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函数f(x)=(x^4-10)/(x^2+3),x∈R最小值
人气:141 ℃ 时间:2020-08-27 08:48:57
解答
f(x)=(x^4-10)/(x^2+3) = { (x^4-9) - 1} /(x^2+3) =x^2-3 -1/(x^2+3) = x^2 -1/(x^2+3) - 3
x∈(-∞,0)时,x^2单调减,-1/(x^2+3)单调减,∴f(x)单调减
x∈(-∞,0)时,x^2单调增,-1/(x^2+3)单调增,∴f(x)单调增
∴x=0时有最小值:f(0)=0-1/3-3=-10/3
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