Question

求y=（x+1）/(x²+3x+5) (x>-1) 的最值

Answer 1

由x>-1得出x+1>0
y=(x+1)/(x²+3x+5)
=(x+1)/[(x+1)²+(x+1)+3]
=1/[(x+1)+1+3/(x+1)]≤1/(1+2√3)
当且仅当x=√3-1时取到最大值。

Answer 2

令g(x)=
1/y
=(x²+3x+5)/(x+1)=
[(x+1)2+x+1+4]/

(x+1)

=
x+1
+1+
4/

(x+1)>=1+2根号下[

x+1*

4/
(x+1)]=1+4=5

当且仅当
x+1= 4/ (x+1)[x>-1],即
x=1时gmin=5

则
y=f(x)=1/
g(x)<=1/5

则当且仅当
x+1=

4/
(x+1)[
x>-1],即x=1时，ymax=1/5