已知x>1,求函数y=3x+4⼀(x-1)+1最小值 是x取多少

2025-05-14 12:39:07
推荐回答(1个)
回答1:

y=3x+4/(x-1)+1
→3x²-(y+2)x+y+3=0.
△=(y+2)²-12(y+3)≥0
→y²-8y-32≥0
→y≥4+4√3,y≤4-4√3.
而x>1,故所求最小值为:
y|min=4+4√3.
此时易得,x=(3+2√3)/3.