第1个回答 2017-08-08
-3≤x≤2
则0≤x²≤9
2≤x²+2≤11
所以1/11≤1/(x²+2)≤1/2
2/11≤2/(x²+2)≤1
3+2/11≤3+2/(x²+2)≤3+1
即35/11≤y≤4本回答被提问者采纳
则0≤x²≤9
2≤x²+2≤11
所以1/11≤1/(x²+2)≤1/2
2/11≤2/(x²+2)≤1
3+2/11≤3+2/(x²+2)≤3+1
即35/11≤y≤4本回答被提问者采纳
第2个回答 2017-08-08
y= 3 + 2/(x^2+2)
y' = -4x/(x^2+2)^2
y'=0
=>x=0
y'|x=0+ <0
y'|x=0- >0
x=0 (max)
y(0) = 4
y(2)=3 + 2/6 = 10/3
y(-3)=3 + 2/11 = 35/11
y的取值范围 = [ 35/11, 4]
y' = -4x/(x^2+2)^2
y'=0
=>x=0
y'|x=0+ <0
y'|x=0- >0
x=0 (max)
y(0) = 4
y(2)=3 + 2/6 = 10/3
y(-3)=3 + 2/11 = 35/11
y的取值范围 = [ 35/11, 4]
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