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y=(x²+2x+3)/(x²+1)的值域,
人气:164 ℃ 时间:2020-04-23 07:44:59
解答
y=(x²+2x+3)/(x²+1)=1+2(x+1)/(x²+1)
求导可得y'=-2(x²+2x-1)/(x²+1)²
令y'=0可得x=√2-1或-√2-1
lim(x→∞)(x²+2x+3)/(x²+1)=1
当x=√2-1时y=(3√2+6)/4取得最大值
当x=-√2-1时y=(6-3√2)/4取得最小值
∴y=(x²+2x+3)/(x²+1)的值域为[(6-3√2)/4,(3√2+6)/4]
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