Question

高中数学函数求值域题

1.求y=x²＋x＋1／x＋1的值域 (请写出详细过程）

2.求y＝x／x²＋x＋1的值域(请写出详细过程）

Answer 1

请在此输入您的回答

Answer 2

(1)
y=(x^2＋x＋1)／(x＋1)

= x + 1/(x+1)
y' = 1 - 1/(x+1)^2
y'=0
(x+1)^2 -1 =0
x(x+2)=0
x=0 or -2

y'' = 2/(x+1)^3
y''(0) = 2 >0 (min)
y''(-2) = -2 <0 (max)

max y = y(-2) = -2 + 1/(-2+1) = -1
min y = y(0) = 0 + 1/(0+1) =1
值域=[-1,1]

(2)
y＝x／(x^2＋x＋1)
y' =[ (x^2＋x＋1) - x(2x+1) ]/(x^2＋x＋1)^2
=(-x^2 +1)/(x^2＋x＋1)^2
y'=0
1-x^2=0
x= 1 or -1

y'|1+ >0
y'|1- < 0
x=1 (max)

y'|-1+ <0
y'|-1- > 0
x=-1 (min)

max y = y(1) = 1／(1＋1＋1) =1/3
min y =y(-1) = -1／(1-1＋1) =-1
值域=[-1,1/3]