Question

函数y=2x^2+1/x，(x>0)的最小值为...

如题。要详细解答过程~ 有图的最好了~~

Answer 1

这个需用导数求了。
y = 2x² + 1/x
y' = 4x - 1/x²
y' = 0
4x - 1/x² = 0
4x³ - 1 = 0
x = (1/4)^(1/3) = 1 / 2^(2/3)
∴当x = 1 / 2^(2/3)时，y = 2 * 1 / [2^(2/3)]² + 1 / [1 / 2^(2/3)]
= 2^(-4/3+1) + 2^(2/3)
= 1 / 2^(1/3) + 2^(2/3)
= 3 / [2^(1/3)]
∴最小值为3 / [2^(1/3)]

Answer 2

y=2x^2+1/x
=2x^2+1/(2x)+1/(2x)
>=3*三次根号[2x^2*1/(2x)*1/(2x)
=3*三次根号(1/2) 当且仅当 2x^2=1/(2x) 即 4x^3=1 x=三次根号(1/4)时成立

三次根号(1/2)=(三次根号4)/2
三次根号(1/4)= 三次根号(2)/2