Question

求函数f(x)=(x+5)(x+2)/(x+1),(x＜-1)的最大值

Answer 1

f(x)=(x+5)(x+2)/(x+1)
= (x+2)*[(x+1) + 4]/(x+1)
= (x+2)*[1 + 4/(x+1)]
= (x+2) + 4(x+2)/(x+1)
= (x+2) + 4[(x+1) + 1]/(x+1)
= (x+2) + 4[1 + 1/(x+1)]
= (x+2) + 4 + 4/(x+1)
= (x+1) + 4/(x+1) + 5
= -[-(x+1) - 4/(x+1)] + 5
x < -1, -(x+1) > 0 ，所以
-(x+1) + 4/[-(x+1)] ≥ 2 √[-(x+1)] * √[-4/(x+1)] = 4
-[-(x+1) - 4/(x+1)] ≤ -4
原式 ≤ -4 + 5 = 1
原式最大值为 1。
当 (x+1) = 4/(x+1) 时，取得最大值，即
(x+1)^2 = 4,
x+1 = -2
x = -3

Answer 2

令t=-(x+1)>0
x+1=-t
x+5=4-t
x+2=1-t
(x+5)(x+2)/(x+1)
=(4-t)(1-t)/-t
=(4-5t+t^2)/-t
=-[4/t-5+t]
=-[4/t+t-5]
=5-(4/t+t)
要想值最大，即4/t+t最小
由基本不等式
4/t+t>=2根号(4/t*t)=4
等号成立时4/t=t
t=2(舍去负值)
所以x+1=-2
x=-3时
f(x)最大值为5-4=1

Answer 3

令t=x+1，则t<0
f(x)=(t+4)(t+1)/t
=(t²+5t+4)/t
=t+4/t+5
因为(-t)+(-4/t)≥4
所以t+4/t≤-4
所以f(x)max=1