Question

当x>1时,函数y=x-1/x^2-3x+5的最大值是？

当x>1时,函数y=x-1/x^2-3x+5的最大值？？

Answer 1

x>1 x-1>0
y=(x-1)/(x²-3x+5)
=(x-1)/(x²-x-2x+2+3)
=(x-1)/[x(x-1)-2(x-1)+3]
=1/[x-2 +3/(x-1)]
=1/[(x-1) +3/(x-1) -1]
由均值不等式得(x-1)+3/(x-1)≥2√3
(x-1)+ 3/(x-1) -1≥2√3-1
0<1/[(x-1)+3/(x-1) -1]≤1/(2√3-1)
0<1/[(x-1)+3/(x-1) -1]≤(2√3+1)/11
0<y≤(2√3+1)/11
函数的最大值为(2√3+1)/11

Answer 2

x>1
x-1>0
y=(x-1)/(x²-3x+5)
=(x-1)/(x²-x-2x+2+3)
=(x-1)/[x(x-1)-2(x-1)+3]
=1/[x-2 +3/(x-1)]
=1/[(x-1) +3/(x-1) -1]
由均值不等式得：(x-1)+3/(x-1)≥2√3
(x-1)+ 3/(x-1) -1≥2√3-1
所以：0<1/[(x-1)+3/(x-1) -1]≤1/(2√3-1)
即：0<1/[(x-1)+3/(x-1) -1]≤(2√3+1)/11
因此：0<y≤(2√3+1)/11
故，函数的最大值为(2√3+1)/11

Answer 3

大于1还是大于等于1？
你求导。