x)=x+1/(x+1)
=x+1 + 1/(x+1) -1
≥2√(x+1)/(x+1) -1
=1
y=x(1-3x)
=3 x(1/3-x)
≤3 [(x + 1/3 - x)/2]^2
=3 * 1/36
=1/12